HARMONIC ANALYSIS
PIOTR HAJLASZ
1. Preliminaries
Let us first fix notation that will be used throughout the book. We will be working
mostly in the Euclidean spaces and with the Lebesgue measure. The Lebesgue measure of a
measurable set E will be denoted by |E|. The volume of a ball of radius r is |B(0, r)| = ωn rn .
Then it is well known that the (n − 1)-dimensional measure of the sphere S n−1 (0, r) equals
nωn rn−1 . The integral average over a set E of positive measure will be denoted by
Z
Z
1
f (x) dx.
f (x) dx =
|E| E
E
The Lebesgue measure on a surface will be denoted by dσ so the integration in the spherical
coordinates will take the form
Z ∞Z
Z
f (sθ) dσ(θ) ds.
f (x) dx =
Rn
S n−1
0
Unless we will explicitly state otherwise, all functions and all function spaces will be complex valued.
A generic open set in Rn will usually be denoted by Ω and C0∞ (Ω) will stand for a
space of smooth functions with compact support in Ω. The space of continuous functions
vanishing at infinity will be denote by C0 (Rn ), i.e. f ∈ C0 (Rn ) if f is continuous and
lim f (x) = 0.
|x|→∞
C0 (Rn ) is a Banach space with respect to the supremum norm k · k∞ and C0∞ (Rn ) is a
dense subset of C0 (Rn ).
We will use the multiindex notation
Dα u =
∂ |α|
,
∂xα1 1 . . . ∂xαnn
xα = xα1 1 · . . . · xαnn
where α = (α1 , . . . , αn ), |α| = α1 + . . . + αn . Then the product rule takes the form
X α!
Dα (f g) =
Dβ f Dγ g, where α! = α1 ! · . . . · αn !.
β!γ!
β+γ=α
Date: February 27, 2017.
1
2
PIOTR HAJLASZ
The characteristic function of a set E will be denoted by χE . By δa we will denote the
Dirac measure centered at a i.e.,
Z
f (x) dδa (x) = f (a).
Rn
By an increasing function we will mean a non-decreasing function and increasing functions that are actually increasing, will be called strictly increasing. Similar terminology
applies to decreasing and strictly decreasing functions.
By sgn x we will denote the sign of the number x, i.e. sgn x = 1 for positive x, sgn x = −1
for negative x and sgn 0 = 0.
Various constants will usually be denoted by C. By writing C(n, m) we will indicate the
the constant C(n, m) depends on n and m only. We adopt the rule that a constant C may
change its value within one string of estimates. This will allow us to minimize the use of
indices.
HARMONIC ANALYSIS
3
2. The Maximal function
2.1. The maximal theorem.
Definition 2.1. For a locally integrable function f ∈ L1loc (Rn ) the Hardy-Littlewood maximal function is defined by
Z
|f (y)| dy, x ∈ Rn .
Mf (x) = sup
r>0
B(x,r)
The operator M is not linear but it is subadditive. We say that an operator T from a
space of measurable functions into a space of measurable functions is subadditive if
|T (f1 + f2 )(x)| ≤ |T f1 (x)| + |T f2 (x)| a.e.
and
|T (kf )(x)| = |k||T f (x)| for k ∈ C.
The following integrability result, known also as the maximal theorem, plays a fundamental
role in many areas of mathematical analysis.
Theorem 2.2 (Hardy-Littlewood-Wiener). If f ∈ Lp (Rn ), 1 ≤ p ≤ ∞, then Mf < ∞
a.e. Moreover
(a) For f ∈ L1 (Rn )
(2.1)
5n
|{x : Mf (x) > t}| ≤
t
Z
|f | for all t > 0.
Rn
(b) If f ∈ Lp (Rn ), 1 < p ≤ ∞, then Mf ∈ Lp (Rn ) and
1/p
p
n/p
kMf kp ≤ 2 · 5
kf kp for 1 < p < ∞,
p−1
kMf k∞ ≤ kf k∞ .
1
n
1
n
Remark
R 2.3. Note that if f ∈ L (R ) is a non-zero function, then Mf 6∈ L (R ). Indeed,
if λ = B(0,R) |f | > 0, then for |x| > R, B(0, R) ⊂ B(x, R + |x|) so
Z
λ
Mf (x) ≥
|f | ≥
,
ωn (R + |x|)n
B(x,R+|x|)
Since the function on the right hand side is not integrable on Rn , the statement (b) of the
theorem is not true for p = 1.
4
PIOTR HAJLASZ
Remark 2.4. If g ∈ L1 (Rn ), then the Chebyschev inequality
Z
1
(2.2)
|{x : |g(x)| > t}| ≤
|g| for t > 0
t Rn
is easy to prove. Thus if an (not necessarily linear) operator T is bounded in L1 , kT f k1 ≤
Ckf k1 , it immediately follows from Chebyschev’s inequality that
Z
C
(2.3)
|{x : |T f (x)| > t}| ≤
|f (x)| dx.
t Rn
The maximal function M is not bounded in L1 , but it still satisfies the weaker estimate
(2.3). For this reason estimate (2.1) is called the weak type estimate.
In the proof of Theorem 2.2 we will need the following two results.
Theorem 2.5 (Cavalieri’s principle). If µ is a σ-finite measure on X and Φ : [0, ∞) →
[0, ∞) is increasing, absolutely continuous and Φ(0) = 0, then
Z ∞
Z
Φ0 (t)µ({|f | > t}) dt .
Φ(|f |) dµ =
0
X
Proof. The result follows immediately from the equality
Z
Z Z |f (x)|
Φ(|f (x)|) dµ(x) =
Φ0 (t) dt dµ(x)
X
X
0
and the Fubini theorem.
Corollary 2.6. If µ is a σ-finite measure on X and 0 < p < ∞, then
Z ∞
Z
p
tp−1 µ({|f | > t}) dt .
|f | dµ = p
0
X
The next result has many applications that go beyond the maximal theorem. Here and
in what follows by 5B we denote the ball concentric with B and with five times the radius.
Theorem 2.7 (5r-covering lemma). Let B be a family of balls in a metric space such that
sup{diam B : B ∈ B} < ∞. Then there is a subfamily of pairwise disjoint balls B 0 ⊂ B
such that
[
[
B⊂
5B .
B∈B0
0
B∈B
More precisely, for every B ∈ B there is B 0 ∈ B such that B ∩ B 0 6= ∅ and B ⊂ 5B 0 .
If the metric space is separable, then the family B 0 is countable (or finite) and we can
arrange it as a (possibly finite) sequence B 0 = {Bi }∞
i=1 so
[
B∈B
B⊂
∞
[
5Bi .
i=1
Remark 2.8. Here B can be either a family of open balls or closed balls. In both cases
the proof is the same.
HARMONIC ANALYSIS
5
Proof. Let sup{diam B : B ∈ B} = R < ∞. Divide the balls in the family B according to
their diameters:
n
R
R o
Fj = B ∈ B : j < diam B ≤ j−1 .
2
2
S∞
Clearly B = j=1 Fj .
First we select balls as large as possible so we choose balls from the family F1 and we
take as many as we can: we define B1 ⊂ F1 to be a maximal family of pairwise disjoint
balls.
Suppose the families B1 , . . . , Bj−1 are already defined.
In the construction of Bj we want to select balls form Fj , but we have to make sure that
balls that we choose do not intersects with any of the previously selected balls. Thus we
have to select balls from
j−1
n
[ o
0
0
F̂j = B ∈ Fj : B ∩ B = ∅ for all B ∈
Bi .
i=1
Then we define Bj to be the maximal family of pairwise disjoint balls in F̂j
S
Finally we define B 0 = ∞
j=1 Bj . It follows immediately from the definition of Bj that
S
S
every ball B ∈ Fj intersects with a ball in ji=1 Bj . If B ∩ B 0 6= ∅, B 0 ∈ ji=1 Bi , then
diam B ≤
R
2j−1
=2·
R
< 2 diam B 0
j
2
and hence B ⊂ 5B 0 .
Proof of Theorem 2.2. (a) Let f ∈ L1 (Rn ) and let Et = {x : Mf (x) > t}. For x ∈ Et ,
there is rx > 0 such that
Z
Z
−1
|f | > t so |B(x, rx )| < t
|f | .
B(x,rx )
B(x,rx )
Observe that supx∈Et rx < ∞, because f ∈ L1 (Rn ). The family of balls {B(x, rx )}x∈Et
forms a covering of the set Et so applying the 5r-covering lemma
there is a sequence of
S∞
pairwise disjoint balls B(xi , rxi ), i = 1, 2, . . . such that Et ⊂ i=1 B(xi , 5rxi ) and hence
Z
∞
∞ Z
X
5n X
5n
n
|Et | ≤ 5
|B(xi , rxi )| ≤
|f | ≤
|f | .
t
t
n
B(x
,r
)
R
x
i
i
i=1
i=1
The proof is complete.
(b) Let f ∈ Lp (Rn ), 1 < p ≤ ∞. Since kMf k∞ ≤ kf k∞ we can assume that 1 < p < ∞.
Let f = f1 + f2 , where
f1 = f χ{|f |>t/2} ,
f2 = f χ{|f |≤t/2}
be a decomposition of |f | into its ‘lower’ and ‘upper’ parts. It is easy to check that f1 ∈
L1 (Rn ) and clearly f2 ∈ L∞ (Rn ).
The idea now if to apply the weak type estimate from (a) to f1 and the estimate
kMf k∞ ≤ kf k∞ to f2 . This will give us an estimate for the size of the level set |{Mf > t}|
6
PIOTR HAJLASZ
and the result will follow from Corollary 2.6. The same idea is the core of the so called real
interpolation method and we will see it again in the proof of the Marcinkiewcz Interpolation
Theorem 8.9.
Since |f | ≤ |f1 | + t/2 we have Mf ≤ Mf1 + t/2 and hence
{Mf > t} ⊂ {Mf1 > t/2} .
Thus
Z
2 · 5n
|Et | = |{Mf > t}| ≤
|f1 (x)| dx
t
Rn
Z
2 · 5n
=
|f (x)| dx .
t
{|f |>t/2}
(2.4)
Cavalieri’s principle gives
Z
Z ∞
p
|Mf (x)| dx = p
tp−1 |{Mf > t}| dt
n
R
Z
Z0 ∞
2 · 5n
p−1
|f (x)| dx dt
t
≤ p
t
{|f |>t/2}
0
Z
Z 2|f (x)|
n
= 2·5 p
|f (x)|
tp−2 dt dx
Rn
0
Z
p
= 2p · 5n
|f (x)|p dx
p − 1 Rn
and the results follows.
Remark 2.9. Note that we proved in (2.4) the following inequality
Z
2 · 5n
(2.5)
|{x : Mf (x) > t}| ≤
|f (x)| dx
t
{|f |>t/2}
which is slightly stronger than (2.1). Later we will see that the measure of the set |{Mf >
t}| has a similar bound from below, see Proposition 2.33.
Remark 2.10. For a positive measure µ on Rn we define the maximal function by
Mµ(x) = sup
r>0
µ(B(x, r))
.
|B(x, r)|
A minor modification of the proof of Theorem 2.2(a) leads to the following result.
Proposition 2.11. If µ is a finite positive Borel measure on Rn , then
5n
µ(Rn ) for all t > 0.
t
Remark 2.12. It is natural to inquire whether in the Euclidean case, there is a version of
Theorem 2.7 for families of cubes. The answer is in the positive and it follows immediately
from Theorem 2.7 and the fact that cubes with sides parallel to coordinate axes are balls
in Rn with respect to the metric d∞ (x, y) = maxi |xi − yi |. In the following result 5Q will
denote a cube concentric with Q and with five times the diameter.
|{x : Mµ(x) > t}| ≤
HARMONIC ANALYSIS
7
Corollary 2.13. Let F be a family of open or closed cubes in Rn with sides parallel to
coordinate directions such that sup{diam Q : Q ∈ F} < ∞. Then there is a subfamily
F 0 ⊂ F of pairwise disjoint cubes such that
[
[
Q⊂
5Q.
Q∈F
Q∈F 0
More precisely for every Q ∈ F there is Q0 ∈ F 0 such that Q ∩ Q0 6= ∅ and Q ⊂ 5Q0 .
Although we are interested mostly in the Euclidean setting, it was important to formulate
Theorem 2.7 in a the setting of metric spaces as we could conclude Corollary 2.13 directly
from Theorem 2.7.
In the next two sections we will show applications of the maximal theorem.
2.2. Fractional integration theorem. As a first application of the maximal theorem
we will prove a result about integrability of the Riesz potentials.
Definition 2.14. For 0 < α < n and n ≥ 2 we define the Riesz potential by
n
Z
π 2 2α Γ α2
1
f (y)
.
(Iα f )(x) =
dy , where γ(α) =
γ(α) Rn |x − y|n−α
Γ n−α
2
At this moment the particular value of the constant γ(α) is not important to us. We
could even replace this constant by 1.1
Theorem 2.15 (Hardy-Littlewood-Sobolev). Let α > 0, 1 < p < ∞ and αp < n. Then
there is a constant C = C(n, p, α) such that
kIα f kp∗ ≤ Ckf kp
for all f ∈ Lp (Rn ) ,
where p∗ = np/(n − αp).
Remark 2.16. It is not difficult to show directly2 that if Iα : Lp → Lq is bounded, then
q = p∗ . The idea is to use the change of variables x 7→ tx, t > 0, and see how both sides of
the inequality kIα f kq ≤ Ckf kp change. This is so called a scaling argument.
We precede the proof with a technical lemma.
Lemma 2.17. If 0 < α < n, and δ > 0, then there is a constant C = C(n, α) such that
Z
|f (y)|
dy ≤ Cδ α Mf (x) for all x ∈ Rn .
n−α
|x
−
y|
B(x,δ)
Proof. For x ∈ Rn and δ > 0 consider the annuli
δ
δ
A(k) = B x, k − B x, k+1 .
2
2
1It
will play an important role later: I2 f is the convolution with the fundamentals solution of −∆, see
Theorem 5.60. Also Riesz potentials will be studied in Section 7 in connection with fractional powers of
the Laplace operator.
2A nice exercise.
8
PIOTR HAJLASZ
We have
Z
B(x,δ)
∞ Z
X
|f (y)|
|f (y)|
dy
=
dy
n−α
|x − y|n−α
|x
−
y|
A(k)
k=0
α−n Z
∞ X
δ
|f (y)| dy
≤
k+1
2
A(k)
k=0
α−n n Z
∞ X
δ
δ
≤ ωn
|f (y)| dy
k+1
k
2
2
k)
B(x,δ/2
k=0
!
α−n X
∞
1
1
≤ ωn δ α
Mf (x) .
kα
2
2
k=0
The proof is complete.
Proof of Theorem 2.15. Fix δ > 0. Hölder’s inequality and integration in polar coordinates
yield
Z
1/p0
Z
|f (y)|
dy
dy ≤ kf kp
n−α
(n−α)p0
Rn \B(x,δ) |x − y|
Rn \B(x,δ) |x − y|
1/p0
Z ∞
n−1−(n−α)p0
= kf kp nωn
s
ds
= C(n, p, α)δ
δ
α−(n/p)
kf kp ,
because nωn equals the (n − 1)-dimensional measure of the unit sphere S n−1 and n − (n −
α)p0 < 0. This and the lemma give
|Iα f (x)| ≤ C δ α M f (x) + δ α−(n/p) kf kp .
Taking3
δ=
Mf (x)
kf kp
−p/n
yields
αp
αp
|Iα f (x)| ≤ C(Mf (x))1− n kf kpn
which is equivalent to
∗
αp
p∗
|Iα f (x)|p ≤ C(Mf (x))p kf kpn .
Integrating both sides over Rn and applying boundedness of the maximal function in Lp
gives the result.
2.3. The Lebesgue differentiation theorem. We will show now how to prove the
Lebesgue differentiation theorem from the maximal theorem.
For h ∈ Rn we define τh f (x) = f (x + h).
Lemma 2.18. If f ∈ Lp (Rn ), 1 ≤ p < ∞, then kτh f − f kp → 0 as h → 0.
3We
apply here a standard trick: we take δ > 0 such that δ α M f (x) = δ α−(n/p) kf kp .
HARMONIC ANALYSIS
9
This result is obvious if f is a compactly supported smooth function, because in that
case τh f converges uniformly to f . The general case follows from the density of C0∞ in Lp .
R
Lemma 2.19. Let f ∈ L1 (Rn ) and let fr (x) = B(x,r) f (y) dy. Then fr → f in L1 as
r → 0. In particular, there is a sequence ri → 0 such that
Z
f (y) dy = f (x) a.e.
lim
i→∞
B(x,ri )
Proof. We have
Z
Z
Z
|fr (x) − f (x)| dx ≤
|f (y) − f (x)| dy dx
Rn
Rn
Z
B(x,r)
Z
|f (x + y) − f (x)| dy dx
=
Rn
B(0,r)
Z
(2.6)
kτy f − f k1 dy .
=
B(0,r)
Since τy f → f in L1 as y → 0, the right hand side of (2.6) converges to 0 as r → 0. The
existence of a sequence ri follows from the fact that from a sequence convergent in L1 we
can extract a subsequence convergent a.e.
The Lebesgue differentiation theorem gives much more.
Theorem 2.20 (Lebesgue differentiation theorem). If f ∈ L1loc (Rn ), then
Z
lim
f (y) dy = f (x) a.e.
r→0 B(x,r)
1
n
RProof. Since the theorem is local in nature we can assume that f ∈ L (R ). Let fr (x) =
f (y) dy and define
B(x,r)
Ωf (x) = lim sup fr (x) − lim inf fr (x) .
r→0
r→0
It suffices to prove that Ωf = 0 a.e. Indeed, this property means that fr → g converges
a.e. to a measurable function g. Since by Lemma 2.19, fri → f a.e. we get g = f a.e.
Observe that Ωf ≤ 2Mf , hence for any ε > 0, Theorem 2.2(a) yields
Z
C
|{x : Ωf (x) > ε}| ≤
|f | .
ε Rn
Let h be a continuous function such that kf − hk1 < ε2 . Continuity of h implies Ωh = 0
everywhere and hence
Ωf ≤ Ω(f − h) + Ωh = Ω(f − h) ,
so
Z
C
|{Ωf > ε}| ≤ |{Ω(f − h) > ε}| ≤
|f − h| ≤ Cε .
ε Rn
Since ε > 0 can be arbitrarily small we conclude Ωf = 0 a.e.
10
PIOTR HAJLASZ
If f ∈ L1loc (Rn ), then we can define f at every point by the formula
Z
(2.7)
f (x) := lim sup
f (y) dy .
r→0
B(x,r)
According to the Lebesgue differentiation theorem this is a representative of f in the class
of functions that coincide with f a.e.
Definition 2.21. Let f ∈ L1loc (Rn ). We say that x ∈ Rn is a Lebesgue point of f if
Z
lim
|f (y) − f (x)| dy = 0 ,
r→0 B(x,r)
where f (x) is defined by (2.7).
Theorem 2.22. If f ∈ L1loc (Rn ), then the set of points that are not Lebesgue points of f
has measure zero.
Proof. For c ∈ Q let Ec be the set of points for which
Z
(2.8)
lim
|f (y) − c| dy = |f (x) − c|
r→0 B(x,r)
does not hold. Clearly |Ec | = 0 and hence the set E =
for x ∈ Rn \ E and all c ∈ Q, (2.8) is satisfied.
S
c∈Q
Ec has measure zero too. Thus
If x ∈ Rn \ E and f (x) ∈ R, approximating f (x) by rational numbers one can easily
check that
Z
lim
|f (y) − f (x)| dy = |f (x) − f (x)| = 0 .
r→0 B(x,r)
Indeed, if ε > 0 is arbitrary and c ∈ Q is such that |f (x) − c| < ε/3, then there is δ > 0
such that for 0 < r < δ
Z
Z
|f (y) − f (x)| dy ≤
|f (y) − c| dy + |c − f (x)|
B(x,r)
B(x,r)
ε
+ |c − f (x)| < ε.
< |f (x) − c| +
3
The proof is complete.
Definition 2.23. We say that x ∈ Rn is a p-Lebesgue point of f ∈ Lploc , 1 ≤ p < ∞ if
Z
lim
|f (y) − f (x)|p dy = 0 .
r→0 B(x,r)
The same method as above leads to the following result that we leave as an exercise.
Theorem 2.24. If f ∈ Lploc (Rn ), 1 ≤ p < ∞, then the set of points that are not p-Lebesgue
points of f has measure zero.
Definition 2.25. Let E ⊂ Rn be a measurable set. We say that x ∈ Rn is a density point
of E if
|B(x, r) ∩ E|
lim
= 1.
r→0
|B(x, r)|
HARMONIC ANALYSIS
11
Applying the Lebesgue theorem to the characteristic function of the set E, i.e., f = χE
we obtain
Theorem 2.26. Almost every point of a measurable set E ⊂ Rn is its density point and
a.e. point of Rn \ E is not a density point of E.
The Lebesgue theorem shows that for almost all x, the averages of f over balls centered
as x converge to f (x). It is natural to inquire whether we can replace balls by other sets
like cubes or even balls, but not centered at x.
Definition 2.27. We say that a family F of measurable subsets of Rn is regular at x ∈ Rn
if
(a) The sets are bounded and have positive measure.
(b) There is a sequence {Si } ⊂ F with |Si | → 0 as i → ∞.
(c) There is a constant C > 0 such that for all S ∈ F we have |S| ≥ C|BS |, where BS
is the smallest closed ball centered at x that contains S.
Note that if |Si | → 0, then the radius of BSi approaches to zero so the sets Si approach
to x.
Examples of families regular at x include:
• Family of all balls that contain x.
• Family of all cubes centered at x.
• Family of all cubes that contain x.
Theorem 2.28. If f ∈ L1loc (Rn ), x is a Lebesgue point of f , and if F is a family regular
at x, then
Z
lim
f (y) dy = f (x) .
S∈F
|S|→0
S
Proof. We have
Z
Z
Z
1
f (y) dy − f (x) ≤
|f (y) − f (x)| dy ≤
|f (y) − f (x)| dy
|S| BS
S
S
Z
Z
|BS |
1
|f (y) − f (x)| dy ≤
|f (y) − f (x)| dy → 0
≤
|S| BS
C BS
as |S| → 0.
Note that if F is a family regular at 0 and if we define the maximal function associated
with F by
Z
MF f (x) = sup |f (x − y)| dy ,
S∈F
S
then a routine calculation shows that
(2.9)
MF f (x) ≤ CMf (x) ,
so MF satisfies the claim of Theorem 2.2 (with different constants). In particular MF is
a bounded operator in Lp , 1 < p ≤ ∞.
12
PIOTR HAJLASZ
Let F be the family of all rectangular boxes in Rn that contain the origin and have sides
parallel to the coordinate axes. With the family we can associate the maximal function
Z
f (x) = sup |f (x − y)| dy .
Mf
S∈F
S
f in Lp , 1 <
Note that the family F is not regular at 0 and hence the boundedness of M
p ≤ ∞ cannot be concluded from (2.9). However, we have
Theorem 2.29 (Zygmund). For 1 < p < ∞ there is a constant C = C(n, p) > 0 such
that
f kp ≤ Ckf kp .
kMf
(2.10)
Moreover, if f ∈ Lploc , 1 < p < ∞, then
Z
f (x − y) dy = f (x) a.e.
(2.11)
lim
diam S→0
S∈F
S
Proof. First we will prove how to conclude (2.11) from (2.10). Note that since the family
is not regular at 0, (2.11) is not a consequence of Theorem 2.28 (see also Theorem 2.30).
However,
Z
Z
f (x)
0 ≤ lim sup f (x − y) dy − lim inf
f (x − y) dy ≤ 2Mf
diam S→0
S∈F
diam S→0
S∈F
S
S
and hence (2.11) follows from (2.10) by almost the same argument as the one used to
deduce the Lebesgue Differentiation Theorem from Theorem 2.2. We leave details to the
reader.
Thus it remains to prove (2.10). In dimension one
Z x+b
f
Mf (x) = sup
|f (y)| dy
a,b>0 x−a
f ≤ CMf
is the so called non-centered Hardy-Littlewood maximal function. Clearly Mf
which gives (2.10). For simplicity of notation we will prove the higher dimensional result
for n = 2 only. The proof in the case of general n > 1 is similar. We have
Z x2 +b2 Z x1 +b1
f
Mf (x1 , x2 ) = sup
|f (y1 , y2 )| dy1 dy2
a1 ,b1 >0
a2 ,b2 >0
x2 −a2
Z
≤
x1 −a1
x2 +b2 sup
a2 ,b2 >0 x2 −a2
Z
x1 +b1
sup
|f (y1 , y2 )| dy1
dy2 .
a1 ,b1 >0 x1 −a1
On the right hand side we have iteration of one dimensional non-centered maximal functions. First we apply the maximal function to variable y1 and evaluate it at x1 and then we
apply the maximal function to the variable y2 and evaluate it at x2 . It is easy to see now
that inequality (2.10) follows from the one dimensional case applied twice and the Fubini
theorem.
HARMONIC ANALYSIS
13
ff does
Surprisingly (2.11) does not hold for p = 1 and hence the maximal function M
not satisfy the weak type estimate (2.1).4 Namely one can prove the following result that
we leave without a proof.
Theorem 2.30 (Saks). Let F be the family of all rectangular boxes in Rn that contain the
origin and have sides parallel to the coordinate axes. Then the set of functions f ∈ L1 (Rn )
such that
Z
lim sup f (x − y) dy = ∞ for all x ∈ Rn
diam S→0
S∈F
1
S
n
is a dense Gδ subset of L (R ). In particular it is not empty.
2.4. The Calderón-Zygmund decomposition. The following result will play an important role in what follows.
Theorem 2.31 (Calderón-Zygmund decomposition). Suppose f ∈ L1 (Rn ), f ≥ 0 and
α > 0. Then there is an open set Ω and a closed set F such that
(a) Rn = Ω ∪ F , Ω ∩ F = ∅;
(b) f ≤ α a.e. in F ;
S
(c) Ω can be decomposed into cubes Ω = ∞
k=1 Qk with pairwise disjoint interiors such
that
Z
α<
f (x) dx ≤ 2n α , k = 1, 2, 3, . . .
Qk
Proof. Decompose Rn into a grid of identical cubes, large enough to have
Z
f (x) dx ≤ α
Q
for each cube in the grid. Take a cube Q from the grid and divide it into 2n identical cubes.
Let Q0 be one of the cubes from this partition. We have two cases:
Z
Z
f (x) dx > α
or
f (x) dx ≤ α .
Q0
Q0
If the fist case holds we include the open cube Q0 to the family {Qk }. Note that
Z
Z
Z
n
−1
n
α<
f = 2 |Q|
f ≤2
f ≤ 2n α
Q0
Q0
Q
so the condition (c) is satisfied. If the second case holds we divide Q0 into 2n identical
cubes and proceed as above. We continue this process infinitely manyStimes or until it is
terminated. We apply it to all the cubes of the original grid. Let Ω = ∞
k=1 Qk , where the
cubes are defined by the first case of the process. It remains to prove that f ≤ α a.e. in
the set F = Rn \ Ω. The set F consists of faces of the cubes (this set has measure zero)
and points x such
that there is a sequence of cubes Q̃i with the property Rthat x ∈ Q̃i ,
R
diam Q̃i → 0, Q̃i f ≤ α. According to Theorem 2.28 for a.e. such x we have Q̃i f → f (x)
and hence f ≤ α a.e. in F .
Corollary 2.32. Let f , α and Ω be as in Theorem 2.31. Then |Ω| < α−1 kf k1 .
4Such
estimate would imply (2.11).
14
PIOTR HAJLASZ
Proof. We have
|Ω| =
∞
X
|Qk | ≤
k=1
∞
X
α
−1
Z
|f | ≤ α−1 kf k1 .
Qk
k=1
The proof is complete.
In Remark 2.9 we obtained an upper bound for the measure of the level set of the maximal
function which improves the weak type estimate from Theorem 2.2. We will prove now a
similar lower bound.
Proposition 2.33. If f ∈ L1 (Rn ), then
Z
Z
2−n C(n)−1
2 · 5n
|f (x)| dx ≤ |{x : Mf (x) > t}| ≤
|f (x)| dx.
t
t
{|f |>C(n)t}
{|f |>t/2}
where we can take C(n) = ωn nn/2 .
Proof. For the proof of the right inequality see (2.4). In order to prove the left inequality
we apply
S the Calderón-Zygmnud decomposition to the function |f | and α = s > 0. If
Ω = k Qk is an open set as in Theorem 2.31, then
Z
|f (x)| dx ≤ 2n s for all k.
s<
Qk
Let `k be the edge-length of the cube Qk . Then for any x ∈ Qk , Qk ⊂ Bx = B(x,
Z
Z
n
|Bx |
|f (y)| dy ≤
|f (y)| dy ≤ ωn n 2 Mf (x).
s<
|Qk | Bx
Qk
√
n`k ) so
Hence
|{x : Mf (x) >
n
sωn−1 n− 2 }|
Z
2−n X
|f (x)| dx
≥
|Qk | ≥
s
Q
k
k
Z
2−n
≥
|f (y)| dy
s {|f |>s}
X
where the last inequality follows from the fact
means
S that |f | ≤ s for almost all x 6∈ Ω whichn/2
that the set {|f | > s} is contained in Ω = k Qk . Now it suffices to take s = tωn n . 2.5. Integrability of the maximal function. In Remark 2.3 we observed that if f ∈
L1 (Rn ), then Mf 6∈ L1 (Rn ). However, we showed that the function cannot be globally
integrable. It turns out that, in general, the maximal function need not be even locally
integrable. We will actually characterize all functions such that the maximal function is
locally integrable.
Definition 2.34. We say that a measurable function f belongs to the Zygmund space
L log L if |f | log(e + |f |) ∈ L1 .
It is easy to see that for a space with finite measure we have
\
Lp ⊂ L log L ⊂ L1 ,
p>1
HARMONIC ANALYSIS
15
so the Zygmund space is an intermediate space between all Lp for p > 1 and L1 .
Theorem 2.35 (Stein). Suppose that a measurable function f defined in Rn equals zero
outside a ball B. Then Mf ∈ L1 (B) if and only if f ∈ L log L(B).
Proof. Suppose that f ∈ L log L(B).5 Cavalieri’s principle (Theorem 2.5) and weak type
estimates from Proposition 2.33 give
Z
Z ∞
Mf (x) dx =
|{x ∈ B : Mf (x) > t}| dt
B
0
Z ∞
|{x ∈ B : Mf (x) > t}| dt
≤ |B| +
Z1 ∞ Z
2 · 5n
≤ |B| +
|f (x)| dx dt
t
1
{|f |>t/2}
Z Z max{2|f (x)|,1} dt
= |B| + C(n)
|f (x)| dx
t
1
ZB
= |B| + C(n) |f (x)| max{0, log(2|f (x)|)} dx
ZB
≤ |B| + 2C(n) |f (x)| log(e + |f (x)|) dx < ∞.
B
To prove the other implication we assume Mf ∈ L1 (B) and we want to prove that
f ∈ L log L(B). In the proof of the first implication we used the right inequality from
Proposition 2.33, but now we want to use the left inequality. However, the following calculations seem to disprove integrability of Mf on B.
Z ∞ Z
Z ∞
Z
C1
|{x ∈ B : Mf (x) > t}| dx ≥
Mf (x) dx =
|f (x)| dx dt
t {|f (x)|>C2 t}
0
0
B
Z Z |f (x)|/C2
dt = C1
|f (x)| dx = +∞.
t
B
0
|
{z
}
= +∞ if f (x) 6= 0.
These calculations are erroneous. Although the function f is supported in the ball B only,
the maximal function Mf is positive on all of Rn and in the left estimate in Proposition 2.33
we deal with the measure of the set
{x ∈ Rn : Mf (x) > t} and not of the set {x ∈ B : Mf (x) > t}.
We will show not how to overcome this difficulty.
Note that Mf ∈ L1loc (Rn ). Indeed, Mf is integrable in B and locally bounded outside
the closed ball B, so we need to verify integrability of Mf in a neighborhood of the
boundary of B.6 But it is easy to see that if x is near the boundary and outside the ball,
5The
proof of integrability of Mf will be similar to the proof of Lp integrability of the maximal function
when f ∈ Lp , p > 1.
6Integrability on B and local boundedness outside B does not imply local integrability in Rn . The
function f (x) = 0 for x ≤ 0 and f (x) = 1/x for x > 0 is integrable on (−∞, 0] and is locally bounded on
(0, ∞).
16
PIOTR HAJLASZ
we can estimate the value of Mf (x) by (constant times) the value of the maximal function
in a point being the reflection of x across the boundary. Thus the integrability of Mf near
the boundary follows from the integrability of Mf in B. Note also that the set {Mf > 1}
is bounded, because Mf (x) decays to zero as x → ∞. Thus local integrability of Mf
and boundedness of the set {Mf > 1} implies that the function Mf in integrable in
{Mf > 1}. With these remarks we can complete the proof as follows.
Z
∞ >
Mf (x) dx
{Mf >1}
Z ∞
|{Mf > t}| dt
≥
1
Z ∞ −n −1 Z
2 C
≥
|f (x)| dx dt
t
1
{|f |>Ct}
Z max{|f (x)|/C,1} !
Z
dt
−n −1
= 2 C
|f (x)| dx
t
1
B
Z
−n −1
= 2 C
|f (x)| max{0, log(|f (x)|/C)} dx .
B
This easily implies that f ∈ L log L.
Remark 2.36. A more careful analysis leads to the following version of Stein’s theorem.
In an open set Ω ⊂ Rn we define the local maximal function by
o
nZ
|f | : x ∈ Q ⊂ Ω ,
MΩ f (x) = sup
Q
where the supremum is taken over all cubes Q in Ω that contain x.
Theorem 2.37 (Stein). Let Q ⊂ Rn be a cube. Then MQ f ∈ L1 (Q) if and only if
f ∈ L log L(Q). Moreover
Z
Z
Z
|f |
−(n+1)
n+2
5
MQ f ≤
|f | log e +
≤2
MQ f ,
|f |Q
Q
Q
Q
where
Z
|f |Q =
|f | .
Q
2.6. The Minkowski integral inequality. Let Fk ∈ Lp (Y, ν) and denote the variable
by y. If we write F (k, y) instead of Fk (y), then the classical Minkowski inequality k|F1 | +
. . . + |Fm |kp ≤ kF1 kp + . . . + kFm kp can be rewritten as
!1/p
1/p
Z X
p
X Z
p
(2.12)
|F (k, y)| dν(y)
≤
|F (k, y)| dν(y)
.
Y
k
k
Y
Since the sum over k can be regarded as an integral with respect to the counting measure,
the following result is a natural generalization of the Minkowski inequality.
HARMONIC ANALYSIS
17
Theorem 2.38 (Minkowski’s integral inequality). If µ and ν are σ-finite measures on X
and Y respectively and if F : X × Y → R is measurable, then for 1 ≤ p < ∞ we have
Z Z
p
1/p Z Z
1/p
p
|F (x, y)| dµ(x) dν(y)
≤
|F (x, y)| dν(y)
dµ(x) .
Y
X
X
Y
Remark 2.39. It would be natural to try approximate F by simple functions in x and
apply the discrete version of the inequality (2.12). The proof given below uses however, a
very different and much more elegant argument.
Proof. If p = 1 then we actually have equality by Fubini’s theorem. ThusR assume that 1 <
p < ∞. We will use the fact that the Lp (Y, ν) norm of the function y 7→ X |F (x, y)| dµ(x)
equals to the norm of the corresponding functional on Lq (Y, ν), p−1 + q −1 = 1.
Z Z
p
1/p
|F (x, y)| dµ(x) dν(y)
Y
X
Z
Z
|F (x, y)| dµ(x) dν(y)
= sup
h(y)
h∈Lq (ν)
khkq =1
X
Y
h(y)|F (x, y)| dν(y) dµ(x)
Z Z
=
sup
h∈Lq (ν)
khkq =1
≤
X
Y
Z Z
sup
h∈Lq (ν)
khkq =1
X
q
|h(y)| dν(y)
|
{z
}
Y
1/q Z
1/p
|F (x, y)| dν(y)
dµ(x)
p
Y
1
Z Z
=
X
1/p
|F (x, y)| dν(y)
dµ(x) .
p
Y
The proof is complete.
2.7. Convergence of averages. In this section we will show that a large class of averages
of f converge to f . Our approach will be based on the maximal function estimates. This
will lead to far reaching generalizations of the Lebesgue Differentiation Theorem 2.20. Note
that in the proofs of Theorem 2.20 and its generalizations discussed above (Theorems 2.28
and 2.29), maximal function also plays an important role. The results of this section will
be used later in Section 3.3
Since many averages are constructed with the help of convolution, let us recall that the
convolution of measurable functions f and g is defined as
Z
Z
(f ∗ g)(x) =
f (x − y)g(y) dy =
f (y)g(x − y) dy.
Rn
Rn
The equality of the integrals follows from a linear change of variables. This equality actually
means that f ∗ g = g ∗ f . One can also easily check that (f ∗ g) ∗ h = f ∗ (g ∗ h).
Theorem 2.40. Let g ∈ L1 (Rn ). If f ∈ Lp (Rn ), 1 ≤ p < ∞, then f ∗ g ∈ Lp (Rn ) and
(2.13)
kf ∗ gkp ≤ kf kp kgk1 .
18
PIOTR HAJLASZ
If f ∈ C0 (Rn ), then f ∗ g ∈ C0 (Rn ) and
kf ∗ gk∞ ≤ kf k∞ kgk1 .
(2.14)
Proof. Let f ∈ Lp (Rn ), 1 ≤ p < ∞. The inequality can be obtained from the Minkowski
Integral Inequality (Theorem 2.38) as follows
p 1/p
Z Z
kf ∗ gkp =
f (x − y)g(y) dy dx
Rn
Rn
1/p
p
Z Z
dx
≤
|f (x − y)| |g(y)| dy
| {z } |{z}
Rn
Rn
dν
dµ
Z
≤
Z
Rn
|f (x − y)|p dx
1/p
|g(y)| dy
Rn
= kf kp kgk1 .
The proof in the case of f ∈ C0 (Rn ) is left to the reader as an exercise.
Remark 2.41. In the above proof one could use the classical Minkowski inequality instead
of the integral one via the following estimate
Z
Z
|f (x − y)| |g(y)| dy =
|f (x − y)| |g(y)|1/p |g(y)|1/q dy
Rn
Rn
Z
p
|g(y)| dy
|f (x − y)| |g(y)| dy
≤
Rn
1/q
1/p Z
.
Rn
We leave details to the reader.
Theorem 2.40 is a special case of a more general Young’s inequality. We leave the proof
as an exercise.
Theorem 2.42 (Young’s inequality). If 1 ≤ p, q, r ≤ ∞ and q −1 = p−1 + r−1 − 1, then
kf ∗ gkq ≤ kf kp kgkr .
For ϕ ∈ L1 (Rn ) and t > 0 we define
x
ϕt (x) = t−n ϕ
.
t
Note that
R
Rn
ϕt =
R
Rn
ϕ.
Example 2.43. If ϕ = ωn−1 χB(0,1) , then ϕt = (ωn tn )−1 χB(0,t) so
Z
Z
1
f (x − y) dy =
f (y) dy.
f ∗ ϕt (x) =
ωn tn B(0,t)
B(x,t)
Thus7 in that case f ∗ ϕt → f a.e. and in L1 , provided f ∈ L1 . Note also that
(2.15)
sup(|f | ∗ ϕt )(x) = Mf (x).
t>0
7See
Lemma 2.19 and Theorem 2.20.
HARMONIC ANALYSIS
19
Actually this identity8 plays a crucial role in the proof of the pointwise convergence f ∗ϕt →
f a.e. In what follows we will study convergence of f ∗ ϕt for more general functions ϕ.
Not surprisingly we will be looking for conditions that will guarantee an estimate similar
to (2.15).9
Theorem 2.44. Let ϕ ∈ L1 (Rn ) and let ϕt (x) = t−n ϕ(x/t) for t > 0. If f ∈ Lp (Rn ),
1 ≤ p < ∞ or f ∈ C0 (Rn ) and p = ∞, then
Z
(2.16)
lim+ kf ∗ ϕt − af kp = 0, where a =
ϕ(x) dx.
t→0
Rn
Remark 2.45. The two most interesting cases are when a = 1 and when a = 0.
Remark 2.46. If f ∈ C0 (Rn ), then f ∗ ϕt ∈ C0 (Rn )10 and (2.16) with p = ∞ means that
f ∗ ϕt converges to af uniformly.
Remark 2.47. If a = 1, then (2.16) means that f ∗ ϕt → f in Lp so the operators
f 7→ f ∗ ϕt converge pointwise, i.e. for every f , to the identity operator f 7→ f . For that
reason the family of functions {ϕt : t > 0} is often called an approximation of identity
(provided a = 1).
Proof. We will prove the result when f ∈ Lp (Rn ), 1 ≤ p < ∞, leaving the case of f ∈
C0 (Rn ) to the reader. We have
Z
(f (x − y) − f (x)) ϕt (y) dy .
|f ∗ ϕt (x) − af (x)| = Rn
Given δ > 0, the Minkowski Integral Inequality yields
Z Z
1/p
kf ∗ ϕt − af kp ≤
|f (x − y) − f (x)|p dx
|ϕt (y)| dy
Rn
Rn
Z
=
kτ−y f − f kp |ϕt (y)| dy
Rn
Z
kτ−ty f − f kp |ϕ(y)| dy,
=
Rn
where the last equality follows from a simple linear change of variables.
Note that kτ−ty f − f kp |ϕ(y)| ≤ 2kf kp |ϕ(y) ∈ L1 and for every y ∈ Rn , kτ−ty f − f kp → 0
as t → 0 (Lemma 2.18) so
Z
kf ∗ ϕt − af kp ≤
kτ−ty f − f kp |ϕ(y)| dy → 0 as t → 0
Rn
by the Dominated Convergence Theorem.
Under assumptions of Theorem 2.44, if f ∈ Lp (Rn ), 1 ≤ p < ∞, there is a sequence
ti → 0 such that f ∗ ϕti → af a.e. as i → ∞, but a more challenging question is to
find conditions that would guarantee f ∗ ϕt → af a.e. as t → 0. That would be a true
generalization of the Lebesgue Differentiation Theorem.
8Along
with weak type estimates for the maximal function, see Theorem 2.20.
Theorem 2.49.
10See Theorem 2.40.
9See
20
PIOTR HAJLASZ
Definition 2.48. Let ϕ ∈ L1 (Rn ). We say that Ψ is a radially decreasing majorant of ϕ if
(a) Ψ(x) = η(|x|) for some11 η : [0, ∞) → [0, ∞].12
(b) η is decreasing.13.
(c) |ϕ(x)| ≤ Ψ(x) a.e.
Every function ϕ ∈ L1 has the least radially decreasing majorant14
Ψ0 (x) = ess sup |ϕ(y)|.
|y|≥|x|
Thus the existence of an integrable radially decreasing majorant Ψ of ϕ is equivalent to
Ψ0 ∈ L1 .
Theorem 2.49. Suppose that ϕ ∈ L1 (Rn ) has an integrable radially decreasing majorant
Ψ(x) = η(|x|) ∈ L1 (Rn ) .
Then for f ∈ L1loc (Rn ) and all x ∈ Rn
sup(f ∗ ϕt )(x) ≤ kΨk1 Mf (x) .
(2.17)
t>0
R
If in addition f ∈ Lp (Rn ), 1 ≤ p < ∞, and Rn ϕ(x) dx = a, then
f ∗ ϕt → af
(2.18)
a.e. as t → 0.
Proof. In order to prove (2.17) it suffices to prove that for a radially decreasing and integrable function Ψ we have
|f | ∗ Ψ(x) ≤ kΨk1 Mf (x).
(2.19)
Indeed, since Ψt is also radially decreasing and integrable, applying (2.19) to Ψt will give
|f ∗ ϕt (x)| ≤ |f | ∗ Ψt (x) ≤ kΨt k1 Mf (x) = kΨk1 Mf (x).
Thus it remains to prove (2.19).
For a positive integer k let
k
k
k·2
k·2
X
X
1
1
χ
(x)
=
χ (x) .
sk (x) =
k
{Ψ≥i/2 }
k
k Bi
2
2
i=1
i=1
Since Ψ is radially symmetric the sets
n
io
Bi = x : Ψ(x) ≥ k
2
are balls15 centered at zero.
11Functions
of this form, i.e. functions constant on spheres S n−1 (0, r) are called radially symmetric.
may happen that η(t) = +∞ for all t.
13i.e. non-increasing
14However, it may happen that Ψ = +∞ everywhere.
0
15Open or closed – find an example such that for some t the balls are open while for other values of t
they are closed.
12It
HARMONIC ANALYSIS
21
The sequence {sk } is increasing and convergent to Ψ almost everywhere. Indeed, for
x 6= 0, Ψ(x) < ∞16 so Ψ(x) < k0 , for some k0 ∈ N. Then for k ≥ k0
`
`+1
≤ Ψ(x) < k , for some ` = 0, 1, 2, . . . k · 2k − 1.
k
2
2
k
Since the inequality Ψ ≥ i/2 is satisfied for i = 1, 2, . . . , `, the definition of the function
sk gives that sk (x) = `/2k so
1
Ψ(x) − k < sk (x) ≤ Ψ(x).
2
That means sk (x) → Ψ(x) for all all x 6= 0. Observe also that the sequence {sk } is
increasing, sk ≤ sk+1 . Indeed, if sk (x) = `/2k , then Ψ(x) ≥ `/2k = (2`)/(2k+1 ) so the
definition of sk+1 gives
1
`
sk+1 (x) ≥ k+1 · 2` = k = sk (x).
2
2
Thus if we can prove
|f | ∗ sk (x) ≤ kΨk1 Mf (x),
(2.20)
inequality (2.19) will follow from the Monotone Convergence Theorem.
We have
k
|f | ∗ sk (x) =
k·2
X
|Bi |
i=1
2k
|f | ∗
χBi
(x).
|Bi |
17
Since
χBi
|f | ∗
(x) =
|Bi |
k
Z
|f (x − y)| dy ≤ Mf (x) and
Bi
k·2
X
|Bi |
i=1
2k
= ksk k1 ≤ kΨk1 ,
inequality (2.20) follows. The proof of (2.17) is complete.
It remains now to prove convergence (2.18).
If p = 1, then
(2.21)
Ωf (x) := lim sup f ∗ ϕt − lim inf
f ∗ ϕt ≤ 2kΨk1 Mf (x)
+
t→0+
t→0
and almost the same argument as in the proof of Theorem 2.20 yields that18 Ωf = 0 a.e.
so f ∗ ϕt converges a.e. to a measurable function. Since f ∗ ϕt → af in L1 (Theorem 2.44),
(2.18) follows.
The case 1 < p < ∞ uses a similar argument. If f ∈ Lp (Rn ), 1 < p < ∞, then
Chebyschev’s inequality (2.2) and Theorem 2.2(b) yield
Z
1
p
p
|{x : Mf (x) > t}| = |{x : |Mf (x)| > t }| ≤ p
|Mf (x)|p dx
t Rn
Z
C(n, p)
≤
|f (x)|p dx.
tp
n
R
16Because
Ψ ∈ L1 .
the balls Bi are centered at the origin.
18We will actually repeat this argument in the case of 1 < p < ∞.
17Because
22
PIOTR HAJLASZ
Hence for ε > 0, inequality (2.21) yields
C
|{x : Ωf (x) > ε}| ≤ p
ε
Z
|f (x)|p dx,
Rn
where the constant C depends on n, p and kΨk1 . If h ∈ C0 (Rn ), then Ωh=0.19 Taking
h ∈ C0 (Rn ) such that kf − hkpp ≤ εp+1 we conclude that
|{x : Ωf (x) > ε}| = |{x : Ω(f − h)(x) > ε}| < Cε
so Ωf = 0 a.e. and exactly as in the case of p = 1, (2.18) follows.
19Because
h ∗ ϕt converges uniformly (and hence pointwise) to h.
HARMONIC ANALYSIS
23
3. The Fourier transform
3.1. Fourier transform.
Definition 3.1. For f ∈ L1 (Rn ) we define the Fourier transform as
Z
n
X
f (x)e−2πix·ξ dx where x · ξ =
xj ξj .
F(f )(x) = fˆ(ξ) =
Rn
j=1
Theorem 3.2. The Fourier transform has the following properties
(a)
∧
: L1 (Rn ) → C0 (Rn ) is a bounded linear operator with kfˆk∞ ≤ kf k1 .
(b) (f ∗ g)ˆ = fˆĝ for f, g ∈ L1 (Rn ).
(c) If f, g ∈ L1 (Rn ), then fˆg, f ĝ ∈ L1 (Rn ) and
Z
Z
f (x)ĝ(x) dx.
fˆ(x)g(x) dx =
Rn
Rn
(d) If τh f (x) = f (x + h), then
(τh f )ˆ(ξ) = fˆ(ξ)e2πih·ξ
and
(e) If ft (x) = t−n f (x/t), then
(ft )ˆ(ξ) = fˆ(tξ) and
(f (x)e2πih·x )ˆ(ξ) = fˆ(ξ − h).
(f (tx))ˆ(ξ) = (fˆ)t (ξ) .
(f) If ρ ∈ O(n) is an orthogonal transformation, then
(f (ρ ·))ˆ(ξ) = fˆ(ρξ).
Proof. (a) Clearly,
Indeed,
∧
: L1 (Rn ) → L∞ (Rn ) is a bounded linear mapping with kfˆk∞ ≤ kf k1 .
|fˆ(ξ)| ≤
Z
f (x)e−2πix·ξ dx = kf k1 .
Rn
The Dominated Convergence Theorem implies that the function fˆ is continuous. It remains
to prove that fˆ(ξ) → 0 as |ξ| → ∞. Let ξ 6= 0. Since eπi = −1 we have
Z
Z
−2πix·ξ
fˆ(ξ) =
f (x)e
dx = −
f (x)e−2πix·ξ eπi dx
Rn
Rn
Z
ξ = −
f (x) exp − 2πi x −
· ξ dx
2|ξ|2
Rn
24
PIOTR HAJLASZ
Z
f x+
= −
Rn
Hence
Z
1
fˆ(ξ) =
2
and thus Lemma 2.18 yields
ξ −2πix·ξ
e
dx .
2|ξ|2
f (x) − f x +
Rn
1
f − τ ξ 2
2|ξ|
2
|fˆ(ξ)| ≤
ξ −2πix·ξ
e
dx
2|ξ|2
f → 0 as |ξ| → ∞.
1
(b)
Z
Z
(f ∗ g)ˆ(ξ) =
n
f (x − y)g(y) dy e−2πix·ξ dx
n
ZR ZR
=
Rn
f (x − y)e−2πi(x−y)·ξ dx g(y)e−2πiy·ξ dy
Rn
= fˆ(ξ)ĝ(ξ) .
(c) fˆg, f ĝ ∈ L1 , because the functions fˆ, ĝ are bounded and the equality of the integrals
easily follows from the Fubini theorem.
(d)
Z
Z
f (x)e−2πi(x−h)·ξ dx
f (x + h)e
dx =
n
R
Z
= e2πih·ξ
f (x)e−2πix·ξ dx = e2πih·ξ fˆ(ξ) .
−2πix·ξ
(τh f )ˆ(ξ) =
Rn
Rn
The second equality follows from a similar argument.
(e)
Z
−n
t
(ft )ˆ(ξ) =
ZR
n
f
x
e
t
−2πix·ξ
Z
f (y)e−2πi(ty)·ξ dy
dx =
Rn
f (y)e−2πiy·(tξ) dy = fˆ(tξ) .
=
Rn
The second formula follows from the first one if we replace t by t−1 .
(f)20
Z
(f (ρ ·))ˆ(ξ) =
−2πix·ξ
f (ρx)e
Rn
Z
=
Z
dx =
f (y)e−2πi(ρ
−1 y)·ξ
dy
Rn
f (y)e−2πiy·(ρξ) dy = fˆ(ρξ) .
Rn
Equality (ρ−1 y) · ξ = y · (ρξ) follows from the fact that the mapping x 7→ ρx is an
isometry.
20This
property is in some sense obvious: the Fourier transform does not depend on the choice of the
coordinate system since it depends on the scalar product x · ξ. Thus the Fourier transform should stay the
same if we rotate the coordinate system by ρ.
HARMONIC ANALYSIS
25
Theorem 3.3. Suppose f ∈ L1 (Rn ) and xk f (x) ∈ L1 (Rn ), where xk is the k-th coordinate
function. Then fˆ is differentiable with respect to ξk and
∂ fˆ
(−2πixk f (x))ˆ =
(ξ) .
∂ξk
Proof. Let ek be the unit vector along the k-th coordinate. Then the second part of Theorem 3.2(d) gives
−2πi(hek )·x
∧
fˆ(ξ + hek ) − fˆ(ξ)
e
−1
=
f (x) (ξ) → (−2πixk f (x))ˆ(ξ) .
h
h
The convergence follows from the continuity of the Fourier transform in L1 .
Definition 3.4. We say that f ∈ L1loc (Rn ) is Lp -differentiable with respect to xk if there is
g ∈ Lp (Rn ) such that
p
Z f (x + hek ) − f (x)
dx → 0 as h → 0.
−
g(x)
h
Rn
The function g is called the Lp -partial derivative of f with respect to xk and is denoted by
g = ∂f /∂xk .
Theorem 3.5. If f ∈ L1 (Rn ) is L1 -differentiable with respect to xk , then
∧
∂f
= 2πiξk fˆ(ξ) .
∂xk
Proof. The first part of Theorem 3.2(d) and the L1 -differentiability of f give
∧
∧
2πi(hek )·ξ
∂f
e
−
1
f
(x
+
he
)
−
f
(x)
∂f
k
=
− fˆ(ξ)
−
→0
∂xk
h
∂xk
h
as h → 0, so
∧
e2πi(hek )·ξ − 1
∂f
= 2πiξk fˆ(ξ) .
(ξ) = lim fˆ(ξ)
h→0
∂xk
h
The proof is complete.
Remark 3.6. With each polynomial
P (x) =
X
aα x α
|α|≤m
of variables x1 , . . . , xn we associate a differential operator
X
X
∂ |α|
P (D) =
aα D α =
aα α 1
.
∂x1 . . . ∂xαnn
|α|≤m
|α|≤m
Then under suitable assumptions Theorems 3.3 and 3.5 have the following higher order
generalizations
(3.1)
P (D)fˆ(ξ) = (P (−2πix)f (x))ˆ(ξ), (P (D)f )ˆ(ξ) = P (2πiξ)fˆ(ξ).
The next result is our first (and actually one of the most important) example of the
Fourier transform.
26
PIOTR HAJLASZ
2
2
Theorem 3.7. If f (x) = e−π|x| , then fˆ(ξ) = e−π|ξ| .
Proof. First observe that it suffices to prove the result in dimension 1. Indeed, assuming it
for n = 1, the case of general n follows from Fubini’s theorem
Z
n
n Z ∞
Y
Y
2
2
−π|x|2 −2πix·ξ
−πx2k −2πixk ξk
ˆ
dxk =
e
e
dx =
e
e
e−πξk = e−π|ξ| .
f (ξ) =
Rn
−∞
k=1
k=1
In the case n = 1 we will show two different proofs. The first one will use the contour
integration while the second one will use differential equations.
2
Thus in the remaining part of the proof we will assume that f (x) = e−πx is a function
of one variable.
2
Proof by contour integration. The function e−πz is holomorphic and hence its integral
along the following curve equals zero.
Letting R → ∞ we obtain
Z
∞
−πx2
e
Z
∞
dx =
−∞
2
e−π(x+iξ) dx .
−∞
The left hand side equals 1, so
∞
Z
2
2
e−πx e−2πixξ eπξ dx .
1=
−∞
Hence
−πξ 2
e
Z
∞
=
2
e−πx e−2πixξ dx = fˆ(ξ) .
−∞
The proof is complete.
Proof by differential equations. We have
f 0 (x) = −2πxf (x),
−i(−2πixf (x)) = f 0 (x)
so Theorems 3.3 and 3.5 yield
∧
−i − 2πixf (x) (ξ) = fb0 (ξ) ,
| {z }
{z
}
|
(fˆ)0 (ξ)
2πiξ fˆ(ξ)
(fˆ)0 (ξ) = −2πξ fˆ(ξ).
HARMONIC ANALYSIS
27
Solving this differential equation for the unknown function ξ 7→ fˆ(ξ) yields
Z ∞
2
−πξ 2
−πξ 2
ˆ
ˆ
ˆ
f (ξ) = f (0)e
=e
, because f (0) =
e−πx dx = 1.
−∞
3.2. Measures of finite total variation. Let us first recall basic facts regarding measures of finite total variation.
If µ is a complex (Borel) measure on Rn , then there is a unique positive measure |µ|
called the total variation of measure µ and a Borel function h : Rn → C, |h(x)| = 1 for all
x ∈ Rn such that
Z
µ(E) =
h(x) d|µ|(x) for all Borel sets E ⊂ Rn .
E
n
The space B(R ) of complex measures of finite total variation kµk := |µ|(Rn ) is a Banach
space with the norm k · k.
Note that f ∈ L1 (Rn ) defines a measure of finite total variation by
Z
µ(E) =
f (x) dx.
E
In that case
Z
|µ|(E) =
Z
|f (x)| dx so kµk =
|f (x)| dx = kf k1 < ∞.
Rn
E
This proves that L1 (Rn ) is a closed subspace of B(Rn ).
In a special case when µ is a real valued measure, called signed measure, we have h :
Rn → {±1} so if we define h+ = hχ{h=1} and h− = hχ{h=−1} , then
Z
±
h± (x) d|µ|(x)
µ (E) =
E
are positive measures such that
(3.2)
µ = µ+ − µ−
and
|µ| = µ+ + µ− .
The representation (3.2) is called the Hahn decomposition of µ. Note that the measures
µ+ and µ− are supported on disjoint sets.
Theorem 3.8 (Riesz representation theorem). The dual space to C0 (Rn ) is isometrically isomorphic to the space of measures of finite total variation. More precisely, if
Φ ∈ (C0 (Rn ))∗ , then there is a unique measure µ of finite total variation such that
Z
Φ(f ) =
f dµ for f ∈ C0 (Rn ).
Rn
Moreover kΦk = kµk = |µ|(Rn ).
28
PIOTR HAJLASZ
This result allows us to define convolution of measures.
If f, g ∈ L1 (Rn ), then f ∗ g ∈ L1 (Rn ) ⊂ B(Rn ) and hence it acts as a functional on
C0 (Rn ) by the formula
Z
Z
Z
Φ(h) =
h(x)(f ∗ g)(x) dx =
h(x)
f (x − y)g(y) dy dx
n
Rn
Rn
ZR Z
=
h(x)f (x − y) dx g(y) dy
Rn
Rn
Z Z
=
h(x + y)f (x)g(y) dx dy .
Rn
Rn
This suggests how to define convolution of measures.
If µ1 , µ2 ∈ B(Rn ), then
Z
Z
Φ(h) =
h(x + y) dµ1 (x) dµ2 (y)
Rn
Rn
defines a functional on C0 (Rn ) and hence there is a unique measure µ ∈ B(Rn ) such that
Z Z
Z
h(x + y) dµ1 (x) dµ2 (y) =
h(x) dµ(x) for all h ∈ C0 (Rn ).
Rn
Rn
Rn
Definition 3.9. We denote the measure µ by
µ = µ1 ∗ µ2
and call it convolution of measures.
Clearly
µ1 ∗ µ2 = µ2 ∗ µ1
If dµ1 = f dx, dµ2 = g dx, then
and kµ1 ∗ µ2 k ≤ kµ1 k kµ2 k.
d(µ1 ∗ µ2 ) = (f ∗ g) dx,
so the convolution of measures extends the notion of convolution of functions. If dµ1 = f dx
and µ ∈ B(Rn ), then
Z Z
Z Z
h(x + y) dµ1 (x) dµ(y) =
h(x + y)f (x) dx dµ(y)
Rn Rn
Rn Rn
Z Z
=
h(x)f (x − y) dx dµ(y)
Rn Rn
Z
Z
=
h(x)
f (x − y) dµ(y) dx.
Rn
Rn
Thus µ1 ∗ µ can be identified with a function
Z
x 7→
f (x − y) dµ(y) ∈ L1 (Rn ),
Rn
so we can write
Z
f ∗µ=
f (x − y) dµ(y),
Rn
kf ∗ µk1 ≤ kf k1 kµk.
HARMONIC ANALYSIS
29
Theorem 3.10. If 1 ≤ p < ∞, f ∈ Lp (Rn ) and µ ∈ B(Rn ), then
kf ∗ µkp ≤ kf kp kµk.
Proof is almost the same as that for Theorem 2.40 as we leave it to the reader.
Definition 3.11. The Fourier transform of a measure of finite total variation µ ∈ B(Rn )
is defined by
Z
µ̂(x) =
e−2πix·ξ dµ.
Rn
Proposition 3.12. If µ ∈ B(Rn ), then µ̂ is a bounded and continuous function such that
kµ̂k∞ ≤ kµk.
Proof. Boundedness of µ̂ is obvious and continuity follows from the Dominated Convergence Theorem.
Proposition 3.13. If µ1 , µ2 ∈ B(Rn ), then µ\
1 ∗ µ2 = µ̂1 · µ̂2 .
We leave the proof as an exercise.
Remark 3.14. Convolution of measures and the Fourier transform of measures play a
fundamental role in probability.
3.3. Summability methods. The answer to an important question of how to reconstruct
a function f from its Fourier transform fˆ is provided by the inversion formula21
Z
(3.3)
f (x) =
fˆ(ξ) e2πix·ξ dξ .
Rn
It seems that this formula requires both f and fˆ to be integrable. Integrability of f is
needed for the existence of fˆ and integrability of fˆ is needed for the integral in (3.3) to
make sense. We will actually prove in Corollary 3.26 that if f, fˆ ∈ L1 , then (3.3) is true a.e.
There is, however, a problem here. Equality (3.3) means that f (x) is the Fourier transform
of fˆ ∈ L1 evaluated at −x. Since the Fourier transform of an integrable function fˆ ∈ L1
belongs to C0 (Rn ), it follows that f ∈ C0 (Rn ). That means, if f ∈ L1 \ C0 , then fˆ cannot
be integrable so the right hand side of (3.3) does not make much sense. Is there any chance
to give meaning to the integral at (3.3), beyond the class of integrable Fourier transforms
so that the equality at (3.3) would be true for all f ∈ L1 ? The answer is in the positive
but we need to interpret the integral at (3.3) using suitable summability methods.22
21Compare
it with the formula for Fourier series:
Z 1
ˆ
f (n) =
f (x)e−2πixn dx,
0
22An
f (x) =
X
fˆ(n)e2πixn .
n∈Z
example of a summability method for sequences is given by the so called Cesàro method: with
each sequence {an } we associate the limit of (a1 + . . . + an )/n. This extends the notion of limit far beyond
the class of convergent sequences.
30
PIOTR HAJLASZ
If Φ ∈ C0 (Rn ) is such that Φ(0) = 1, then for t > 0 we define the Φ-mean of a function
f as
Z
MΦ,t f = Mt f =
f (x)Φ(tx) dx.
n
R
R
If f ∈ L1 (Rn ), then MΦ,t f → Rn f (x) dx as t → 0+ , but the limit of Mt f might also exist
for some non-integrable functions f .
R
Definition 3.15. We say that Rn f is Φ-summable to ` ∈ R if limt→0+ MΦ,t f = `.
2
In the case of Φ(x) = e−|x| and Φ(x) = e−|x| we obtain the so called the Abel and the
Gauss summability methods. More precisely we have
Definition 3.16. For each t > 0 the Abel mean and the Gauss mean of a function f are
Z
Z
2
−t|x|
At (f ) =
f (x)e
dx and Gt (f ) =
f (x) e−t|x| dx .
Rn
Rn
R
23
We say that Rn f (x) dx is Abel summable (Gauss summable) to ` ∈ R if limt→0 At (f ) = `
(limt→0 Gt (f ) = `).
2
2
2
Remark 3.17. If Φ(x) = e−|x| , then Φ(tx) = e−t |x| so MΦ,t f = Gt2 (f ). This however,
does not cause any problems when you compare notions of the Φ-summability and the
Gauss summability because limt→0+ MΦ,t f = limt→o Gt2 (f ).
Remark 3.18. Clearly, in the case of integrable functions, the Abel and the Gauss methods
give the usual integral. However, the means At (f ) and Gt (f ) areR finite whenever f is
bounded so for some bounded, but non-integrable f , the integral Rn f (x) dx might still
be Abel or Gauss summable to a finite value. For example if f ∈ L1 (Rn ) \ C0 (Rn ), the
the function ξ 7→ fˆ(ξ)e2πix·ξ is in C0 (Rn ) but it is not integrable.
However, as we will see
R
n
ˆ
(Theorems 3.23 and 3.30) for almost all x ∈ R , the integral Rn f (ξ)e2πix·ξ dξ is both Abel
and Gauss summable to f (x). This gives some meaning of the inversion formula (3.3) for
generic f ∈ L1 (Rn ).
We want to find a reasonably large class of Φ-means (including the Abel and the Gauss
means) such that
Z
fˆ(ξ)e2πix·ξ dξ
Rn
is Φ-summable to f (x) i.e.
Z
Mt (x) =
fˆ(ξ)e2πix·ξ Φ(tξ) dξ → f (x) as t → 0+
Rn
and we can consider both, convergence Mt → f almost everywhere and in L1 (with respect
to variable x). The main result reads as follows
R
Theorem 3.19. Let Φ ∈ L1 (Rn ) be such that ϕ = Φ̂ ∈ L1 (Rn ) and Rn ϕ(x) dx = 1. For
f ∈ L1 (Rn ) let
Z
Mt (x) =
fˆ(ξ)e2πix·ξ Φ(tξ) dξ
Rn
23Exercise.
Prove that if lima→∞
Ra
0
f (x) dx = `, then At =
R∞
0
f (x)e−tx dx converges to ` as t → 0.
HARMONIC ANALYSIS
31
be the Φ-mean of the integral (3.3). Then
Mt → f in L1 as t → 0.
(3.4)
If in addition ϕ = Φ̂ has an integrable radially decreasing majorant, then
Mt → f a.e. as t → 0.
(3.5)
Proof. The lemma below is a link between the Φ-mean and the approximation by convolution discussed in Section 2.7.
Lemma 3.20. If f, Φ ∈ L1 (Rn ) and ϕ = Φ̂, then
Z
Z
2πix·ξ
ˆ
f (ξ)e
Φ(tξ) dξ =
f (x − y)ϕt (−y) dy = f ∗ ϕ̃t (x),
Rn
where ϕ̃(y) = ϕ(−y).
Rn
Proof. For h ∈ Rn define
g (h) (x) = e2πix·h Φ(tx) ∈ L1 (Rn ) .
Recall that24
(w(x)e2πih·x )∧ (ξ) = ŵ(ξ − h) and (w(tx))∧ (ξ) = (ŵ)t (ξ).
Hence
(g (h) )∧ (ξ) = (Φ̂)t (ξ − h) = ϕt (ξ − h)
so replacing h by x we get
(g (x) )∧ (ξ) = ϕt (ξ − x).
We have
Z
fˆ(ξ)e2πix·ξ Φ(tξ) dξ =
Z
fˆ(ξ)g (x) (ξ) dξ =
Rn
Rn
Z
=
Z
f (ξ)(g (x) )∧ (ξ) dξ
Z
f (ξ)ϕt (ξ − x) dξ =
f (x − y)ϕt (−y) dx.
Rn
Rn
Rn
We used here Theorem 3.2(c) and in the last equality the change of variables −y = ξ−x. Now (3.4) follows directly from Theorem 2.44 and (3.5) follows from Theorem 2.49. Remark 3.21. Theorem 3.19 is not easy to apply since given Φ ∈ L1 , we have to compute
2
Φ̂ and show its integrability. In the case of the Gauss mean Φ(x) = e−|x| it can be easily
done with the help of Theorem 3.7, but the case of the Abel mean Φ(x) = e−|x| is much
harder due to difficulty of computing the Fourier transform of Φ(x) = e−|x| .
In Theorems 3.23 and 3.30 we will investigate the case of the the Gauss and the Abel
and methods.
Theorem 3.22. Let f (x) = e−4π
2 t|x|2
, t > 0. Then
(a) W (x, t) := fˆ(x) = (4πt)−n/2 e−|x| /(4t) .
(b) The function W has the following scaling property with respect to t: if ϕ(x) =
W (x, 1), then W (x, t) = ϕt1/2 (x).
2
24see
Theorem 3.2.
32
PIOTR HAJLASZ
(c)
Z
W (x, t) dx = 1 for all t > 0.
Rn
Proof. We can write
f (x) = e−4π
2 t|x|2
= u((4πt)1/2 x),
2
where u(x) = e−π|x| .
2
Since (ψ(tx))∧ (ξ) = (ψ̂)t (ξ) and û(ξ) = u(ξ) = e−π|ξ| we obtain25
2
W (ξ, t) = fˆ(ξ) = (û)(4πt)1/2 (ξ) = u(4πt)1/2 (ξ) = (4πt)−n/2 e−|ξ| /(4t)
which is (a). In particular
ϕ(ξ) = W (ξ, 1) = u(4π)1/2 (ξ).
Since (ψt1 )t2 (x) = ψt1 t2 (x) we get
W (ξ, t) = u(4πt)1/2 (ξ) = u(4π)1/2 t1/2 (ξ) = ϕt1/2 (ξ)
R
R
which is (b). Finally equality Rn ψt1 = Rn ψt2 , t1 , t2 > 0 along with W (ξ, t) = ϕt1/2 (ξ)
yields (c):
Z
Z
Z
1
2
W (ξ, t) dξ =
W ξ,
dξ =
e−π|ξ| dξ = 1.
4π
Rn
Rn
Rn
As an application we obtain
Theorem 3.23 (The Gauss-Weierstrass summability method). If f ∈ L1 (Rn ), then
Z
Z
2πix·ξ −4π 2 t|ξ|2
ˆ
f (ξ)e
e
f (y)W (x − y, t) dy → f as t → 0+
(3.6)
dξ =
Rn
Rn
both in L1 (Rn ) and almost everywhere.
2
2
Proof. First we will prove equality. Let Φ(x) = e−4π |x| . Using notation from Theorem 3.22
Φ̂(x) = W (x, 1) = ϕ(x). Since ϕ(x) = ϕ(−x), ϕ̃t1/2 (x) = ϕt1/2 (x) = W (x, t). Thus
Lemma 3.20 yields
Z
Z
2πix·ξ −4π 2 t|ξ|2
ˆ
f (ξ)e
e
dξ =
fˆ(ξ)e2πix·ξ Φ(t1.2 ξ) dξ
n
n
R
R
Z
W (x − y, t)f (y) dy.
= f ∗ ϕ̃t1/2 (x) =
Rn
Theorem 3.22 shows that the function Φ satisfies the assumptions of Theorem 3.19 so both
convergences in L1 and almost everywhere follow.
Remark 3.24. The function W (x, t) = ϕt1/2 (x) satisfies the assumptions of Theorem 2.49
with a = 1 so for f ∈ Lp , 1 ≤ p < ∞
Z
(3.7)
f (y)W (x − y, t) dy → f a.e. and in Lp as t → 0+ .
Rn
25The
function W is defined as a Fourier transform so it will be more convenient for us to prove (a)-(c)
with the variable x replaced by ξ.
HARMONIC ANALYSIS
33
Remark 3.25. W (x, t) is called the Gauss-Weierstrass kernel. Under suitable assumptions
about f , differentiation under the sign of the integral shows that the function
Z
Z
|x−y|2
1
− 4t
w(x, t) =
W (x − y, t)f (y) dy =
e
f (y) dy
(4πt)n/2 Rn
Rn
= ∆x w in the interior of Rn+1
satisfies the heat equation ∂w
+ . As we showed, the function
∂t
w(x, t) converges to f (x) as t → 0, so w is a solution to
∂w
= ∆x w
on Rn+1
+ ,
∂t
w(x, 0) = f (x), x ∈ Rn .
Corollary 3.26. If both f and fˆ are integrable26, then
Z
(3.8)
f (x) =
fˆ(ξ)e2πix·ξ dξ a.e.
Rn
Proof. Since fˆ ∈ L1 , the left hand side of (3.6) converges to
follows from Theorem 3.23.
R
Rn
fˆ(ξ)e2πix·ξ dξ so the result
The above inversion formula motivates the following definitions.
Definition 3.27. The inverse Fourier transform is defined by
Z
−1
ˇ
f (ξ)e2πix·ξ dξ.
f (x) = F (f )(x) =
Rn
Corollary 3.28. If f1 , f2 ∈ L1 (Rn ) and fˆ1 = fˆ2 on Rn , then f1 = f2 a.e.
Proof. Let f = f1 − f2 . Then f ∈ L1 and fˆ = 0 ∈ L1 so (3.8) yields that f = 0 a.e.
The following result provides another example of a function that satisfies the assumptions
of Theorem 3.19.
Theorem 3.29. Let f (x) = e−2π|x|t , t > 0. Then
(a)
P (x, t) := fˆ(x) = cn
where
t
,
+ |x|2 )(n+1)/2
n+1
(t2
Γ 2
.
π (n+1)/2
(b) The function P has the following scaling property with respect to t: if ϕ(x) =
P (x, 1), then P (x, t) = ϕt (x).
(c)
Z
cn =
P (x, t) dx = 1 for all t > 0.
Rn
By the same arguments as before we obtain.
assumption about integrability of fˆ is very strong. As already observed, the equality (3.8) implies
that f equals a.e. to a function in C0 (Rn ).
26The
34
PIOTR HAJLASZ
Theorem 3.30 (The Abel summability method). If f ∈ L1 (Rn ), then
Z
Z
2πix·ξ −2π|ξ|t
ˆ
f (ξ)e
e
dξ =
f (y)P (x − y, t) dy → f as t → 0+
1
Rn
n
Rn
both in L (R ) and almost everywhere.
Remark 3.31. The function P (x, t) = ϕt (x) satisfies the assumptions of Theorem 2.49
with a = 1 so for f ∈ Lp , 1 ≤ p < ∞
Z
f (y)P (x − y, t) dy → f a.e. and in Lp as t → 0+ .
Rn
Remark 3.32. P (x, t) is called the Poisson kernel. Under suitable assumptions about f ,
differentiation under the sign of the integral shows that the function
Z
P (x − y, t)f (y) dy
u(x, t) =
Rn
satisfies the Laplace equation ∆(x,t) u = 0 in the interior of Rn+1
+ . As we showed u(x, t)
converges to f as t → 0+ so u(x, t) is a solution to the Dirichlet problem in the half-space
∆(x,t) u = 0
on Rn+1
+ ,
u(x, 0) = f (x), x ∈ Rn
The proof of Theorem 3.29 is substantially more difficult than that of Theorem 3.22.
Since the formula for the Fourier transform involves the Γ function we need to recall its
basic properties.
Definition 3.33. For 0 < x < ∞ we define
Z ∞
Γ(x) =
tx−1 e−t dt .
0
Theorem 3.34.
(a) Γ(x + 1) = xΓ(x) for all 0 < x < ∞.
(b) Γ(n + 1) =
√n! for all n = 0, 1, 2, 3, . . .
(c) Γ(1/2) = π.
Proof. (a) follows from the integration by parts. Since Γ(1) = 1, (b) follows from (a) by
induction. The substitution t = s2 gives
Z ∞
Z ∞
Z ∞
2
x−1 −t
2(x−1) −s2
Γ(x) =
t e dt =
s
e 2s ds = 2
s2x−1 e−s ds
0
and hence
0
0
Z ∞
√
1
2
Γ
=2
e−s ds = π .
2
0
The formula Γ(x) = Γ(x + 1)/x allows us to define Γ(x) for all negative non-integer
values of x. For example
√
3 Γ(−1/2)
Γ(1/2)
4 π
Γ −
=
=
=
.
2
−3/2
(−3/2)(−1/2)
3
HARMONIC ANALYSIS
35
Definition 3.35. For x ∈ (−∞, 0) \ {−1, −2, −3, . . .}, the function Γ(x) is defined by
Γ(x) =
Γ(x + k)
,
x(x + 1) · . . . · (x + k − 1)
where k is a positive integer such that x + k ∈ (0, 1).
Lemma 3.36.
Z
√
π/2
n
sin θ dθ =
0
πΓ
nΓ
n+1
2
n
2
for n = 1, 2, 3, . . .
Proof. Denote the left hand side by an and the right hand side by bn . Easy one time
integration by parts27 shows that
n+1
an .
an+2 = (n + 1)(an − an+2 ), an+2 =
n+2
Also elementary properties of the Γ function show that
n+1
bn+2 =
bn
n+2
and now it is enough to observe that a1 = 1 = b1 = 1, a2 = π/4 = b2 .
Lemma 3.37.
(a) The volume of the unit ball in Rn equals
π n/2
2π n/2
.
=
ωn =
nΓ(n/2)
Γ n2 + 1
(3.9)
(b) The (n − 1)-dimensional measure of the unit sphere in Rn equals nωn
Proof.
It follows from the picture and the Fubini theorem that the volume of the upper half of
the ball equals
Z 1
1
ωn−1 r(h)n−1 dh .
ωn =
2
0
The substitution
h = 1 − cos θ, dh = sin θ dθ, r(h) = sin θ
27Use
sinn+2 Θ = (− cos θ)0 sinn+1 Θ.
36
PIOTR HAJLASZ
and Lemma 3.36 give
1
ωn = ωn−1
2
so
π/2
Z
sinn θ dθ = ωn−1
0
π 1/2 Γ
nΓ
n+1
2
n
2
,
2π 1/2 Γ n+1
2
ωn =
ωn−1 .
nΓ n2
If
2π n/2
nΓ n2
then a direct computation shows that a1 = 2 = ω1 and that an satisfies the same recurrence relationship as ωn , so ωn = an for all n. The second equality in (3.9) follows from
Theorem 3.34(a).
an =
(b) follows from the fact that the (n−1)-dimensional measure of a sphere of radius r equals
to the derivative with respect to r of the volume of an n-dimensional ball of radius r. 2
We will also need the following result.
Lemma 3.38. For β > 0 we have
−β
e
1
=√
π
Z
∞
0
e−u −β 2 /(4u)
√ e
du .
u
Proof. Applying the theory of residues to the function eiβz /(1 + z 2 ) one can easily prove
that
Z
2 ∞ cos βx
−β
e =
dx .
π 0 1 + x2
This and an obvious identity
Z ∞
1
2
=
e−(1+x )u du
2
1+x
0
yields
Z
2 ∞ cos βx
−β
e
=
dx
π 0 1 + x2
Z
Z ∞
2 ∞
2
=
cos βx
e−u e−ux du dx
π 0
Z 0
Z
2 ∞ −u ∞ −ux2
=
e
e
cos βx dx du
π 0
0
Z
Z
2 ∞ −u 1 ∞ −ux2 iβx =
e
e
e dx du
π 0
2 −∞
Z ∞
Z ∞
2
2
(x=−2πy) 2
−u
=
e
π
e−4π uy e−2πiβy dy du
π 0
{z
}
| −∞
W (β,u)
=
2
π
Z
0
∞
e−u
1rπ
2
u
e−β
2 /(4u)
du
HARMONIC ANALYSIS
=
1
√
π
Z
∞
0
37
e−u −β 2 /(4u)
√ e
du .
u
The proof is complete.
Proof of Theorem 3.29. (a) By a change of variables formula it suffices to prove the formula
for t = 1. We have
Z
Z Z ∞ −u
1
e
−2π|x| −2πix·ξ
−4π 2 |x|2 /(4u)
√
√ e
e
e
dx =
du e−2πix·ξ dx
π 0
u
Rn
Rn
Z ∞ −u
Z
e
1
2
2
√
e−4π |x| /(4u) e−2πix·ξ dx du
= √
π 0
u Rn
|
{z
}
W (ξ,(4u)−1 )
r n
∞ −u e
1
u
2
√
= √
e−u|ξ| du
π
π 0
u
Z ∞
1
2
e−u(1+|ξ| ) u(n−1)/2 du
=
(n+1)/2
π
0
Z ∞
1
1
=
e−s s(n−1)/2 ds .
π (n+1)/2 (1 + |ξ|2 )(n+1)/2 0
|
{z
}
Z
Γ[(n+1)/2]
(b) is obvious.
(c) Because of the scaling property (b) it suffices to consider t = 1. We have
Z
Z
dx
P (x, 1) dx
=
cn
2 (n+1)/2
n (1 + |x| )
Rn
ZR∞ Z
dσ
polar
rn−1 dr
=
cn
2 )(n+1)/2
(1
+
r
n−1
0
S
(0,1)
Z ∞
rn−1
=
cn n ωn
dr
(1 + r2 )(n+1)/2
0
Z π/2
r=tan θ
=
cn n ωn
sinn−1 θ dθ
0
n+1
Γ 2
π 1/2 Γ n2
2π n/2
=
n
π (n+1)/2 n Γ n2 (n − 1)Γ n−1
2
=
1.
The proof is complete.
Corollary 3.26 applied to f (x) = e−4π
2 t|x|2
Corollary 3.39.
Z
2
2
W (ξ, t)e2πix·ξ dξ = e−4π t|x|
Rn
n
for all x ∈ R .
and to f (x) = e−2πt|x| yields
Z
and
Rn
P (ξ, t)e2πix·ξ dξ = e−2π|x|t
38
PIOTR HAJLASZ
The Weierstrass and Poisson kernels have the following semigroup property.
Corollary 3.40. If Wt (x) = W (x, t) and Pt (x) = P (x, t), then for t1 , t2 > 0
(Wt1 ∗ Wt2 )(x) = Wt1 +t2 (x)
and
(Pt1 ∗ Pt2 )(x) = Pt1 +t2 (x)
for all x ∈ Rn .
Proof. It follows from Corollary 3.39 with x replaced by −x that
Ŵt (x) = e−4π
2 t|x|2
and P̂t (x) = e−2π|x|t .
Hence Theorem 3.2(b) yields
(Wt1 ∗ Wt2 )ˆ(x) = Ŵt1 (x)Ŵt2 (x) = Ŵt1 +t2 (x)
(Pt1 ∗ Pt2 )ˆ(x) = P̂t1 (x)P̂t2 (x) = P̂t1 +t2 (x)
2
and the result follows from Corollary 3.28.
3.4. The Schwarz class and the Plancherel theorem.
Definition 3.41. We say that f belongs to the Schwarz class S (Rn ) = Sn if f ∈ C ∞ (Rn )
and for all multiindices α, β
sup |xα Dβ f (x)| = pα,β (f ) < ∞.
x∈Rn
That means all derivatives of f rapidly decrease to zero as |x| → ∞, faster than the inverse
of any polynomial.
Clearly C0∞ (Rn ) ⊂ Sn , but also e−|x| ∈ Sn so functions in the Schwarz class need not
be compactly supported.
2
{pα,β } is a countable family of norms in Sn and we can use it to define a topology in
Sn .
Definition 3.42. We say that a sequence (fk ) converges to f in Sn if
lim pα,β (fk − f ) = 0 for all multiindices α, β.
k→∞
This convergence is metrizable. Indeed, dα,β (f, g) = pα,β (f − g) is a metric and if we
arrange all these metrics in a sequence d01 , d02 , . . ., then
d(f, g) =
∞
X
k=1
2−k
d0k (f, g)
1 + d0k (f, g)
defines a metric in Sn such that fn → f in Sn if and only if fn → f in the metric d.
Proposition 3.43. The space Sn has the following properties.
(a) Sn equipped with the metric d is a complete metric space.
(b) C0∞ (Rn ) is dense in Sn .
(c) If ϕ ∈ Sn , then τh ϕ → ϕ in Sn as h → 0, where τh ϕ(x) = ϕ(x + h).
HARMONIC ANALYSIS
39
(d) The mapping
Sn 3 ϕ 7→ xα Dβ ϕ(x) ∈ Sn
is continuous.
(e) If ϕ ∈ Sn , then
ϕ(x + hek ) − ϕ(x)
∂ϕ
→
(x)
h
∂xk
in the topology of Sn .
(f) If ϕ, ψ ∈ Sn , then ϕ ∗ ψ ∈ Sn and
as h → 0.
Dα (ϕ ∗ ψ) = (Dα ϕ) ∗ ψ = ϕ ∗ (Dα ψ)
for any multiindex α.
We leave the proof as an exercise.
Theorem 3.44. The Fourier transform is a continuous, one-to-one mapping of Sn onto
Sn such that
(a)
∂f
∂xj
∧
(ξ) = 2πiξj fˆ(ξ),
(b)
(−2πixj f )ˆ(ξ) =
∂ fˆ
(ξ) ,
∂ξj
(c)
(f ∗ g)ˆ = fˆĝ,
(d)
Z
Z
f (x)ĝ(x) dx =
Rn
fˆ(x)g(x) dx ,
Rn
(e)
Z
f (x) =
fˆ(ξ)e2πix·ξ dξ .
Rn
Proof. We already proved (a), (b), (c) and (d). Now we will prove that the Fourier
transform is a continuous mapping from Sn into Sn . Formulas (a) and (b) imply that
ξ α Dβ fˆ(ξ) = C(Dα (xβ f ))ˆ(ξ).
Since kĝk∞ ≤ kgk1 we get
pα,β (fˆ) = kξ α Dβ fˆ(ξ)k∞ ≤ CkDα (xβ f )k1 .
An application of the Leibnitz rule28 implies that Dα (xβ f ) equals a finite sum of expressions
of the form xβi Dαi f . Since
Z
βi αi
kx D f k1 =
|(1 + |x|2 )n xβi Dαi f (x)|(1 + |x|2 )−n dx
Rn
28Product
rule.
40
PIOTR HAJLASZ
≤ C(n) sup |(1 + |x|2 )n xβi Dαi f (x)| < ∞
x∈Rn
it follows that fˆ ∈ Sn . One can also easily deduce that the mapping ˆ : Sn → Sn is
continuous.
If f ∈ Sn than fˆ ∈ Sn and hence both f and fˆ are integrable, so (e) follows from the
inversion formula. This formula also shows that the Fourier transform applied four times
is an identity on Sn and hence the Fourier transform is a bijection on Sn .
2
Theorem 3.45 (Plancherel). The Fourier transform is an L2 isometry on a dense subset
Sn of L2
kfˆk2 = kf k2 , f ∈ Sn ,
and hence it uniquely extends to an isometry of L2
kfˆk2 = kf k2 , f ∈ L2 (Rn ) .
Moreover for f ∈ L2 (Rn )
Z
fˆ(ξ) = lim
R→∞
f (x)e−2πix·ξ dx
|x|<R
2
in the L sense, i.e.
Z
ˆ
f −
|x|<R
f (x)e−2πix·ξ dx → 0
as R → ∞
2
and similarly
Z
fˆ(ξ)e2πix·ξ dξ
f (x) = lim
R→∞
|ξ|<R
2
in the L sense.
¯
Proof. Given f ∈ Sn let g = fˆ, so ĝ = f¯. Indeed,
Z
Z
¯ˆ
−2πix·ξ
ĝ(ξ) =
f (x)e
dx =
fˆ(x)e2πix·ξ dx = f¯(x) .
Rn
Rn
Hence Theorem 3.44(d) gives
Z
Z
¯
kf k2 =
ff =
Rn
Z
f ĝ =
Rn
fˆg =
Rn
Z
¯
fˆfˆ = kfˆk2 .
Rn
Thus the Fourier transform is an L isometry on Sn . Since Sn is a dense subset of L2 it
uniquely extends to an isometry of L2 . Now for f ∈ L2 (Rn ) we have
2
L2
L1 3 f χB(0,R) −→ f
and hence
Z
(f χB(0,R) )ˆ(ξ) =
as R → ∞
2
L
f (x)e−2πix·ξ dx −→ fˆ(ξ)
|x|≤R
as R → ∞. Similarly
Z
2
L
fˆ(ξ)e2πix·ξ dξ −→ f (x)
as R → ∞.
|ξ|<R
HARMONIC ANALYSIS
Proposition 3.46. If f, g ∈ L2 (Rn ), then
Z
Z
ˆ
f (x)g(x) dx =
Rn
41
f (x)ĝ(x) dx .
Rn
Proof. Approximate f and g in L2 by functions in Sn , apply Theorem 3.44(d) and pass to
the limit.
Consider the class L1 (Rn ) + L2 (Rn ) consisting of functions of the form f = f1 + f2 ,
f1 ∈ L1 , f2 ∈ L2 . Then we define
fˆ = fˆ1 + fˆ2 .
In order to show that the Fourier transform is well defined in the class L1 + L2 we need to
show that it does not depend on the particular choice of the representation f = f1 + f2 .
Indeed, if we also have f = g1 + g2 , g1 ∈ L1 , g2 ∈ L2 , then f1 − g1 = g2 − f2 ∈ L1 ∩ L2 and
hence
fˆ1 − ĝ1 = (f1 − g1 )ˆ = (g2 − f2 )ˆ = ĝ2 − fˆ2 ,
fˆ1 + fˆ2 = ĝ1 + ĝ2 .
It is an easy exercise to show that
Lp (Rn ) ⊂ L1 (Rn ) + L2 (Rn ),
for 1 ≤ p ≤ 2,
p
and hence the Fourier transform is well defined on L (Rn ), 1 ≤ p ≤ 2 and
ˆ: Lp (Rn ) → L2 (Rn ) + C0 (Rn ),
for 1 ≤ p ≤ 2.
Later we will prove the Hausdorff-Young Inequality which implies that
0
ˆ: Lp (Rn ) → Lp (Rn ),
0
where p is the Hölder conjugate to p.
for 1 ≤ p ≤ 2,
42
PIOTR HAJLASZ
4. Miscellaneous results about the Fourier transform
In this chapter we will show some interesting results about the Fourier transform. Some
of them, but not all, will be used later.
4.1. The Poisson summation formula.
Theorem 4.1 (Poisson summation formula). If f ∈ C 1 (R) and
C
(4.1)
|f (x)| + |f 0 (x)| ≤
for x ∈ R,
1 + x2
then
∞
∞
X
X
fˆ(n).
f (n) =
n=−∞
n=−∞
Remark 4.2. Here fˆ(n) stand for the Fourier transform of f evaluated at n – do not get
confused with the Fourier series.
P
Note that it follows from the growth condition (4.1) that the series n f (n) converges
and that any function f ∈ Sn satisfies (4.1).
Proof. It follows from (4.1) that
(4.2)
g(x) =
∞
X
f (x + k)
k=−∞
is a periodic function of class C 1 with period 1. Therefore g is can be represented as a
Fourier series29
∞
X
ĝ(n)e2πinx
g(x) =
n=−∞
where
Z
ĝ(n) =
1
−2πinx
g(x)e
0
Z
dx =
∞ Z
X
k=−∞
0
1
−2πinx
f (x + k)e
dx =
∞ Z
X
k=−∞
k+1
f (x)e−2πinx dx
k
∞
=
f (x)e−2πinx dx = fˆ(n)
−∞
29Now
ĝ(n) are Fourier series coefficients – do not get confused with the Fourier transform!
HARMONIC ANALYSIS
43
so
∞
X
∞
X
f (k) = g(0) =
ĝ(n) =
n=−∞
k=−∞
∞
X
fˆ(n).
n=−∞
As an application we will prove the Jacobi identity.
Definition 4.3. The Jacobi ‘theta’ function is defined by
ϑ(t) =
∞
X
2
e−πn t ,
t > 0.
n=−∞
Clearly ϑ ∈ C ∞ (R).
1
Corollary 4.4 (Jacobi identity). For t > 0 we have ϑ(t) = t− 2 ϑ(1/t).
2
Proof. The function f (x) = t−1/2 e−πx /t , t > 0 can be written as
2
f (x) = t−1/2 u(xt−1/2 ) where u(x) = e−πx .
Hence30
2
fˆ(ξ) = t−1/2 (û)t−1/2 (ξ) = t−1/2 ut−1/2 (ξ) = u(ξt1/2 ) = e−πξ t .
Thus the Poisson summation formula yields
t
−1/2
∞
X
e
−πn2 /t
=
∞
X
2t
e−πn
i.e., t−1/2 ϑ(1/t) = ϑ(t).
n=−∞
n=−∞
Corollary 4.5. For t > 0 we have
∞
X
1
π 1 + e−2πt
=
.
t2 + n2
t 1 − e−2πt
n=−∞
Remark 4.6. It is easy to show by taking the limit as t → 0+ in the above identity that
∞
X
1
π2
=
.
2
n
6
n=1
Proof. In Theorem 3.29 we showed that the Fourier transform of f (x) = e−2π|x|t equals31
fˆ(x) =
30We
31We
t
π(t2 + x2 )
use here two facts: (ϕ(sx))∧ (ξ) = (ϕ̂)s (ξ) = s−n ϕ̂(ξ/s) with n = 1 and û(ξ) = u(ξ).
are in dimension n = 1 and c1 = 1.
44
PIOTR HAJLASZ
so the Poisson summation formula gives32
∞
∞
X
t X
1
1 + e−2πt
−2π|n|t
=
e
=
π n=−∞ t2 + n2 n=−∞
1 − e−2πt
from which the theorem follows.
4.2. The Heisenberg inequality. Celebrated Heisenberg’s uncertainty principle asserts
that position and momentum of a particle cannot be measured at the same time and it
can be written as an inequality
h
σx σp ≥
4π
where σx and σp are standard deviations of position and momentum and h is the Plank
constant. Translating it into our language the above inequality can be formulated as
Theorem 4.7 (Heisenberg’s inequality). For any f ∈ L2 (R) and a, b ∈ R,
sZ
sZ
∞
∞
kf k22
(x − a)2 |f (x)|2 dx
(ξ − b)2 |fˆ(ξ)|2 dξ ≥
.
4π
−∞
−∞
2
Moreover the equality holds if and only if f (x) = Ce2πibx e−k(x−a) for some C ∈ C and
k > 0.
Proof. We will prove the result under the assumption that f ∈ S1 . The case of general
f ∈ L2 is left to the reader as an exercise. Let f (x) = e2πibx g(x − a). Since
fˆ(ξ) = ĝ(ξ − b)e−2πa(ξ−b)
we easily obtain that the Heisenberg inequality is equivalent to
1/2
1/2 Z ∞
Z ∞
kgk22
2
2
2
2
|ξ| |ĝ(ξ)| dξ
≥
|x| |g(x)| dx
(4.3)
.
4π
−∞
−∞
Note that (|g|2 )0 = (gg)0 = 2 re g 0 g so integration by parts gives
Z β
Z β
Z β
β
0
2
2
2
xg 0 (x)g(x) dx.
|g(x)| dx =
x |g(x)| dx = x|g(x)| − 2 re
α
α
α
α
Letting α → −∞, β → +∞ and using the fact that g ∈ S1 yields
Z ∞
Z ∞
2
|g(x)| dx = −2 re
xg 0 (x)g(x) dx
−∞
Z ∞ −∞
(4.4)
≤ 2
|x| |g(x)| |g 0 (x)| dx
−∞
Z
(4.5)
∞
≤ 2
2
1/2 Z
∞
|x| |g(x)| dx
−∞
32In
2
1/2
|g (x)| dx
0
2
−∞
the Poisson summation formula we assumed that f ∈ C 1 (R) and it satisfies (4.1). However, our
function f is not C 1 . The condition (4.1) was required for the function (4.2) to be represented as a Fourier
series. Although the condition (4.1) is not satisfied any longer, it is easy to check that the function defined
by (4.2) is periodic and Lipschitz continuous so it is still equal to its Fourier series and the proof carries
on.
HARMONIC ANALYSIS
Z
1/2 Z
∞
2
45
2
|gb0 (ξ)|2
|x| |g(x)| dx
= 2
−∞
Z
dξ
−∞
1/2 Z
∞
2
2
∞
|x| |g(x)| dx
= 2
1/2
∞
2
1/2
|2πiξĝ(ξ)| dξ
−∞
−∞
from which (4.3) follows. Finally, let us investigate the case of equality. For the equality
in the Schwarz inequality (4.5) we need |g 0 (x)| to be proportional to |xg(x)|, say |g 0 (x)| =
k|xg(x)|, for some k > 0 so g 0 (x) = keiϕ(x) xg(x) for some real valued function ϕ(x). But
then equality in (4.4) implies that
−re e−ϕ(x) = 1 so eiϕ(x) = −1
2
and we have g 0 (x) = −kxg(x). Solving this differential equation yields g(x) = Ce−kx and
2
hence f (x) = Ce2πibx e−k(x−a) .
4.3. Eigenfunctions of the Fourier transform. The Fourier transform applied four
times is identity on L2 (Rn ) so if f is an eigenfunction of the Fourier transform,
fˆ = λf,
then f = |λ|4 f
so λ ∈ {−1, 1, −i, i}.
2
We know already one eigenfunction with the eigenvalue 1, namely f (x) = e−πx and now
we will find now all eigenfunctions in dimension n = 1.
2
It is natural to search among functions of the form p(x)e−πx , where p(x) is a polynomial.33
Definition 4.8. Hermite functions are defined by the formula
(−1)n πx2 d n −2πx2
hn (x) =
e
e
, for all integers n ≥ 0.
n!
dx
2
Note that h0 (x) = e−πx .
2
Lemma 4.9. hn (x) = pn (x)e−πx for a polynomial pn (x) of order n of the form
pn (x) =
(4π)n n
x + lower order terms.
n!
In particular hn ∈ S1 .
This lemma easily follows from the definition of hn .
Lemma 4.10. h0n (x) − 2πxhn (x) = −(n + 1)hn+1 (x).
Proof. Applying the product rule to
h 2 i h (−1)n d n
i
πx
−2πx2
hn (x) = e
·
e
n!
dx
33Note
e−πξ
2
2
2
that evaluation of the Fourier transform of p(x)e−πx is straightforward: since F(e−πx )(ξ) =
it suffices to apply (3.1).
46
PIOTR HAJLASZ
gives
h0m (x)
h
i h
= 2πxhn (x) + e
·
πx
(−1)n
n! }
| {z
−(n+1)
d n+1
i
−2πx2
= 2πxhn (x) − (n + 1)hn+1 (x).
e
dx
(−1)n+1
(n+1)!
Lemma 4.11. If f ∈ C ∞ (R) and n is a positive integer, then
d n
d n
d n−1
(4πx)
f (x) −
(4πxf (x)) = −4πn
f (x)
dx
dx
dx
for all x ∈ R.
The result follows from a simple induction.
Lemma 4.12. h0n (x) + 2πxhn (x) = 4πhn−1 (x), where h−1 (x) = 0.
Proof. For n = 0 we check the equality directly. For n ≥ 1 Lemma 4.10 gives
h0n (x) + 2πxhn (x) = 4πxhn (x) − (n + 1)hn+1 (x)
(−1)n+1 πx2 d n+1 −2πx2
(−1)n πx2 d n −2πx2
= 4πx
e
e
e
−(n + 1)
e
n!
dx
(n + 1)!
dx {z
|
}
|
{z
}
2
+(−1)n /n!
d n
( dx
) (−4πxe−2πx )
d n
d n
i
(−1)n πx2 h
−2πx2
−2πx2
=
e
4πx
e
−
(4πxe
)
n!
dx
|
{z dx
}
d n−1 −2πx2
−4πn( dx
)
e
by Lemma 4.11
= 4π
(−1)n−1 πx2 d n−1 −2πx2
e
e
= 4πhn−1 (x).
(n − 1)!
dx
Lemma 4.13. ĥn = (−i)n hn and ȟn = in hn .
Remark 4.14. The lemma yields
ĥ4n+2 = −h4n+2 ,
ĥ4n = h4n ,
ĥ4n+3 = ih4n+3 ,
ĥ4n+1 = −ih4n+1
Thus the lemma provides examples of eigenfunctions corresponding to all possible eigenvalues 1, −1, i, −i.
Proof. Since hn ∈ S1 we can compute the Fourier transform directly. Lemma 4.10 gives
h0n − 2πxhn = −(n + 1)hn+1 ,
c0 − 2πxh
\n = −(n + 1)ĥn+1 .
h
n
Since (−2πixf )∧ = (fˆ)0 and fb0 = 2πiξ fˆ(ξ) we have
2πiξ ĥn (ξ) − i(ĥn )0 (ξ) = −(n + 1)ĥn+1 (ξ)
(4.6)
We will prove the equality ĥn = (−i)n hn by induction.
2
For n = 0, h0 = e−πx so ĥ0 = h0 = (−i)0 h0 .
HARMONIC ANALYSIS
47
Suppose now that the equality holds for n and we will prove it for n + 1. The equality
(4.6) yields.
2πiξ(−i)n hn (ξ) − i((−i)n hn (ξ))0 = −(n + 1)ĥn+1 (ξ),
−(−i)n+1 2πξhn (ξ) + (−i)n+1 h0n (ξ) = −(n + 1)ĥn+1 (ξ),
0
= −(n + 1)ĥn+1 (ξ).
(−i)n+1
hn (ξ) − 2πξhn (ξ)
{z
}
|
−(n+1)hn+1 (ξ) by Lemma 4.10
This proves that ĥn+1 = (−i)n+1 hn+1 (ξ) which completes the proof of the claim that
ĥn = (−i)n hn (ξ) fro all n. Finally
hn = (ĥn )∨ = ((−i)n hn )∨ = (−i)n ȟn ,
ȟn = in hn .
Let Kf = f 00 − 4π 2 x2 f .
Lemma 4.15. Khn = −4π(n + 12 )hn
Proof. Lemmas 4.10 and 4.12 yield
h0n − 2πxhn = −(n + 1)hn+1 ,
h00n − 2πhn − 2πxh0n = −(n + 1)h0n+1 ,
h00n = 2πhn + 2πxh0n − (n + 1)h0n+1 .
Hence
Khn = h00n − 4π 2 x2 hn = 2πhn + 2πxh0n − (n + 1)h0n+1 − 4π 2 x2 hn
= 2πhn − (n + 1)h0n+1 + 2πx (h0n − 2πxhn )
|
{z
}
−(n+1)hn+1
1)(h0n+1
= 2πhn − (n +
1
hn .
= −4π n +
2
+ 2πxhn+1 )
Lemma 4.16. For real valued functions f, g ∈ S1 we have34
hKf, gi = hf, Kgi
Proof. The integration by parts yields
Z
Z
Z
00
2 2
00
hKf, gi =
(f − 4π x f )g =
f g − 4π 2 x2 f g
R
R
ZR
Z
Z
=
f g 00 − 4π 2 x2 f g = f (g 00 − 4π 2 x2 g) = hf, Kgi.
R
R
Lemma 4.17. The Hermite functions hn are orthogonal,
hhn , hm i = 0,
34Here
h·, ·i denotes the L2 scalar product.
for n 6= m.
48
PIOTR HAJLASZ
Proof. We have
1
1
−4π n +
hhn , hm i = hKhn , hm i = hhn , Khm i = −4π m +
hhn , hm i
2
2
and the result easily follows.
Lemma 4.18.
(4π)n
khn k22 = √
.
2n!
Proof. Since by Lemma 4.10 we have (n + 1)hn+1 = 2πxhn − h0n , integration by parts yields
Z
Z
Z
Z
0
2
(n + 1)khn+1 k2 = 2π xhn hn+1 − hn hn+1 = 2π xhn hn+1 + hn h0n+1
R
R
R
Z R
0
2
hn (hn+1 + 2πxhn+1 ) = 4πkhn k2 .
=
|
{z
}
R
4πhn
Replacing n by n − 1 yields
khn k22 =
4π
khn−1 k22
n
and a simple induction argument yields
(4π)n
kh0 k22 .
n!
Now the result follows from a simple observation that
Z
Z
Z
1
1
2
2
−πx2 2
−2πx2
e−πy dy = √ .
kh0 k2 = (e
e
) dx =
dx = √
2 R
2
R
R
khn k22 =
The last two lemmas show that the scaled Hermite functions
(4π)n −1/2
hn
en = √
2n!
form an orthonormal family in L2 (R).
2
n
Theorem 4.19. The fmaily {en }∞
n=0 is an orthonormal basis of L (R ).
R
Proof. It suffices toRprove that if f ∈ L2 (R) and R f hn = 0 for all n ≥ 0, then f = 0 a.e.
Thus suppose that R f hn = 0 for all n ≥ 0. Since
(4π)2
2
n
hn (x) =
x + . . . e−πx ,
n!
2
it easily follows that the functions xn e−πx , n ≥ 0, are linear combinations of the functions
h0 , h1 , . . . , hn . Hence
Z
2
f (x)xn e−πx dx = 0 for all n ≥ 0.
R
HARMONIC ANALYSIS
49
2
Since by the Schwarz inequality f (x)e−πx ∈ L1 we can compute the Fourier transform
directly
Z
Z
∞
X
(−2πixξ)n
−πx2 ∧
−πx2 −2πixξ
−πx2
f (x)e
=
f (x)e
e
dx =
f (x)e
dx
n!
R
R
n=0
Z
∞
X
(−2πiξ)n
2
=
f (x)xn e−πx dx = 0.
n!
R
n=0
2
2
Vanishing of the Fourier transform of f (x)e−πx implies that f (x)e−πx = 0 a.e. and hence
f (x) = 0 a.e.
2
2
Since {en }∞
n=0 is an orthonormal basis of L (R) any f ∈ L can be written as
f=
∞
X
hf, en i en
n=0
n
This and the fact that ên = (−i) en yield
Theorem 4.20 (Wiener’s definition of the Fourier transform). For f ∈ L2 (R) we have
fˆ =
∞
X
hf, en i(−i)n en .
n=0
Let Hk , k = 0, 1, 2, 3 be subspaces of L2 (R) defined by
Hk = span {e4n+k }∞
n=0 .
The subspaces Hk are orthogonal and
L2 (R) = H0 ⊕ H1 ⊕ H2 ⊕ H3 .
Clearly, the subspaces Hk are eigenspaces of the Fourier transform:
fˆ = (−i)k f
for f ∈ Hk and k = 0, 1, 2, 3.
4.4. The Bochner-Hecke formula. As was pointed out in Section 4.3 computing the
2
Fourier transform of P (x)e−π|x| is straightforward due to formula (3.1) and Theorem 3.7,
but the computations are often algebraically complicated and it is not always obvious how
to write the answer in a compact and nice form. We have see examples of such computations
in Section 4.3 when we found eigenfunctions of the Fourier transform and now we will see
another example of this type. The results of this section will be used later in Section 6.3
Definition 4.21. We say that Pk (x) is a homogeneous harmonic polynomial of degree k if
X
Pk (x) =
aα xα and ∆Pk = 0 .
|α|=k
Theorem 4.22 (Bochner-Hecke). If Pk (x) is a homogeneous harmonic polynomial of degree k, then
2
2
F Pk (x)e−π|x| (ξ) = (−i)k Pk (ξ)e−π|ξ| .
50
PIOTR HAJLASZ
Proof. Since the polynomial Pk is homogeneous of degree k,
X
Pk (−2πix) =
aα (2πix)α = (−2πi)k Pk (x).
|α|=k
2
Hence (3.1) applied to f (x) = e−π|x| yields35
Z
2
2
(4.7)
Pk (x)e−π|x| e−2πix·ξ dx = F Pk (x)e−π|x| (ξ)
Rn
2
= (−2πi)−k Pk (D)e−π|ξ| = Q(ξ)e−π|ξ|
2
for some polynomial Q and it remains to prove that Q(ξ) = Pk (−iξ).
Lemma 4.23. If P (x) is a polynomial36 in Rn , then
Z
Z
P
2
−π j (xj +iξj )2
P (x)e
(4.8)
dx =
P (x − iξ)e−π|x| dx .
Rn
Rn
Proof. By the same argument involving the same contour integration as in the proof of
Theorem 3.7 for any polynomial P of one variable we have
Z ∞
Z ∞
2
−π(x+iξ)2
(4.9)
P (x)e
dx =
P (x − iξ)e−πx dx .
−∞
−∞
If P is a polynomial in n variables, then (4.8) follows from (4.9) and the Fubini theorem. 2
Multiplying both sides of (4.7) by eπ|ξ| and applying (4.8) we obtain
Z
Z
P
2
−π j (xj +iξj )2
Pk (x − iξ)e−π|x| dx .
Pk (x)e
dx =
Q(ξ) =
Rn
Rn
We can write
Pk (x − η) =
X
η α Pα (x)
α
and then
Z
−π|x|2
Pk (x − η)e
dx =
Rn
X
η
α
Z
α
2
Pα (x)e−π|x| dx ,
Rn
so clearly the integral is a polynomial is η. With this notation we obtain
Z
X
2
α
Q(ξ) =
(iξ)
Pα (x)e−π|x| dx
Rn
α
and hence
Q(ξ/i) =
X
α
35Instead
ξ
α
Z
−π|x|2
Pα (x)e
Rn
Z
dx =
2
Pk (x − ξ)e−π|x| dx .
Rn
of using (3.1) we could apply the differential operator Pk (Dξ ) to both sides of the equality
Z
2
2
e−π|x| e−2πix·ξ dx = e−π|ξ|
Rn
and differentiate under the sign of the integral.
36Any polynomial, not necessarily harmonic or homogeneous.
HARMONIC ANALYSIS
51
Since Pk is a harmonic function, it has the mean value property37
Z
Pk (sθ − ξ) dσ(θ) = |S n−1 |Pk (−ξ) .
S n−1
Thus integration in polar coordinates gives
Z
Z ∞
2
n−1
s
Pk (sθ − ξ) dσ(θ) e−πs ds
Q(ξ/i) =
n−1
0
Z ∞S
2
sn−1 |S n−1 |e−πs ds
= Pk (−ξ)
Z0
2
= Pk (−ξ)
e−π|x| dx = Pk (−ξ)
Rn
and hence Q(ξ) = Pk (−iξ). The proof is complete.
Using homogeneity of Pk and the second part of Theorem 3.2(e) one can easily deduce
from the Bochner-Hecke formula the folloiwing result.
Corollary 4.24. If Pk is a homogeneous harmonic polynomial of degree k, then for any
t > 0 we have
n
2
2
F Pk (x)e−πt|x| (ξ) = t−k− 2 (−i)k Pk (ξ)e−π|ξ| /t .
We leave details as an easy exercise.
4.5. Symmetry of the Fourier transform. If f is radially symmetric i.e., f = f ◦ ρ for
all ρ ∈ O(n), then fˆ ◦ ρ = f[
◦ ρ = fˆ for all ρ ∈ O(n) so fˆ is radially symmetric too. Now
we will prove a more delicate result about the symmetry of the Fourier transform.
Proposition 4.25. Let f ∈ L1 (Rn ) be a real valued function of the form
xj
f (x) =
g(|x|).
|x|
Then there is a continuous function h : [0, ∞) → R, h(0) = 0, h(t) → 0 as t → ∞ such
that
ξj
fˆ(ξ) = i h(|ξ|), ξ ∈ Rn .
|ξ|
While the function f is not radially symmetric a more symmetric object is the map
x
F (x) =
g(|x|) ∈ L1 (Rn , Rn ), x ∈ Rn
|x|
so f is the jth component of F . We will prove that
ξ
F̂ (ξ) = i h(|ξ|), ξ ∈ Rn
|ξ|
from which Proposition 4.25 will readily follow.
In the proof we will use the following result which is interesting on its own. It ill be used
later one more time in a proof of Theorem 6.2.
37We
integrate here Pk over the sphere centered at −ξ and of radius s.
52
PIOTR HAJLASZ
Lemma 4.26. If m : Rn → Rn is a measurable function that is homogeneous of degree 0,
i.e. m(tx) = m(x) for t > 0, and commutes with the orthogonal transformations, i.e.
(4.10)
m(ρ(x)) = ρ(m(x))
for all x ∈ Rn and ρ ∈ O(n), then there is a constant c such that
x
(4.11)
m(x) = c
for all x =
6 0.
|x|
Proof. Let e1 , e2 , . . . , en be the standard orthogonal basis of Rn . If [ρjk ] is the matrix
representation of ρ ∈ O(n), then the condition (4.10) reads as
(4.12)
mj (ρ(x)) =
n
X
ρjk mk (x),
j = 1, 2, . . . , n,
k=1
where m(x) = (m1 (x), . . . , mn (x)).
Let m1 (e1 ) = c. Consider all ρ ∈ O(n) such that ρ(e1 ) = e1 . This condition means that
the first column of the matrix [ρjk ] equals e1 , i.e. ρ11 = 1, ρj1 = 0, for j > 1. Since columns
are orthogonal, for k > 1 we have
0=
n
X
ρj1 ρjk = ρ1k .
j=1
Thus
1 0
 0 ρ22
ρ=
 ... ...
0 ρn2


... 0
. . . ρ2n 
,
. . . .. 
. 
. . . ρnn
where [ρjk ]nj,k=2 is the matrix of an arbitrary orthogonal transformation in the (n − 1)dimensional subspace orthogonal to e1 .
For x = e1 = ρ(e1 ) = ρ(x) and j ≥ 2 identity (4.12) yields
mj (e1 ) =
n
X
k=1
ρjk mk (e1 ) =
n
X
ρjk mk (e1 ) ,
k=2
and hence
 
m2 (e1 )
..

=
.

mn (e1 )


ρ22 . . . ρ2n
m2 (e1 )
.. . .
.
..
.
. ..  
.
.
ρn2 . . . ρnn
mn (e1 )
That means the vector [m2 (e1 ), . . . , mn (e1 )]T is fixed under an arbitrary orthogonal transformation of Rn−1 , so it must be a zero vector, i.e.
m2 (e1 ) = . . . = mn (e1 ) = 0 .
Now formula (4.12) for any ρ ∈ O(n) and x = e1 , takes the form
mj (ρ(e1 )) = ρj1 m1 (e1 ) = cρj1 .
HARMONIC ANALYSIS
53
By homogeneity it suffices to prove (4.11) for |x| = 1. Let ρ ∈ O(n) be such that ρ(e1 ) = x.
Then ρj1 = xj , j = 1, 2, . . . , n and hence
xj
mj (x) = cρj1 = cxj = c
.
|x|
This completes the proof of the lemma.
Proof of Proposition 4.25. Since the function F is odd
Z
Z
iF (x)(cos(2πx · ξ) − i sin(2πx · ξ)) dx =
iF̂ (ξ) =
Rn
sin(2πx · ξ)F (x) dx
Rn
so the mapping ξ 7→ iF̂ (ξ) takes values in Rn . Fix k > 0 and define mk : S n−1 (0, k) → Rn
by mk (ξ) = iF̂ (ξ) for |ξ| = k. Extend mk to mk : Rn \{0} → Rn as a function homogeneous
of degree 0, i.e.
kξ mk (ξ) = i F̂
for ξ 6= 0.
|ξ|
We claim that
(4.13)
mk (ρ(ξ)) = ρ(mk (ξ)) for ρ ∈ O(n) and ξ 6= 0.
Since mk is homogeneous of degree 0, it suffices to check (4.13) for |ξ| = k. We have38
Z
Z
x
x
sin(2πx · ρ(ξ)) g(|x|) dx =
sin(2πρ−1 (x) · ξ) g(|x|) dx
mk (ρ(ξ)) =
|x|
|x|
n
Rn
ZR
Z
ρ(x)
ρ(x)
sin(2πx · ξ)
sin(2πx · ξ)
=
g(|ρ(x)|) dx =
g(|x|) dx
|ρ(x)|
|x|
Rn
Rn
Z
x
sin(2πx · ξ) g(|x|) dx = ρ(mk (ξ)).
= ρ
|x|
Rn
According to Lemma 4.26 there is a constant h(k) ∈ R such that mk (ξ) = −h(k)ξ/|ξ|. In
particular for |ξ| = k we have
ξ
iF̂ (ξ) = mk (ξ) = − h(|ξ|).
|ξ|
Clearly h is continuous, h(0) = 0 and h(t) → 0 as t → ∞, because iF̂ ∈ C0 (Rn , Rn ).
38First we
use equality x·ρ(ξ) = ρ−1 (x)·ξ which follows from the fact that the orthogonal transformation
ρ preserves the scalar product and then we use the change of variables x = ρ(y) whose absolute value
of the Jacobian equals 1.
−1
54
PIOTR HAJLASZ
5. Tempered distributions
5.1. Basic constructions.
Definition 5.1. The space Sn0 of all continuous linear functionals on Sn is called the
space of tempered distributions. The evaluation of u ∈ Sn0 on ϕ ∈ Sn will usually be
denoted by u[ϕ]. The space Sn0 is equipped with the weak-∗ convergence (usually called
weak convergence or just convergence). Namely uk → u in Sn0 if uk [ψ] → u[ψ] for every
ψ ∈ Sn .
Here are examples:
1. If f ∈ Lp (Rn ), 1 ≤ p ≤ ∞, then
Z
f (x)ϕ(x) dx defines Lf ∈ Sn0 .
Lf [ϕ] =
Rn
2. If µ is a measure of finite total variation, then
Z
ϕ dµ defines Lµ ∈ Sn0 .
Lµ [ϕ] =
Rn
3. We say that a function f is a tempered Lp function if f (x)(1 + |x|2 )−k ∈ Lp (Rn ) for some
nonnegative integer k. If p = ∞ we call f a slowly increasing function. Then
Z
Lf [ϕ] =
f (x)ϕ(x) dµ defines Lf ∈ Sn0 for all 1 ≤ p ≤ ∞.
Rn
Note that slowly increasing functions are exactly measurable functions bounded by polynomials.
4. A tempered measure is a Borel measure µ such that
Z
(1 + |x|2 )−k dµ < ∞
Rn
for some integer k ≥ 0. As before Lµ ∈ Sn0 .
5. L[ϕ] = Dα ϕ(x0 ) is a tempered distribution L ∈ Sn0 .
The distributions generated by a function or by a measure will often be denoted by
Lf [ϕ] = f [ϕ], Lµ [ϕ] = µ[ϕ].
HARMONIC ANALYSIS
55
Suppose that u ∈ Sn0 . If there is a tempered Lp function f such that u[ϕ] = f [ϕ]
for ϕ ∈ Sn , then we can identify u with the function f and simply write u = f . The
identification is possible, because the function f is uniquely defined (up to a.e. equivalence).
This follows from a well known result.
R
Lemma 5.2. If Ω ⊂ Rn is open and f ∈ L1loc (Ω) satisfies Ω f ϕ = 0 for all ϕ ∈ C0∞ (Ω),
then f = 0 a.e.
Note that not every function f ∈ C ∞ (Rn ) defines a tempered distribution, because it
may happen that
R form some ϕ ∈ Sn the function f ϕ is not integrable and hence the
integral f [ϕ] = Rn f (x)ϕ(x) dx does not make sense.
Theorem 5.3. A linear functional L on Sn is a tempered distribution if and only if there
is a constant C > 0 and a positive integer m such that
X
|L(ϕ)| ≤ C
pα,β (ϕ) for all ϕ ∈ Sn .
|α|,|β|≤m
Proof. If a linear functional L satisfies the given estimate, then clearly it is continuous
on Sn , so it remains to prove the converse implication. Let L ∈ Sn0 . We claim that there
is a positive integer m such that |L(ϕ)| ≤ 1 for all
n
X
1o
ϕ ∈ ϕ ∈ Sn :
pα,β (ϕ) ≤
:= Nm .
m
|α|,|β|≤m
To the contrary suppose that there is a sequence ϕk ∈ Sn such that |L(ϕk )| > 1 and
X
1
(5.1)
pα,β (ϕk ) ≤ , k = 1, 2, 3, . . .
k
|α|,|β|≤k
Note that (5.1) implies that ϕk → 0 in Sn , so the inequality |L(ϕk )| > 1 contradicts
continuity of L. This proves the claim. Denote
X
kϕk =
pα,β (ϕ) .
|α|,|β|≤m
Observe that k · k is a norm. For an arbitrary 0 6= ϕ ∈ Sn , ϕ̃ = ϕ/(mkϕk) satisfies
kϕ̃k ≤ 1/m, so ϕ̃ ∈ Nm and hence
|L(ϕ)| = mkϕk|L(ϕ̃)| ≤ mkϕk
2
which proves the theorem.
For any function g on Rn we define g̃(x) = g(−x). Then it easily follows from the Fubini
theorem that for u, ϕ, ψ ∈ Sn
Z
Z
(u ∗ ϕ)(x)ψ(x) dx =
u(x)(ϕ̃ ∗ ψ)(x) dx .
Rn
Rn
Regarding the functions u and u ∗ ϕ as distributions we can rewrite this equality as
(u ∗ ϕ)[ψ] = u[ϕ̃ ∗ ψ]
If u ∈ Sn0 and ϕ ∈ Sn , then ψ 7→ u[ϕ̃ ∗ ψ] is a tempered distribution so it motivates the
following definition.
56
PIOTR HAJLASZ
Definition 5.4. If u ∈ Sn0 and ϕ ∈ Sn , then the convolution of u and ϕ is a tempered
distribution defined by the formula
(u ∗ ϕ)[ψ] := u[ϕ̃ ∗ ψ] .
The following two results are left as an easy exercise.
Proposition 5.5. If u ∈ Sn0 , ϕ, ψ ∈ Sn , then
(u ∗ ϕ) ∗ ψ = u ∗ (ϕ ∗ ψ) .
Proposition 5.6. If µ ∈ B(Rn ) and ϕ ∈ Sn , then µ ∗ ϕ = ϕ ∗ µ where µ ∗ ϕ is the
convolution of a distribution µ ∈ Sn0 with a function ϕ ∈ Sn and ϕ ∗ µ is the convolution
of a function with a measure.
Theorem 5.7. If u ∈ Sn0 and ϕ ∈ Sn , then the tempered distribution u∗ϕ can be identified
with a slowly increasing function f defined by
f (x) = u[τ−x ϕ̃] = u[ϕ(x − ·)]
for all x ∈ Rn .
Moreover f ∈ C ∞ (Rn ) and all its derivatives are slowly increasing.
Proof. Note that the equality u[τ−x ϕ̃] = u[ϕ(x − ·)] is obvious. First we will prove that
the function f (x) = u[τ−x ϕ̃] = u[ϕ(x − ·)] is C ∞ and f , as well as all its derivatives, are
slowly increasing.
It follows from Theorem 3.43(c) that f is continuous. Observe that for a fixed x ∈ Rn
f (x + hek ) − f (x)
ϕ(x + hek − ·) − ϕ(x − ·)
=u
→ u[(∂k ϕ)(x − ·)] as h → 0
h
h
because it is easy to check using Theorem 3.43(e) that for every x ∈ Rn
ϕ(x + hek − ·) − ϕ(x − ·)
→ (∂k ϕ)(x − ·) as h → 0
h
in the topology of Sn . Thus the partial derivatives
∂k f (x) = u[(∂k ϕ)(x − ·)]
exist and are continuous by Theorem 3.43(c). Since ∂k ϕ ∈ Sn , iterating the above process
for any multiindex α we obtain
(5.2)
Dα f (x) = u[(Dα ϕ)(x − ·)] .
This implies that f ∈ C ∞ (Rn ). Since u is a tempered distribution, Theorem 5.3 gives the
estimate
X
(5.3)
|f (x)| = |u[τ−x ϕ̃]| ≤ C
pα,β (τ−x ϕ̃) ≤ C 0 (1 + |x|)m .
|α|,|β|≤m
Indeed,
pα,β (τ−x ϕ̃) = sup |z α (Dβ ϕ̃)(z − x)| = sup |(z + x)α (Dβ ϕ̃)(z)|
z∈Rn
z∈Rn
≤ C 1 + |x||α| sup 1 + |z||α| |Dβ ϕ̃(z)| ≤ C 0 (1 + |x|)m .
z∈Rn
HARMONIC ANALYSIS
57
Hence f is slowly increasing, and so it defines a tempered distribution. Since the derivatives
of f satisfy (5.2), which is an expression of the same type as the one in the definition of
f , we conclude that all derivatives Dα f are slowly increasing.
It remains to prove that f = u ∗ ϕ in the sense of tempered distributions i.e.,
(u ∗ ϕ)[ψ] = f [ψ] for all ψ ∈ Sn .
(5.4)
Using f (y) = u[ϕ(y − ·)], (5.4) reads as
hZ
i Z
u
ϕ(y − ·)ψ(y) dy =
Rn
u[ϕ(y − ·)]ψ(y) dy.
Rn
The idea of proving this equality is to approximate the integral on on the left hand side by
Riemann sums, use linearity of u to go under the sign of the sum and to pass to the limit.
However, the details of this approximation argument are not obvious. They are elementary and tedious, but not obvious and so in most of the textbooks they are summarized
as ‘easy to see’. For the sake of completeness we decided to include details.
First of all observe that is suffices to prove (5.4) under the assumption that ϕ, ψ ∈ C0∞ .
Indeed, suppose we know (5.4) for compactly supported functions, but now ϕ, ψ ∈ Sn .
Let39 C0∞ 3 ϕk → ϕ in Sn and C0∞ 3 ψ` → ψ in Sn . Observe that fk (x) = u[τ−x ϕ̃k ] → f (x)
pointwise. It follows from the proof of (5.3) that all the functions fk have a common
estimate independent of k
|fk (x)| ≤ C(1 + |x|)m
and hence fk → f in Sn0 by the Dominated Convergence Theorem. It is also clear from
the definition of the convolution that u ∗ ϕk → u ∗ ϕ in Sn0 , so
(u ∗ ϕ)[ψ` ] = lim (u ∗ ϕk )[ψ` ] = lim fk [ψ` ] = f [ψ` ] .
k→∞
k→∞
The second equality follows from assuming (5.4) for all ϕk , ψ` ∈ C0∞ . Now letting ` → ∞
yields (5.4).
If the support of ψ is contained in a cube Q with integer edge-length we divide Q into
cubes {Qki }i of edge-length 2−k and centes yki . Then the Riemann sums converge in the
topology of Sn (as a function of x) to the integral
Z
X
ϕ(yki − x)ψ(yki )|Qki | →
ϕ(y − x)ψ(y) dy
Rn
i
and hence
Z
hZ
i
X
ϕ(y − ·)ψ(y) dy = lim
u
u[ϕ(yki − ·)]ψ(yki )|Qki | =
Rn
k→∞
i
u[ϕ(y − ·)]ψ(y) dy.
Rn
Boring (but elementary) details of the convergence of the Riemann sums in the topology
of Sn are moved to Appendix 5.7.
Note that formula (5.2) implies
Theorem 5.8. If u ∈ Sn0 and ϕ ∈ Sn , then for any multiindex α we have
Dα (u ∗ ϕ)(x) = (u ∗ (Dα ϕ))(x) .
39See
Theorem 3.43(b).
58
PIOTR HAJLASZ
The next result will require the use of the Banach-Steinhaus theorem.
Theorem 5.9. If uk → u is Sn0 and ϕk → ϕ in Sn , then uk [ϕk ] → u[ϕ].
Proof. In order to be able to use triangle inequality to prove the convergence we need to
know uniform estimates for the sequence uk . Convergence uk → u in Sn0 means pointwise
convergence on a complete metric space Sn so it is natural to use the Banach-Steinhaus
Theorem to get uniform estimates for uk . We state it as a lemma.
Lemma 5.10 (Banach-Steinhaus). Let X be a complete metric space and let {fi }i∈I be a
family of continuous real-valued functions on X. If the functions in the family are pointwise
bounded, i.e.
sup |fi (x)| < ∞ for x ∈ X,
i∈I
then there is an open set U ⊂ X and a constant M > 0 such that
sup |fi (x)| ≤ M
for all x ∈ U .
i∈I
It easily follows from the linearity that it suffices to consider the case uk → 0 and ϕk → 0.
The functions ϕ 7→ |uk [ϕ]| are continuous on the complete metric space (Sn , d). They are
pointwise bounded on Sn since |uk [ϕ]| → 0 for every ϕ ∈ Sn . Thus Lemma 5.10 implies
that
sup |uk [ϕ]| ≤ M for all ϕ ∈ B(ϕ0 , r0 ).
k
Fix ε > 0. There is k1 such that |uk [ϕ0 ]| < ε for k ≥ k1 . Since ϕk → 0 in Sn , then also
ε−1 ϕk → 0 and hence
d(ϕ0 + ε−1 ϕk , ϕ0 ) = d(ε−1 ϕk , 0) → 0 ,
so there is k2 such that for k ≥ k2 , ϕ0 + ε−1 ϕk ∈ B(ϕ0 , r0 ). Hence for k ≥ max{k1 , k2 } we
have
|uk [ε−1 ϕk ]| ≤ |uk [ϕ0 + ε−1 ϕk ]| + |uk [ϕ0 ]| ≤ M + ε ,
so |uk [ϕk ]| ≤ ε(M + ε).
Theorem 5.11. If uk → u in Sn0 , then there is a constant C > 0 and a positive integer
m such that
X
sup |uk [ϕ]| ≤ C
pα,β (ϕ) for all ϕ ∈ Sn .
k
|α|,|β|≤m
Proof. Let vk = uk − u. Clearly vk → 0 in Sn0 . It suffices to prove that there is C > 0 and
a positive integer m such that
X
sup |vk [ϕ]| ≤ C
pα,β (ϕ) for all ϕ ∈ Sn .
k
|α|,|β|≤m
Indeed, this estimate, the inequality
sup |uk [ϕ]| ≤ |u[ϕ]| + sup |vk [ϕ]|
k
and Theorem 5.3 readily yield the result.
k
HARMONIC ANALYSIS
59
As in the proof of Theorem 5.3 it suffices to prove that there is a positive integer m such
that supk |vk [ϕ]| ≤ 1 for
n
X
1o
ϕ ∈ Nm := ϕ ∈ Sn :
pα,β (ϕ) ≤
.
m
|α|,|β|≤m
To the contrary suppose that for any positive integer i we can find ϕi ∈ Ni such that
supk |vk [ϕi ]| > 1, i.e.
|vki [ϕi ]| > 1 for some ki .
(5.5)
Clearly ϕi → 0 in Sn . If the sequence ki is bounded, (5.5) contradicts continuity of
functionals vk . If ki is unbounded, then there is a subsequence kij → ∞ and again (5.5)
contradicts Theorem 5.9.
Theorem 5.12. C0∞ (Rn ) is dense in Sn0 , i.e. if u ∈ Sn0 , then there is a sequence wk ∈
C0∞ (Rn ) such that
wk [ψ] → u(ψ) for all ψ ∈ Sn .
R
Proof. Let u ∈ Sn0 . If ϕ ∈ C0∞ (Rn ), Rn ϕ = 1, then for every ψ ∈ Sn , ϕ̃ε ∗ ψ → ψ in Sn 40
so
(u ∗ ϕε )[ψ] = u[ϕ̃ε ∗ ψ] → u[ψ] .
∞
Since uε = u ∗ ϕε ∈ C we obtain a sequence of smooth functions that converge to u in
Sn0 . If we take a cut-off function η ∈ C0∞ (Rn ), 0 ≤ η ≤ 1, η(x) = 1 for |x| ≤ 1 and η(x) = 0
for |x| ≥ 2, then one can easily prove that for ψ ∈ Sn
η(x/k)ψ(x) → ψ(x) in Sn as k → ∞.
Using this fact and Theorem 5.941 we obtain that wk (x) = η(x/k)(u∗ϕk−1 ) ∈ C0∞ converges
to u in Sn0 . Indeed,
wk [ψ] = (u ∗ ϕk−1 )[η(·/k)ψ(·)] → u[ψ].
If ϕ, ψ ∈ Sn , then the integration by parts gives
Z
Z
α
|α|
D ϕ(x)ψ(x) dx = (−1)
ϕ(x)Dα ψ(x) dx,
Rn
For h ∈ Rn we have
Z
Z
(τh ϕ)(x)ψ(x) dx =
Rn
Moreover
Dα ϕ[ψ] = ϕ[(−1)α ψ].
Rn
ϕ(x)(τ−h ψ)(x) dx,
τh ϕ[ψ] = ϕ[τ−h ψ].
Rn
Z
Z
ϕ̃(x)ψ(x) dx =
ZRn
ϕ̂(x)ψ(x) dx =
ZRn
ϕ̃[ψ] = ϕ[ψ̃],
ϕ(x)ψ̂(x) dx,
ϕ̂[ψ] = ϕ[ψ̂],
ϕ(x)ψ̌(x) dx,
ϕ̌[ψ] = ϕ[ψ̌].
ZRn
ϕ̌(x)ψ(x) dx =
Rn
ϕ(x)ψ̃(x) dx,
ZRn
Rn
40Why?
41Instead
of using Theorem 5.9 which is somewhat not constructive as it is based on the BanachSteinhaus theorem, we could prove convergence by a brute force estimates.
60
PIOTR HAJLASZ
If η ∈ C ∞ is slowly increasing and all derivatives of η are slowly increasing, i.e. every
derivative Dα η is bounded by a polynomial, then for ψ ∈ Sn , ηψ ∈ Sn and the mapping
ψ 7→ ηψ is continuous in Sn . In particular xα ψ(x) ∈ Sn . Moreover
Z
Z
(ηϕ)[ψ] =
η(x)ϕ(x) ψ(x) dx =
ϕ(x) η(x)ψ(x) dx = ϕ[ηψ].
Rn
Rn
This motivates the following definition.
Definition 5.13. For u ∈ Sn0 we define
• The distributional partial derivative Dα u is a tempered distribution defined by the
formula
Dα u[ψ] = u[(−1)|α| Dα ψ].
• The translation τh u ∈ Sn0 is defined by
(τh u)[ψ] = u[τ−h ψ].
• The reflection ũ ∈ Sn0 is defined by
ũ[ψ] = u[ψ̃].
• The Fourier transform F(u) = û ∈ Sn0 and the inverse Fourier transform ǔ ∈ Sn0
are
û[ψ] = u[ψ̂] and ǔ[ψ] = u[ψ̌].
• If η ∈ C ∞ is slowly increasing and all derivatives of η are slowly increasing, then
we define
(ηu)[ψ] = u[ηψ]. In particular (xα u)[ψ] = u[xα ψ].
The formulas preceding the definition show that on the subclass Sn ⊂ Sn0 the partial
derivative, the translation, the reflection, the Fourier transform and the multiplication by
a function defined in the distributional sense coincide with those defined in the classical
way.
If f ∈ L1 (Rn )+L2 (Rn ), in particular, if f ∈ Lp (Rn ), 1 ≤ p ≤ 2, then the classical Fourier
transform coincides with the distributional one. That easily follows from Theorem 3.2(c)
and Proposition 3.46.
The basic properties of the Fourier transform, distributional derivative and convolution
in Sn0 are collected in the next result whose easy proof is left to the reader.
Theorem 5.14. The Fourier transform in a homeomorphism of Sn0 onto itself.42 Moreover
for u ∈ Sn0 and ϕ ∈ Sn we have
(a)
(b)
(c)
(d)
(e)
(û)∨ = u,
(u ∗ ϕ)ˆ = ϕ̂û,
Dα (u ∗ ϕ) = Dα u ∗ ϕ = u ∗ Dα ϕ,
(Dα u)ˆ = (2πix)α û,
((−2πix)α u)ˆ = Dα û.
42With
respect to the weak convergence in Sn0 .
HARMONIC ANALYSIS
61
Just for the illustration we will prove the property (d). For any ψ ∈ Sn we have
(Dα u)∧ [ψ] = u[(−1)|α| Dα ψ̂] = u[(−1)|α| ((−2πix)α ψ)∧ ]
= u[((2πix)α ψ)∧ ] = (2πix)α û [ψ].
Note that in the case (c), u ∗ ϕ ∈ Sn0 and Dα (u ∗ ϕ) is understood in the distributional
sense. On the other hand u ∗ ϕ ∈ C ∞ and Theorem 5.8 shows that the distributional
derivative of u ∗ ϕ coincides with the classical one.
Example 5.15. The Dirac measure δa defines a tempered distribution so
Z
δ̂a [ψ] = δa [ψ̂] = ψ̂(a) =
ψ(x)e−2πix·a dx
Rn
and hence δ̂a = e−2πix·a .
Example 5.16. Let µ be a measure on R defined by µ(E) = #(E ∩ Z), where #A stands
for the cardinality of a set A. In other words µ is the counting measure on Z trivially
extended to R. Clearly, µ is a tempered measure so it defines a tempered distribution.
Moreover µ̂ = µ in S10 . Indeed, this is a simple consequence of the Poisson summation
formula Theorem 4.1 that
∞
∞
X
X
µ̂[ϕ] = µ[ϕ̂] =
ϕ̂(n) =
ϕ(n) = µ[ϕ].
n=−∞
n=−∞
5.2. Tempered distributions as derivatives of functions. Using the notion of distributional derivative we can provide a new class of examples of a distributions in Sn0 . If fα ,
|α| ≤ m are slowly increasing functions and aα ∈ C for |α| ≤ m, then
X
(5.6)
u=
aα Dα fα ∈ Sn0 .
|α|≤m
where the derivative Dα fα is understood in the distributional sense.
The aim of this section is to prove a surprising result that every distribution in Sn0 can
be represented in the form (5.6), see Theorem 5.21.
Proposition 5.17. Let N be a positive integer, then the operators43
(I − ∆)N : Sn → Sn
and
(I − ∆)N : Sn0 → Sn0
are homeomorphisms. The inverse operator in both cases is given by the operator
(I − ∆)−N := F −1 (1 + 4π 2 |ξ|2 )−N F .
Remark 5.18. That means for every v ∈ Sn (v ∈ Sn0 ) the equation (I − ∆)N u = v has
a unique solution u ∈ Sn (u ∈ Sn0 ) given by
u = (I − ∆)N v = F −1 (1 + 4π 2 |ξ|2 )−N F(v) .
43I
− ∆ stands for the identity minus the Laplace operator.
62
PIOTR HAJLASZ
Proof. Suppose that u, v ∈ Sn satisfy (I − ∆)N u = v. Taking the Fourier transform yields
(1 + 4π 2 |ξ|2 )N F(u) = F(v)
and hence
F(u) = (1 + 4π 2 |ξ|2 )−N F(v).
Since F(u) ∈ Sn , the right hand side is also in Sn so we can take the inverse Fourier
transform and we obtain
(5.7)
u = F −1 (1 + 4π 2 |ξ|2 )−N F(v) .
Thus if a solution u ∈ Sn exists, it is unique and given by (5.7). To see that the formula
indeed, defines a solution u ∈ Sn observe that the function (1 + 4π 2 |ξ|2 )−N and all its
derivatives are slowly increasing so if f ∈ Sn , then (1 + 4π 2 |ξ|2 )−N f ∈ Sn . Hence for any
v ∈ Sn , the right hand side of (5.7) defines a function in Sn and backward computations
shows that (I − ∆)N u = v.
Similarly, if u, v ∈ Sn0 , then (I − ∆)N u = v if and only if the Fourier transforms are
equal i.e.,
(1 + 4π 2 |ξ|2 )N F(u) = F(v)
which is equivalent to
F(u) = (1 + 4π 2 |ξ|2 )−N F(v), u = F −1 (1 + 4π 2 |ξ|2 )−N F(v) .
Note that we were allowed to multiply both sides by (1 + 4π 2 |ξ|2 )−N because this function
and all its derivatives are slowly increasing.
The operator (I −∆)N maps functions of class C k to functions of class C k−2N so it lowers
regularity of functions. That means the inverse operator (I − ∆)−N increases regularity.
Hence we should expect that it also increases regularity of distributions. Since every tempered distribution has finite order in a sense that it can be estimated by a finite sum as in
Theorem 5.3, one could expect that for u ∈ Sn0 and sufficiently large N , the distribution
(I − ∆)−N u is actually a regular function. As we will see this intuition is correct.
Remark 5.19. The operator (I − ∆)−N is called a Bessel potential and one can prove that
it can be represented as an integral operator. In Section ?? we will study Bessel potentials
in detail. We will find an explicit integral formula for the Bessel potentials and we will
show how use them to characterize Sobolev spaces.
k,1
Definition 5.20. By Cloc
we will denote the class of k times continuously differentiable
0,1
function whose derivatives of order k are locally Lipschitz continuous. In particular Cloc
denotes the class of locally Lipschitz continuous functions.
Theorem 5.21. Suppose u ∈ Sn0 satisfies the estimate44
X
|u(ϕ)| ≤ C
pα,β (ϕ) .
|α|,|β|≤m
If k is a nonnegative integer and N > (n + m + k)/2, then the tempered distribution
v = (I − ∆)−N u
has the following properties:
44See
Theorem 5.3.
HARMONIC ANALYSIS
63
(a) If k = 0, then v is a slowly increasing function.
k−1,1
. Also all deriva(b) If k ≥ 1, then v is a slowly increasing function of the class Cloc
tives of v or order |α| ≤ k are slowly increasing.
Proof. We will need
Lemma 5.22. If P (x) is a polynomial in Rn of degree p and N > (n + p)/2, then all
derivatives of
P (x)
f (x) =
(1 + |x|2 )N
belong to L1 (Rn ).
Proof. Since 2N − p > n, f ∈ L1 (Rn ). We have
Q(x)
∂f
=
,
∂xi
(1 + |x|2 )2N
deg Q = 2N + p − 1.
The function on the right hand side is of the same form as f . Since
n + (2N + p − 1)
2
1
we conclude that ∂f /∂xi ∈ L . The integrability of higher order derivatives follows by
induction.
2
2N >
We will prove that the distributional derivatives of v of orders |γ| ≤ k are slowly increasing functions. We have
(−1)|γ| Dγ v[ψ] = u F (1 + 4π 2 |ξ|2 )−N F −1 (Dγ ψ) .
Hence
|α|,|β|≤m
X
= C
|α|,|β|≤m
≤ C
sup xα Dβ F (1 + 4π 2 |ξ|2 )−N F −1 (Dγ ψ) X
|Dγ v[ψ]| ≤ C
x∈Rn
ξ β+γ
α
−1
F (ψ) Cα,β,γ sup F D
2
2
N
(1 + 4π |ξ| )
x∈Rn
β+γ
X x
α
−1
D
F (ψ) .
2
2
N
(1 + 4π |x| )
0
|α|,|β|≤m
1
Note that
xβ+γ
−1
D
F (ψ)
(1 + 4π 2 |x|2 )N
X
α!
xβ+γ
αi
=
D
Dβi (F −1 (ψ)) .
2
2
N
α1 !βi !
(1 + 4π |x| )
α +β =α
α
i
i
Since deg xβ+γ ≤ m + k and N > (n + m + k)/2, Lemma 5.22 gives
xβ+γ
αi
D
∈ L1 (Rn ) ,
(1 + 4π 2 |x|2 )N
64
PIOTR HAJLASZ
so
|Dγ v[ψ]| ≤ C
X
≤ C
X
kF −1 (xβi ψ)k∞
i
kxβi ψk1
i
0
≤ C k(1 + |x|2 )m/2 ψ(x)k1 .
This proves that for |γ| ≤ k the functional
ψ 7→ Dγ v[ψ]
is bounded on L1 ((1 + |x|2 )m/2 dx). Thus there are functions gγ ∈ L∞ such that
Z
γ
ψ(x)gγ (x)(1 + |x|2 )m/2 dx ,
D v[ψ] =
Rn
i.e.
Dγ v = gγ (x)(1 + |x|2 )m/2
in the distributional sense. In particular, if γ = 0, v(x) = g0 (x)(1 + |x|2 )m/2 is slowly
increasing which proves (a). The part (b) follows from the following result whose proof is
postponed to Section ??, see Theorem ??
Proposition 5.23. If a function u is slowly increasing and its distributional derivatives
k−1,1
of order less than or equal k, k ≥ 1, are slowly increasing functions, then u ∈ Cloc
.
5.3. Tempered distributions with compact support. We will investigate now properties of the Fourier transform of tempered distributions with compact support.
Lemma 5.24. Let Ω ⊂ Rn be open and u ∈ Sn0 . If u[ϕ] = 0 for all ϕ ∈ C0∞ (Ω), then
u[ϕ] = 0 for all ϕ ∈ Sn such that supp ϕ ⊂ Ω.
Proof. Let ϕ ∈ Sn , supp ϕ ⊂ Ω and let η be a cut-off function, i.e. η ∈ C0∞ (Rn ), 0 ≤ η ≤ 1,
η(x) = 1 for |x| ≤ 1 and η(x) = 0 for |x| ≥ 2. One can easily prove that η(x/R)ϕ(x) → ϕ(x)
in Sn as R → ∞. Since η(x/R)ϕ(x) ∈ C0∞ (Ω), the lemma follows.
Definition 5.25. Let u ∈ Sn0 . The support of u (supp u) is the intersection of all closed
sets E ⊂ Rn such that
ϕ ∈ C0∞ (Rn \ E) ⇒ u[ϕ] = 0.
Proposition 5.26. If u ∈ Sn0 and ϕ ∈ C0∞ (Rn \ supp u), then u[ϕ] = 0.
Proof. Indeed, complement of supp u is the union of all openSsets Ui such that if ϕ ∈
C0∞ (Ui ), then u[ϕ] = 0, and we need to prove that if ϕ ∈ C0∞ ( i Ui ), then still u[ϕ] = 0.
Since the support of ϕ is compact we can select a finite subfamily of open sets Ui covering
the support and the result follows from a simple argument involving partition of unity. Thus the support of u is the smallest closed set such that the distribution vanishes on
C0∞ functions supported outside that set.
The lemma shows that we can replace ϕ ∈ C0∞ (Rn \E) in the above definition by ϕ ∈ Sn
with supp ϕ ⊂ Rn \ E.
HARMONIC ANALYSIS
65
Before we state the next result we need some facts about analytic and holomorphic
functions in several variables.
Definition 5.27. We say that a function f : Ω → C defined in an open set Ω ⊂ Rn is
R-analytic, if in a neighborhood on any point x0 ∈ Ω it can be expanded as a convergent
power series
X
(5.8)
f (x) =
aα (x − x0 )α , aα ∈ C,
α
i.e. if in a neighborhood of any point f equals to its Taylor series.
We say that a function f : Ω → C defined in an open set Ω ⊂ Cn is C-analytic if in a
neighborhood of any point z0 ∈ Ω it can be expanded as a convergent power series
X
aα (z − z0 )α .
f (z) =
α
Rn has a natural embedding into Cn , just like R into C.
Rn 3 x = (x1 , . . . , xn ) 7→ (x1 + i · 0, . . . , xn + i · 0) = x + i · 0 ∈ Cn .
It is easy to see that an R-analytic function f in Ω ⊂ Rn extends to a C-analytic function
f˜ in an open set Ω̃ ⊂ Cn , Ω ⊂ Ω̃. Namely, if f satisfies (5.8), we set
X
f˜(z) =
aα (z − z0 )α , z0 = x0 + i · 0.
α
On the other hand, if f˜ is C-analytic in Ω̃ ⊂ Cn , then the restriction f of f˜ to Ω = Ω̃ ∩ Rn
is R-analytic.
For example for any ξ ∈ Rn , f (x) = ex·ξ is R-analytic and f˜(z) = ez·ξ is its C-analytic
extension.
Definition 5.28. We say that a continuous function f : Ω → C defined in an open set
Ω ⊂ Cn is holomorphic if
∂f
= 0 for i = 1, 2, . . . , n.
∂z i
It is easy to see that C-analytic function are holomorphic, but the converse implication
is also true.
Lemma 5.29 (Cauchy). If f is holomorphic in45
Dn (w, r) = D1 (w1 , r1 ) × . . . × D1 (wn , rn ) ⊂ Cn
and continuous in the closure Dn (x, r), then
Z
Z
1
f (ξ) dξ1 . . . dξn
(5.9)
f (z) =
...
n
(2πi) ∂D1 (w1 ,r1 )
∂D1 (wn ,rn ) (ξ1 − z1 ) . . . (ξn − zn )
for all z ∈ Dn (w, r).
45Product
of one dimensional discs.
66
PIOTR HAJLASZ
Proof. The function f is holomorphic in each variable separately, so (5.9) follows from
one dimensional Cauchy formulas and the Fubini theorem.
2
Just like in the case of holomorphic functions of one variable one can prove that the
integral on the right hand side of (5.9) can be expanded as a power series and hence
defines a C-analytic function. We proved
Theorem 5.30. A function f is holomorphic in Ω ⊂ Cn if and only if it is C-analytic.
Now we can state an important result about compactly supported distributions.
Theorem 5.31. If u ∈ Sn0 has compact support, then û is a slowly increasing C ∞ function
and all derivatives of û are slowly increasing. Moreover û is R-analytic on Rn and has a
holomorphic extension to Cn .
Proof. Let η ∈ C0∞ (Rn ) be such that η(x) = 1 in a neighborhood of supp u. Then u = ηu
in Sn0 and hence for ψ ∈ Sn we have
Z
−2πix·(·)
η(·)ψ(x)e
dx
û[ψ] = (ηu)ˆ[ψ] = u
Rn
Z
=
u η(·)e−2πix·(·) ψ(x) dx .
Rn
We could pass with u under the sign of the integral, because of an argument with approximation of the integral by Riemann sums.46
Note that the function
F (x1 , . . . , xn ) = u η(·)e−2πix·(·)
is C ∞ smooth and
Dα F (x1 , . . . , xn ) = u (−2πi(·))α η(·)e−2πix·(·) .
Indeed, we could differentiate under the sign of u because the corresponding difference
quotients converge in the topology of Sn . It also easily follows from Theorem 5.3 that F
and all its derivatives are slowly increasing. Thus we may identify û with F , so û ∈ C ∞ .
Moreover F has a holomorphic extension to Cn by the formula
F (z1 , . . . , zn ) = u η(·)e−2πiz·(·)
so in particular F (x1 , . . . , xn ) is R-analytic.
2
Remark 5.32. Note that if u ∈ Sn0 has compact support, then we can reasonably define
u[e−2πix·(·) ] by the formula
u e−2πix·(·) := u η(·)e−2πix·(·) ,
where η ∈ C0∞ (Rn ) is such that η = 1 in a neighborhood of supp u. Note that this construction does not depend on the choice of η. In the same way we can define u[f ] for any
f ∈ C ∞ (Rn )
46Compare
with the proof of Theorem 5.7.
HARMONIC ANALYSIS
67
Theorem 5.33. If u ∈ Sn0 and supp u = {x0 }, then there is an integer m ≥ 0 and complex
numbers aα , |α| ≤ m such that
X
u=
aα D α δx 0 .
|α|≤m
Proof. Without loss of generality we may assume that x0 = 0. According to Theorem 5.3
the distribution u satisfies the estimate
X
|u[ϕ]| ≤ C
pα,β (ϕ) .
|α|,|β|≤m
First we will prove that for ϕ ∈ Sn we have
Dα ϕ[0] = 0 for |α| ≤ m
(5.10)
=⇒
u[ϕ] = 0 .
Indeed, it follows from Taylor’s formula that
ϕ(x) = O(|x|m+1 ) as x → 0
and hence also
Dβ ϕ(x) = O(|x|m+1−|β| ) as x → 0 for all |β| ≤ m.
(5.11)
Let η ∈ C0∞ (Rn ) be a cut-off function, i.e. 0 ≤ η ≤ 1, η(x) = 1 for |x| ≤ 1, η(x) = 0 for
|x| ≥ 2 and define ηε (x) = η(x/ε). The estimate (5.11) easily implies that
pα,β (ηε ϕ) → 0 as ε → 0
47
for all |α|, |β| ≤ m.
hence
Note that ϕ − ηε ϕ = 0 in a neighborhood of 0, so u[ϕ − ηε ϕ] = 0 and
X
|u[ϕ]| ≤ |u[ϕ − ηε ϕ]| + |u[ηε ϕ]| ≤ 0 +
pα,β (ηε ϕ) → 0 as ε → 0.
|α|,|β|≤m
This completes the proof of (5.10).
Let now ψ ∈ Sn be arbitrary and let
h(x) = ψ(x) −
X Dα ψ(0)
xα .
α!
|α|≤m
Clearly
Dα h(0) = 0 for |α| ≤ m.
(5.12)
We have


X Dα ψ(0)
ψ(x) = η(x) 
xα  + η(x)h(x) + (1 − η(x))ψ(x) .
α!
|α|≤m
Since (1 − η)ψ vanishes in a neighborhood of 0 we have u[(1 − η)ψ] = 0. The equality (5.12)
implies that ϕ = ηh ∈ C0∞ (Rn ) satisfies the assumptions of (5.10), so u[ηh] = 0. Hence
h
X Dα ψ(0) i
u[ψ] = u η(x)
xα
α!
|α|≤m
47Check
it!
68
PIOTR HAJLASZ
η(x)xα
=
(−1) u
(−1)|α| Dα ψ(0) .
|
{z
}
α!
|α|≤m |
{z
}
Dα δ0 [ψ]
X
|α|
aα
The proof is complete.
2
Corollary 5.34. Let u ∈ Sn0 . If supp û = {ξ0 }, then u = P (x)e2πix·ξ0 for some polynomial
P (x) in Rn . In particular if supp û = {0}, then u is a polynomial.
We leave the proof as an exercise.
Corollary 5.35. If u ∈ Sn0 satisfies ∆u = 0, then u is a polynomial.
Proof. We have
−4π 2 |ξ|2 û = (∆u)ˆ = 0 in Sn0 .
This implies that supp û = {0}, so u is a polynomial by Corollary 5.34.
2
5.4. Homogeneous distributions. A very important class of distributions is provided
by so called homogeneous distributions that we will study now.
Definition 5.36. A function f ∈ C(R \ {0}) is homogeneous of degree a ∈ R if for all
x 6= 0 and all t > 0 we have f (tx) = ta f (x). More generally a measurable function on Rn
is homogeneous of degree a ∈ R if
f (tx) = ta f (x) for almost all (x, t) ∈ Rn × (0, ∞).
If f ∈ C(Rn \ {0}) is homogeneous of degree a > −n, then it defines a tempered
distribution. Indeed,
a
x |f (x)| = |x| f
≤ C|x|a
|x| and |x|a is integrable in a neighborhood of the origin (because a > −n).
If f is a measurable homogeneous function of degree a we can assume (by changing f
on a set of measure zero) that f is continuous on almost all open rays going out of 0. Then
on the remaining rays (which form a set of measure zero) we can make f homogeneous.
This is to say that we can find a representative of f (in class of functions that are equal
a.e.) such that
f (tx) = ta f (x) for all x 6= 0 and t > 0.
If a measurable function that is not a.e. equal to zero is homogeneous of degree a ≤ −n,
then the integration in the spherical coordinate system shows that it is not integrable in
any neighborhood of 0 and hence it cannot define a tempered distribution.
Lemma 5.37. Suppose that f ∈ L1loc (Rn ) defines a tempered distribution. Then the function f is homogeneous of degree a ∈ R if and only if 48
(5.13)
f [ϕt ] = ta f [ϕ] for all ϕ ∈ Sn and all t > 0.
Remark 5.38. If f is not equal to zero a.e., then necessarily a > −n.
48As
always ϕt (x) = t−n ϕ(x/t).
HARMONIC ANALYSIS
Proof. For ϕ ∈ Sn we have
Z
Z
f [ϕt ] =
f (x)ϕt (x) dx =
Rn
69
a
Z
f (tx)ϕ(x) dx and t f [ϕ] =
Rn
ta f (x)ϕ(x) dx.
Rn
Hence condition (5.13) is equivalent to the equality
Z
Z
f (tx)ϕ(x) dx =
ta f (x)ϕ(x) dx for all ϕ ∈ Sn and all t > 0
(5.14)
Rn
Rn
which, in turn, is equivalent to the fact that for all t > 0, equality f (tx) = ta f (x) holds
a.e.
The above result motivates the following definition.
Definition 5.39. We say that a distribution u ∈ Sn0 is homogeneous of degree a ∈ R if
49
u[ϕt ] = ta u[ϕ] for all ϕ ∈ Sn and all t > 0.
Proposition 5.40. u ∈ Sn0 is homogeneous of degree a ∈ R if and only if û ∈ Sn0 is
homogeneous of degree −n − a.
Proof. Let ϕ ∈ Sn . Then
ϕ
bt (ξ) = ϕ̂(tξ) = t−n (t−1 )−n ϕ̂(ξ/t−1 ) = t−n (ϕ̂)t−1 (ξ)
and hence
û[ϕt ] = u[ϕ
bt ] = t−n u[(ϕ̂)t−1 ] = t−n−a u[ϕ̂] = t−n−a û[ϕ].
If u ∈ Sn0 is homogeneous of degree a, the above equality shows that û is homogeneous of
degree −n − a. Now if û is homogeneous of degree −n − a, then its Fourier transform i.e.,
ũ is homogeneous of degree −n − (−n − a) = a and hence also u is homogeneous of degree
a.
Example 5.41. u(x) = |x|a , a ≥ 0, is homogeneous of degree a and it defines a tempered
distribution. However, û is homogeneous of degree −n − a ≤ −n and hence it cannot be
represented as a function.
The notion of a homogeneous distribution will play a significant role in the proof of the
next result.
Theorem 5.42. For 0 < a < n,
n−a
a− n
2Γ
π
2 |ξ|a−n .
F |x|−a (ξ) =
Γ a2
Remark 5.43. |x|−a is homogeneous of degree −a > −n and |ξ|a−n is homogeneous of
degree a − n > −n so both |x|−a and |ξ|a−n are tempered distributions represented as
functions.
49Now
a can be any real number and we do not impose restriction a > −n that was necessary in the
case of functions.
70
PIOTR HAJLASZ
Proof. First we will prove the theorem under the assumption that n/2 < a < n. In that
case
|x|−a = |x|−a χ{|x|≤1} + |x|−a χ{|x|>1} ∈ L1 + L2 .
Thus the Fourier transform of |x|−a is a function in C0 + L2 .
The Fourier transform commutes with orthogonal transformations
f[
◦ ρ = fˆ ◦ ρ,
for ρ ∈ O(n) and f ∈ L1 + L2 .
If f is radially symmetric i.e., f = f ◦ ρ for all ρ ∈ O(n), then fˆ ◦ ρ = f[
◦ ρ = fˆ for all
ρ ∈ O(n) so fˆ is radially symmetric too. Thus the Fourier transform of |x|−a ∈ L1 + L2 is
a radially symmetric function in C0 + L2 , homogeneous of degree a − n as a distribution.
Hence Lemma 5.37 yields
F |x|−a (ξ) = ca,n |ξ|a−n
2
and it remains to compute the coefficient ca,n . Employing the fact that ϕ(x) = e−π|x| is a
fixed point of the Fourier transform we have
ca,n |x|a−n [ϕ] = F(|x|−a )[ϕ] = |x|−a [ϕ̂] = |x|−a [ϕ],
i.e.,
Z
(5.15)
ca,n
e
−π|x|2
a−n
|x|
Z
2
e−π|x| |x|−a dx.
dx =
Rn
Rn
The integrals in this identity are easy to compute.
Lemma 5.44. For γ > −n we have
Z
n+γ
Γ
2
e−π|x| |x|γ dx = γ 2 n .
π2Γ 2
Rn
Proof. For γ > −n we have50
Z
2
e−π|x| |x|γ dx
∞
Z
=
2
e−πs sγ+n−1 ds
nωn
Rn
0
t=πs2
=
=
nωn
Z
n+γ
2
2π
nωn
2π
n+γ
2
∞
e−t t
n+γ
−1
2
dt
0
Γ
n+γ
2
=
n+γ
2
γ
π 2 Γ n2
Γ
.
Applying the lemma to both sides of (5.15) yields
Γ a2
Γ n−a
Γ n−a
a− n
2 2
.
ca,n a−n n = − a n
so ca,n = π 2
Γ a2
π 2Γ 2
π 2 Γ 2
This completes the proof when n/2 < a < n.
50Recall
that the volume of the unit sphere in Rn equals nωn so the first equality is just integration in
2π n/2
the spherical coordinates. In the last equality we use ωn = nΓ(n/2)
, see (3.9).
HARMONIC ANALYSIS
71
Suppose now that 0 < a < n/2. Since n/2 < n − a < n we have
n
π 2 −a Γ a2
−(n−a)
|ξ|−a
(ξ) =
F |x|
n−a
Γ 2
and applying the Fourier transform to this equality yields
n
π 2 −a Γ a2
a−n
F |ξ|−a (x).
| − x|
=
n−a
Γ 2
Replacing x by ξ and ξ by x yields the result when 0 < a < n/2.
We are left with the case of a = n/2. Take a sequence n/2 < ak < n, ak → a = n/2.
Since the result is true for ak we have
Z
Z
n−ak
ak − n
2Γ
π
2
|x|−ak ϕ̂(x) dx =
|x|ak −n ϕ(x) dx.
ak
n
n
Γ
R
R
2
Passing to the limit ak → a = n/2 yields the result for a = n/2.
Remark 5.45. Clearly |ξ|ak −n → |ξ|a−n and |x|−ak → |x|−a in Sn0 . The last convergence
and the continuity of the Fourier transform yields that F(|x|−ak ) → F(|x|−a ) in Sn0 . Hence
the result for a = n/2 follows by passing to the limit in the equality
n
k
π ak − 2 Γ n−a
−ak
2
F |x|
(ξ) =
|ξ|ak −n in Sn0 .
ak
Γ 2
Remark 5.46. In the case a = n/2 we obtain a particularly simple formula
Z
Z
n
−n
−n
−n
2
2
2
|x| ϕ̂(x) dx =
|x|− 2 ϕ(x) dx
F(|x| ) = |x| , i.e.,
Rn
Rn
for all ϕ ∈ Sn . This formula can be regarded as a version of the Poisson summation
formula.51
5.5. The principal value.
Proposition 5.47. If u ∈ Sn0 is homogeneous of degree a, then Dα u is homogeneous of
degree a − |α|.
We leave the proof as an (easy) exercise.
In particular if f ∈ C ∞ (Rn \ {0}) is homogeneous of degree 1 − n, then it defines
a distribution, but its first order distributional partial derivatives are homogeneous of
degree −n and hence they cannot be represented as functions. It is however, still possible
to represent these distributional derivatives as improper integrals, the so called principal
value of the integral.
Definition 5.48. If for any ε > 0, a function f is integrable in Rn \ B(0, ε), then we define
the principal value of the integral as
Z
Z
p.v.
f (x) dx = lim
f (x) dx
Rn
provided the limit exists.
51cf.
Theorem 4.1.
ε→0
|x|≥ε
72
PIOTR HAJLASZ
If a function f is slowly increasing, but has singularity at 0, then we can try to define a
distribution p.v. f by
Z
Z
(p.v. f )[ϕ] = p.v.
f (x)ϕ(x) dx = lim
f (x)ϕ(x) dx for ϕ ∈ Sn .
ε→0
Rn
|x|≥ε
In many interesting cases p.v. f defines a tempered distribution, but not always.
Theorem 5.49. Let
KΩ (x) =
Ω(x/|x|)
,
|x|n
x 6= 0,
where Ω ∈ L1 (S n−1 ) satisfies
Z
Ω(θ) dσ(θ) = 0.
S n−1
Then
p.v. KΩ ∈ Sn0
and
|(p.v. KΩ )[φ]| ≤ C(n)kΩkL1 (S n−1 ) k∇ϕk∞ + k|x|ϕ(x)k∞
for all ϕ ∈ Sn .
Remark 5.50. Distributions p.v. KΩ ∈ Sn0 described in the above result will be called
distributions of the p.v. type or simply p.v. distributions.
Remark 5.51. Note that the function KΩ is homogeneous of degree −n so the function
KΩ does not define a tempered distribution.
R
Remark 5.52. It is not difficult to verify that if S n−1 Ω(θ) dσ(θ) 6= 0, then p.v. KΩ
does not define a tempered distribution because p.v. KΩ [ϕ] will diverge for some functions
ϕ ∈ Sn .
Proof. For ϕ ∈ Sn we have
Z
Z
Ω(x/|x|)
Ω(x/|x|)
(p.v. KΩ )[ϕ] = lim
(ϕ(x) − ϕ(0)) +
ϕ(x) dx
ε→0
n
n
|x|
|x|
|x|≥1
ε≤|x|≤1
Z
Z
|Ω(x/|x|)|
|Ω(x/|x|)|
dx + kϕ(y)|y|k∞
dx
≤ k∇ϕk∞
n−1
|x|
|x|n+1
|x|≤1
|x|≥1
and the result follows upon integration in spherical coordinates. We could subtract ϕ(0)
R
in the first integral because ε≤|x|≤1 Ω(x/|x|)
dx = 0. In the last step used the following
|x|n
estimates
Z 1
Z 1
d
|ϕ(x) − ϕ(0)| = ϕ(tx) dt = ∇ϕ(tx) · x dt ≤ |x| k∇ϕk∞
dt
0
0
and
|ϕ(x)| =
|ϕ(x)| |x|
kϕ(y)|y|k∞
≤
.
|x|
|x|
HARMONIC ANALYSIS
73
The next result illustrates a simple situation when the principal value appears in a
natural way. Note that the function xj /|x|n+1 satisfies the assumptions of Theorem 5.49.
However, the proof of Proposition 5.53 will be straightforward and we will not use Theorem 5.49.
Proposition 5.53. For n ≥ 2, the distributional partial derivatives of |x|1−n ∈ Sn0 satisfy
∂
xj
1−n
|x|
= (1 − n) p.v.
∈ Sn0 ,
n+1
∂xj
|x|
i.e.,
(5.16)
Z
xj
∂
1−n
|x|
[ϕ] = (1 − n) lim
ϕ(x) dx. for ϕ ∈ Sn .
ε→0 |x|≥ε |x|n+1
∂xj
Before we prove the proposition let us recall the integration by parts formula for functions
defined in a domain in Rn . If Ω ⊂ Rn is a bounded domain with piecewise C 1 boundary
and f, g ∈ C 1 (Ω), then
Z
Z
(5.17)
(∇f (x)g(x) + f (x)∇g(x)) dx =
f g~ν dσ ,
Ω
∂Ω
where ~ν = (ν1 , . . . , νn ) is the unit outer normal vector to ∂Ω. Comparing jth components
on both sides of (5.17) we have
Z
Z
Z
∂f
∂g
(5.18)
(x) g(x) dx = − f (x)
(x) dx +
f g νj dσ .
∂xj
Ω ∂xj
Ω
∂Ω
Proof of Proposition 5.53. Let A(ε, R) = {x : ε ≤ |x| ≤ R}. We have
Z
∂
∂ϕ
1−n
|x|
|x|1−n dx
[ϕ] = −
∂xj
Rn ∂xj
Z
∂ϕ
1−n
|x|
dx
= lim lim −
ε→0 R→∞
ε≤|x|≤R ∂xj
Z
Z
∂
1−n
ϕ(x)
= lim lim
|x|
dx −
ϕ(x)|x|1−n νj dσ
ε→0 R→∞
∂x
ε≤|x|≤R
∂A(ε,R)
| j {z }
(1−n)xj /|x|n+1
Z
xj
= lim (1 − n)
ϕ(x) n+1 −
ε→0
|x|
|x|≥ε
Z
ϕ(x)|x|1−n νj dσ
|x|=ε
Indeed, we could pass to the limit as R → ∞, because the part of the integral over ∂A(ε, R)
corresponding to {|x| = R} converges to zero. Since the integral of the function |x|1−n νj
over the sphere |x| = ε equals zero we have
Z
Z
1−n
1−n
ϕ(x)|x|
νj dσ = (ϕ(x) − ϕ(0))|x|
νj dσ ≤ Cε → 0 as ε → 0,
|x|=ε
|x|=ε
because
ϕ(x) − ϕ(0) |x|1−n νj ≤ Cε · ε1−n = ε2−n
and the surface area of the sphere {|x| = ε} equals Cεn−1 .
74
PIOTR HAJLASZ
Remark 5.54. We proved that p.v. (xj /|x|n+1 ) defines a tempered distribution somewhat
indirectly by showing that for ϕ ∈ Sn equality (5.16) is satisfied. Since ∂|x|1−n /∂xj ∈ Sn0 ,
equality (5.16) implies that p.v. (xj /|x|n+1 ) ∈ Sn0 .
Remark 5.55. When n = 1 the counterpart of Theorem 5.53 is that
d
x
1
(log |x|) = p.v. 2 = p.v.
dx
|x|
x
in Sn0 .
We leave the proof as an exercise.
We plan to generalize Proposition 5.53 to a class of more general functions. Let’s start
with the following elementary result.
Theorem 5.56. Suppose that K ∈ C 1 (Rn \ {0}) is such that both K and |∇K| have
polynomial growth52 for |x| ≥ 1 and there are constants C, α > 0 such that
|K(x)| ≤
C
|x|n−1−α
for 0 < |x| < 1,
C
for 0 < |x| < 1.
|x|n−α
Then K ∈ Sn0 and the distributional partial derivatives ∂K/∂xj , 1 ≤ j ≤ n, coincide with
pointwise derivatives, i.e. for ϕ ∈ Sn
Z
Z
∂ϕ
∂K
∂K
K(x)
[ϕ] := −
(x) dx =
(x) ϕ(x) dx .
∂xj
∂xj
Rn
Rn ∂xj
|∇K(x)| ≤
Proof. Let A(ε) = {x : ε ≤ |x| ≤ ε−1 }. From (5.18) we have
Z
∂K
∂ϕ
[ϕ] = lim −
K(x)
(x) dx
ε→0
∂xj
∂xj
ε≤|x|≤ε−1
Z
Z
∂K
= lim
(x) ϕ(x) dx −
K(x)ϕ(x) νj dσ(x) .
ε→0
ε≤|x|≤ε−1 ∂xj
∂A(ε)
Since
Z
lim
ε→0
ε≤|x|≤ε−1
∂K
(x) ϕ(x) dx =
∂xj
Z
Rn
∂K
(x) ϕ(x) dx
∂xj
it remains to show that
Z
lim
ε→0
K(x)ϕ(x) νj dσ = 0 .
∂A(ε)
The integral on the outer sphere |x| = ε−1 converges to 0 since K has polynomial growth
and ϕ rapidly converges to 0 as |x| → ∞ and on the inner sphere |x| = ε we have
Z
C
K(x)ϕ(x) νj dσ ≤ n−1−α εn−1 = Cεα → 0 as ε → 0.
ε
|x|=ε
The proof is complete.
52i.e.,
|K(x)| + |∇K(x)| ≤ C(1 + |x|)m for some C > 0, m ≥ 1 and all |x| ≥ 1.
HARMONIC ANALYSIS
75
An interesting problem is the case α = 0, i.e. when K and ∇K satisfy the estimates
(5.19)
|K(x)| ≤
C
,
|x|n−1
|∇K(x)| ≤
C
|x|n
for x 6= 0.
One such situation was described in Proposition 5.53.
Here we make an additional assumption about K. We assume that K ∈ C 1 (Rn \ {0}) is
homogenelus of degree 1 − n, i.e.
K(x) =
K(x/|x|)
|x|n−1
for x 6= 0.
Since K is bounded in {|x| ≥ 1} and integrable in {|x| ≤ 1} we have K ∈ Sn0 and there
is no need to interpret K through the principal value of the integral. The first estimate at
(5.19) is satisfied. To see that the second estimate if satisfied too we observe that ∇K is
homogeneous of degree −n. Indeed, for 1 ≤ j ≤ n and t > 0
∂
∂
∂K
(tx)t =
(K(tx)) = t1−n
K(x)
∂xj
∂xj
∂xj
and hence
(∇K)(tx) = t−n ∇K(x) .
Thus
(∇K)(x/|x|)
,
|x|n
from which the second estimate at (5.19) follows.
∇K(x) =
x 6= 0
Observe that ∇K(x) is not integrable at any neighborhood of 0, but we may try to consider the principal value of ∇K(x), i.e. the principal value of each of the partial derivatives
∂K/∂xj .
Theorem 5.57. Suppose that K ∈ C 1 (Rn \ {0}) is homogeneous of degree 1 − n. Then
∇K(x) is homogeneous of degree −n. Moreover
Z
(5.20)
∇K(θ) dσ(θ) = 0
S n−1
and
p.v. ∇K(x) ∈ Sn0
is a well defined tempered distribution, i.e. for each 1 ≤ j ≤ n
p.v.
∂K
(x) ∈ Sn0 .
∂xj
Finally the distributional gradient ∇K satisfies
(5.21)
∇K = cδ0 + p.v. ∇K(x) ,
|{z}
| {z }
dist.
pointwise
where
Z
c=
K(x)
S n−1
x
dσ(x) .
|x|
76
PIOTR HAJLASZ
In other words for ϕ ∈ Sn and 1 ≤ j ≤ n we have
Z
Z
∂K
∂ϕ
∂K
[ϕ] := −
K(x)
(x) dx = cj ϕ(0) + lim
(x)ϕ(x) dx ,
ε→0 |x|≥ε ∂xj
∂xj
∂xj
Rn
where
Z
cj =
K(x)
S n−1
xj
dσ(x) .
|x|
Remark 5.58. Note
that Proposition 5.53 follows from this result: since K(x) = |x|1−n ,
R
the constant c = S n−1 |x|−n x dσ(x) = 0.
Remark 5.59. The fact that p.v. ∇K ∈ Sn0 is a straightforward consequence of (5.20) and
Theorem 5.49. However, our argument will be direct without referring to Theorem 5.49.
Proof. We already checked that ∇K(x) is homogeneous of degree −n. For r > 1 let
A(1, r) = {x : 1 ≤ |x| ≤ r}. From the integration by parts formula (5.17) we have
Z
Z
K(x) ~ν (x) dσ(x)
∇K(x) dx =
∂A(1,r)
1≤|x|≤r
Z
Z
x
x
K(x)
= −
K(x)
dσ(x) +
dσ(x) = 0 .
|x|
|x|
|x|=1
|x|=r
Indeed, the last two integrals are equal by a simple change of variables and homogeneity of
K. Thus the integral on the left hand side equals 0 independently of r. Hence its derivative
with respect to r is also equal zero.
Z
Z
d 0= ∇K(x) dx =
∇K(θ) dσ(θ) .
dr r=1+ 1≤|x|≤r
|x|=1
This proves (5.20). As we will see (5.20) implies that p.v. ∇K(x) ∈ Sn0 is a well defined
tempered distribution as well as that the distributional gradient ∇K satisfies (5.21). Let
ϕ ∈ Sn . We have
Z
∇K[ϕ] := −
K(x)∇ϕ(x) dx
Rn
Z
= lim lim −
K(x)∇ϕ(x) dx
ε→0 R→∞
ε≤|x|≤R
Z
Z
= lim lim
∇K(x) ϕ(x) dx −
K(x)ϕ(x) ~ν (x) dσ(x)
ε→0 R→∞
ε≤|x|≤R
∂A(ε,R)
Z
Z
x
dσ(x) .
= lim
∇K(x) ϕ(x) dx +
K(x)ϕ(x)
ε→0
|x|
|x|≥ε
|x|=ε
It remains to prove that
Z
Z
x
x
lim
K(x)ϕ(x)
dσ(x) = ϕ(0)
K(x)
dσ(x) .
ε→0 |x|=ε
|x|
|x|
|x|=1
Let
Z
x
c=
dσ(x) =
K(x)
|x|
|x|=1
Z
K(x)
|x|=ε
x
dσ(x) .
|x|
HARMONIC ANALYSIS
77
The last equality follows from a simple change of variables and homogeneity of K. We have
Z
x
(5.22)
K(x)ϕ(x)
dσ(x)
|x|
|x|=ε
Z
x
K(x) ϕ(x) − ϕ(0)
= cϕ(0) +
dσ(x)
|x|
|x|=ε
→ cϕ(0)
as ε → 0. Indeed, for |x| = ε
K(x) ϕ(x) − ϕ(0) x ≤ Cε1−n ε = Cε2−n .
|x| Since the surface area of the sphere {|x| = ε} is nωn εn−1 , the integral on the right hand
side of (5.22) converges to 0 as ε → 0.
5.6. Fundamental solution of the Laplace equation. A straightforward application
of Theorem 5.57 gives a well known formula for the fundamental solution to the Laplace
equation.
Theorem 5.60. For n ≥ 2 we have53
∆Φ = δ0 ,
where
Φ(x) =


1
2π
log |x|
 −
1
n(n−2)ωn
1
|x|n−2
if n = 2,
if n ≥ 3.
Proof. We will prove the theorem for n ≥ 3, but a similar argument works for n = 2.
According to Theorem 5.5654
1 x
(5.23)
∇Φ(x) =
in Sn0 .
nωn |x|n
Note that the function Φ(x) is harmonic in Rn \ {0} and hence
n
1 X ∂ xj
for x 6= 0.
0 = ∆Φ(x) = div ∇Φ(x) =
nωn j=1 ∂xj |x|n
Now Theorem 5.57 gives a formula for the distributional Laplacian
n
X
1
∂ xj
∆Φ = div ∇Φ = cδ0 +
p.v.
= cδ0 ,
nωn
∂xj |x|n
j=1
|
{z
}
0
where
c=
n Z
X
j=1
|x|=1
1 xj xj
1
dσ(x)
=
nωn |x|n |x|
nωn
Z
dσ(x) = 1 .
|x|=1
The proof is complete.
53Note
that when n ≥ 3, then Φ ∗ f = −I2 f , where I2 is the Riesz potential, see Definition 2.14.
(5.23) is also true for n = 2.
54Formula
78
PIOTR HAJLASZ
For ϕ ∈ Sn let u(x) = (Φ ∗ ϕ)(x). Then55 u ∈ C ∞ (Rn ) and
∆u(x) = ∆(Φ ∗ ϕ)(x) = (∆Φ) ∗ ϕ (x) = (δ0 ∗ ϕ)(x) = ϕ(x) .
Hence convolution with the fundamental solution of the Laplace operator provides an
explicit solution of the Poisson equation
∆u = ϕ .
This explains the importance of the fundamental solution in partial differential equations.
Observe that the above calculation gives also
ϕ(x) = ∆(Φ ∗ ϕ)(x)
n
X
∂2
(Φ ∗ ϕ)(x)
=
2
∂x
j
j=1
n X
∂Φ ∂ϕ
=
∗
(x)
∂x
∂x
j
j
j=1
Z
∇Φ(x − y) · ∇ϕ(y) dy
=
Rn
Z
1
(x − y) · ∇ϕ(y)
=
dy
nωn Rn
|x − y|n
for every x ∈ Rn . In the last equality we employed (5.23). Thus we proved
Theorem 5.61. For ϕ ∈ Sn , n ≥ 2 we have
Z
1
(x − y) · ∇ϕ(y)
ϕ(x) =
dy
nωn Rn
|x − y|n
for all x ∈ Rn .
From this theorem we can conclude a similar result for higher order derivatives.
Theorem 5.62. For ϕ ∈ Sn , n ≥ 2 and m ≥ 1 we have
Z X α
m
D ϕ(y) (x − y)α
ϕ(x) =
dy
nωn Rn
α! |x − y|n
for all x ∈ Rn .
|α|=m
Proof. Fix x ∈ Rn and define
ψ(y) =
X
Dβ ϕ(y)
|β|≤m−1
(x − y)β
.
β!
Then ψ(x) = ϕ(x) and56
∂ψ
(y) =
∂yj
55As
56We
X
|β|=m−1
D
β+δj
(x − y)β
ϕ(y)
β!
a convolution of Φ ∈ Sn0 with ϕ ∈ Sn .
compute ∂ψ/∂yj using the Leibniz rule and observe that the lower order terms cancel out.
HARMONIC ANALYSIS
79
where57 δj = (0, . . . , 1, . . . , 0). Hence Theorem 5.60 applied to ψ yields
ϕ(x) = ψ(x)
1
=
nωn
=
1
nωn
=
m
nωn
n
X
xj − yj
∂ψ
(y)
dy
|x − y|n
Rn j=1 ∂yj
n
X
X Z
(x − y)β xj − yj
β+δj
D
ϕ(y)
dy
β!
|x − y|n
n
j=1 |β|=m−1 R
X Z Dα ϕ(y) (x − y)α
dy ,
α!
|x − y|n
Rn
Z
|α|=m
because for α with |α| = m
X
j,β: β+δj
m
1
=
.
β!
α!
=α
The proof is complete.
5.7. Appendix. We will present here details of the last step in the proof of Theorem 5.7.
Namely we will show that if u ∈ Sn0 and ϕ, ψ ∈ C0∞ (Rn ), then
i Z
hZ
u[ϕ(y − ·)ψ(y)] dy.
ϕ(y − ·)ψ(y) dy =
u
Rn
Rn
R
To approximate the integral Rn η(y) dy, η ∈ C0∞ by Riemann sums, we fix a cube
k
Q = [−N, N ]n so large that supp η ⊂ Q and divide the cube into cubes {Qki }m
i=1 of
sidelength 2−k . Denote the centers of these cubes by yki . Clearly
Z
mk
X
η(yki )|Qki | →
η(y) dy as k → ∞,
Rn
i=1
and
mk Z
X
(5.24)
i=1
Suppose ϕ, ψ ∈
|η(y) − η(yki )| dy → 0 as k → ∞.
Qki
C0∞ (Rn ),
mk
X
so ϕ̃ ∗ ψ ∈ C0∞ (Rn ). Hence for every x ∈ Rn
Z
ϕ̃(x − yki )ψ(yki )|Qki | →
ϕ̃(x − y)ψ(y) dy as k → ∞.
wk (x) =
Rn
i=1
The functions wk (x) belong to Sn .58 We will prove that
Z
wk (x) →
ϕ̃(x − y)ψ(y) dy = (ϕ̃ ∗ ψ)(x) as k → ∞
Rn
in the topology of Sn . First let us show that
sup |wk (x) − (ϕ̃ ∗ ψ)(x)| → 0 as k → ∞.
x∈Rn
571
on jth coordinate, 0 otherwise.
finite linear combinations of functions from Sn .
58As
80
PIOTR HAJLASZ
We have
mk
X
sup |wk (x) − (ϕ̃ ∗ ψ)(x)| ≤
x∈Rn
i=1
Z
|ϕ̃(x − yki )ψ(yki ) − ϕ̃(x − y)ψ(y)| dy
sup
x∈Rn
≤ kϕ̃k∞
Qki
mk Z
X
mk
X
+
i=1
|ψ(yki ) − ψ(y)| dy
Qki
i=1
Z
|ϕ̃(x − yki ) − ϕ̃(x − y)| |ψ(y)| dy → 0.
sup
x∈Rn
Qki
The convergence of the first sum follows from (5.24) and the convergence of the second
one follows from uniform continuity of ϕ̃. Since
mk
X
β
D wk (x) =
Dβ ϕ̃(x − yki )ψ(yki )|Qki | ,
i=1
exactly the same argument as above shows that
sup |Dβ wk (x) − (Dβ ϕ̃ ∗ ψ)(x) | → 0 as k → ∞.
|
{z
}
x∈Rn
Dβ (ϕ̃∗ψ)(x)
Therefore
pα,β (wk − ϕ̃ ∗ ψ) → 0 as k → ∞,
because |x| is bounded as the functions have compact support.
α
This proves that wk → ϕ̃ ∗ ψ in Sn .
Since the function f (y) = u[ϕ̃(· − y)] is smooth, u[ϕ̃(· − y)]ψ(y) ∈ C0∞ and hence
mk
i
hZ
X
u[ϕ̃(· − yki )]ψ(yki )|Qki |
ϕ(y − ·)ψ(y) dy = u[ϕ̃ ∗ ψ] = lim u[wk ] = lim
u
k→∞
Rn
k→∞
u[ϕ(y − ·)]ψ(y) dy.
u[ϕ̃(· − y)]ψ(y) dy =
=
Rn
i=1
Z
Z
Rn
HARMONIC ANALYSIS
81
6. Convolution with principal value distributions
In Theorem 5.49 we proved that if Ω ∈ L1 (S n−1 ) satisfies
Z
Ω(θ) dσ(θ) = 0,
S n−1
then
Ω(x/|x|)
|x|n
defines a tempered distribution. In this chapter we will study convolution operators of the
form
Z
Ω(y/|y|)
TΩ ϕ(x) = WΩ ∗ ϕ(x) = lim
ϕ(x − y) dy.
ε→0 |y|≥ε
|y|n
∨
d
Since TΩ ϕ = (ϕ̂W
Ω ) for ϕ ∈ Sn , the main objective will be to compute the Fourier
d
d
transform W
Ω and conclude properties of TΩ from those of WΩ . We will focus on the
2
setting of tempered distributions and L spaces only.
WΩ = p.v. KΩ ,
where KΩ (x) =
Operators TΩ are called singular integrals and the main objective is to study their boundedness in Lp spaces for 1 < p < ∞. We will return to this problem in Chapters ?? and ??.
First we will consider the Riesz transforms which are amongst the most important singular integrals, but have a very simple structure. Finally we will consider WΩ in its general
form.
6.1. Riesz transforms: L2 theory. In Proposition 5.53 we proved that for n ≥ 2
∂
xj
1−n
|x|
= (1 − n) p.v.
∈ Sn0 .
n+1
∂xj
|x|
In this section we will investigate the Riesz transforms which are convolutions with the
distributions
Γ n+1
xj
2
Wj = cn p.v.
, cn =
, n ≥ 2.
n+1
|x|n+1
π 2
The choice of the coefficient cn will be justified in Theorem 6.2; with this choice the Fourier
transform of Wj has a particularly simple form.
Note that when n = 1, Wj becomes59
1
x
1
1
W = p.v. 2 = p.v.
π
|x|
π
x
59cf.
Remark 5.55.
82
PIOTR HAJLASZ
and convolution with W is known as the Hilbert transform. Thus the Riesz transforms are
higher dimensional generalizations of the Hilbert transform. The Hilbert transform will
be carefully studied in Section 10. The reader may choose to read Section 10.1 now. This
section contains elementary properties of the Hilbert transform and it fits well into the
content of Chapter 6.
Definition 6.1. For 1 ≤ j ≤ n the Riesz transform Rj of a function ϕ ∈ Sn is defined by
(Rj ϕ)(x) = (Wj ∗ ϕ)(x)
Z
yj
= cn p.v.
ϕ(x − y) dy
n+1
Rn |y|
Z
xj − y j
ϕ(y) dy .
= cn lim
ε→0 |x−y|≥ε |x − y|n+1
Theorem 6.2.
cj (ξ) = −i ξj .
W
|ξ|
Proof. Applying Proposition 5.53 and then Theorem 5.42 we have
∧
∧
xj
1
∂
1−n
p.v. n+1
=
|x|
|x|
1 − n ∂xj
∧
2πiξj
=
|x|1−n
1−n
n−(n−1)
Γ
2
2πiξj n−1− n
2
|ξ|−1
=
π
n−1
1−n
Γ 2
=
π
Γ
n+1
2
n+1
2
−iξj
.
|ξ|
We used here equalities
Γ(1/2) = π
1/2
and
n−1
Γ
2
n−1
2
=Γ
n+1
2
.
The proof is complete.
Corollary 6.3. For ϕ ∈ Sn and 1 ≤ j ≤ n
∨
iξj
(6.1)
(Rj ϕ)(x) = − ϕ̂(ξ) (x) .
|ξ|
Rj ϕ ∈ C ∞ is slowly increasing and all its derivatives are slowly increasing. Moreover
kRj ϕk2 ≤ kϕk2
for ϕ ∈ Sn .
ξ
Remark 6.4. One has to be careful here. Although ϕ ∈ Sn , in general −i |ξ|j ϕ̂ 6∈ Sn so
the inverse Fourier transform in (6.1) has to be understood in the distributional or in the
L2 sense.
HARMONIC ANALYSIS
83
Proof. Equality (6.1) follows from
ξj
ϕ̂.
|ξ|
Since Rj ϕ is a convolution with a tempered distribution, it is smooth, slowly increasing and
all its derivatives are slowly increasing, see Theorem 5.7. Finally the L2 estimate follows
from the fact that the Fourier transform is an isometry on L2 and the multiplication by
the function −iξj /|ξ| is bounded in L2 with norm 1 (since the supremum of the function
is 1).60
(Rj ϕ)∧ = (Wj ∗ ϕ)∧ = ϕ̂Ŵj = −i
The above result allows us to extend the Riesz transforms to bounded operators in L2 .
Definition 6.5. The Riesz transforms are bounded operators on L2 (Rn ), n ≥ 2,
for f ∈ L2 (Rn ) and j = 1, 2, . . . , n,
kRj f k2 ≤ kf k2
defined by the formula
(Rj f )(x) =
iξj
− fˆ(ξ)
|ξ|
∨
(x) for f ∈ L2 (Rn ).
Remark 6.6. The Riesz transform on L2 was defined as an extension of a bounded operator
Rj defined on a subspace Sn ⊂ L2 by the formula Rj ϕ = cn (p.v. xj /|x|n+1 ) ∗ ϕ. An
interesting question is whether the L2 extension can also be defined by the same p.v.
integral formula and if yes, whether the limit in the definition of the p.v. integral exists
in the L2 sense or in the a.e. sense. These highly non-trivial questions will be answered in
Section ??.
Corollary 6.7. The Riesz transforms satisfy
n
X
Rj2 = −I on L2 (Rn ).
j=1
Proof. For f ∈ L2 we have
n
X
Rj2 f
∧
(ξ) =
j=1
n X
j=1
iξj
−
|ξ|
2
fˆ(ξ) = −fˆ(ξ)
and the result follows upon taking the inverse Fourier transform.
The next general and elementary result applies to Riesz transforms.
Theorem 6.8. If m ∈ L∞ (Rn ), then the operator
Tm f = (mfˆ)∨ , f ∈ L2
is bounded in L2 ,
kTm f k2 ≤ kmk∞ kf k2 .
Moreover, there is a tempered distribution u ∈ Sn0 such that
Tm ϕ = u ∗ ϕ
60cf.
Theorem 6.8 and estimates (6.3).
for all ϕ ∈ Sn .
84
PIOTR HAJLASZ
Hence Tm ϕ is smooth, slowly increasing, all its derivatives are slowly increasing and
(6.2)
Dα (Tm ϕ) = Tm (Dα ϕ)
for all multiindices α.
If m1 , m2 ∈ L∞ , then Tm1 ◦ Tm2 = Tm2 ◦ Tm1 = Tm1 m2 .
Proof. Boundedness of Tm on L2 is an obvious consequence of the Plancherel theorem
(6.3)
ˆ
ˆ
kTm f k2 = kTd
m f k2 = kmf k2 ≤ kmk∞ kf k2 = kmk∞ kf k2 .
To prove existence of u ∈ Sn0 , observe that m ∈ L∞ is a tempered distribution and
u = m̌ ∈ Sn0 satisfies û = m. Thus (u ∗ ϕ)∧ = ϕ̂m = (Tm ϕ)∧ so Tm ϕ = u ∗ ϕ.
Smoothness of Tm ϕ = u ∗ ϕ along with equality (6.2) follow from Theorems 5.7 and 5.8.
Finally, the composition formula Tm1 ◦ Tm2 = Tm2 ◦ Tm1 = Tm1 m2 follows directly from
the definition of Tm .
An amazing property of the Riesz transforms is that they allow us to compute mixed
partial derivatives ∂j ∂k u if we only know ∆u. More precisely we have
Proposition 6.9. If ϕ ∈ Sn , then for 1 ≤ j, k ≤ n we have
∂ 2ϕ
= −Rj Rk ∆ϕ(x) = −∆(Rj Rk ϕ)(x), .
∂xj ∂xk
Proof. For ϕ ∈ Sn we have
(∂j ∂k ϕ)∧ (ξ) = (2πiξj )(2πiξk )ϕ̂(ξ)
−iξj
−iξk
= −
(−4π 2 |ξ|2 )ϕ̂(ξ)
|ξ|
|ξ|
−iξj
−iξk d
∆ϕ(ξ)
= −
|ξ|
|ξ|
= (−Rj Rk ∆ϕ)∧ (ξ).
Thus taking the inverse Fourier transform in L2 yields
∂ 2ϕ
= −Rj Rk ∆ϕ(x).
∂xj ∂xk
The composition of the Riesz transforms Rj Rk is understood as composition of bounded
operators in L2 and according to Theorem 6.8 there is u ∈ Sn0 such that
∨
−iξj
−iξk
Rj Rk ϕ =
ϕ̂
= u ∗ ϕ.
|ξ|
|ξ|
so Rj Rk ϕ ∈ C ∞ is slowly increasing and all its derivatives are slowly increasing. Also
∂ 2ϕ
= −Rj Rk ∆ϕ(x) = −u ∗ ∆ϕ = −∆(u ∗ ϕ) = −∆(Rj Rk ϕ).
∂xj ∂xk
The proof is complete.
HARMONIC ANALYSIS
85
Remark 6.10. One can prove that if j 6= k, then Rj Rk can be written as a convolution
with a p.v. distribution. However, Rj2 cannot be written as a convolution with a p.v.
distribution. Fortunately, it is still a convolution with a tempered distribution, but not of
a p.v. type.61
As a direct consequence of Proposition 6.9 and of boundedness of the Riesz transforms
in L2 we obtain
Corollary 6.11.
∂ϕ ∂xj ∂xk ≤ k∆ϕk2
for ϕ ∈ Sn .
2
Remark 6.12. The same estimate can be proved directly using integration by parts. Later
we will see that a similar estimate is also true for the Lp norm, 1 < p < ∞. But this is a
much harder result and a simple integration by parts is not enough.
Remark 6.13. It is quite convincing that the proof of Proposition 6.9 applied to u ∈ Sn0
such that ∆u = f ∈ L2 , gives ∂j ∂k u = −Rj Rk f . However this is not true. For example
if u = xy, then, as a slowly increasing function, u ∈ S20 . Clearly ∆u = 0 ∈ L2 , but
∂x ∂y u = 1 6= 0 = −Rx Ry 0. The problem is of a very delicate nature and it is important to
understand it well. Remember that û is a tempered distribution, not a function so in the
equality
−iξj
−iξk
(2πiξj )(2πiξk )û(ξ) = −
[(−4π 2 |ξ|2 )û(ξ)]
|ξ|
|ξ|
the left hand side is well defined since we are allowed to multiply distributions by
(2πiξj )(2πiξk ). However, the right hand side is not well defined: (−4π 2 |ξ|2 )û(ξ) is a distribution and we are not allowed to multiply this distribution by a non-smooth function.
c = fˆ ∈ L2 so why are we not
It is actually still confusing. After all, (−4π 2 |ξ|2 )û = ∆u
allowed to multiply it by a bounded function? We are, but if we do, the above equality
is no longer true, because we switch from the distributional context to the L2 context.
In fact, the distribution û may have a non-trivial part that is supported at 0 such that
the multiplication by (2πiξj )(2πiξk ) does not ‘kill’ this part. However, multiplication by
−4π 2 |ξ|2 does ‘kill’ it so it may happen that
−iξj
−iξk ˆ
(2πiξj )(2πiξk )û(ξ) = −
f +v
|ξ|
|ξ|
where v is a distribution supported at 0. This will be carefully explained in the next result.
Theorem 6.14. If u ∈ Sn0 satisfies
∆u = f ∈ L2 (Rn ) ,
(6.4)
then for any 1 ≤ j, k ≤ n
where Pjk
61c.f.
∂ 2u
= −Rj Rk f + Pjk ,
∂xj ∂xk
is a harmonic polynomial.
Problems ?? and ?? and Example 6.41.
86
PIOTR HAJLASZ
Proof. We will only prove that Pjk are polynomials and we leave the proof that they are
harmonic to the reader. According to Corollary 5.34 it suffices to prove that the tempered
distribution
2
∧
∂ u
+ Rj Rk f
∂xj ∂xk
has support contained in {0}. To this end we need to show that if ϕ ∈ Sn , supp ϕ ⊂
Rn \ {0}, say ϕ(x) = 0 for |x| ≤ r, then
2
∧
∂ u
+ Rj Rk f
[ϕ] = 0, i.e., (∂j ∂k u)∧ [ϕ] = (−Rj Rk f )∧ [ϕ] .
∂xj ∂xk
Let η ∈ C ∞ (Rn ) be such that η(x) = 0 for |x| ≤ r/2 and η(x) = 1 for |x| ≥ r. The function
η and its derivatives are slowly increasing and ηϕ = ϕ. Hence
(∂j ∂k u)∧ [ϕ] = (2πiξi )(2πiξj )û [ηϕ]
= η(ξ)(2πiξi )(2πiξj )û [ϕ] = ♥.
Observe that
η(ξ)(2πiξi )(2πiξj ) = − η(ξ)
|
−iξk
−iξj
(−4π 2 |ξ|2 )
|ξ|
|ξ|
{z
}
λ(ξ)
∞
and λ ∈ C and its derivatives are slowly increasing since η vanishes in a neighborhood
of ξ = 0. Thus62
♥ =
− λ(ξ)(−4π 2 |ξ|2 )û [ϕ]
= (−λ(ξ)fˆ(ξ))[ϕ]
−iξj
−iξk ˆ
(6.5)
f (ξ) [ϕ]
=
−η(ξ)
|ξ|
|ξ|
−iξj
−iξk ˆ
=
−
f (ξ) [ηϕ]
|ξ|
|ξ|
−iξj
−iξk ˆ
=
−
f (ξ) [ϕ]
|ξ|
|ξ|
= (−Rj Rk f )∧ [ϕ] .
The proof is complete.
6.2. The Beurling-Ahlfors transform. We proved that
∂ 2ϕ
= −Rj Rk ∆ϕ = −∆(Rj Rk ϕ) for ϕ ∈ Sn
∂xj ∂xk
the Fourier transform of (6.4) yields −4π 2 |ξ|2 û = fˆ in Sn0 . We should understand this equality
in
and not in L2 , because despite the fact that −4π 2 |ξ|2 û = fˆ ∈ L2 , it may happen that the distribution
û is not a function. If λ ∈ C ∞ (Rn ) and all its derivatives are slowly increasing, then −λ(ξ)(−4π 2 |ξ|2 )û =
−λ(ξ)fˆ in Sn0 . Now −λfˆ ∈ L2 and fˆ ∈ L2 are functions. Functions that are equal a.e. define the same
distribution. Hence equality (6.5) is true because the functions that define distributions on the left hand
side and on the right hand side of (6.5) are equal for all ξ 6= 0 so they are equal a.e.
62Taking
Sn0
HARMONIC ANALYSIS
87
so the Riesz transforms, in some sense, can change the direction of partial derivatives. Now
we will find an operator S such that
∂ϕ
∂ϕ
S
=
for ϕ ∈ S (R2 ) = S (C).
∂ z̄
∂z
We will be using complex notation for points in R2 = C
z = x + iy
and ξ = ξ1 + iξ2 ,
and also for the Lebesgue measure
dz = dz + idy, dz̄ = dx − idy
dz̄ ∧ dz
= dx ∧ dy = dx dy.
2i
so
Definition 6.15. The operator
S = (iR1 + R2 )2
defined on L2 (C).
is called the Beurling-Ahlfors transform.
The next result collects basic properties of the Beurling-Ahlfors transform.
Theorem 6.16. The Beurling-Ahlfors transform has the following properties
(1) S : L2 (C) → L2 (C) is a surjective isometry,
kSf k2 = kf k2
for f ∈ L2 (C).
(2) The inverse operator is
S −1 f = S f¯.
(3) For f ∈ L2 (C) we have
∨
¯
ξˆ
f (ξ) ,
Sf (ξ) =
ξ
S
−1
f (ξ) =
∨
ξˆ
f (ξ) .
ξ¯
(4)
S
∂ϕ
∂ z̄
=
∂ϕ
∂
(Sϕ) =
∂ z̄
∂z
for ϕ ∈ S (R2 ) = S (C).
Proof. Since
¯ 2 2 2
ξ¯
ξ
ξ1 − iξ2
iξ1
iξ2
=
=
= i −
+ −
ξ
|ξ|
|ξ|
|ξ|
|ξ|
we conclude that
¯ ∨
ξ ˆ
Sf = (iR1 + R2 ) f =
f
ξ
2
for f ∈ L2 (C).
Clearly
∨
ξ ξ¯
ξ ˆ
−1
·
=
1
so
S
f
=
f
ξ¯ ξ
ξ¯
which proves (3). The fact that S is an isometry easily follows from the Plancherel theorem
ξ¯ c
ˆ
ˆ
kSf k2 = kSf k2 = ξ f = kf k2 = kf k2 .
2
88
PIOTR HAJLASZ
Since S has the inverse operator, it maps L2 (C) onto L2 (C). This establishes (1). A simple
computation based on the fact that
F(f¯(ξ)) = F(f )(−ξ) and F −1 (f¯(ξ)) = F −1 (f )(−ξ)
yields
S f¯ =
∨
ξˆ
f
= S −1 f
ξ¯
which is (2). Equality
S
∂ϕ
∂ z̄
=
∂
(Sϕ) for ϕ ∈ S (R2 ) = S (C)
∂ z̄
follows from Theorem 6.8, see (6.2). Recall that
∂
1 ∂
∂
∂
1 ∂
∂
=
+i
=
−i
and
∂ z̄
2 ∂x
∂y
∂z
2 ∂x
∂y
so
∧
∂ϕ
1
= (2πiξ1 + i(2πiξ2 ))ϕ̂ = πiξ ϕ̂
∂ z̄
2
and
∧
1
∂ϕ
= (2πiξ1 − i(2πiξ2 ))ϕ̂ = πiξ¯ϕ̂.
∂z
2
Therefore
∧
∧
ξ¯
∂ϕ
∂ϕ
= πiξ ϕ̂ = πiξ¯ϕ̂ =
.
S
∂ z̄
ξ
∂z
This completes the proof of (4).
Corollary 6.17. For ϕ ∈ S (C) we have
∂ϕ = ∂ϕ .
∂ z̄ ∂z 2
2
Proof. This is an immediate consequence of parts (4) and (1) of Theorem 6.16.
Similarly as in the case of the Riesz transforms, the Beurling-Ahlfors transform can
be represented as a convolution with a principal value distribution. In order to find a
formula we need to investigate a connection between the Beurling-Ahlfors transform and
the fundamental solution to the Laplace equation. Recall that63
1
1
Φ(z) =
log |z| =
log |z|2
2π
4π
satisfies
2
∂
∂2
∂2
∆Φ =
+
Φ
=
4
Φ = δ0 .
∂x2 ∂y 2
∂ z̄∂z
Hence
2
1 ∂
∂ϕ
1 ∂
∂
1
∂
(6.6)
(4πΦ) ∗
=
(4πΦ) ∗ ϕ =
(4πΦ) ∗ ϕ = ϕ.
π ∂z
∂ z̄
π ∂ z̄ ∂z
π ∂ z̄∂z
63Theorem
5.60.
HARMONIC ANALYSIS
89
An argument similar to that used in the proof of Theorem 5.56 allows us to differentiate
Φ once in the pointwise sense without necessity of taking the principal value distribution
so64
∂
∂
1 ∂
z̄
1
(6.7)
(4πΦ(z)) =
log |z|2 = 2 (z z̄) = 2 =
for z 6= 0
∂z
∂z
|z| ∂z
|z|
z
and hence
∂
1
∂
∂
=
(4Φ) = ∆Φ = δ0 .
∂ z̄ πz
∂ z̄ ∂z
We proved
1
Proposition 6.18. The function G(z) = πz
is a fundamental solution of the operator
∂/∂ z̄, i.e.
∂
1
= δ0 in S (C).
∂ z̄ πz
Also (6.7) allows us to rewrite (6.6) as65
Z
Z
1 ∂
ϕz̄ (ξ) ¯
ϕ(ξ) ¯
1
dξ ∧ dξ =
dξ ∧ ξ = ϕ(z).
(6.8)
2πi C z − ξ
2πi ∂ z̄ C z − ξ
This is a representation formula similar to that in Theorem 5.61. It allows us to reconstruct
the function ϕ from its derivative ϕz̄ . In paricular it follows immediately that the only
holomorphic function in S (C) is the zero function.
Definition 6.19. The Cauchy transform of a function f is defined by
Z
f (ξ) ¯
1
dξ ∧ ξ.
Cf (z) = (G ∗ f )(z) =
2πi C z − ξ
Thus (6.8) reads as
Corollary 6.20.
∂
(Cϕ) = C
∂ z̄
∂ϕ
∂ z̄
= ϕ for ϕ ∈ S (C).
The next result allows us to identify the Beurling-Ahlfors transform with the convolution
with a principal value distribution. A different proof will be presented in Section 6.3, see
Example 6.27.
Theorem 6.21. For ϕ ∈ S (C) we have
(6.9)
Z
∂ϕ
1
ϕ(ξ)
1
1
∂
¯
dξ ∧ dξ = −
S(ϕ) =
(Cϕ) = C
=−
lim
p.v. 2 ∗ ϕ.
∂z
∂z
2πi ε→0 |ξ−z|≥ε (z − ξ)2
π
z
Proof. First we will prove that for ϕ ∈ S (C) we have
Z
∂
∂ϕ
1
ϕ(ξ)
1
1
¯
(6.10)
(Cϕ) = C
=−
lim
dξ ∧ dξ = −
p.v. 2 ∗ ϕ.
∂z
∂z
2πi ε→0 |ξ−z|≥ε (z − ξ)2
π
z
64If
f is holomorphic and g is smooth in the real sense, then
1
that dξ1 dξ2 = 2i
dξ¯ ∧ dξ.
65Recall
∂
∂z (f
◦ g)(z) =
∂f
∂g
∂z (g(z)) ∂z (z).
90
PIOTR HAJLASZ
We could actually conclude this equality from Theorem 5.57 but we prefer to provide a
direct proof. Note that
∂ϕ
∂
(6.11)
C
=
(Cϕ), for ϕ ∈ S (C)
∂z
∂z
is a consequence of a general rule how we differentiate convolution of a tempered distribution with ϕ ∈ S (C).
It is easy to check that in the complex notation Green’s theorem can be rewritten as
follows
Lemma 6.22 (Green’s theorem). If Ω ⊂ C is a bounded domain with the piecewise C 1
boundary and f, g ∈ C 1 (Ω), then
Z Z
∂f
∂g
+
dz̄ ∧ dz =
g dz − f dz̄.
∂z ∂ z̄
Ω
∂Ω
Here
Z
Z
Z
f (dx − idy) =
f dz̄ =
∂Ω
∂Ω
f¯ dz.
∂Ω
In particular
Z
Z
∂
f g dz̄
(f g) dz̄ ∧ dz = −
Ω ∂z
∂Ω
i.e. we have the following version of the integration by parts formula
Z
Z
Z
∂f
∂g
(6.12)
g dz̄ ∧ dz = − f
dz̄ ∧ dz −
f g dz̄.
∂z
Ω ∂z
Ω
∂Ω
Applying (6.12) to our situation yields66
Z
∂ϕ
∂ϕ(ξ)/∂ξ ¯
2πi C
dξ ∧ dξ
=
∂z
z−ξ
C
Z
∂ϕ(ξ)/∂ξ ¯
= lim
dξ ∧ dξ
ε→0 |ξ−z|≥ε
z−ξ
Z
Z
∂
1
ϕ(ξ) ¯
¯
= lim −
ϕ(ξ)
dξ ∧ dξ + lim
dξ = ♥.
ε→0
ε→0
∂ξ z − ξ
|ξ−z|≥ε
|ξ−z|=ε z − ξ
The sign ‘+’ in the last limit is opposite to the sign ‘-’ in the boundary integral in (6.12).
This is because the positive orientation of the boundary |ξ − z| = ε of the exterior domain
|ξ − z| ≥ ε is clockwise.
Note that
Z
|ξ−z|=ε
and hence
dξ¯
=−
z−ξ
Z
lim
ε→0
66In
|ξ−z|=ε
Z
|ξ|=ε
dξ¯
=−
ξ
ϕ(ξ) ¯
dξ = lim
ε→0
z−ξ
Z
|ξ|=ε
dξ
= −ε−2
ξ¯
Z
|ξ−z|=ε
Z
ξ dξ = 0
|ξ|=ε
ϕ(ξ) − ϕ(z) ¯
dξ = 0.
z−ξ
the integration by parts we actually should consider the domain ε ≤ |ξ − z| ≤ R and let ε → 0 and
R → ∞ since (6.12) applies to bounded domains. We simply skipped one step and passed to the limit as
R → ∞, ‘in mind’, cf. the proof of Proposition 5.53.
HARMONIC ANALYSIS
91
Since
∂
∂ξ
1
z−ξ
=
1
(z − ξ)2
we obtain
Z
♥ = − lim
ε→0
|ξ−z|≥ε
ϕ(ξ)
dξ¯ ∧ dξ.
(z − ξ)2
This completes the proof of (6.10).
Note that (6.10) was a quite straightforward consequence of Green’s formula. It remains
∂
(Cϕ) and this is not obvious.67 Let us start with an interesting
to prove that S(ϕ) = ∂z
remark.
Remark 6.23. The space
Sz̄ = {ϕz̄ : ϕ ∈ S (C)}
is dense in L2 (C). For if not, we would find 0 6= f ∈ L2 which is orthogonal to Sz̄ , i.e.,
Z
ϕz̄ f¯ dz̄ ∧ dz = 0 for all ϕ ∈ S (C).
C
That means f¯z̄ = 0 in S (C) and hence ∆f¯ = 4f¯z̄z = 0 in S 0 (C). According to Corollary 5.35, f¯ is a polynomial. Since f¯ ∈ L2 , we conclude that f = 0 which is a contradiction.
Observe also that
∂
S(ϕz̄ ) = ϕz and
(Cϕz̄ ) = ϕz .
∂z
This shows that equality (6.9) holds on the subspace Sz̄ ⊂ S (C) which is dense in L2 .
∂
Since the operator S is an isometry on L2 , we conclude that ϕ 7→ ∂z
C(ϕ) is an isometry on
∂
2
C(ϕ)
a dense subset Sz̄ of L . We could conclude the theorem if we could prove that ϕ 7→ ∂z
2
is bounded in the L norm on S (C). After all, boundedness of this operator follows from
Theorem 6.21, but we cannot use it before we prove Theorem 6.21. Vicious circle! In our
proof given below we will follow the idea described in this remark by constructing a suitable
L2 approximation of ϕ ∈ S (C) by functions of the class Sz̄ (C).
0
Let’s return to the proof. Let ϕ ∈ S (C) and let ψ = Cϕ. As we know68 ψz̄ = ϕ. However,
in general ψ 6∈ S (C).69 Since the Cauchy transform is bounded by the Riesz potential I1
and ϕ ∈ Lp for all p, the Fractional Integration Theorem 2.15 implies that ψ ∈ Lp for all
2 < p < ∞, but in general ψ 6∈ L2 .70
¯ ∈ L∞ defines a tempered distribution. If u = (ξ/ξ)
¯ ∨ ∈ S 0 (C), then S(ϕ) = u ∗ ϕ
function ξ/ξ
(c.f. Theorem 6.8) and it remains to prove that u is the p.v. distribution given by (6.10). However, finding
¯ is not easy, see Example 6.27. On the other hand since S = (iR1 + R2 )2 is a
the Fourier transform of ξ/ξ
composition of convolutions with p.v. distributions, S should be easy to represent as a convolution with
a p.v. distribution. However, finding a formula for the composition is not easy either. For example Rj2
cannot be represented as a composition with a p.v. distribution (see Problem ?? and Example 6.41.) We
will discuss this topic in detail in Section 6.6.
68Corollary 6.20.
69cf. Problem ??
70cf. Problem ??
67The
92
PIOTR HAJLASZ
Let η ∈ C ∞ (C), η(z) = 1 for |z| ≤ 1 and η(z) = 0 for |z| ≥ 2 be a standard cutoff function. Let ψ R (z) = ψ(z)η(z/R). Since the function has compact support we have
S(ψz̄R ) = ψzR . We claim that
ψz̄R → ϕ in L2 as R → ∞.
(6.13)
The chain rule gives
ψz̄R = ψz̄ η
z
R
+ ψ(z)ηz̄
z 1
z
z 1
= ϕ(z)η
+ ψ(z)ηz̄
.
R R
R
R R
Clearly
z
→ ϕ in L2 .
R
The function ηz̄ (z/R) is bounded an supported in B(0, 2R). Hence for any 2 < p < ∞ we
have71
1/2
Z
1/p
Z
2
p
ψ(z)ηz̄ z 1 ≤ C
|ψ|
≤
C
|ψ|
R R
ϕ(z)η
B(0,2R)
2
≤ CR
−2/p
B(0,2R)
kψkp → 0 as R → ∞.
This proves (6.13) and thus72
S(ψz̄R ) → S(ϕ) in L2 .
(6.14)
On the other hand by the arguing in the same way as in the proof of (6.13) we have
z 1
z
∂
(6.15) S(ψz̄R ) = ψzR = ψz (z)η
+ ψ(z)ηz
→ ψz =
(Cϕ) in L2 as R → ∞.
R
R R
∂z
∂
Comparing (6.14) and (6.15) yields S(ϕ) = ∂z
(Cϕ). The proof is complete.
6.3. Higher Riesz transforms.
Definition 6.24. Higher Riesz transforms of degree k are defined as a convolution with
(6.16)
p.v.
Pk (x)
,
|x|n+k
where Pk (x) is a homogeneous harmonic polynomial of degree k ≥ 1.
If k = 1 and P1 (x) = cn xj , we obtain the Riesz transform Rj .
If Pk is a homogeneous harmonic polynomial of degree k ≥ 1 we can write
Pk (x)
Pk (x)|x|−k
Pk (x/|x|)
Ω(x/|x|)
=
=
=
|x|n+k
|x|n
|x|n
|x|n
and in order to prove that (6.16) defines a tempered distribution we need to prove that
Z
Z
(6.17)
Ω(θ) dσ(θ) =
Pk (θ) dσ(θ) = 0 .
S n−1
71In
S n−1
the first inequality we use the fact that the area of the disc B(0, 2R) equals CR2 . We also use here
Hölder’s inequality and the fact that ψ ∈ Lp .
72S is bounded in L2 .
HARMONIC ANALYSIS
93
If ~ν is an outward normal vector to the unit sphere, then
∂Pk
d = Pk (tx) = ktk−1 Pk (x) = kPk (x)
∂~ν
dt t=1
t=1
73
and hence Green’s formula yields
Z
Z
Z
∂Pk
k
Pk (θ) dσ(θ) =
(θ) dσ(θ) =
∆Pk dx = 0 .
ν
S n−1
S n−1 ∂~
Bn
Thus according to Theorem 5.49
Z
Pk (x)
WΩ [ϕ] = p.v.
ϕ(x) dx
n+k
Rn |x|
is a well defined tempered distribution.
Our aim is to prove the following result.
Theorem 6.25. If Pk is a homogeneous harmonic polynomial of degree k ≥ 1, then
∧
Pk (x)
Pk (ξ)
,
(ξ) = γk
p.v. n+k
|x|
|ξ|k
where
k n/2
γk = (−i) π
Γ k2
.
Γ k+n
2
Example 6.26. This result immediately implies Theorem 6.2.
Example 6.27. We will show now that the formula for the Beurling-Ahlfors transform
1
1
(6.18)
S(ϕ) = −
p.v. 2 ∗ ϕ for ϕ ∈ S (C)
π
z
originally proved in Theorem 6.21 is a straightforward consequence of Theorem 6.25.
Let n = 2 and let74 P2 (ξ) = ξ¯2 = ξ12 − 2iξ1 ξ2 − ξ22 . Clearly P2 is a homogeneous harmonic
polynomial of degree 2. Since γ2 = −π we obtain
∧
∧
1
P2 (z)
P2 (ξ)
ξ¯
p.v. 2
(ξ) = p.v.
(ξ)
=
−π
=
−π
z
|z|4
|ξ|2
ξ
so
∧
1
1
ξ¯
[
−
p.v. 2 ∗ ϕ
= ϕ̂ = S(ϕ)
π
z
ξ
which proves (6.18).
We will deduce Theorem 6.25 from the following result.
Theorem 6.28. If Pk is a homogeneous harmonic polynomial of degree k and 0 < α < n,
then
∧
Pk (x)
Pk (ξ)
(ξ) = γk,α k+α ,
k+n−α
|x|
|ξ|
73
R
∂g
∂f
(f ∆g − g∆f ) dx = ∂Ω (f ∂~
ν is the outward normal.
ν − g ∂~
ν ) dσ where ~
74We use complex notation in R2 = C.
R
Ω
94
PIOTR HAJLASZ
where
k
γk,α = (−i) π
n
−α
2
Γ k+α
2
.
Γ k+n−α
2
Remark 6.29. Note that the function
Pk (x)
|x|k+n−α
is a tempered L1 function, so it defines a tempered distribution without necessity of taking
the principal value of the integral.
Proof. For any t > 0 and ϕ ∈ Sn , Corollary 4.24 gives75
Z
Z
n
2
−πt|x|2
k
Pk (x)e
Pk (x)e−π|x| /t t−k− 2 ϕ(x) dx .
ϕ̂(x) dx = (−i)
Rn
Rn
Now we multiply both sides by
tβ−1 ,
where β =
k+n−α
>0
2
and integrate with respect to 0 < t < ∞. Since
Z ∞
2
e−πt|x| tβ−1 dt = (π|x|2 )−β Γ(β)
0
the integral on the left hand side will be equal to
∧
Z
Γ k+n−α
Pk (x)
Pk (·)
Γ(β)
2
ϕ̂(x) dx =
[ϕ] .
(6.19)
k+n−α
2β
πβ
| · |k+n−α
π 2
Rn |x|
Similarly
∞
Z
β−1
−π|x|2 /t −k− n
2
e
t
t
0
s=π|x|2 /t
=
(π|x|2 )−
k+α
2
Z
dt =
Z ∞
∞
2
e−π|x| /t t−
k+α
−1
2
dt
0
e−s s
k+α
−1
2
ds
0
=
2 − k+α
2
(π|x| )
Γ
k+α
2
.
Thus the integral on the right hand side equals
k+α Z
Pk (x)
kΓ
2
(6.20)
(−i)
ϕ(x) dx .
k+α
k+α
π 2
Rn |x|
Since integrals at (6.19) and (6.20) are equal one to another, the theorem follows.
Proof of Theorem 6.25. For ϕ ∈ Sn we take the identity
Z
Z
Pk (x)
Pk (x)
ϕ̂(x) dx = γk,α
ϕ(x) dx
k+n−α
k+α
Rn |x|
Rn |x|
75
R
Rn
f (x)ϕ̂(x) dx =
R
Rn
fˆ(x)ϕ(x) dx.
HARMONIC ANALYSIS
95
and let α → 0. The right hand side converges to
Z
k
Γ
n
Pk (x)
2 (−i)k π 2
ϕ(x) dx .
k+n
k
Γ 2
Rn |x|
To compute the limit on the left hand side we observe that the integral of Pk (x)|x|−(k+n−α)
on the unit ball equals zero, see (6.17), and hence
Z
Pk (x)
ϕ̂(x) dx
k+n−α
Rn |x|
Z
Z
Pk (x)
Pk (x)
=
(
ϕ̂(x)
−
ϕ̂(0))
dx
+
ϕ̂(x) dx
k+n−α
k+n−α
|x|≤1 |x|
|x|≥1 |x|
Z
Z
Pk (x)
Pk (x)
α→0+
(ϕ̂(x) − ϕ̂(0)) dx +
ϕ̂(x) dx
−→
k+n
k+n
|x|≥1 |x|
|x|≤1 |x|
Z
Z
Pk (x)
Pk (x)
= lim
(ϕ̂(x) − ϕ̂(0)) dx +
ϕ̂(x) dx
k+n
ε→0 ε≤|x|≤1 |x|k+n
|x|≥1 |x|
Z
Pk (x)
= lim
ϕ̂(x) dx
ε→0 |x|≥ε |x|k+n
∧
Pk (x)
=
p.v. k+n
[ϕ] .
|x|
Comparing the above limits yields the result.
6.4. Another proof of Theorem 6.2. Two proof presented above76 are somewhat indirect since they were based on other results: Theorem 5.42 and Theorem 6.25. In this
section we will present a different, more straightforward proof. A one dimensional version
of this argument will be presented again in Section 10 in the proof of Theorem 10.2.
For ϕ ∈ Sn we have
Z
cj [ϕ] = Wj [ϕ̂] = lim cn
W
ε→0
Z
ϕ̂(ξ)
|ξ|≥ε
Z
= lim cn
ε→0
ξj
dξ
|ξ|n+1
−2πix·ξ
ϕ(x)e
ε≤|ξ|≤ε−1
Rn
Z
= lim cn
ε→0
Z
ϕ(x)
e−2πix·ξ
n
−1
R
ε≤|ξ|≤ε
|
{z
I
ξj
dξ
|ξ|n+1
ξj
dξ
dx = ♥ .
|ξ|n+1
}
dx
Expressing the integral I is the spherical coordinates yields
Z
Z ε−1
n−1
−2πix·(sθ) sθj
I =
s
e
dσ(θ) ds
sn+1
ε
S n−1
Z ε−1 Z
ds
=
cos(2πsx · θ) − i sin(2πsx · θ) θj dσ(θ)
s
ε
S n−1
76The
second proof was given in Remark 6.26.
96
PIOTR HAJLASZ
Z
ε−1
= −i
ε
Z
= −i
S n−1
Z
= −i
S n−1
Thus
Z
ds
sin(2πsx · θ)θj dσ(θ)
s
S n−1
!
Z ε−1
sin(2πs|x · θ|)
ds sgn (x · θ)θj dσ(θ)
s
ε
!
Z 2π|x·θ|ε−1
sin t
dt sgn (x · θ)θj dσ(θ) .
t
2π|x·θ|ε
Z
π
ϕ(x) −icn
sgn (x · θ)θj dσ(θ) dx .
♥=
2 S n−1
Rn
Indeed, we could pass to the limit under the sign of the integral using the Dominated
Convergence Theorem and we employed the equality77
Z ∞
π
sin t
dt = .
t
2
0
Thus it remains to prove that
Z
π
xj
(6.21)
cn
sgn (x · θ)θj dσ(θ) =
.
2 S n−1
|x|
It is obvious that the function m : Rn → Rn defined as
Z
m(x) =
sgn (x · θ)θ dσ(θ)
Z
S n−1
is homogeneous of degree 0. We will use Lemma 4.26 so we need to check that m commutes
with orthogonal transformations. It does because
Z
sgn (ρ(x) · θ)θ dσ(θ)
m(ρ(x)) =
S n−1
Z
=
sgn (x · ρ−1 (θ))θ dσ(θ)
n−1
ZS
(6.22)
=
sgn (x · θ)ρ(θ) dσ(θ)
S n−1
Z
(6.23)
= ρ
sgn (x · θ)θ dσ(θ)
S n−1
= ρ(m(x)) .
Note that (6.22) follows from the fact that ρ induces a volume preserving change of variables
on S n−1 , while (6.23) is a direct consequence of linearity of ρ. Thus the Lemma 4.26 yields
Z
x
sgn (x · θ)θ dσ(θ) = c
|x|
S n−1
and hence looking at the jth component we have
Z
xj
(6.24)
sgn (x · θ)θj dσ(θ) = c
.
|x|
S n−1
77Recall
that this is understood as an improper integral which is consistent with our setting since we
pass to the limit as ε → 0.
HARMONIC ANALYSIS
97
Thus it remains to prove that
π −1 2π (n−1)/2
= 2ωn−1 .
c = cn
=
2
Γ n+1
2
Taking x = ej in (6.24) we have
Z
|θj | dσ(θ) = c .
S n−1
The unit ball B n−1 in coordinates perpendicular to xj split the sphere S n−1 into two half
spheres S±n−1 . Thus
Z
c=2
θj dσ(θ) .
n−1
S+
Recall that if M ⊂ Rn is a graph of a C 1 function f : Ω → R, Ω ⊂ Rn−1 , then for a
measurable function g on M we have
Z
Z
p
g dσ =
g(x, f (x)) 1 + |∇f (x)|2 dx .
M
Ω
In our situation we parametrize S+n−1 as a graph of the function
p
f (x) = 1 − |x|2 , x ∈ B n−1 .
Form the picture
98
PIOTR HAJLASZ
we conclude
Z
Z
Z
θj dσ(θ) =
n−1
S+
p
(1 − h) 1 + tan2 α dx
(1 − h) dσ(θ) =
n−1
S+
B n−1
Z
=
dx = ωn−1 ,
B n−1
because
p
1 + tan2 α =
1
1
=
cos α
1−h
and the result follows.
2
The proof of the next result is based on an argument similar to that used to prove (6.24);
we leave details to the reader as an exercise.
Lemma 6.30. Let K be a function of one variable, then for n ≥ 2 we have
Z
Z 1
n−3
K(x · θ) dσ(θ) = (n − 1)ωn−1
K(s|x|)(1 − s2 ) 2 ds
S n−1
−1
n
for all x ∈ R \ {0}.
6.5. The general case. In this section Rwe will consider operators TΩ ϕ = WΩ ∗ ϕ in the
most general form where Ω ∈ L1 (S n−1 ), S n−1 Ω(θ) dσ(θ) = 0, and
WΩ = p.v. KΩ ,
where KΩ (x) =
Ω(x/|x|)
.
|x|n
The following lemma will play an important role.
Lemma 6.31. If F is a function homogeneous of degree zero and F S n−1 ∈ L1 (S n−1 ), then
F is a tempered L1 function
Z
|F (x)|(1 + |x|n+1 )−1 dx < ∞.
Rn
Hence F ∈ Sn0 . Moreover
|F [ϕ]| ≤ C(n)kF kL1 (S n−1 ) sup (1 + |x|n+1 )|ϕ(x)|.
x∈Rn
Proof. We have
Z
|F [ϕ]| ≤
Rn
Z
=
Z
s
0
Z
0
∞
n−1
n−1
|F (x)ϕ(x)| dx =
s
|F (θ)ϕ(sθ)| dσ(θ) ds
0
S n−1
Z ∞
|F (θ)|
sn−1 |ϕ(sθ)| ds dσ(θ) = ♥
S n−1
Z
∞
Z
|ϕ(sθ)| ds ≤
0
∞
sn−1
sup (1 + |x|n+1 )|ϕ(x)| ds
1 + sn+1 |x|=s
≤ C(n) sup (1 + |x|n+1 )|ϕ(x)|.
x∈Rn
Hence
♥ ≤ C(n)kF kL1 (S n−1 ) sup (1 + |ξ|n+1 )|ϕ(ξ)|.
ξ∈Rn
HARMONIC ANALYSIS
99
In particular taking78 ϕ(x) = (1 + |x|n+1 )−1 yields
Z
|F (x)|(1 + |x|n+1 )−1 ≤ CkF kL1 (S n−1 ) < ∞
Rn
so F is a tempered L1 function.
The main result of this section generalizes Theorem 6.2.
Theorem 6.32. Let n ≥ 2 and Ω ∈ L1 (S n−1 ) be such that
Z
Ω(θ) dσ(θ) = 0 .
S n−1
Then the Fourier transform of the distribution WΩ is a tempered L1 function, homogeneous
of degree zero such that
Z
d
d
W
(6.25)
kWΩ kL1 (S n−1 ) ≤ CkΩkL1 (S n−1 ) and
Ω (ξ) dσ(ξ) = 0.
S n−1
d
Therefore W
Ω is a tempered distribution satisfying
n+1
d
(6.26)
|W
sup (1 + |ξ| )|ϕ(ξ)| .
Ω [ϕ]| ≤ CkΩkL1 (S n−1 )
ξ∈Rn
Moreover
1
iπ
d
W
Ω(θ) log
−
sgn (ξ · θ) dσ(θ)
Ω (ξ) =
|ξ · θ|
2
S n−1
Z
1
iπ
0
=
Ω(θ) log 0
−
sgn (ξ · θ) dσ(θ) ,
|ξ · θ|
2
S n−1
Z
(6.27)
where ξ 0 = ξ/|ξ|.
Finally, if Ω is odd, then
(6.28)
iπ
d
W
Ω (ξ) = −
2
Z
Ω(θ) sgn (ξ 0 · θ) dσ(θ) is odd
S n−1
and if Ω is even, then
Z
(6.29)
d
W
Ω (ξ) =
S n−1
Ω(θ) log
1
|ξ 0 · θ|
dσ(θ) is even.
We will precede the proof with some technical lemmas.
Lemma 6.33. For θ ∈ S n−1 define a function on S n−1 by fθ (ξ 0 ) = log |ξ01·θ| . Then fθ ∈
Lp (S n−1 ) for all 1 ≤ p < ∞ and
sup kfθ kLp (S n−1 ) = C(n, p) < ∞.
θ∈S n−1
78ϕ(x)
= (1 + |x|n+1 )−1 does not belong to Sn , but the above estimates work for this function too.
100
PIOTR HAJLASZ
Proof. According to Lemma 6.3079
Z
Z 1
1
1
p
p
0
2 n−3
(6.30)
log
dσ(ξ
)
=
(n
−
1)ω
(1
−
s
) 2 dx < ∞
log
n−1
|ξ 0 · θ|
|s|
S n−1
−1
The last integral has three singularities80
n−3
n−3
1
at s = 0, (1 + s) 2 at s = −1, (1 − s) 2 at s = 1
logp
|s|
and all these singularities are integrable.
Lemma 6.34. If h : [0, ∞) → R is continuous, bounded and the improper integral
Z ∞
h(s)
ds
s
1
converges, then for µ > λ > 0 and N > ε > 0 we have
Z N
h(λs)
−
h(µs)
≤ 2khk∞ log µ .
ds
(6.31)
s
λ
ε
Moreover
Z
(6.32)
lim
N →∞
ε→0
Proof. We have
Z
ε
N
ε
N
µ
h(λs) − h(µs)
ds = h(0) log
.
s
λ
h(λs) − h(µs)
ds =
s
Z
λN
λε
µε
Z
=
λε
Z µN
h(s)
h(s)
ds −
ds
s
s
µε
Z µN
h(s)
h(s)
ds −
ds .
s
s
λN
Estimating the absolute value of the last two integrals gives (6.31). Since
Z µε
µ
h(s)
ds → h(0) log
as ε → 0
s
λ
λε
and
Z µN
h(s)
ds → 0 as N → ∞
s
λN
(6.32) follows.
Corollary 6.35. For a 6= 0 we have
Z N −isa
e
− cos s
1
π
lim
ds = log
− i sgn a
N →∞
s
|a|
2
ε
ε→0
and
N
e−isa − cos s 1
ds ≤ 2 log + C,
s
|a|
ε
where C is some positive constant independent of a, ε and N .
Z
79Since
80The
|ξ 0 · θ| ≤ 1, log |ξ01·θ| ≥ 0 and we do not have to take the absolute value of the log term.
last two are singularities only when n = 2.
2
HARMONIC ANALYSIS
101
Proof. We have
Z N −isa
Z N
Z N
e
− cos s
cos(sa) − cos s
sin(sa)
ds =
ds − i
ds
s
s
s
ε
ε
ε
Z N
Z N
cos(s|a|) − cos s
sin(s|a|)
=
− i sgn (a)
ds
s
s
ε
ε
Z N
Z N |a|
cos(s|a|) − cos s
sin s
=
− i sgn (a)
ds
s
s
ε
ε|a|
and the result follows from Lemma 6.34.
Proof of Theorem 6.32. The structure of the proof is as follows
(1) We will prove that the two integrals at (6.27) are equal.
(2) Define
Z
1
iπ
0
Ω(θ) log 0
F (ξ) =
−
sgn (ξ · θ) dσ(θ) ,
|ξ · θ|
2
S n−1
Clearly F is homogeneous of degree zero. We will prove that
Z
kF kL1 (S n−1 ) ≤ CkΩkL1 (S n−1 ) ,
F (ξ) dσ(ξ) = 0.
S n−1
Hence according to Lemma 6.31, F is a tempered L1 function so F ∈ Sn0 and
n+1
|F [ϕ]| ≤ CkΩkL1 (S n−1 ) sup (1 + |ξ| )|ϕ(ξ)| .
ξ∈Rn
(3) Since ξ 7→ log |ξ01·θ| is even and ξ 7→ sgn (ξ 0 · θ) is odd, it follows that if Ω is odd or
even, then
Z
Z
1
iπ
0
Ω(θ) sgn (ξ · θ) dσ(θ) or F (ξ) =
Ω(θ) log 0
F (ξ) = −
dσ(θ)
2 S n−1
|ξ · θ|
S n−1
are odd and even respectively.
d
(4) Finally we will prove that W
Ω (ξ) = F (ξ).
First observe that the equality of the two integrals in (6.27) follows from
Z
1
1
1
1
log
= log
+ log 0
and
Ω(θ) log
dσ(θ) = 0 .
|ξ · θ|
|ξ|
|ξ · θ|
|ξ|
S n−1
This proves (1).
Let
1
iπ
0
F (ξ) =
Ω(θ) log 0
−
sgn (ξ · θ) dσ(θ)
|ξ · θ|
2
S n−1
be the function defined by the integral in the second line of (6.27).
Z
It is clear that F (ξ) is homogeneous of degree zero. We will show that it is integrable
on the unit sphere and
kF kL1 (S n−1 ) ≤ CkΩkL1 (S n−1 ) .
102
PIOTR HAJLASZ
Note that
Z
iπ
sgn (ξ 0 · θ) dσ(θ)
2
n−1
S
is a bounded by π2 kΩkL1 (S n−1 ) , so this component of F (ξ) does not cause any troubles and
hence we only need to estimate
Z
1
G(ξ) =
Ω(θ) log 0
dσ(θ) .
|ξ · θ|
S n−1
Lemma 6.33 with p = 1 yields
Z
Z
1
dσ(ξ 0 )
kGkL1 (S n−1 ) ≤
Ω(θ)
log
dσ(θ)
n−1
|ξ 0 · θ|
S n−1
S
Z
Z
1
0
≤
|Ω(θ)|
dσ(ξ ) dσ(θ) = C(n)kΩkL1 (S n−1 ) .
log 0
|ξ · θ|
S n−1
S n−1
R
It easily follow now from the Fubini theorem that S n−1 F (ξ) dσ(ξ) = 0. That completes
the proof of (2).
ξ 7→
Ω(θ)
Since F ∈ L1 (S n−1 ), the step (3) is obvious.
d
To prove (4) we need to show that W
Ω = F in the sense of distributions i.e.,
Z
Z
1
iπ
0
d
Ω(θ) log 0
ϕ(ξ)
W
−
sgn (ξ · θ) dσ(θ) dξ.
Ω [ϕ] =
|ξ · θ|
2
S n−1
Rn
We have
d
W
Ω [ϕ] = WΩ [ϕ̂]
Z
Ω(x/|x|)
= lim
ϕ̂(x) dx
ε→0
|x|n
ε≤|x|≤N
N →∞
Z
Z
Ω(x/|x|) −2πix·ξ
= lim
e
dx dξ
ϕ(ξ)
ε→0
|x|n
Rn
ε≤|x|≤N
N →∞
Z
Z
Z N
−2πisθ·ξ ds
= lim
ϕ(ξ)
Ω(θ)
e
dσ(θ) dξ
ε→0
s
Rn
S n−1
ε
N →∞
Z
Z
Z N
ds
−2πisθ·ξ
ϕ(ξ)
Ω(θ)
e
−
cos(2πs|ξ|)
= lim
dσ(θ) dξ
ε→0
s
n
n−1
R
S
ε
N →∞
= ♦.
The last equality follows from the fact that the integral of Ω over the sphere vanishes. We
have
Z N
Z 2π|ξ|N −isθ·ξ0
ds
e
− cos s
−2πisθ·ξ
e
− cos(2πs|ξ|)
=
ds
s
s
2π|ξ|ε
ε
1
π
−→ log
− i sgn (θ · ξ 0 )
0
|θ · ξ |
2
by Corollary 6.35. Also Corollary 6.35 and integrability of G easily imply that
Z N
ds −2πisθ·ξ
ϕ(ξ)Ω(θ)
e
− cos(2πs|ξ|)
s
ε
HARMONIC ANALYSIS
≤ C|ϕ(ξ)| |Ω(θ)| 1 + log
|ξ 0
1
· θ|
103
∈ L1 (Rn × S n−1 )
so the Dominated Convergence Theorem gives
Z
Z
π
1
0
− i sgn (θ · ξ ) dσ(θ) dξ .
♦ =
ϕ(ξ)
Ω(θ) log
|θ · ξ 0 |
2
Rn
S n−1
2
The proof is complete.
If Ω is an odd function, then
iπ
d
W
Ω (ξ) = −
2
Z
Ω(θ) sgn (ξ 0 · θ) dσ(θ) .
S n−1
n−1
d
In particular the Fourier transform W
Ω is bounded. More generally any function Ω on S
can be decomposed into its even and odd parts
1
Ωe (θ) = (Ω(θ) + Ω(−θ)),
2
1
Ωo (θ) = (Ω(θ) − Ω(−θ)) .
2
Corollary 6.36. Let Ω ∈ L1 (S n−1 ) be such that
Z
Ω(θ) dσ(θ) = 0 .
S n−1
If Ωo ∈ L1 (S n−1 ) and Ωe ∈ Lq (S n−1 ) for some q > 1, then the Fourier transform of WΩ
is a bounded function. In particular the operator TΩ ϕ = WΩ ∗ ϕ for ϕ ∈ Sn extends to a
bounded operator in L2 .
0
Proof. Lemma 6.33 implies that log(1/|ξ 0 · θ|) belongs to Lq (S n−1 ) and hence
Z
1
n−1 Ωe (θ) log |ξ 0 · θ| dσ(θ) ≤ CkΩe kLq (S n−1 ) ,
S
with a constant C independent of ξ. Now the formulas (6.28) and (6.29) yield
Z
Z
1
iπ
0
d
W
Ωo (θ) sgn (ξ · θ) dσ(θ) +
Ωe (θ) log 0
dσ(θ)
Ω (ξ) = −
2 S n−1
|ξ · θ|
S n−1
from which we have
d
kW
Ω k∞ ≤ C(kΩo kL1 (S n−1 ) + kΩe kLq (S n−1 ) ) .
∨
d
Since TΩ ϕ = (ϕ̂W
Ω ) and multiplication by a bounded function constitutes a bounded
operator in L2 it follows that
d
kTΩ ϕk2 ≤ kW
Ω k∞ kϕk2
for ϕ ∈ Sn .
d
It is natural to expect that if Ω in Theorem 6.32 is smooth, then W
Ω is smooth away
from the origin. As we will see this it true. The next result is quite surprising since it gives
d
a complete characterization of such Fourier transforms W
Ω.
104
PIOTR HAJLASZ
Theorem 6.37. If Ω ∈ C ∞ (S n−1 ) satisfies
Z
(6.33)
Ω(θ) dσ(θ) = 0,
S n−1
then
∞
n
d
m=W
Ω ∈ C (R \ {0})
(6.34)
is homogeneous of degree zero and
Z
(6.35)
m(ξ) dσ(ξ) = 0.
S n−1
Conversely if m ∈ C ∞ (Rn \ {0}) is homogeneous of degree zero and it satisfies (6.35), then
d
there is Ω ∈ C ∞ (S n−1 ) satisfying (6.33) such that m = W
Ω.
Lemma 6.38. If Ω ∈ C ∞ (S n−1 ) and
Z 1
n−3
(6.36)
|K(s)|(1 − s2 ) 2 ds < ∞
−1
∞
then m ∈ C (S
n−1
) where
Z
Ω(θ)K(ξ · θ) dσ(θ).
m(ξ) =
S n−1
Remark 6.39. We do not assume here that the integral of Ω vanishes.
Remark 6.40. According to Lemma 6.30 inequality (6.36) implies that for every ξ ∈ S n−1
Z
Z 1
n−3
|K(ξ · θ)| dσ(θ) = (n − 1)ωn−1
|K(s)|(1 − s2 ) 2 ds < ∞.
S n−1
−1
Proof. Observe that it suffices to prove that m is smooth in a neighborhood of e1 . Indeed,
if ξ0 ∈ S n−1 and ρ(e1 ) = ξ0 for some ρ ∈ O(n), then
Z
Ω(θ)K(ρ(ξ) · θ) dσ(θ)
m̃(ξ) := (m ◦ ρ)(ξ) =
S n−1
Z
Z
−1
=
Ω(θ)K(ξ · ρ (θ)) dσ(θ) =
Ω(ρ(θ))K(ξ · θ) dσ(θ)
S n−1
∞
S n−1
n−1
and Ω ◦ ρ ∈ C (S ). Thus smoothness of m̃ is a neighborhood of e1 will imply smoothness of m in a neighborhood of ξ0 , because m = m̃ ◦ ρ−1 and ρ−1 : S n−1 → S n−1 is a
diffeomorphism that maps a neighborhood of ξ0 onto a neighborhood of e1 , where m̃ is
smooth.
Thus we will prove that m is smooth in a neighborhood of e1 .
If ξ is in the right hemisphere generated by e1 , we can represent it in a local coordinate
system
v

u
n
X
u
ξ = q(ξ2 , . . . , ξn ) = t1 −
ξj2 , ξ2 , . . . , ξn  , (ξ2 , . . . , ξn ) ∈ B n−1 (0, 1).
j=2
It suffices to prove that (m ◦ q)(ξ2 , . . . , ξn ) is smooth in B n−1 (0, 1).
HARMONIC ANALYSIS
105
Given ξ = q(ξ2 , . . . , ξn ) let
ρ(ξ2 , . . . , ξn ) ∈ SO(n),
ρ(ξ2 , . . . , ξn )(ξ) = e1
be a rotation that rotates ξ to e1 in the plane span {e1 , ξ} and fixes the orthogonal complement of that plane. Such a rotation is unique and it is easy to see that components of
the matrix ρ(ξ2 , . . . , ξn ) smoothly depend in (ξ2 , . . . , ξn ). By an argument similar to the
one used at the beginning of the proof
Z
Z
(m ◦ q)(ξ2 , . . . , ξn ) =
Ω(θ)K(ξ · θ) dσ(θ) =
(Ω ◦ ρ−1 (ξ2 , . . . , ξn ))(θ)K(θ1 ) dσ(θ).
S n−1
S n−1
Now it is easy to see that we can differentiate with respect to ξ2 , . . . , ξn under the sign of
the integral infinitely many times.81
Proof of Theorem 6.37. Let Ω ∈ C ∞ (S n−1 ) satisfy (6.33). For ξ 6= 0,
Z
d
Ω(θ)K(ξ · θ) dσ(θ)
m(ξ) = WΩ (ξ) =
S n−1
where
K(s) = log
1
iπ
− sgn (s)
|s|
2
and we already proved in (6.30) that
Z 1
n−3
|K(s)|(1 − s2 ) 2 ds < ∞.
−1
∞
n−1
Thus m ∈ C (S ) by Lemma 6.38 and since m is homogeneous of degree zero, m ∈
C ∞ (Rn \ {0}). The fact that the integral of m over the sphere vanishes was already proved
in Theorem 6.32, see (6.25).
Suppose now that m ∈ C ∞ (Rn \ {0}) is homogeneous of degree zero and it satisfies
(6.35). Since m is a tempered distribution, m̂ is a tempered distribution too and
n ∧
∂ m
(ξ) = (2πiξj )n m̂(ξ).
∂xnj
The function ∂ n m/∂xnj is homogeneous of degree −n. Since
!
∂
∂ n−1 m
∂ nm
=
∂xnj
∂xj ∂xn−1
j
it is the derivative of a function that is homogeneous of degree 1 − n, Theorem 5.57 yield
Z
∂ n m(θ)
∂ nm
∂ nm
dσ(θ)
=
0
and
=
cδ
+
p.v.
.
0
∂xnj
∂xnj
∂xnj
S n−1
81Perhaps
it will be easier to see it if we rewrite the integral in a way that it does not involve integration
of functions on manifolds (sphere). Let F ∈ C ∞ (Rn ) be a smooth extension of Ω from S n−1 to Rn . Then
Z
(m ◦ q)(ξ2 , . . . , ξn ) =
F (ρ−1 (ξ2 , . . . , ξn )(x))dµ(x)
Rn
n−1
where µ is a measure concentrated
. Namely µ is the extension of K(θ1 )dσ(θ) from
R on the sphere S
n−1
n
S
to R by zero, i.e., µ(E) = S n−1 ∩E K(θ1 ) dσ(θ).
106
PIOTR HAJLASZ
Taking the Fourier transform yields
(2πi)n ξjn m̂(ξ)
∧
∂ nm
= c + p.v.
(ξ).
∂xnj
Note that the function on the right hand side is of class C ∞ (Rn \ {0}), homogeneous of
degree zero (by the first part of the proof).
The above equality yields
∧ n
n
X
X
∂ nm
2n
n
n
(ξj )m̂(ξ) =
ξj c + p.v.
(ξ) .
(6.37)
(2πi)
∂xnj
j=1
j=1
We claim that the distribution m̂ coincides in Rn \ {0} with the function that is smooth
in Rn \ {0}, homogeneous of degree −n
∧ Pn n
∂nm
c + p.v. ∂xn
(ξ)
j=1 ξj
j
P
(6.38)
m̂(ξ) =
in Rn \ {0}.
(2πi)n nj=1 (ξj2n )
That means, if ϕ ∈ Sn with supp ϕ ⊂ Rn \ {0}, then
∧ Pn n
∂nm
Z
c + p.v. ∂xn
(ξ)
j=1 ξj
j
P
ϕ(x) dx.
(6.39)
m̂[ϕ] =
(2πi)n nj=1 (ξj2n )
Rn
To see this observe that since ϕ vanishes in a neighborhood of 0, the function
positive on the support of ϕ so
ϕ
Pn
ψ=
∈ Sn
2n
n
(2πi)
j=1 (ξj )
Pn
2n
j=1 ξj
is
is in the Schwarz class.82 Now it is easy to see that if we evaluate both sides of the
distributional equality (6.37) on ψ we obtain equality (6.39).
We proved that in Rn \ {0}, m̂ coincides with a smooth function, homogeneous of degree
−n which is given by the formula on the right hand side of (6.38). We still do not know
what happens at 0. There is a possibility that m̂ has a part supported at 0.
Let Ω̃(ξ) = m̂(ξ) for ξ ∈ S n−1 . Thus
m̂(ξ) =
Ω̃(ξ/|ξ|)
|ξ|n
for ξ 6= 0.
We claim that
Z
(6.40)
Ω̃(θ) dσ(θ) = 0,
S n−1
Indeed, let ϕ ∈ Sn be a radial function supported in the annulus 1 ≤ |x| ≤ 2 and positive
in the interior of the annulus. Then the integration in the spherial coordinates gives
Z
Z
Ω̃(x/|x|)
(6.41)
m̂[ϕ] =
ϕ(x) dx = C
Ω̃(θ) dσ(θ), where C > 0.
|x|n
Rn
S n−1
82Why?
HARMONIC ANALYSIS
107
On the other hand the Fourier transform commutes with orthogonal transformations and
hence ϕ̂(ξ) is also a radial function ϕ̂(ξ) = f (|ξ|) so
Z
Z
Z ∞
(6.42)
m̂[ϕ] = m[ϕ̂] =
m(ξ)ϕ̂(ξ) dξ =
m(θ)dσ(θ)
sn−1 f (s) ds = 0.
Rn
S n−1
0
Now (6.40) follows from the equality between (6.41) and (6.42).
We proved that m̂(x) coincides with Ω̃(x/|x|)/|x|n in Rn \ {0}. Hence the distribution
(6.43)
m̂ − p.v.
Ω̃(x/|x|)
|x|n
is supported at the origin. According to Corollary 5.34 its Fourier transform is a polynomial.
However, the Fourier transform of (6.43) is a bounded function so the polynomial must
be constant. The constant must be zero because m and (p.v. Ω̃(x/|x|)/|x|n )∧ have the zero
integrals over S n−1 . Hence
∧
Ω̃(x/|x|)
Ω(x/|x|)
m̂ = p.v.
so m(ξ) = p.v.
(ξ)
|x|n
|x|n
where Ω(θ) = Ω̃(−θ).
6.6. Algebra of singular integrals. In this section we will investigate compositions of
singular integrals
Z
Z
Ω(y/|y|)
f (x − y) dy where
Ω(θ) dσ(θ) = 0.
TΩ f (x) = WΩ ∗ f (x) = lim
ε→0 |y|≥ε
|y|n
S n−1
We will assume that Ω ∈ C ∞ (S n−1 ).
In general we cannot expect that the composition of singular integrals is a singular
integral. This can be seen in the following example
Example 6.41. There is a function Ω ∈ C ∞ (S n−1 ) with the vanishing integral such that
that the square of a Riesz transform satisfies83
1
(6.44)
Rj2 = − I + TΩ
n
Indeed,
∧ ξj2
1
1
2
=
−
ϕ̂(ξ) = m(ξ)ϕ̂(ξ)
Rj + I [ϕ]
n
n |ξ|2
and
!
Z
Z
Z
n
X
1
1
m(ξ) dσ(ξ) =
− ξj2 dσ(ξ) =
1−
ξk2 dσ(ξ) = 0
n
n
S n−1
S n−1
S n−1
k=1
so the existence of Ω satisfying (6.44) follows from Theorem 6.37.
However, the class of operators of the form aI + TΩ is closed under compositions.
83I
stands for the identity.
108
PIOTR HAJLASZ
Theorem 6.42. If m ∈ C ∞ (Rn \ {0}) is homogeneous of degree zero, then there is a ∈ C
and Ω ∈ C ∞ (S n−1 ) with vanishing integral such that
T ϕ := (mϕ̂)∨ = aϕ + TΩ ϕ
Proof. Let a ∈ C be such that
Ω such that T f − af = TΩ .
R
S n−1
for ϕ ∈ Sn .
(m(θ) − a) dσ(θ) = 0. Then by Theorem 6.37 there is
Corollary 6.43. The class A of operators of the form
aI + TΩ
where a ∈ C and Ω ∈ C ∞ (S n−1 ) has vanishing integral coincides with the class of operators
of the form
T ϕ = (mϕ̂)∨ ,
where m ∈ C ∞ (Rn \{0}) is homogeneous of degree zero. Hence A is a commutative algebra.
d
An operator in the class A is invertible if and only if m = a + W
Ω does not vanish at any
n−1
point of S .
HARMONIC ANALYSIS
109
7. Riesz potentials and fractional powers of the Laplacian
Recall that for 0 < α < n the Riesz potantial Iα is defined by84
Z
1
f (y)
(Iα f )(x) = (Uα ∗ f )(x) =
dy ,
γ(α) Rn |x − y|n−α
where
n
n−α
α− n
2Γ
π 2 2α Γ α2
π
2 |x|α−n .
γ(α) =
so Uα = (2π)−α
α
Γ n−α
Γ
2
2
Clearly, Iα ϕ is smooth, slowly increasing, and all its derivatives are slowly increasing when
ϕ ∈ Sn .85
A relationship between Riesz potentials and Riesz transforms is explained in the next
result
Proposition 7.1. If n ≥ 2 and ϕ ∈ Sn , then for 1 ≤ j ≤ n
∂
(I1 ϕ)(x) = −Rj ϕ(x) .
∂xj
Proof. This result easily follows from Proposition 5.53
Γ n−1
∂U2
∂
∂
1−n
2
(I1 ϕ) =
|x|
∗ϕ=
∗ϕ
n+1
∂xj
∂xj
∂xj
2π 2
Γ n−1
xj
2
= (1 − n)
p.v. n+1 ∗ ϕ = −Rj ϕ.
n+1
|x|
2π 2
cα (ξ) = (2π)−α |ξ|−α = (4π 2 |ξ|2 )−α/2 . Since (Iα ϕ)∧ = ϕ̂U
cα
According to Theorem 5.42, U
we get
Theorem 7.2. For 0 < α < n and ϕ ∈ Sn ,86
(7.1)
84c.f.
(Iα ϕ)∧ (ξ) = (4π 2 |ξ|2 )−α/2 ϕ̂(ξ),
Definition 2.14.
in general Iα ϕ 6∈ Sn , see Remark 7.5.
86The Fourier transform in (7.1) is understood as a Fourier transform of a tempered distribution U ∗ ϕ.
α
On the other hand (4π 2 |ξ|2 )−α/2 ϕ̂ ∈ L1 so the inverse Fourier transform in (7.2) is just a regular inverse
Fourier transform of an integrable function.
85However,
110
PIOTR HAJLASZ
i.e.,
∨
Iα ϕ = (4π 2 |ξ|2 )−α/2 ϕ̂(ξ) .
(7.2)
Recall that when n ≥ 3 the fundamental solution of the Laplace operator is87
Φ(x) = −
1
1
= −U2 (x).
n(n − 2)ωn |x|n−2
That means
(7.3)
−∆(I2 ϕ) = −∆(U2 ∗ ϕ) = U2 ∗ ((−∆)ϕ) = I2 ((−∆)ϕ) = ∆(Φ ∗ ϕ) = δ0 ∗ ϕ = ϕ.
Hence in some sense the Riesz potential I2 is the inverse of −∆ so it is reasonable to use
notation
∨
(−∆)−1 ϕ = I2 ϕ = (4π 2 |ξ|2 )−1 ϕ̂(ξ) .
With this notation (7.3) reads as
(−∆)(−∆)−1 ϕ = (−∆)−1 (−∆)ϕ = ϕ.
Also for ϕ ∈ Sn and a positive integer k we have
∨
∨
−∆ϕ = 4π 2 |ξ|2 ϕ̂(ξ)
so (−∆)k ϕ = (4π 2 |ξ|2 )k ϕ̂(ξ) .
This suggests how to define fractional powers of the Laplace operator, including powers
with negative exponents
Definition 7.3. For α > −n and ϕ ∈ Sn we define
∨
(−∆)α/2 ϕ = (4π 2 |ξ|2 )α/2 ϕ̂(ξ) .
Thus
∨
Iα ϕ = (4π 2 |ξ|2 )−α/2 ϕ̂(ξ) = (−∆)−α/2 ϕ for 0 < α < n.
Proposition 7.4. For any α > −n there is u ∈ Sn0 such that (−∆)α/2 ϕ = u ∗ ϕ for
ϕ ∈ Sn . In particular (−∆)α/2 ϕ is smooth, slowly increasing, and all its derivatives are
slowly increasing.
Proof. If 0 > α > −n, then (−∆)α/2 ϕ = I−α ϕ = U−α ∗ ϕ so it remains to consider the case
α ≥ 0. However, our argument will work for all α > −n. Let88
v(ξ) = ṽ(ξ) = (4π 2 |ξ|2 )α/2 ∈ Sn0
and u = v̂
so û = ṽ = v.
Then
∧
(u ∗ ϕ)∧ = ϕ̂û = ϕ̂v = (4π 2 |ξ|2 )α/2 ϕ̂(ξ) = (−∆)α/2 ϕ (ξ).
87Theorem
5.60.
α > −n, the singularity of v at ξ = 0 is integrable and hence the function v defines a tempered
distribution.
88Since
HARMONIC ANALYSIS
111
Remark 7.5. Although (−∆)α/2 ϕ is always smooth, it is not always in the Schwarz class
Sn . Namely for every α 6∈ {0, 2, 4, 6, . . .}, α > −n, there is ϕ ∈ Sn such that (−∆)α/2 ϕ 6∈
Sn . Suppose to the contrary that for some α 6∈ {0, 2, 4, 6, . . .}, α > −n, and all ϕ ∈ Sn we
have (−∆)α/2 ϕ ∈ Sn . Then the Fourier transform of this function also belongs to Sn i.e.,
(4π 2 |ξ|2 )α/2 ϕ̂(ξ) ∈ Sn .
(7.4)
However |ξ|α is not C ∞ smooth at ξ = 0 so if ϕ̂(0) 6= 0, the function (7.4) is not C ∞
smooth at ξ = 0 and hence it cannot belong to Sn .
On the other hand if α = 2k ∈ {0, 2, 4, 6, . . .}, then
(−∆)α/2 = (−∆)k : Sn → Sn
since the Laplace operator maps Sn to Sn and so does the k-fold composition of (−∆).
Proposition 7.6. If α > −n, then for ϕ ∈ Sn we have
(−∆)(−∆)α/2 ϕ = (−∆)α/2 (−∆)ϕ = (−∆)
α+2
2
ϕ.
Remark 7.7. We need to understand this result as follows. The function (−∆)α/2 ϕ is
smooth so we can apply the classical Laplace operator −∆ to it. Also since (−∆)α/2 ϕ is
slowly increasing and all its derivatives are slowly increasing, the classical Laplace operator
coincides with the distributional one (since there is no problem with the integration by
parts). This is how we understand the left hand side. In the middle term (−∆)ϕ ∈ Sn so
we can apply (−∆)α/2 to it. And finally the the term on the right hand side is well defined
because ϕ ∈ Sn .
Proof. According to Proposition 7.4, (−∆)α/2 ϕ = u ∗ ϕ for some u ∈ Sn0 . Hence
(−∆)((−∆)α/2 ϕ) = −∆(u ∗ ϕ) = u ∗ (− ∆ϕ ) = (−∆)α/2 (−∆ϕ)
|{z}
in Sn
=
d
(4π 2 |ξ|2 )α/2 (−∆ϕ)
∨
∨
α+2
= (4π 2 |ξ|2 )α/2 (4π 2 |ξ|2 )ϕ̂ = (−∆) 2 ϕ.
Remark 7.8. Formally we can extend the above result to composition of operators
(−∆)α/2 (−∆)β/2 as follows
∧
(−∆)α/2 (−∆)β/2 ϕ)
= (4π 2 |ξ|2 )α/2 (4π 2 |ξ|2 )β/2 ϕ̂(ξ)
∧
α+β
α+β
= (4π 2 |ξ|2 ) 2 ϕ̂(ξ) = (−∆) 2 ϕ
so formally
(7.5)
(−∆)α/2 (−∆)β/2 = (−∆)
α+β
2
on Sn
provided α, β > −n and α + β > −n.
The problem is that in general if ϕ ∈ Sn , then (−∆)β/2 ϕ 6∈ Sn so in the composition
(−∆)α/2 (−∆)β/2 ϕ
we apply the operator (−∆)α/2 to a function which is not in Sn and Definition 7.3 requires
the function to which we apply (−∆)α/2 to be in the space Sn . Well, the formula that
112
PIOTR HAJLASZ
defines (−∆)α/2 actually applies to a much larger class of function than just Sn and with
this extension in mind we can justify the above composition formula. There is however, a
price we have to pay for it. If ϕ ∈ Sn , then89 (−∆)α/2 ϕ = uα ∗ ϕ and (−∆)β/2 ϕ = uβ ∗ ϕ
for some uα , uβ ∈ Sn0 and we would like to write
(−∆)α/2 (−∆)β/2 ϕ = uα ∗ (uβ ∗ ϕ)
(7.6)
and this does not make any sense if uβ ∗ ϕ = (−∆)β/2 ϕ 6∈ Sn . An example where this
creates a problem will be seen in the next result. A formal proof, similar to the arguments
used above will be very short and elegant, but the actual rigorous proof will be quite long
and boring.
Theorem 7.9. If α, β > 0, α + β < n, then
(7.7)
Iα (Iβ ϕ) = Iα+β ϕ
for ϕ ∈ Sn .
Remark 7.10. Formally we can write this equality as
(−∆)−α/2 (−∆)−β/2 ϕ = (−∆)−
(7.8)
α+β
2
ϕ
when α, β > 0, α + β < n and ϕ ∈ Sn . This looks like a special case of the situation
considered in Remark 7.8 so, does it complete the proof of Theorem 7.9? Not really, because
we run into the problem described in (7.6). The equality can be written as
Uα ∗ (Uβ ∗ ϕ) = Uα+β ∗ ϕ
If we would know that Iβ ϕ = Uβ ∗ ϕ ∈ Sn we could apply the Fourier transform and easily
prove the identity, but in general, without this information, we have no reason to claim
c
that (Uα ∗ (Uβ ∗ ϕ))∧ = U\
β ∗ ϕ Uα .
Proof.
Lemma 7.11. If α, β > 0, α + β < n, then there is a constant90 C0 = C0 (α, β, n) such
that
Z
dy
C0
= n−(α+β) .
n−α
n−β
|y|
|x|
Rn |x − y|
Proof. First observe that the integral on the left hand side is finite whenever x 6= 0. To see
this let R = |x| and divide the integral over Rn into four integrals over mutually disjoint
regions
Z
Z
Z
Z
... dy,
... dy,
... dy,
... dy.
B(0, R
)
2
B(x, R
)
2
B(0,2R)\(B(0, R
)∪B(x, R
))
2
2
Rn \B(0,2R)
It is very easy to see that the first three integrals are finite. Indeed, in the first integral
we have a singularity |y|n−β at y = 0 only and this singularity is integrable. In the second
integral we have a singularity |x − y|n−α at y = x only which is integrable again and in the
third integral we do not have singularities at all. It remains to prove finiteness of the last
integral.
89Proposition
90In
7.4.
Corollary 7.12 we will find the exact value of C0 .
HARMONIC ANALYSIS
113
If |y| ≥ 2R, then |x − y| ≥ |y| − |x| ≥ |y| − |y|/2 = |y|/2 so
Z
Z
dy
dy
n−α
≤2
< ∞,
n−α
n−β
2n−(α+β)
|y|
Rn \B(0,2R) |x − y|
Rn \B(0,2R) |y|
because 2n − (α + β) > n.
Note that the integral from the lemma is invariant under rotations and hence
Z
dy
|y|n−β = f (|x|).
n−α
|x
−
y|
n
R
For x 6= 0 let x0 = x/|x| and t = |x| so x = tx0 . A simple change of variables gives
Z
Z
dy
dy
=
n−α
n−β
n−α |y|n−β
|y|
Rn |x − y|
Rn |tx0 − y|
Z
dy
= |x|α+β−n f (1).
= tα+β−n
n−α |y|n−β
|x
−
y|
n
0
R
We have
Iα (Iβ ϕ) =
=
=
=
=
Z
Z
1
1
ϕ(z)
dz dy
n−β
γ(α)γ(β) Rn |x − y|n−α
Rn |y − z|
Z Z
dy
1
ϕ(z) dz
n−α |y − z|n−β
γ(α)γ(β) Rn
Rn |x − y|
Z Z
1
dy
ϕ(z) dz
n−α |y|n−β
γ(α)γ(β) Rn
Rn |(x − z) − y|
Z
C0
ϕ(z)
dz
γ(α)γ(β) Rn |x − z|n−(α+β)
C0 γ(α + β)
Iα+β ϕ(x).
γ(α)γ(β)
We could use the Fubini theorem here because the function that we integrated over Rn ×Rn
was integrable. To see the integrability, repeat the above computations with ϕ replaced by
|ϕ|.
It remains to show that
γ(α)γ(β)
.
γ(α + β)
To prove this it suffices to verify that Iα (Iβ ϕ) = Iα+β ϕ for just one non-zero function ϕ.
To this end let ϕ ∈ Sn be such that ϕ̂ = 0 in a neighborhood of 0. Then
∨ ∨
Iα (Iβ ϕ) = Iα (4π 2 |ξ|2 )−β/2 ϕ̂
= (4π 2 |ξ|2 )−α/2 (4π 2 |ξ|2 )−β/2 ϕ̂ = Iα+β ϕ.
|
{z
}
C0 =
∈Sn
The proof is complete.
As a corollary we obtain
114
PIOTR HAJLASZ
Corollary 7.12. If α, β > 0, α + β < n, then
Z
1
dy
γ(α)γ(β)
=
.
n−α
n−β
n−(α+β)
|y|
γ(α + β) |x|
Rn |x − y|
In the next result we will prove an interesting formula for the fractional Laplacian
(−∆)α/2 when 0 < α < 2.
Theorem 7.13. For 0 < α < 2 and ϕ ∈ Sn we have
Z
1
ϕ(y) − ϕ(x)
α/2
(−∆) ϕ = lim
dy
ε→0 γ(−α) |x−y|≥ε |y − x|n+α
Z
ϕ(x + y) + ϕ(x − y) − 2ϕ(x)
1
dy,
=
2γ(−α) Rn
|y|n+α
where91
n
π 2 2−α Γ − α2
.
γ(−α) =
Γ n+α
2
Proof. First we will prove equality of the integrals
Z
Z
1
ϕ(x + y) + ϕ(x − y) − 2ϕ(x)
ϕ(y) − ϕ(x)
(7.9)
dy = lim
dy.
n+α
ε→0 |x−y|≥ε |y − x|n+α
2 Rn
|y|
It easily follows from Taylor’s formula that
|ϕ(x + y) + ϕ(x − y) − 2ϕ(x)|
kD2 ϕk∞
≤
C
,
|y|n+α
|y|n+α−2
where
kD2 ϕk∞ = sup
x∈Rn
2 !1/2
n X
∂ 2ϕ
(x)
.
∂xj ∂xk j,k=1
Since n + α − 2 < n, both sides of this inequality are integrable over the ball {|y| ≤ 1}.
They are also integrable in {|y| > 1} because of rapid decay of ϕ. Thus the first integral
at (7.9) is well defined and finite. For the second integral we have
Z
Z
Z
ϕ(y) − ϕ(x)
ϕ(x + y) − ϕ(x)
ϕ(x − y) − ϕ(x)
dy =
dy =
dy
n+α
n+α
|y|
|y|n+α
|x−y|≥ε |y − x|
|y|≥ε
|y|≥ε
so
Z
Z
ϕ(y) − ϕ(x)
1
ϕ(x + y) + ϕ(x − y) − 2ϕ(x)
dy =
dy
n+α
2 |y|≥ε
|y|n+α
|x−y|≥ε |y − x|
and hence passing to the limit as ε → 0+ yields (7.9).
Observe that
(7.10)
91We
(x, y) 7→
ϕ(x + y) + ϕ(x − y) − 2ϕ(x)
∈ L1 (Rn × Rn ).
|y|n+α
use here Γ evaluated at a negative number. Actually, Γ(− α2 ) = − α2 Γ(1 − α2 ), where 1 − α2 > 0, see
Definition 3.35. Note also that γ(−α) is given by the same formula as the constant in the definition of the
Riesz potential.
HARMONIC ANALYSIS
115
Indeed, for (x, y) ∈ Rn × B(0, 1) we have
2
−2n
ϕ(x + y) + ϕ(x − y) − 2ϕ(x) ≤ C kD ϕkL∞ (B(x,1)) ≤ C (1 + |x|)
∈ L1 (Rn × B(0, 1))
|y|n+α
|y|n+α−2
|y|n+α−2
and
Z
Z
|y|≥1
Rn
|ϕ(x + y) + ϕ(x − y) − 2ϕ(x)|
dx dy ≤ 3kϕk1
|y|n+α
Z
|y|≥1
dy
< ∞.
|y|n+α
This allows us to compute the Fourier transform of the integrable function
Z
ϕ(x + y) + ϕ(x − y) − 2ϕ(x)
(7.11)
f (x) =
dy
|y|n+α
Rn
by changing the order of integration. We have
Z
Z
ϕ(x + y) + ϕ(x − y) − 2ϕ(x)
−2πix·ξ
ˆ
f (ξ) =
e
dy dx
|y|n+α
Rn
Rn
Z
Z
1
−2πix·ξ
e
(ϕ(x + y) + ϕ(x − y) − 2ϕ(x)) dx dy
=
n+α
Rn
Rn |y|
Z
e2πiy·ξ + e−2πiy·ξ − 2
dy
= ϕ̂(ξ)
|y|n+α
Rn
Z
1 − cos(2πy · ξ)
= −2ϕ̂(ξ)
dy.
|y|n+α
Rn
Let ρ ∈ O(n) be such that ρ(ξ) = |ξ|e1 . Then the change of variables ỹ = ρ(y) gives
Z
Z
Z
1 − cos(2πy · ξ)
1 − cos(2πρ−1 (y) · ξ)
1 − cos(2πy · ρ(ξ))
dy =
dy =
dy
n+α
−1
n+α
|y|
|ρ (y)|
|y|n+α
Rn
Rn
Rn
Z
1 − cos(2π|ξ|y1 )
dy
=
|y|n+α
Rn
Z
1 − cos y1
α
= (2π|ξ|)
dy
|y|n+α
Rn
where in the last equality we used another change of variables ỹ = 2π|ξ|y. Actually it is
a consequence of the above calculation that the function (1 − cos y1 )/|y|n+α is integrable
on Rn , but one can see it more directly by using the estimate |1 − cos y1 | ≤ C|y|2 in a
neighborhood of the origin. Let
Z
1 − cos y1
C(α) =
dy.
|y|n+α
Rn
We proved that
∧
fˆ(ξ) = −2ϕ̂(ξ)(2π|ξ|)α C(α) = −2C(α) (−∆)α/2 ϕ (ξ).
Hence
(−∆)
α/2
Z
1
1
ϕ(x + y) + ϕ(x − y) − 2ϕ(x)
ϕ(x) = −
f (x) = −
dy
2C(α)
2C(α) Rn
|y|n+α
Z
1
ϕ(x + y) − ϕ(x)
= −
lim
dy.
C(α) ε→0 |y|≥ε
|y|n+α
116
PIOTR HAJLASZ
It remains to prove that
n
π 2 2−α Γ − α2
C(α) = −
.
Γ n+α
2
(7.12)
2
Let ϕ(x) = e−π|x| . Since ϕ̂ = ϕ we have
Z
Z
2
ϕ(0 + y) − ϕ(0)
1
1
e−π|y| − 1
α/2
lim
(7.13) (−∆) ϕ(0) = −
dy = −
dy.
C(α) ε→0 |y|≥ε
|y|n+α
C(α) Rn |y|n+α
Both sides of this equality are easy to compute. Recall that according to Lemma 5.44 for
γ > −n we have
Z
Z ∞
n+γ
Γ
2
2
e−π|x| |x|γ dx = nωn
e−πs sn+γ−1 ds = γ 2 n .
n
π2Γ 2
R
0
R
Since for g ∈ L1 , ǧ(0) = Rn g(x) dx, we obtain
Z
∨
2
α/2
2
2 α/2
(−∆) ϕ(0) = (4π |ξ| ) ϕ̂(ξ) (0) =
(2π|ξ|)α e−π|ξ| dξ
(7.14)
Rn
n+α
n+α
α α
2 π2Γ 2
α Γ
2 .
= (2π) α n =
Γ n2
π2Γ 2
On the other hand
Z
2
e−π|y| − 1
dy
|y|n+α
Rn
Z
=
∞
nωn
0
=
γ=2−α−n
=
=
1
Z
ds = nωn
∞
−πs2
(e
0
− 1)
s−α
−α
0
ds
Z
nωn ∞ −πs2 1−α
(e
− 1) s ds = −2π
e
s
ds
α 0
0
Z ∞
n+γ
2π
2π Γ 2
−πs2 n+γ−1
− nωn
e
s
ds = −
γ
α
α π 2 Γ n2
0
α
α+n Γ −
2
.
π 2
Γ n2
nωn
α
Z
∞
−πs2
−
n−1 e
s
n+α
s
−πs2
0 −α
This identity, (7.13) and (7.14) yield
α
α
2α π 2 Γ n+α
α+n Γ −
1
2
2
=−
π 2
C(α)
Γ n2
Γ n2
which easily implies (7.12).
Corollary 7.14. For 0 < α < 2
n
Z
π 2 2−α Γ − α2
1 − cos y1
dy = −
.
|y|n+α
Γ n+α
Rn
2
Corollary 7.15. If 0 < α < 2 and ϕ ∈ Sn , then (−∆)α/2 ϕ ∈ L1 (Rn ).
Proof.
Z
1
ϕ(x + y) + ϕ(x − y) − 2ϕ(x)
1
(−∆) ϕ =
dy =
f (x) ∈ L1
n+α
2γ(−α) Rn
|y|
2γ(−α)
since we proved in (7.10) that the function f defined in (7.11) is integrable.
α/2
HARMONIC ANALYSIS
117
It follows from Proposition 7.6 that is 0 < α < 2 and k ≥ 0 is an integer, then
α
α
α
(−∆) 2 +k ϕ = (−∆) 2 (−∆)k ϕ = (−1)k (−∆) 2 (∆k ϕ)
and hence Theorem 7.13 can be generalized to the case of higher powers as follows
Corollary 7.16. If 0 < α < 2, k ≥ 0 is an integer and ϕ ∈ Sn , then
Z
α
(−1)k
∆k ϕ(y) − ∆k ϕ(x)
+k
2
(−∆)
ϕ = lim
dy
ε→0 γ(−α) |x−y|≥ε
|y − x|n+α
Z
∆k ϕ(x + y) + ∆k ϕ(x − y) − 2∆k ϕ(x)
(−1)k
dy,
=
2γ(−α) Rn
|y|n+α
where
n
π 2 2−α Γ − α2
.
γ(−α) =
Γ n+α
2