HARMONIC ANALYSIS
T.KOMOROWSKI
1. Fourier Series
1.1. Definition. We assume that T := R/∼ with identification relation x ∼ y iff x−y = 2πn,
n ∈ Z. Let f ∈ L1 (T) we let
Z
ˆ
f (x)e−inx dx
f (n) :=
T
for all n ∈ Z. The measure on T is dx := (2π)−1 dx, where dx is the standard Lebesgue
measure on R. The formal series (needs not even be convergent)
X
fˆ(n)einx
n
shall be called the Fourier series of f .
Basic properties:
P 1)
Z
einx dx = δn0
T
P 2) if f is a trigonometric polynomial, i.e.
X
f (x) =
an einx
|n|≤N
then an = fˆ(n).
P 3) if f ∈ L1 (T)
|fˆ(n)| ≤ kf kL1 .
1.2. Why do we consider Fourier Series? Consider a periodic, initial value problem for
the heat eqt.

2 u(t, x), t > 0, x ∈ R
 ∂t u(t, x) = ∂xx

u(0, x) = f (x), u(t, x + 2π) = u(t, x)
How can we solve it? Look for the solutions of the form u(t, x) = ekt+inx Note, k = −n2 and
the fact that u is to be periodic implies n ∈ Z.
P
2
The superposition principle yields that u(t, x) = n a(n)e−n t+inx is a solution to the heat
eqt. The initial condition yields
X
f (x) =
a(n)einx .
n
So the best candidates for the coefficients are the Fourier ones...
1
2
T.KOMOROWSKI
1.3.
Convolution of two functions. Suppose that f, g ∈ L1 (T). We define f ∗ g(x) :=
R
T f (x − y)g(y)dy.
Proposition 1.1. P 1) For any f, g ∈ L1 (T) we have f ∗g is defined as an element of L1 (T).
P 2) We have f ∗ g = g ∗ f
P 3) f[
∗ g(k) = fˆ(k)ĝ(k)
P 4) kf ∗ gkL1 (T) ≤ kf kL1 (T) kgkL1 (T) , in fact for nonnegative f, g we have equality.
P 5) kf ∗ gkL∞ (T) ≤ kf kL∞ (T) kgkL1 (T) .
P 6) For any f ∈ L1 (T) and g ∈ C(T) we have f ∗ g ∈ C(T).
Proof. Exercise.
1.4. Auxiliary result.
Proposition 1.2.
sin[(n + 21 )x]
,
sin x2
|k|≤n
n
1
sin[(n + 1)x/2] 2
1 X
Dn (x) =
An (x) :=
,
n+1
n+1
sin x2
k=0
X
Sn (x; f ) :=
fˆ(k)eikx = f ∗ Dn (x),
Dn (x) :=
X
eikx =
|k|≤n
n
X
Fn (x; f ) :=
Sk (x; f ) = f ∗ An (x),
k=0
1.5. The question of the convergence of the Fourier series. Does the Fourier series
converge? If so, in what sense?
1.5.1. Uniform convergence.
Theorem 1.3. (Fejér) Suppose that f ∈ C(T). Then
lim kFn (f ) − f k∞ = 0.
n→=∞
Proof. The theorem follows from the following lemma.
Lemma 1.4. We have
An (x) ≥ 0,
Z
An (x)dx = 1,
T
for any δ > 0
lim sup An (x) = 0.
n→+∞ |x|≥δ
Proof. Exercise.
Note that for any x ∈ T and δ > 0
Z
|Fn (x; f ) − f (x)| ≤ An (y)[f (x − y) − f (x)]dy T
Z
Z
=
An (y)[f (x − y) − f (x)]dy + An (y)[f (x − y) − f (x)]dy [|y|≤δ]
[|y|<δ]
HARMONIC ANALYSIS
3
Z
≤ 2kf k∞
An (y)dy + ωf (δ).
[|y|≤δ]
Hence,
lim sup kFn (f ) − f k∞ ≤ ωf (δ)
n→+∞
for arbitrary δ > 0 and the conclusion of the theorem follows.
Remark. The argument above in fact shows that if we have any family of functions
Aα ∈ L1 (T) such that
A (x) ≥ 0,
Z α
Aα (x)dx = 1,
T
for any δ > 0
lim sup Aα (x) = 0.
α |x|≥δ
then
lim kAα ∗ f − f k∞ = 0.
α
Note that we also have
lim kAα ∗ f − f kL1 (T) = 0.
(1.1)
α
L1 (T).
for f ∈
can write then
Indeed, for any ε > 0 there exists f1 ∈ C(T) such that kf − f1 kL1 < ε we
kAα ∗ f − f kL1 ≤ kAα ∗ (f − f1 )kL1 + kAα ∗ f1 − f1 kL1 + kf − f1 kL1
≤ kAα kL1 kf − f1 kL1 + kAα ∗ f1 − f1 kL1 + kf − f1 kL1
| {z }
=1
≤ 2ε + kAα ∗ f1 − f1 kL1 .
Letting δ → 0+ we obtain
lim sup kAα ∗ f − f kL1 ≤ 2ε.
δ→0+
We shall return to the issue of a.s. convergence and the Lp –convergence later on...
Example 1. (Poisson kernel). For any r ∈ (0, 1) we have
X
1 − r2
Pr (x) :=
r|n| einx =
.
ix |2
|1
−
re
n
Exercise. The kernel satisfies the assumptions made in the remark.
We have therefore for any f ∈ C(T)
X
r|n| fˆ(n)einx = Pr ∗ f (x) → f (x), as r → 1−
n
uniformly on T and in L1 (T) if f ∈ L1 (T).
Example 2. (Mollifiers). Another important
example is provided by the following conR
struction. Suppose that c ∈ (0, π) g ≥ 0, R gdx = 1 (density) and g(x) = 0 for |x| ≥ c > 0.
We define g on T by periodization. Let gδ := δg(x/δ). Then for any f ∈ C(T) we have
lim kgδ ∗ f − f k∞ = 0. and for any f ∈ L1 (T) we have lim kgδ ∗ f − f kL1 (T) = 0.
δ→0+
δ→0+
Corollary 1.5. (Weierstrass) The set of trigonometric polynomials is dense in C(T). The
set of polynomials is dense in C[−1, 1].
4
T.KOMOROWSKI
Proof. The first part is obvious. As for the other one suppose that f ∈ C[−1, 1]. Then
g(x) := f (cos x) belongs to C(T). For any ε > 0 one can find a trig. polynomial
X
an einx
g̃(x) :=
|n|≤N
such that kg̃ − gk∞ < ε. But, of course also g̃(−x) satisfies this estimate and
N
X
1
h̃(x) := [g̃(x) + g̃(−x)] =
(an + a−n ) cos nx
| {z }
2
n=0
=bn
satisfies the estimate. Choose
h(x) =
N
X
bn cos(n arccos x).
n=0
Excercise. Prove that cos n arccos x, n ≥ 0 are polynomials.
Corollary 1.6. (Uniqueness of Fourier series) Suppose that fˆ(n) = 0 for all n and f ∈ L1 (T).
Then, f = 0.
Proof. Suppose first that f ∈ C(T). Note that Fn (f ) = 0. According to Fejér theorem f is a
limit of Fn (f ), so f = 0. If f ∈ L1 (T) then for a mollifier g ∈ C(T) we have kf −f ∗gδ kL1 (T) → 0
as δ → 0+. But f\
∗ gδ (n) = fb(n)gbδ (n) = 0 so f ∗ gδ = 0 (note that f ∗ gδ ∈ C(T)), thus
f = 0.
1
Suppose that f ∈ C (T). We have
Z
Z
parts 1
−inx
ˆ
f 0 (x)e−inx dx.
f (n) =
f (x)e
dx =
n
T
T
Hence,
kf 0 kL1 (T)
ˆ
|f (n)| ≤
, ∀ n.
n
More generally, if f ∈ C k (T) one gets
kf (k) kL1 (T)
, ∀ n.
nk
Excercise. We say that f ∈ Lipα (T) for α ∈ (0, 1), if ∃ M > 0 such that |f (x) − f (y)| ≤
M |x − y|α . Suppose that f ∈ Lipα (T) then
C
|fˆ(n)| ≤ α , ∀ n.
n
Note that
Z
Z
Z 1
−1
−1
π −inx
−inx
iπ−inx
ˆ
f (n) =
f (x)e
dx =
f (x)e
dx =
f x+
e
dx
2π T
2π T
2π T
n
hence,
Z h
i
kf kLipα
1
π
−inx
|fˆ(n)| =
f (x) − f x +
e
dx ≤
.
4π T
n
2nα
Question: Can we characterize the condition f ∈ Lipα (T) in terms of the asymptotic
behavior of fˆ(n)?
|fˆ(n)| ≤
HARMONIC ANALYSIS
5
Theorem 1.7. (Riemann-Lebesgue) Suppose that f ∈ L1 (T). Then
lim
|n|→+∞
fˆ(n) = 0.
Proof. If f ∈ C 1 (T) the theorem is obvious. Suppose that f ∈ L1 (T). For any ε > 0 one can
find g ∈ C 1 (T) such that kf − gkL1 < ε. We have therefore
ε
+ |ĝ(n)|.
|fˆ(n)| ≤ |fˆ(n) − ĝ(n)| + |ĝ(n)| ≤
2π
Hence,
ε
lim sup |fˆ(n)| ≤
2π
n→+∞
thus it equals to 0.
One can observe that the more regular a function coefficients the faster its Fourier coefficients decay at infinity. In fact, we can easily conclude the following.
Theorem 1.8. Suppose that f ∈ C 1+α (T). Then,
lim kSn (f ) − f k∞ = 0.
n→+∞
Proof. From the above estimates we get
|fˆ(n)| ≤
C
n1+α
,
∀ n 6= 0.
Using Weierstrass convergence criterion for function series we conclude that Sn (f ) ⇒ g. But
n
Fn (f ) =
1X
Sk (f ) ⇒ f.
n
k=0
Hence g = f .
Theorem 1.9. Suppose that f ∈ C 1 (T). Then,
fˆ0 (k) = ik fˆ(k)
for all k ∈ Z.
Proof. Exercise.
1.5.2. Pointwise convergence. Is it true that for f ∈ C(T) we have at least Sn (x; f ) →
f (x), as n → +∞? The answer is in the negative.
Exercise. Show that there exist 0 < C1 < C2 < +∞ such that
Z
C1 log n ≤
|Dn (x)|dx ≤ C2 log n, ∀ n ≥ 2.
T
Define sn : C(T) → R as sn f := Sn (0; f ). It is a bounded linear functional. In fact
Z
|sn (f )| ≤
|Dn (x)f (x)|dx ≤ C2 log nkf k∞ .
T
Choosing
fn :=


1, Dn (x) > 0

−1, Dn (x) < 0,
6
T.KOMOROWSKI
we would find that
Z
|sn (f )| =
|Dn (x)|dx ≥ C1 log n.
T
This is not entirely correct because f 6∈ C(T), but that can be easily remedied by taking
fn = f except close to zeros of Dn (x) and then extrapolate e.g. linearly. We can achieve
easily kfn k∞ and
C1
|sn (fn )| ≥
log n.
2
Note that we have limn→+∞ ksn k = +∞. Hence by the Banach-Steinhaus theorem there must
be f ∈ C(T) for which lim supn→+∞ sn (f ) = +∞, or equivalently lim supn→+∞ Sn (0; f ) =
+∞. In fact the set of those f -s is residual, i.e. its complement is contained in a union of
closed and nowhere dense sets.
Remark 1. It has been shown by Carleson that if f ∈ C(T) then Sn (x; f ) → f (x) for
x ∈ T \ N , where m1 (N ) = 0.
Let ωf (x; δ) := sup[|f (y) − f (x)| : |y − x| ≤ δ] be the modulus of continuity of f at x. As
an application of the Lebesgue–Jordan theorem we obtain immediately the following.
Theorem 1.10. (Dini Theorem) Suppose that for some δ0 > 0
Z δ0
ωf (x; δ)
dδ < +∞.
δ
0
Then, limn→+∞ Sn (x; f ) = f (x).
Proof. With no loss of generality assume that x = 0 (otherwise we consider f˜(y) := f (x + y)).
We have
Z
sin [(n + 1/2)y]
Sn (0; f ) − f (0) = Dn ∗ f (0) − f (0) =
[f (y) − f (0)]dy.
sin(y/2)
T
Let
g(y) :=
f (y) − f (0)
.
sin(y/2)
We have g ∈ L1 (T) for
Z
Z
Z
ωf (0; y)
|f (y) − f (0)|
|g(y)|dy =
dy ≤ C
dy < +∞.
|y|
T
T | sin(y/2)|
T
Using Theorem 1.7 we obtain
Z
lim [Sn (0; f ) − f (0)] = lim
n→+∞
n→+∞
sin [(n + 1/2)y]
[f (y) − f (0)] = 0.
sin(y/2)
T
Theorem 1.11. (Dirichlet-Jordan Theorem) If f ∈ BV (T) (a function of bounded variation)
then Sn (x; f ) → 1/2(f (x−) + f (x+)) for x ∈ T.
Proof. Let Sf (0) := 1/2(f (0−) + f (0+)). We have
Zπ
1
sin [(n + 1/2)y]
Sn (0; f ) =
f (y)dy.
2π
sin(y/2)
−π
HARMONIC ANALYSIS
7
Let
g(y) :=
f (y) sin(y/2)
.
y/2
We need to show that
Zπ
1
Tλ g :=
2π
sin(y(2λ)−1 )
g(y)dy → Sg(0),
y/2
−π
as λ → 0+. We have however
πλ
Z −1
1
Tλ g =
2π
sin(y/2)
g (λy) dy
y/2
−πλ−1
R πλ−1
R0
We divide the above integral into two integrals: 0
and −πλ−1 . Since we deal with both
of them in the same fashion consider only the first one. Let
Z
1 +∞ sin(z/2)
Si(y) :=
dz.
π y
z/2
We have Si(0) = 1. We can write
1
2π
πλ
Z −1
sin(y/2)
[g (λy) − g(0+)]dy
y/2
0
πλ
Z −1
1
= −
2π
parts
Si(y)d[g (λy) − g(0+)]
0
=−
1
2π
Zπ
Si(yλ−1 )dg (y) → 0,
0
because |G(·) is bounded (via Lebesgue dominated convergence theorem).
Example. Gibbs phenomenon.

 1, x ∈ [nπ, (n + 1)π), n even
f (x) =

−1, x ∈ [nπ, (n + 1)π), n odd
Calculate
fˆ(n) =
S2N +1 (x; f ) =


0,

2
iπn ,
n even
n
odd.
N
X 2einx
4 X sin[(2n + 1)x]
=
.
iπn
π
2n + 1
n=0
|n|≤N
Calculate maximum of S2N +1 (x; f ). Note that
0
S2N
+1 (x; f )
N
4X
=
cos[(2n + 1)x]
π
n=0
8
T.KOMOROWSKI
"
0
!"
Figure 1. Function f(x)
use 2 cos[(2n + 1)x] sin x = sin[2(n + 1)x] − sin(2nx) and get
0
S2N
+1 (x; f ) =
2 sin[2(N + 1)x]
.
π
sin x
It equals to 0 at xN = π/[2(N + 1)] we have
2
S2N +1 (xN ; f ) =
(N + 1)π
N sin
X
n=0
2n+1
2(N +1)
2n+1
2(N +1)
2
→
π
Z
0
π
sin x
dx ≈ 1.17....
x
1.5.3. The L2 -convergence. Suppose that f ∈ L2 (T). The space is a Hilbert space when
equipped with the scalar product
Z
(f, g)L2 (T) =
f ḡdx.
T
einx
Note that en (x) :=
are orthonormal. Sn (f ) is the orthogonal projection onto the subspace
of n-th degree orthogonal polynomials, i.e. f − Sn (f ) ⊥ ek , k = 0, ±1, . . . , ±n. As a result
we get
kf k2L2 (T) = kf − Sn (f )k2L2 (T) + kSn (f )k2L2 (T)
X
= kf − Sn (f )k2L2 (T) +
|fˆ(k)|2 .
|k|≤n
Also, for m ≤ n
(1.2)
kSm (f ) − Sn (f )k2L2 (T) =
X
|fˆ(k)|2 .
m<|k|≤n
Corollary 1.12. We have
2
ˆ
n |f (n)|
P
< +∞.
Theorem 1.13. We have limn→+∞ kf − Sn (f )kL2 (T) = 0 for any f ∈ L2 (T).
Proof. Using (1.2) we conclude that Sn (f ) converges in L2 (T). Denote its limit by g. However,
the above means that Fn (f ) converges also to g in L2 (T). According to (1.1) we obtain
Fn (f ) → f , so f = g.
HARMONIC ANALYSIS
9
Corollary 1.14. limn→+∞ kf − Fn (f )kL2 (T) = 0.
Corollary 1.15. (Parseval’s identity)
kf k2L2 (T) =
(1.3)
X
|fˆ(n)|2 .
n
Questions:
1. Can we claim limn→+∞ kf − Sn (f )kLp (T) = 0, or at least limn→+∞ kf − Fn (f )kLp (T) = 0
for f ∈ Lp (T) and p ∈ [1, +∞)?
2. How about the a.s. convergence?
3. Is there a similar duality relation to (1.3) for other Lp spaces?
1.5.4. Filters. Fourier coefficients of Dn (x):

 0, |k| > n
D̂(k) =

1, |k| ≤ n.
n
!n
Figure 2. Filter Dn (x)
Fourier coefficients of An (x):
Â(k) =


0,
|k| > n

n+1−|k|
n+1 ,
|k| ≤ n.
Exercise. Suppose that m ≤ n we let
An,m (x) :=
X
Ân (k)eikx .
|k|≤m
Then,
An,m =
n−m
m+1
Dm +
Am .
n+1
n+1
10
T.KOMOROWSKI
1
n
!n
Figure 3. Filter An (x)
In consequence (see the first example in Section 1.5.2),
n−m
m+1
(1.4)
kAn,m kL1 ≤ C
log m +
,
n+1
n+1
which implies that
(1.5)
sup kAn,m kL1 ≤ C log n.
1≤m≤n
Theorem 1.16. (Bernstein) Suppose that f is a trigonometric polynomial of degree n then
kf 0 k∞ ≤ nkf k∞ .
Proof. We shall present a simple proof with the constant 2n in place of n. The proof with the
proper constant n is a bit more involved, see cite [1]. We have f 0 = Kn+ ∗ f + Kn− ∗ f, where

0, k > 2n, or k < 0





k, 0 ≤ k ≤ n,
K̂n+ (k) =





2n − k, n < k ≤ 2n,
K̂n− (k) = K̂n+ (−k). We show the estimate for Kn+ ∗ f for the other case is analogous. Note
that Kn+ (x) = neinx An (x) so by virtue of Proposition 1.1 P 5) we obtain
kKn+ ∗ f k∞ ≤ kKn+ kL1 kf k∞
and kKn+ kL1 = nkA+
n kL1 = n.
n
Let Vn := 2A2n−1 − An−1 , n ≥ 1, V0 := A0 . Let Wn := V2 − V2n−1 , n ≥ 1, and W0 := V1 .
Let m ≤ 2n+1 and Wn,m = Am ∗ Wn . Using (1.4) we conclude that for m ≤ 2n+1
(1.6)
kWn,m kL1 ≤ kAm kL1 kWn kL1 ≤ 3.
Exercise. Show that for f ∈ C(T)
X
(1.7)
f=
Wn ∗ f = lim V2n ∗ f
n≥0
n→+∞
HARMONIC ANALYSIS
11
2
1
n!1
!(n!1)
!(2n!1)
2n!1
Figure 4. Filters 2A2n−1 (x) and An−1 (x)
1
!2
n
!2
n!1
0
2
n!1
2
n
Figure 5. Filter V2n (x)
and the convergence is uniform.
The Hölder class
Lα := [f : ∃C > 0 |f (x + h) − f (x)| ≤ Chα ],
when α ∈ (0, 1).
Theorem 1.17. Suppose that α ∈ (0, 1). Then, the necessary and sufficient condition for
f ∈ Lα is that
(1.8)
∃ C > 0 ∀ n : kWn ∗ f k∞ ≤ C2−nα .
Proof.
Suppose that (1.8) holds. Then, the series in (1.7) converges uniformly and g =
P
W
n ∗ f ∈ C(T). On the other hand we have
n≥0
ĝ(k) = lim V̂2n (k)fˆ(k) = fˆ(k)
n→+∞
12
T.KOMOROWSKI
1
n+1
!2
!2
n
!2
n!1
0
2
n!1
2
n
n+1
2
Figure 6. Filter Wn (x)
so, in fact, f = g. Suppose also that
h ∈ [2−N −1 , 2−N ).
(1.9)
We can write
|f (x + h) − f (x)| ≤
X
|Wn ∗ f (x + h) − Wn ∗ f (x)| ≤
n≥0
X
|Wn ∗ f (x + h) − Wn ∗ f (x)| +
N ≥n≥0
X
|Wn ∗ f (x + h) − Wn ∗ f (x)|.
N <n
In the case when n > N we use (1.8) and obtain that the second sum can be bounded from
above by
X
X
C
|Wn ∗ f (x + h) − Wn ∗ f (x)| ≤ C
2−nα ≤ N α ≤ Chα .
2
n>N
n>N
As for the first sum we use Bernstein’s theorem 1.16 to estimate by
|Wn ∗ f (x + h) − Wn ∗ f (x)| ≤ k(Wn ∗ f )0 k∞ h ≤ 2n+2 kWn ∗ f k∞ h ≤ C2n(1−α) h
(1.2)
≤ C2(n−N )(1−α) hα
and, in consequence,
X
|Wn ∗ f (x + h) − Wn ∗ f (x)| ≤ Chα
N ≥n≥0
X
2(n−N )(1−α)
N ≥n≥0
≤C
X
2−n(1−α) hα ≤ Chα .
N ≥n≥0
Thus, f ∈ Lα .
Conversely, suppose that f ∈ Lα , then |f (x + h) − f (x)| ≤ Chα . Note that
Wn ∗ f (x) =
X
2n−1 ≤|k|≤2
2 − (eikh + e−ikh )
.
Ŵn (k)fˆ(k)eikx
2 − (eikh + e−ikh )
n+1
HARMONIC ANALYSIS
13
Exercise. Suppose that fˆ(0) = 0, λ−k = λk for all k > 0 and
X
λk fˆ(k)eikx
g=
|k|≤N
Hint: Show that
(1.10)
g = λN SN f + (λN − λN −1 )N FN −1 f +
N
−2
X
(λk + λk+2 − 2λk+1 )(k + 1)Fk f.
k=0
Prove first (1.10) holds for f such that fˆ(k) = 0 for all k < 0. In general case decompose first
P
P
f = f˜+ + f˜− , where f˜+ = N ≥k>0 fˆ(k)eikx and f˜− = −N ≤k<0 fˆ(k)eikx .
ikh
−ikh
−1
Let λk := [2 − (e + e
)] and
X
Mn f (x) =
Ŵn (k)fˆ(k)eikx [2 − (eikh + e−ikh )] = Wn ∗ [f (x + h) + f (x − h) − 2f (x)].
2n−1 ≤|k|≤2n+1
Using formula (1.10) we can obtain
(1.11)
Wn ∗ f (x) = λ2n+1 Mn f (x) + (λ2n+1 − λ2n+1 −1 )2n+1 A2n+1 −1 ∗ Mn f (x)
+
2n+1
X−2
(λk + λk+2 − 2λk+1 )(k + 1)Ak ∗ Mn f (x).
k=2n−1
Exercise. Show that for any complex valued function F : [a, b] → C that belongs to
C 2 [a, b] we have
(1.12)
|F (x + h0 ) − F (x)| ≤ sup |F 0 |h0
(1.13)
|F (x + h0 ) + F (x − h0 ) − 2F (x)| ≤ 2 sup |F 00 |h20
for any x − h0 , x, x + h0 ∈ [a, b], h > 0.
Using (1.13), with F (s) = [2 − (ei(k+1+s)h + e−i(k+1+s)h )]−1 , h0 = 1, we can estimate by
|Wn ∗ f (x)| ≤ |λ2n+1 ||Mn ∗ f (x)|
(1.14)
+2n+1 h sup 2|2 − (ei(2
n+1 +s)h
+ e−i(2
n+1 +s)h
)|−2 |A2n+1 −1 ∗ Mn f (x)|
s∈[0,1]
+2h
2
2n+1
X−2
(k + 1) sup [4|2 − (ei(2
n+1 +s)h
+ e−i(2
n+1 +s)h
)|−3 ]
s∈[−1,1]
k=2n−1
+ sup [2|2 − (ei(2
n+1 +s)h
+ e−i(2
n+1 +s)h
)|−2 ]}|Ak ∗ Mn f (x)|,
s∈[0,1]
Lemma 1.18. For any m ≥ 1
|2 − (ei(k+s)h + e−i(k+s)h )|−m ≤
1
.
(kh)2m
Proof. Exercise.
Since Ak ∗ Mn f = Wk,n ∗ [f (· + h) + f (· − h) − 2f (·)] , according to (1.6), we have
(1.15)
kAk ∗ Mn f k∞ ≤ 2 sup kWn,m kL1 kf (· + h) − f (·)k∞ ≤ Chα
1≤m≤k
14
T.KOMOROWSKI
Let h = 2−n−1 . Note that
(1.16)
i(2n+1 +s)2−n−1
|2 − (e
−i(2n+1 +s)2−n−1
+e
1
s ≥C>0
)| = 4 sin
1 + n+1
2
2
2
for s ∈ [0, 1]
Taking all of the above into account we get from (1.14) (for h = 2−n−1 )
(1.17)
|Wn ∗ f (x)| ≤ CkWn kL1 kf (· + 2−n−1 ) + f (· − 2−n−1 ) − 2f (·)k∞
+2n+1 h sup [2|2 − (ei(2
n+1 +s)2−n−1
+ e−i(2
n+1 +s)2−n−1
)|−2 ]|A2n+1 −1 ∗ Mn f |
s∈[0,1]
+2h
2
2n+1
X−2
(k + 1){ sup [4|2 − (ei(2
k=2n−1
n+1 +s)2−n−1
+ e−i(2
n+1 +s)2−n−1
)|−3 ]
s∈[−1,1]
+ sup [2|2 − (ei(2
n+1 +s)2−n−1
+ e−i(2
n+1 +s)2−n−1
)|−2 ]}|Ak ∗ Mn f |,
s∈[−1,1]
n+1
Lemma 1.18
≤
2X
−2
C
−2(n+1)
+ |A2n+1 −1 ∗ Mn f | + C2
(k + 1)|Ak ∗ Mn f |
2nα
n−1
k=2
(1.15)
≤
2n+1 −2
X (k + 1)
C
C
−nα
n+1 −nα
+
C2
+
C2
2
≤ nα .
nα
3
2
k
2
n−1
k=2
Remark 1. The above result holds also for α = 1. But in this case one has to modify a
little the definition of the space Lip1 := [f : ∃ C > 0 |f (x + h) + f (x − h) − 2f (x)| ≤ Ch2 ],
see [4].
Remark 2. Combining (1.15) and (1.17) we obtain
1
1
kWn ∗ f k∞ ≤ C Wn ∗ f · + n+1 + f · − n+1 − 2f (·) .
2
2
∞
Exercise. Use the above remark to prove that if f 0 ∈ Lipα then
C
∃ C > 0 ∀ n ≥ 0 : kWn ∗ f k∞ ≤ n(1+α) .
2
In fact, for any k ≥ 1 if f (k) ∈ Lipα then
∃ C > 0 ∀ n ≥ 0 : kWn ∗ f k∞ ≤
C
2n(k+α)
.
Exercise. The converse to the above is also true. Namely, if
C
∃ C > 0 ∀ n ≥ 0 : kWn ∗ f k∞ ≤ n(k+α) .
2
then f (k) ∈ Lipα .
Hint: Use the fact that f (k) ∈ Lipα to conclude
∃ C > 0 ∀ n ≥ 0 : kWn ∗ f (k) k∞ ≤
then use Theorem 1.16.
C
.
2nα
HARMONIC ANALYSIS
15
2. The maximal function. Pointwise convergence.
We shall be interested in proving the following theorem.
Theorem 2.1. Suppose that f ∈ L1 (T). Then
lim Fn (f ) = f,
n→+∞
a.e.
We start with the following very useful covering lemma due to Vitali. For any open cube I
and a > 0 denote by aI the cube that has the same center as I and whose volume |aI| = ad |I|.
Lemma 2.2. Suppose that K ⊂ Rd is a compact set and {Iα } is its covering with open cubes.
Then there exists a finite family of cubes {Iαi }i=1,...,N such that
i) Iαi ∩ Iαj = ∅, if i 6= j,
S
ii) K ⊂ N
i=1 3Iαi .
Proof. Since K is compact there is a finite subfamily F1 of {Iα } that covers K. Choose Iα1 as
the cube of F1 that has the largest volume. In case of a tie choose any one of those intervals
with the largest volume. Among the cubes of F1 choose a subfamily F2 of cubes disjoint with
Iα1 . If it is non-empty pick the cube of F2 that has the largest volume, etc. Suppose that N
is the largest natural such that FN 6= ∅. In the described above way we construct a family of
cubes {Iαi }i=1,...,N that clearly satisfy i). We show that it also must satisfy ii). Suppose that
x ∈ K then thereSis I ∈ F1 such that x ∈ I. If I = Iα1 there is nothing to prove. If otherwise,
then either x ∈ F2 , or I ∩ Iα1 6= ∅. In the latter case though we must have |I| ≤ |Iα1 | so
x ∈ 3Iα1 . In the first case we repeat the above procedure replacing F1 with F2 . After a finite
number of steps we convince ourselves that ii) must hold.
Exercise. Was the assumption that K is compact important? Substantiate your answer.
For any cube I denote by I(c) its center.
Definition 2.3. ( Maximal function of Hardy–Littlewood) Suppose that f ∈ L1loc (Rd ). For
any x ∈ Rd define
Z
1
M f (x) := sup
|f (y)|dy.
x=I(c) |I|
I
Theorem 2.4. ( The maximal inequality of Hardy-Littlewood) For any f ∈ L1 (Rd ) and
λ > 0 we have
Z
3d
(2.1)
md (M f > λ) ≤
|f (y)|dy.
λ
[M f >λ]
Proof. Let Eλ := [M f > λ]. Suppose that K ⊂ Eλ is compact. For any x ∈ K there is Ix
such that x = Ix (c) and
Z
|f (y)|dy > λIx .
Ix
Clearly {Ix }x∈K covers K. According to Lemma 2.2 one can find {Ii } a finite subfamily of
S
disjoint cubes such that K ⊂ N
i=1 3Ii . We can write therefore that
Z
N Z
N
X
3d X
3d
d
md (K) ≤ 3
|Ii | ≤
|f (y)|dy ≤
|f (y)|dy.
λ
λ
i=1
i=1 I
i
Rd
16
T.KOMOROWSKI
Since K was chosen arbitrary the conclusion of the theorem follows.
Remark 1. Let us reformulate (2.1). Let
f, if |f | ≥ λ/2
f1 =
0, if otherwise.
Then,
f ≤ f1 +
λ
2
and, in consequence,
M f ≤ M f1 +
λ
.
2
We obtain therefore
3d
md (M f ≥ λ) ≤ md (M f1 ≥ λ/2) ≤
λ
(2.1)
(2.2)
Z
3d
|f1 |dx =
λ
Z
|f |dx.
[|f |≥λ/2]
Remark 2. Logarithmic inequality. Suppose that supp f ⊂ BR (0) for some R > 0. We
have then
Z
Z
M f dx ≤ md (BR (0)) +
(M f − 1)+ dx
BR (0)
BR (0)
+∞
+∞
Z
Z
3d dλ
= md (BR (0)) +
md (M f ≥ λ)dλ ≤ md (BR (0)) +
λ
1
d
= md (BR (0)) + 3
1
2|f
Z |∨1
Z
|f |dy
dλ
≤ md (BR (0)) + 3d
λ
Z
Z
|f |dy
[f ≥λ/2]
(log 2 + log+ |f |)|f |dy.
1
2.1. Applications of the Maximal Inequality.
2.1.1. Lebesgue density theorem. Suppose that f ∈ L1 (R). Let also
Zx
(2.3)
F (x) := f (y)dy.
0
We shall prove the following.
Theorem 2.5. We have
F (x + h) − F (x)
= f (x).
h→0
h
(2.4)
lim
for a.e. x.
Note that when f ∈ Cc (R) the above result is obviously true. For any f ∈ L1 (R) there
exists a sequence (fn ) ⊂ Cc (R) such that limn→+∞ kf − fn kL1 (R) = 0. Denote by Fn the
function corresponding to fn by (2.3). Of course (2.4) would follow if we could write
(2.5)
F (x + h) − F (x)
Fn (x + h) − Fn (x)
= lim lim
h→0
h→0 n→+∞
h
h
lim
HARMONIC ANALYSIS
17
Fn (x + h) − Fn (x)
= lim fn (x) = f (x).
n→+∞ h→0
n→+∞
h
= lim lim
The proof of Theorem 2.5. Let c > 0 and
1
Sc f (x; h) := sup
I |I|
(2.6)
Z
|f (y) − f (x)|dy,
I
where the supremum is taken over all subintervals I ⊂ [x − h, x + h] such that |I| ≥ ch. It
suffices only to show that
(2.7)
lim Sc f (x; h) = 0
h→0+
for a.e. x. Let
1
En (f ) := x : lim sup Sc f (x; h) ≥
.
n
h→0+
We show that
(2.8)
m(En (f )) = 0
for any n ≥ 1, which of course implies (2.5). Note however that for any g ∈ Cc (R) we have
Fn (f ) = En (f − g). Choose ε > 0 arbitrary and find g so that kf − gkL1 (R) < ε. For any
function u ∈ L1 (Rd ) and I as in the definition of we have
Z
Z
Z x+h
1
1
1
(2.9)
|u(y) − u(x)|dy ≤
|u(y)|dy + |u(x)| ≤
|u(y)|dy + |u(x)|
|I| I
|I| I
ch x−h
2
≤ M u(x) + |u(x)|.
c
Using the above estimate we can write This in turn implies that
(2.10)
c 1
m(En (f )) = m(En (f − g)) ≤ m M (f − g) ≥
+ m |f − g| >
.
4n
2n
Using the maximal inequality we obtain that the first term on the right hand side of (2.10)
can be estimated by
12n
12n
kf − gkL1 (R) <
ε.
c
c
The second term can be estimated by Chebyshev’s inequality by
2nkf − gkL2 (R) < 2nε.
Summarizing we showed that
1
m(En (f )) < 2n 6 +
c
for any ε > 0. Thus m(En (f )) = 0 for any n ≥ 1.
ε
18
T.KOMOROWSKI
2.1.2. Almost everywhere convergence of Fejér sums.
Theorem 2.6. Suppose that f ∈ L1 (T). Then
lim Fn (f ) = f,
n→+∞
a.e.
Proof. Again it suffices only to prove that
Z
AN (x − y)|f (y) − f (x)|dy = 0
(2.11)
lim
N →+∞ T
for a.e. x. Let
Z
1
En (f ) := x ∈ T : lim sup AN (x − y)|f (y) − f (x)|dy ≥
.
n
N →+∞ T
As before we need to show that
(2.12)
m(En (f )) = 0.
To this purpose we show the following.
Lemma 2.7. There exists a constant C ∈ (0, +∞) such that
∀ x ∈ T, f ∈ L1 (T), N ≥ 1.
|AN ∗ f (x)| ≤ CM f (x),
Proof. We have
AN (x) ≤
(2.13)

2
 (N + 1)π

π2
.
(N +1)x2
Exercise. Show the above estimate.
With the notation Zk := [y : 2k /(N + 1) ≤ |x − y| ≤ 2k+1 /(N + 1)] we can write
Z
Z
|AN ∗ f (x)| ≤
AN (x − y)|f (y)|dy =
AN (x − y)|f (y)|dy
|x−y|≤1/(N +1)
T
X
+
0≤k:2k ≤π(N +1)
(2.13)
≤ (N + 1)π
2
Z
AN (x − y)|f (y)|dy
Zk
Z
|f (y)|dy +
|x−y|≤1/(N +1)
0≤k:2k ≤π(N +1)
Zk
π2
|f (y)|dy
(N + 1)(x − y)2
Z
N +1
|f (y)|dy
22k
Zk
k
0≤k:2 ≤π(N +1)
Z
X
N +1
|f (y)|dy
22k
[|x−y|≤2k+1 /(N +1)]
k
≤ 2π 2 M f (x) + π 2
≤ 2π 2 M f (x) + π 2
Z
X
X
0≤k:2 ≤π(N +1)
≤ 2π 2 M f (x) + 8π 2 M f (x)
X 1
.
2k
0≤k
Define
(2.14)
F∗ (x; f ) := sup FN (x; f ).
N ≥1
HARMONIC ANALYSIS
19
Corollary 2.8. There exists constants C, c ∈ (0, +∞) such that
Z
C
|f |dx, ∀ f ∈ L1 (T), λ > 0.
(2.15)
m[F∗ (f ) > λ] ≤
λ [|f |>cλ]
The remainder of the proof is exactly the same as in the previous case, as we can L1
approximate f by elements g ∈ C(T), for which Theorem 1.3 holds.
3. Interpolation theorems. The Lp –convergence.
3.1. Marcinkiewicz real interpolation theorem. We start with the following simple observation.
Lemma 3.1. Suppose that (X, Σ, µ) is a measure space and F : X → [0, +∞] is a measurable
function. Then, for any p ≥ 1
Z
(3.1)
+∞
Z
f dµ = p
λp−1 µ[F ≥ λ]dλ.
p
0
Proof. We have
Z
µ[F ≥ λ] =
1[F ≥λ] dµ
hence,
+∞
+∞Z
Z
Z
p−1
p
λ µ[F ≥ λ]dλ = p
λp−1 1[F ≥λ] dλdµ
0
0
 +∞



Z
Z
Z ZF
Z
Fubini
p−1
p−1




= p
λ 1[F ≥λ] dλ dµ = p
λ dλ dµ = F p dµ.
0
0
Denote by L0 (L0 (X, Σ, µ)) the space of all measurable functions f : X → [−∞, +∞].
Suppose that 1 ≤ p1 < p2 < +∞ and T : Lp1 + Lp2 → L0 is a subadditive operator, i.e.
(3.2)
|T (f + g)| ≤ |T f | + |T g|,
∀ f, g ∈ Lp1 + Lp2 .
Definition 3.2. We say that the operator T is of type (p, q) if there exists +∞ > C > 0 such
that
kT f kLq ≤ Ckf kLp ∀ f ∈ Lp .
The infimum of those C for which the above inequality holds shall be denoted by kT kp,q .
Definition 3.3. Suppose that 1 ≤ p ≤ +∞ and 1 ≤ q ≤ +∞. We say that the operator T is
of weak (p, q) type if
a) in case when q < +∞ there exists a constant C > 0 such that
(3.3)
µ[|T f | ≥ λ] ≤
Ckf kqLp
,
λq
∀ f ∈ Lp .
Define the infimum of such C-s as kT kw
p,q .
b) in case when q = ∞: the operator is of type (p, q).
20
T.KOMOROWSKI
Exercise. We define as Lpw the Lorentz space that consists of all functions f that satisfy
Ckf kpLp
,
µ[|f | ≥ λ] ≤
λp
(3.4)
∀λ > 0
and some C > 0.
i) Prove that Lp ⊂ Lpw ,
ii) Is it true that L1w (0, 1) \ L1 (0, 1) 6= ∅?
iii) Define kf kw
Lp as the infimum of C-s such that (3.4) holds. Can you compare it with the
standard Lp norm?
iv) Check that
w
w
kf + gkw
L1 ≤ 2(kf kL1 + kgkL1 ).
v) The triangle inequality though needs not to hold. Consider the example f = |x|−1 ,
g = |x − 1|−1 on L1w (0, 1).
The following result holds.
Theorem 3.4. (Marcinkiewicz interpolation theorem)
Suppose that 1 ≤ p1 < p2 < +∞ and T is a subadditive operator acting on Lp1 + Lp2 that
is of both weakly (p1 , p1 ) and (p2 , p2 ) types. Assume also that p ∈ (p1 , p2 ) and
1
θ
1−θ
=
+
p
p1
p2
p
p
1
for some θ ∈ (0, 1). Then, we have T (L ∩ (L + Lp2 )) ⊂ Lp and
kT f kLp ≤ Ckf kp .
w
The constant C depends only on kT kw
p1 ,p1 , kT kp2 ,p2 p − p1 , p2 − p.
Proof. For any λ > 0 set fλ := f 1[|f |≤λ] and f (λ) := f 1[|f |>λ] . We have
µ[|T f | ≥ 2λ] ≤ µ[|T (fλ )| ≥ λ] + µ[|T (f (λ) )| ≥ λ]
Since the operator T is both weak (p1 , p1 ) and (p2 , p2 ) types we can write that the left hand
side of the above inequality can be further estimated from above by
!
!
Z
Z
C
C1
2
|f |p1 dµ + p2
|f |p2 dµ .
λp1
λ
[|f |>λ]
[|f |≤λ]
We obtain therefore
+∞
Z
Z
p−1
λ µ[|T f | ≥ 2λ]dλ ≤ C1
+∞
λp−p1 −1 dλ
+∞
Z
+C2
λ
0
p−p2 −1
|f |p1 dµ
[|f |>λ]
0
0
!
Z
Z
!
p2
|f | dµ
dλ
[|f |≤λ]
|f |
Z
Z +∞
λp−p1 −1 dλ + C2 |f |p2 dµ
λp−p2 −1 dλ
0
|f |
Z
Z
C1
C
2
≤
|f |p dµ +
|f |p dµ
p − p1
p2 − p
and the theorem follows from Lemma 3.1.
Z
≤ C1
|f |p1 dµ
Z
HARMONIC ANALYSIS
21
Remark. When T is linear it extends continuously from Πp := Lp ∩ (Lp1 + Lp2 ) to the
entire Lp . Indeed, according to the above theorem we have
kT f − T gkLp = kT (f − g)kLp ≤ Ckf − gkLp
thus, T is continuous and allows for extension to the Lp –closure of Πp , which is Lp . A similar
conclusion can be reached for |T f | if T is only sublinear.
Exercise. The above result extends also to the case when p2 = +∞
Applications:
1. For any p ∈ (1, +∞) there exists a constant Cp ∈ (0, +∞) such that
kM f kLp ≤ Ckf kLp ,
for all f ∈ Lp (R).
For the proof it suffices only to note that M (f + g) ≤ M f + M g and that Hardy-Littlewood
inequality guarantees that M is of weak (1, 1) type. Of course, directly from the definition
we conclude that it is also of (+∞, +∞) type so the assertion follows from the Marcinkiewicz
interpolation theorem.
2. As a direct consequence of (2.15) and the Marcinkiewicz interpolation theorem we obtain
that for any p ∈ (1, +∞) there exists a constant Cp ∈ (0, +∞) such that
kF∗ (f )kLp ≤ Ckf kLp ,
(3.5)
for all f ∈ Lp (R).
Let ε > 0 be arbitrary. For any f ∈ Lp (T) there exists g ∈ C(T) such that kf − gkLp < ε.
We can write therefore
kFN (f ) − f kLp ≤ kFN (f − g) − (f − g)kLp + kFN (g) − gkLp
≤ kFN (f − g)kLp + kf − gkLp + kFN (g) − gkL∞ ≤ kF∗ (f − g)kLp + ε + kFN (g) − gkL∞ .
Using (3.5) and Theorem 1.3 we get
lim sup kFN (f ) − f kLp ≤ (C + 1)ε.
N →+∞
Hence, we have proved
Theorem 3.5. For any f ∈ Lp (T) and p ≥ 1 we have
lim kFN (f ) − f kLp = 0.
N →+∞
3.2. Riesz-Thorin complex interpolation theorem. We have the following maximum
principle.
Theorem 3.6. Suppose that Ω := [x + iy : a < x < b] and f : Ω̄ → C is a continuous
function that is holomorphic on Ω. Assume also that there exist α < π(b − a)−1 and A < +∞
such that
(3.6)
|f (z)| ≤ exp {A exp(α|y|)} ,
∀z ∈ Ω
and
(3.7)
|f (a + iy)| ≤ M,
|f (b + iy)| ≤ M,
Then,
(3.8)
|f (z)| ≤ M,
z ∈ Ω.
∀ y ∈ R.
22
T.KOMOROWSKI
Example. Note that one cannot improve on the value of α. Take a = −π/2, b = π/2 and
f (z) = exp(eiz ). We have
f (−π/2 + iy) = exp(e−iπ/2 e−y ) = exp(−ie−y )
and |f (−π/2 + iy)| = 1. Likewise, |f (π/2 + iy)| = 1. On the other hand f (iy) = exp(e−y ) →
+∞ when y → −∞
The proof of Theorem 3.6. Let g(z) := M −1 f ((a+b)/2+(b−a)π −1 z). We have g : Ω̄1 → C,
where Ω1 := [x + iy : −π/2 < x < π/2],
(3.9)
|g(−π/2 + iy)| ≤ 1,
|g(π/2 + iy)| ≤ 1,
∀y ∈ R
and
(3.10)
|g(z)| ≤ exp {A exp(α|y|)} ,
∀ z ∈ Ω1
where α < 1. It suffices only to show that
(3.11)
|g(z)| ≤ 1,
∀ z ∈ Ω1 .
Let
hε (z) := exp{−2ε cos(βz)}
for any ε > 0 and β ∈ (α, 1). Note that
Re[eiβz + e−iβz ] = (eβy + e−βy ) cos(βx) ≥ δ(eβy + e−βy ),
with δ := cos(βπ/2) > 0 (for β < 1). Thus
|hε (z)| ≤ exp{−2εδ cosh(βx)} < 1,
∀ z ∈ Ω̄1 .
hence |ghε | ≤ 1 on ∂Ω1 . Also we have
n
o
|g(z)hε (z)| ≤ exp Aeα|y| − 2εδ cosh(βy) ≤ 1
for |y| sufficiently large. Hence the function |g(z)hε (z)| ≤ 1 on a boundary of a rectangle
[−π/2, π] × [−|y|, |y|] for all sufficiently large y. Thus, by virtue of the maximum principle for
the modulus of an analytic function we conclude that |g(z)hε (z)| ≤ 1 for all z ∈ Ω̄1 . Finally
we obtain |g(z)| ≤ 1/|hε (z)| for any ε > 0. Letting ε → 0+ we conclude the assertion of the
theorem.
The second result gives more information about the behavior of the modulus of a holomorphic function.
Theorem 3.7. Suppose that Ω := [x+iy : a < x < b] and f : Ω̄ → C satisfies the assumptions
of Theorem 3.6. Let
M (x) := sup[|f (x + iy)| : y ∈ R].
Then,
(3.12)
[M (x)]b−a ≤ [M (a)]b−x [M (b)]x−a ,
x ∈ [a, b].
Remark. The above means that x 7→ log M (x) is a convex function.
The proof of Theorem 3.7. Consider g(z) := f (z)etz . The function satisfies the assumptions
of Theorem 3.6. As a consequence of the maximum principle we conclude that
M (x)etx ≤ max{M (a)eta , M (b)etb }.
Hence
(3.13)
M (x) ≤ max{M (a)et(a−x) , M (b)et(b−x) },
∀ t ∈ R.
HARMONIC ANALYSIS
23
We optimize the right hand side of (3.13) over t and find that the smallest value is attained
when M (a)et(a−x) = M (b)et(b−x) , or Thus,
M (a)
= et(b−a)
M (b)
and
M (x) ≤ [M (a)](b−x)/(b−a) [M (b)](x−a)/(b−a) .
(3.14)
Theorem 3.8. (Riesz-Thorin) Suppose that 1 ≤ p1 < p2 ≤ +∞, 1 ≤ q1 < q2 ≤ +∞ and
T : Lp1 + Lp2 → L0 is a linear operator such that it is of both (p1 , q1 ) and (p2 , q2 ) types. Then,
for any p ∈ [p1 , p2 ] such that
1
1−θ
θ
=
+
p
p1
p2
for some θ ∈ [0, 1]
the operator T extends continuously to an operator of (p, q) type, where
1−θ
θ
1
=
+ .
q
q1
q2
In addition,
kT kp,q ≤ kT kp1−θ
kT kθp2 ,q2
1 ,q1
(3.15)
Proof. Let
1
1
+ 0 = 1.
q q
We can write
Z
kT kp,q =
(3.16)
sup
sup
kf kLp =1 kgk
L
T f gdµ
q 0 =1
Z
=
sup
sup
kf kLp =1,f >0 kgk
Lq
0 =1,g>0
sup sup
T (f φ)gψdµ
|φ|=1 |ψ|=1
For any f, g > 0 define
Z
u(z; f, g, φ, ψ) :=
T (f (1−z)/p1 +z/p2 φ)g 1−[(1−z)/p1 +z/p2 ] ψdµ.
Let M (x; f, g, φ, ψ) := supy∈R |u(x + iy; f, g, φ, ψ)|, x ∈ [0, 1]. Note that
kT kp,q =
sup
sup
sup sup M (θ; f, g, φ, ψ).
kf kL1 =1,f >0 kgkL1 =1,g>0 |φ|=1 |ψ|=1
Each function u(z; f, g, φ, ψ) satisfies the hypotheses of Theorem 3.6. Hence we obtain that
M (x; f, g, φ, ψ) ≤ M 1−x (0; f, g, φ, ψ)M x (1; f, g, φ, ψ),
x ∈ [0, 1]. Taking the supremum on both sides over φ, ψ, f, g we conclude estimate (3.15).
Remark. The above result can be also worded as follows.
24
T.KOMOROWSKI
Theorem 3.9. Suppose that T is defined on D a dense subset of all Lr , r ∈ [p1 , p2 ],
kT f kLqi ≤ Mi kf kLpi ,
and
i = 1, 2, f ∈ D
1
θ
1−θ
+ ,
=
p
p1
p2
1
1−θ
θ
=
+
for some θ ∈ [0, 1].
q
q1
q2
Then,
kT f kLq ≤ M11−θ M2θ kf kLp , f ∈ D.
Proof. The proof follows from an application of Theorem 3.8 and obvious estimates kT kpi ,qi ≤
Mi , i = 1, 2.
Applications:
1. Young’s inequality. Suppose that f ∈ Lp (Rd ), g ∈ Lq (Rd ), p, q ≥ 1, 1/p + 1/q − 1 = 1/r
for some r ≥ 1 then f ∗ g ∈ Lr (Rd ) and
(3.17)
kf ∗ gkLr ≤ kf kLp kgkLq .
Suppose that T (f ) := f ∗ g for f, g ∈ S(Rd ). In the same way as in Proposition 1.1, P 4) we
have
kT (f )kL1 ≤ kf kL1 kgkL1
and obviously
kT (f )kL∞ ≤ kf kL∞ kgkL1 .
Using Riesz-Thorin interpolation theorem we obtain
kT (f )kLp ≤ kf kLp kgkL1
for any p ∈ [1, +∞] and all f ∈ Lp . Let S(g) := f ∗ g. We have
kS(g)kLp ≤ kf kLp kgkL1 .
From Hölder inequality
kS(g)kL∞ ≤ kf kLp kgkLp0
so again by interpolation if
1−θ
θ
1−θ
1
=
+
=
r
p
∞
p
we obtain
kS(g)kLr ≤ kf kLp kgkLq ,
where
1
1−θ
θ
1
θ
1 1
=
+ 0 =1−θ+θ 1−
=1− =1− +
q
1
p
p
p
p r
and (3.17) follows.
2. Markov operator with an invariant measure. Suppose that (X, Σ) is a measurable space
and B(X) is the space of bounded measurable functions. An operator P : B(X) → B(X) is
called Markov if P 1 = 1. and P f ≥ 0, if f ≥ 0.
A probability measure µ on (X, Σ) is called invariant if
Z
Z
P f dµ = f dµ, ∀ f ∈ B(X).
HARMONIC ANALYSIS
25
We show that P extends to a contraction on any Lp (µ). This is trivially true for p = ∞ for
−kf kL∞ 1 ≤ f ≤ kf kL∞ 1
so
−kf kL∞ 1 ≤ P f ≤ kf kL∞ 1
and in consequence kP f kL∞ ≤ kf kL∞ .
On the other hand, for any f we have −|f | ≤ f ≤ |f | so |P f | ≤ P |f | and
Z
Z
Z
inv.
|P f |dµ ≤ P |f |dµ =
|f |dµ.
Hence, kP f kL1 ≤ kf kL1 . Using Riesz-Thorin interpolation theorem we have kP f kLp ≤ kf kLp
for any p ∈ [1, +∞].
3. Hausdorff-Young inequality. Recall that the Fourier coefficients are given by
Z
ˆ
e−inx f (x)dx.
f (n) =
T
So, denoting by fˆ the respective complex number sequence (fˆ(n))n∈Z we obtain
kfˆkl∞ ≤ kf kL1 (T) .
On the other hand, Parseval identity shows that
kfˆkl2 = kf kL2 (T) .
Interpolating between these two estimates with the help of Riesz-Thorin theorem we obtain
!1/p
X
p
kfˆklp =
≤ kf k p0
|fˆ(n)|
L (T)
n
where 1/p + 1/p0 = 1.
4. Weakly singular operators. Suppose that T : L∞ [0, 1] → L∞ [0, 1] is given by
Z 1
f (y)dy
T f (x) :=
,
α
0 |x − y|
where α ∈ [0, 1). Note that
Z
kT f kL∞ ≤
sup
x∈[0,1] 0
1
dy
|x − y|α
!
kf kL∞ .
On the other hand,
Z
kT f kL1 ≤
1
Z
dx
0
0
1
|f (y)|dy
|x − y|α
Z
1
≤ kf kL1
−1
dx
.
|x|α
Interpolating between these inequalities we obtain that T extends to a bounded operator
between T : Lp [0, 1] → Lp [0, 1].
4. The definition and basic properties of the Fourier transform.
Throughout this section all the Lp spaces are complex.
26
T.KOMOROWSKI
4.1. The mollifiers and the Schwartz class of functions.
Definition 4.1. The Schwartz class of functions S(Rd ) is defined as the space that contains
all functions f : Rd → C that are of C ∞ (Rd ) class and
sup(1 + |x|2 )n |Dk f (x)| < +∞
for any n ≥ 0 and multi-index k = (k1 , . . . , kd ). Here Dk f (x) := ∂1k1 . . . ∂dkd f (x).
Definition 4.2. The convolution
of functions on Rd . Suppose that f, g ∈ L1 (Rd ). Define
R
f ∗ g ∈ L1 (Rd ) by f ∗ g(x) := Rd f (x − y)g(y)dy.
Proposition 4.3. C 1) For any f, g ∈ L1 (Rd ) the convolution f ∗ g is defined as an element
of L1 (Rd ).
C 2) We have f ∗ g = g ∗ f
C 3) kf ∗ gkL1 (Rd ) ≤ kf kL1 (Rd ) kgkL1 (Rd ) , in fact for nonnegative f, g we have equality.
C 4) kf ∗ gkL∞ (Rd ) ≤ kf kL∞ (Rd ) kgkL1 (Rd ) .
C 5) For any f ∈ L1 (Rd ) and g ∈ Cb (Rd ) ∩ L1 (Rd ) we have f ∗ g ∈ Cb (Rd ) ∩ L1 (Rd ).
Exercise. Prove the above proposition. R
Suppose that φ ∈ Cc∞ (Rd ) and φ ≥ 0, Rd φ(x)dx = 1. For any δ > 0 define φδ (x) :=
δ −d φ(x/δ). Let
Tδ f (x) := φδ ∗ f (x), f ∈ L1loc (Rd ).
Exercise. Prove that
i) Tδ f ∈ C ∞ (Rd ) for any f ∈ L1 (Rd ),
ii) Tδ f ∈ Cc∞ (Rd ) for any f ∈ L1 (Rd ) that is compactly supported.
Exercise. Verify that
i) kTδ f kL1 ≤ kf kL1 for any f ∈ L1 (Rd ),
ii) kTδ f kL∞ ≤ kf kL∞ for any f ∈ L∞ (Rd ),
iii) conclude from i) and ii) that kTδ f kLp ≤ kf kLp for any f ∈ Lp (Rd ) and p ≥ 1,
iv) using the argument contained in the proof of Lemma 1.4 conclude that for any f ∈
Cc (Rd ) we have Tδ f → f , as δ → 0+ uniformly on compact sets,
v) from iii) and iv) show that
(4.1)
lim kTδ f − f kLp = 0
δ→0+
for any f ∈ Lp (Rd ), where p ∈ [1, +∞).
Exercise. Show that Cc∞ (Rd ) (thus also S(Rd )) is dense in any Lp (Rd ), p ∈ [1, +∞).
4.2. The definition of the Fourier transform on L1 (Rd ).
Definition 4.4. Suppose that f ∈ L1 (Rd ). The Fourier transform of the function is defined
as
Z
ˆ
f (ξ) :=
e−ix·ξ f (x)dx, ξ ∈ Rd .
R
Sometimes we shall also denote the Fourier transform by F(f ).
Basic properties of the Fourier transform. Here we assume always that the minimal
assumptions for the existence of the respective Fourier transforms are satisfied. Denote also
by jp (x) := xp , x ∈ Rd , p = 1, . . . , d.
HARMONIC ANALYSIS
27
(F1)
kfˆkL∞ (Rd ) ≤ kf kL1 (Rd ) ,
(F2)
F(Dk f ) = ik j1k1 . . . jdkd fˆ,
(F3)
F(j1k1 . . . jdkd f ) = ik Dk fˆ,
Corollary 4.5. F(S) ⊂ S.
Proof. Suppose that f ∈ S). Note that for any l ≥ 0, l ∈ Z, a multi-index k = (k1 , . . . , kd )
we have
(F3)
(1 + |ξ|2 )l Dk fˆ(ξ) = (−1)k ik (1 + |ξ|2 )l F(j1k1 . . . jdkd f )(ξ)
(F2)
= (−1)k ik F((I − ∆)l (j1k1 . . . jdkd f ))(ξ).
Here, ∆ :=
Pd
2
p=1 ∂p
and, as a consequence of (F1), we obtain
sup[(1 + |ξ|2 )l |Dk fˆ(ξ)|] ≤ k(I − ∆)l (j1k1 . . . jdkd f )kL1 (Rd ) < +∞.
ξ
Exercise. Show (F1) – (F3).
(F4) fˆ ∈ C0 (Rd ), where C0 (Rd ) denotes the class of functions g : Rd → C that are
continuous and satisfy lim|x|→+∞ g(x) = 0.
Proof. According to Corollary 4.5 the above result holds for any f ∈ S(Rd ). Suppose that
f ∈ L1 (Rd ). Choose an arbitrary ε > 0. There exists g ∈ S(Rd ) such that kf − gkL1 < ε.
Note that
|fˆ(ξ) − ĝ(ξ)| ≤ kf − gkL1 < ε,
so fˆ is a limit, uniform on compacts, of continuous functions. Thus, it must be continuous.
In addition,
lim sup |fˆ(ξ) − ĝ(ξ)| = lim sup |fˆ(ξ)| ≤ ε,
|ξ|→+∞
|ξ|→+∞
therefore, since ε > 0 was arbitrary, we must have fˆ ∈ C0 (Rd ).
Theorem 4.6. (F 5) The inversion formula. Suppose that fˆ ∈ L1 (Rd ) then there exists a
version of f that belongs to C0 (Rd ) and it satisfies
Z
1
(4.2)
f (x) =
eix·ξ fˆ(ξ)dξ.
(2π)d Rd
Proof. Since fˆ ∈ L1 (Rd ) we can write that the right hand side of (4.2) equals
Z
Z
Z Z
1
1
ix·ξ
−iy·ξ
i(x−y)·ξ
(4.3)
e
e
f (y)dy dξ =
e
f (y)dy dξ
(2π)d Rd
(2π)d Rd
Rd
Rd
Since the integral in parentheses equals eix·ξ fˆ(ξ) it belongs to L1 (Rd ) so using the Lebesgue
dominated convergence theorem we can write that the right hand side of (4.3) equals
Z
Z
|ξ|2
1
i(x−y)·ξ
(4.4)
lim
exp −
e
f (y)dy dξ
λ
(2π)d λ→+∞ Rd
Rd
28
T.KOMOROWSKI
Fubini
=
1
lim
(2π)d λ→+∞
|ξ|2
exp −
λ
Rd
Z
Z
f (y)
Rd
i(x−y)·ξ
e
dξ dy
Lemma 4.7. For any a ∈ R
2
Z
√
ξ
λa2
iaξ
e dξ = πλ exp −
.
(4.5)
exp −
λ
4
R
Exercise. Prove the above lemma. Hint: Use contour integration.
Using (4.5) we can write that the right hand side of (4.4) equals
d/2 Z
λ
λ|y − x|2
lim
dy.
f (y) exp −
λ→+∞ 4π
4
Rd
The remaining part of the proof of the inversion formula can be concluded from the following
two facts, that are left out as the exercises.
Exercise. Verify that the family
d/2
λ
λ|x|2
eλ (x) :=
exp −
4π
4
(the so called heat kernels) satisfies the assumptions of the remark after Lemma 1.4.
Exercise. Applying an analogue of the argument used to show Lemma 1.4 prove that
lim keλ ∗ f − f kL1 (Rd ) = 0
λ→+∞
for any f ∈ L1 (Rd ).
Corollary 4.8. Suppose that fˆ ≡ 0. Then, f ≡ 0.
Theorem 4.9. (F 6) (The Fourier transform of a convolution) For any f, g ∈ L1 (Rd ) we have
f[
∗ g(ξ) = fˆ(ξ)ĝ(ξ).
(4.6)
Proof. Note that the left hand side of (4.6) equals
Z
Z
Z
Z
Fubini
−iξ·x
−iξ·y
e
f (x − y)g(y)dy dx =
e
g(y)
Rd
Rd
Rd
Z
=
Rd
e−iξ·y g(y)
Z
−iξ·(x−y)
e
f (x − y)dy dx
Rd
e−iξ·(y) f (y)dy dx = fˆ(ξ)ĝ(ξ).
Rd
(F7) Fourier transform of a shift. Suppose that a ∈ Rd and Ta f (x) := f (x−a), f ∈ L1 (Rd ).
Then
F(Ta f )(ξ) = eia·ξ fˆ(ξ).
Exercise: prove (F7).
4.3. The definition of the Fourier transform on L2 (Rd ).
Theorem 4.10. (F8) Suppose that f, g ∈ S(Rd ) then
Z
Z
1
∗
(4.7)
f (x)g (x)dx =
fˆ(ξ)ĝ ∗ (ξ)dξ.
(2π)d Rd
Rd
HARMONIC ANALYSIS
29
Proof. The right hand side of (4.7) equals
Z
Z Z
Z
1
1
Fubini
−iξ·x
∗
−iξ·x ∗
e
f (x)dx ĝ (ξ)dξ =
f (x)dx
e
ĝ (ξ)dξ
(2π)d Rd
(2π)d Rd
Rd
Rd
Z
(F5)
=
f (x)g ∗ (x)dx.
Rd
Corollary 4.11. (F9) The Plancherel identity. In the special case when f = g then
Z
Z
1
2
|fˆ(ξ)|2 dξ.
(4.8)
|f (x)| dx =
(2π)d Rd
Rd
As a consequence of the above equality we conclude that F extends continuously to a
unitary map F : L2 (Rd , dx) → L2 (Rd , (2π)−d/2 dξ).
Exercise Note that for any f ∈ L1 (Rd ) ∩ L2 (Rd ) we have F(f ) = F(f ).
For the above reason we shall unify the notation and call F(f ) the Fourier transform of
f ∈ L2 (Rd ).
Corollary 4.12. Both formulas (4.7) and (4.8) hold for all f, g ∈ L2 (Rd ).
5. Singular operators.
R1
5.1. Examples. Hilbert transform. We have shown that f 7→ T f given by T f (x) := 0 |x −
y|−α f (y)dy is a bounded operator on any Lp [0, 1], p ∈ [1, +∞] when α ∈ (0, 1). One could
inquire whether the above holds for α = 1. This is the so called case of the singular operator.
Note that we cannot even define then T f for constants different from 0 so we have to proceed
with some degree of care. First, we assume some symmetry that allows to claim that T f = 0
for f =const. To simplify matters assume that f : R → R. Let
Z +∞
f (y)dy
(5.1)
T f (x) :=
.
−∞ x − y
Even here we should be a bit careful since we still have not defined the meaning of the integral
on the right hand side of (5.1). We could use for instance the principal value, i.e.
Z
f (y)dy
(5.2)
T f (x) := lim
,
ε→0+ |y−x|≥ε x − y
when the limit of the right hand side exists in an appropriate sense. Such an operator is
called the Hilbert transform of a function f . Note that it exists for any function f that is
locally Hölder and belongs to L1 (R). Indeed, note that for any ε > 0 we can write
Z
dy
= 0.
1≥|y−x|≥ε x − y
On the other hand if |f (y) − f (x)| ≤ C|y − x|α for some α ∈ (0, 1] and C > 0, |y − x| ≤ 1, so
we can write
Z
Z
Z
f (y)dy
[f (y) − f (x)]dy
f (y)dy
=
+
.
x
−
y
x
−
y
|y−x|≥ε
1≥|y−x|≥ε
|y−x|≥1 x − y
30
T.KOMOROWSKI
The absolute value of the integrand under the first integral on the right hand side can be
estimated by
Cdy
|x − y|1−α
so the passage to the limit, when ε → 0+, can be substantiated for any x fixed by virtue
of the Lebesgue dominated convergence theorem. We wish to understand whether the limit
(5.2) exists for f ∈ Lp , p ≥ 1.
2. The Lp limit of Fourier sums. Recall that
Z
sin[(n + 21 )y]
f (x − y)dy.
Sn (f )(x) =
sin y2
T
We are interested in the existence of the Lp limit of Sn (f ) as n → +∞ if f ∈ Lp (T).
5.2. The Calderon-Zygmund decomposition of a function. Suppose that ak ≤ bk ,
Q
Q
k = 1, . . . , d. The set Qk = dk=1 [ak , bk ] ⊂ Rd , is called a cube. We call Q0k = dk=1 (ak , bk )
the interior of the cube and maxk |bk − ak | the mesh size of the cube. Suppose that p =
Q
(p1 , . . . , pd ) ∈ Zd . The cubes Qp,n = dk=1 [pk 2−n , (pk + 1)2−n ] are called the dyadic cubes of
generation n. We denote such a family of cubes by Qn .
Theorem 5.1. Suppose that f ∈ L1 (Rd ), f ≥ 0 and α > 0 is a positive constant. Then,
there exists a decomposition Rd = Ωα ∪ Fα such that
i) Ωα ∩ Fα = ∅,
ii) f ≤ α,S
a.e. on Fα ,
S
iii) Ωα = k Qk , where Qk ∈ n Qn are such that Q0k ∩ Q0l = ∅ for k 6= l and
Z
1
α<
f (x)dx ≤ 2d α.
m(Qk ) Qk
Proof. We can find n0 sufficiently large so that for each Q ∈ Qn0 we have
Z
1
(5.3)
f (x)dx ≤ α.
m(Q) Q
Any cube satisfying (5.3) shall be called good. Now consider Q ∈ Qn0 . Let Q0 ⊂ Q belongs
to Qn0 +1 . If Q0 does not satisfy (5.3) we call it bad. This way we can distinguish the family
of all good cubes of generation n0 + 1. Let G1 be the union of all such cubes. Using the
family of cubes Qn0 +2 that are contained in G1 we can distinguish again the subfamilies of
good and bad cubes and define as G2 the union of all good cubes of the n0 + 2-nd generation.
This process can be continued until there existTany good cubes in a given generation. Define
as Ωα the union of all bad cubes and Fα := n≥1 Gn . Note that obviously i) and ii) hold.
We need only to show iii) Suppose that the cube Q ∈ Qn0 +k was bad. Then, according to its
definition we have
Z
1
α<
f (x)dx.
m(Q) Q
We claim that in addition
Z
1
f (x)dx ≤ 2d α.
m(Q) Q
Indeed, suppose that
Z
1
d
f (x)dx.
2 α<
m(Q) Q
HARMONIC ANALYSIS
31
We know that there exists a good cube Q0 ∈ Qn0 +k−1 containing Q. We have
Z
Z
2d
2d
d
2 α<
f (x)dx ≤
f (x)dx
m(Q0 ) Q
m(Q0 ) Q0
but then
1
α<
m(Q0 )
which contradicts the fact that Q0 is good.
Z
f (x)dx,
Q0
Corollary 5.2. Suppose that the assumptions of Theorem 5.1 are satisfied. Then, using the
notation of that theorem we can write
iv)
1
m(Ωα ) ≤ kf kL1 ,
α
Proof. Note that from iii) of Theorem 5.1 we have
Z
X
1X
1
m(Ωα ) =
m(Qk ) ≤
f (x)dx ≤ kf kL1
α
α
Qk
k
k
so we can easily conclude iv).
5.3. Calderon-Zygmund theorem on singular operators.
Theorem 5.3. Suppose that
i) k ∈ L2 (Rd ),
ii) kk̂kL∞ (Rd ) < +∞,
iii) k ∈ C 1 (Rd \ {0}) and
|∇k(x)| ≤
B
.
|x|d+1
For f ∈ L2 (Rd ) define
Z
(5.4)
k(x − y)f (y)dy,
T f (x) :=
Rd
x ∈ Rd .
Then, there exists a constant A1 depending only on kk̂kL∞ (Rd ) , B and d such that for any
f ∈ L1 (Rd ) ∩ L2 (Rd ) we have
m(|T f | ≥ λ) ≤
(5.5)
A1
kf kL1 ,
λ
∀ λ > 0.
In addition, for any p ∈ (1, +∞) there exists a constant Ap depending only on kk̂kL∞ (Rd ) ,
p, B, d such that
kT f kLp ≤ Ap kf kLp ,
(5.6)
∀ f ∈ Lp (Rd ) ∩ L2 (Rd ).
The operator T can be therefore extends to a bounded operator T : Lp (Rd ) → Lp (Rd ) for any
p ∈ (1, +∞) and an operator of weak type (1, 1).
Proof. Step 1. T is of type (2, 2). Use (F7) and (F8) we obtain
Z
Z
iξ·x ∗
∗
ˆ
e k̂ (ξ)f (ξ)dξ = T f (x) = T f (x) =
e−iξ·x k̂(ξ)fˆ(ξ)dξ
Rd
Rd
32
T.KOMOROWSKI
hence from Corollary 4.8 we obtain
F(T f )(ξ) = k̂(ξ)fˆ(ξ).
In consequence of Plancherel identity we obtain
Z
−d
2
|k̂(ξ)fˆ(ξ)|2 dξ ≤ kk̂k2L∞ kf k2L2 .
kT f kL2 = (2π)
Rd
Step 2. C-Z decomposition
of f . Let f ∈ L1 (Rd ) and α > 0. We apply to |f | Theorem 5.1
S
and find Ωα = Qi and Fα as in that theorem. Let

f (x),
x ∈ Fα ,


g(x) =
R

 1
f (x)dx, x ∈ Q0i
|Qi | Qi
and b := f − g. We have g ≤ α on Fα and g ≤ 2d α on Ωα . Also,
b(x) = 0, x ∈ Fα ,
Z
1
b(x)dx = 0, ∀ Q0i
|Qi | Qi
and
kbkL1 (Rd ) ≤ 2kf kL1 (Rd )
(5.7)
Since T f = T g + T b we have of course
m[x : |T f (x)| > α] ≤ m[x : |T g(x)| > α/2] + m[x : |T b(x)| > α/2]
Step 3. Estimates of T g. Note first that g ∈ L2 (Rd ). Indeed,
Z
Z
Z
Z
|g(x)|2 dx =
|g(x)|2 dx +
|g(x)|2 dx ≤ α
|f (x)|dx + 22d α2 m[Ωα ]
Rd
Fα
iv)
Z
≤α
Ωα
Fα
|f (x)|dx + 22d αkf kL1 ≤ (1 + 22d )αkf kL1 < +∞.
Fα
Thus,
4kT gk2L2
4A2 kgk2L2
kf kL1
≤
≤ 4A2 (1 + 22d )
.
α2
α2
α
P
Step 4. Estimates of T b. Denoting bi = b1Qi we can write T b = i T bi and
Z
(5.9)
T bi (x) =
k(x − y)b(y)dy.
(5.8)
m[x : |T g(x)| > α/2] ≤
Qi
Denoting by yi the center of Qi we can write that the left hand side of (5.9) equals
Z
(5.10)
[k(x − y) − k(x − yi )]b(y)dy.
Qi
Thanks to iii) we conclude that
(5.11)
|k(x − y) − k(x − yi )| ≤
Bδi
,
|x − yi∗ |d+1
HARMONIC ANALYSIS
33
where yi∗ ∈ [y, yi ] and δi denotes the diameter of Qi . Let Q̃i := [x : dist(x, Qi ) ≤ δi ]. On
Fα \ Q̃i we can estimate as follows
Z
Z
Z
dx
(5.12)
|T bi (x)|dx ≤ Bδi
|b(y)|dy
|x|d+1
Qi
|x|≥δi
(Q̃i )c
Z
|b(y)|dy.
= cd B
Qi
Let
Ω∗α
:=
S
i Q̃i
and
Z
Fα∗
:=
Rd
\
Ω∗α .
We have
X Z
XZ
|T bi (x)|dx ≤
|T b(x)|dx ≤
i
i F∗
α
Fα∗
≤ cd B
XZ
|T bi (x)|dx
(Q̃i )c
Z
|b(y)|dy = cd B
i Q
i
|b(y)|dy ≤ cd Bkf kL1 (Rd ) .
Fα
We have therefore
m[x ∈ Fα∗ : |T b(x)| > α/2] ≤
2cd Bkf kL1 (Rd )
α
.
On the other hand,
m[Ω∗α ] ≤ 10d m[Ωα ]
Cor. 5.2
≤
10d
kf kL1 (Rd ) .
α
We have shown therefore that
2cd B + 10d
kf kL1 (Rd ) .
α
Step 5. The Lp estimates. From (5.8) and (5.13) it follows that
C
(5.14)
m[x : |T f (x)| > α] ≤ kf kL1 (Rd )
α
m[x : |T b(x)| > α/2] ≤
(5.13)
for some constant C > 0 depending only on d, B, kk̂kL∞ . The operator T is therefore of weak
type (1, 1). As a consequence of Marcinkiewicz interpolation theorem we obtain that for any
p ∈ (1, 2) there exists Ap ∈ (0, +∞) such that
kT f kLp (Rd ) ≤ Ap kf kLp (Rd ) ,
∀ f ∈ Lp (Rd ).
On the other hand, suppose that p ∈ (2, +∞) and f, g ∈ Cc∞ (Rd ). We can write
Z Z
(f, T g)L2 =
k(x − y)f (x)g(y)dxdy.
Rd Rd
Let T̃ f (y) :=
R
k(x − y)f (x)dx. Since the kernel k̃(x) := k(−x) satisfies the hypotheses of
Rd
the theorem we know from the already part of it that
|(f, T g)L2 | = |(T̃ f, g)L2 | ≤ kT f kLp0 kgkLp ≤ Ap0 kf kLp0 kgkLp .
Taking the supremum over all f such that kf kLp0 = 1 we conclude that
kT gkLp ≤ Ap0 kgkLp
and we can conclude the assertion of the theorem for p ∈ (2, +∞).
34
T.KOMOROWSKI
5.4. Applications.
5.4.1. The Hilbert transform. Note that the argument of the theorem shows in fact the following.
Theorem 5.4. Suppose that there exists B > 0 for which
1)
B
, x ∈ Rd \ {0},
|k(x)| ≤
|x|d
2)
Z
x ∈ Rd ,
|k(y − x) − k(y)|dy ≤ B,
|y|≥2|x|
3) for any 0 < R1 < R2
Z
k(y)dy = 0.
R1 <|y|<R2
For f ∈
Cc∞ (Rd )
define
Z
(5.15)
k(x − y)f (y)dy,
Tε f (x) :=
x ∈ Rd .
|y−x|≥ε
Then, for each p ∈ (1, +∞) there exists a constant Ap depending only on p, B and d such that
kTε f kLp ≤ Ap kf kLp ,
(5.16)
∀ f ∈ Cc∞ (Rd ), ε > 0.
The operator Tε extends to the entire Lp (Rd ) and
(5.17)
T f := lim Tε f
ε→0+
exists for any f ∈ Lp (Rd ). The operator T : Lp (Rd ) → Lp (Rd ) is bounded for any p ∈ (1, +∞).
Proof. Let kε (x) := 1[ε,+∞) (|x|)k(x). Thanks to 1) we conclude that kε ∈ L2 (Rd ). We also
have
Z
k̂1 (ξ) = lim
e−ix·ξ k(x)dx
R→+∞
1≤|x|≤R
Z
e−ix·ξ k(x)dx + lim
=
1≤|x|≤3π|ξ|−1 ∨1
Z
R→+∞
3π|ξ|−1 ∨1≤|x|≤R
e−ix·ξ k(x)dx.
Denote the first and the second integrals by I and II(R) respectively.
For |ξ| > 1/(3π) we have I = 0. We can write for |ξ| ≤ 1/(3π)
Z
Z
−ix·ξ
|x||k(x)|dx
|I| = [e
− 1]k(x)dx ≤ |ξ|
1≤|x|≤3π|ξ|−1
−1
1≤|x|≤3π|ξ|
Z
dx
1
B|ξ|
≤ Bcd |ξ| ×
= Bcd ,
d−1
|ξ|
|x|
1≤|x|≤3π|ξ|−1
where cd depends only on d.
HARMONIC ANALYSIS
35
To estimate II(R) we let η := π|ξ|−2 ξ (reflection w.r.t. the sphere of radius π 1/2 ). Note
that eiη·ξ = −1. For sufficiently large R we can write as follows
Z
1
II(R) =
[e−ix·ξ − e−i(x+η)·ξ ]k(x)dx
2
3π|ξ|−1 ∨1≤|x|≤R
=
Z
1
2
e−ix·ξ [k(x) − k(x − η)]dx + R(R).
3π|ξ|−1 ∨1≤|x|≤R
Here,

R(R) :=

Z
1

2
Z
e−ix·ξ k(x − η)dx −
3π|ξ|−1 ∨1≤|x|≤R

e−ix·ξ k(x − η)dx .
3π|ξ|−1 ∨1≤|x−η|≤R
Let A(r, R) denote the spherical shell [r ≤ |x| ≤ R]. We have |η| = π|ξ|−1 . Denoting
A1 := A(2π|ξ|−1 ∨ 1, 4π|ξ|−1 ∨ 1) and A2 := A(R − π|ξ|−1 ∨ 1, R + π|ξ|−1 ∨ 1) we obtain
Z
Z
1
1
1
|R(R)| ≤
|k(x − η)|dx ≤
dx ≤ Cd
2
2
|x − η|d
A1 ∪A2
A1 ∪A2
for some Cd > 0 depending only on d.
We can estimate therefore
Z
lim II(R) ≤ 1
R→+∞
2
|k(x) − k(x − η)|dx + lim sup |R(R)|
R→+∞
3|η|≤|x|
Z
1
≤
2
2)
|k(x) − k(x − η)|dx + Cd = B + Cd .
2η|≤|x|≤2|η|
We conclude therefore that |k̂1 (ξ)| ≤ B 0 , where B 0 depends only on d and B. Note that
Z
Z
(ε)
−ix·ξ
k̂ε (ξ) =
e
k(x)dx =
e−ix·(εξ) kε (x)dx = K̂1 (ξ)
ε≤|x|
1≤|x|
where K (ε) (x) := εd k(εx). Note that K (ε) satisfies assumptions 1) – 3) with the same constant
as k. So we conclude that |k̂ε (ξ)| ≤ B 0 for all ε > 0.
Analyzing the proof of Theorem 5.3 one can see that the only place where assumption iii)
has been used was (5.11), which in turn led to estimate (5.12). We can use assumption 2) to
obtain (5.12) directly. Note that
Z
Z Z
(5.18)
|Tε bi (x)|dx =
[kε (x − y) − kε (x − yi )]b(y)dy dx
(Q̃ )c
(Q̃ )c Qi
i
i

Z
≤

Z

|b(y)|dy 
Qi
(Q̃i )c

|kε (x − y) − kε (x − yi )| dx
36
T.KOMOROWSKI


Z
Z

|kε (x − y + yi ) − kε (x)| dx .

|b(y)|dy 
=
Qi
(Q̃i −yi )c
But |y − yi | ≤ δi ≤ 1/2|x| for x ∈ (Q̃i − yi )c . We show that kε (·) satisfies
Z
(5.19)
|kε (y − x) − kε (y)|dy ≤ B 0 , x ∈ Rd
|y|≥2|x|
for some B 0 dependent only on B and d. Indeed, the left hand side of (5.19) can be estimated
by
Z
Z
|k(y)||1[ε,+∞) (|y − x|) − 1[ε,+∞) (|y|)|dy
|k(y − x) − k(y)|1[ε,+∞) (|y − x|)dy +
|y|≥2|x|
|y|≥2|x|
Z
cond. 2)
≤
|k(y)||1[ε,+∞) (|y − x|) − 1[ε,+∞) (|y|)|dy.
B+
|y|≥2|x|
The second integral J on the right hand side can be split into two integrals: first, over
D1 := [y : |y − x| ≥ ε, |y| < ε], second over D2 := [y : |y − x| < ε, |y| ≥ ε]. Note that
D1 ⊂ A(ε/2, ε) and D2 ⊂ A(ε, 2ε). As a consequence we obtain
Z
dy
= 2Bωd−1 log 2.
|J| ≤ B
|y|d
A(ε/2,ε)∪A(ε,2ε)
Using condition (5.19). we can estimate the utmost right hand side of (5.18) by
Z
B 0 |b(y)|dy.
Qi
Then repeating the respective part of the proof of Theorem 5.3 we conclude the Lp estimates
(5.16). Note also that for any f ∈ Cc∞ (Rd ) we have
Tε f (x) = T1 f (x) + Gε f (x),
where
Z
3)
Z
k(x − y)f (y)dy =
Gε f (x) :=
1>|y−x|≥ε
k(x − y)[f (y) − f (x)]dy.
1>|y−x|≥ε
The passage to the limit, as ε → 0+, for Gε f is possible under the integral on the utmost
right hand side upon the application of the Lebesgue dominated convergence theorem. We
have
Bk∇f k∞ |y − x|
Bk∇f k∞
|k(x − y)[f (y) − f (x)]| ≤
≤
.
d
|x − y|
|x − y|d−1
Hence limε→0+ Gε f (x) exists for each x ∈ Rd and the convergence is uniform w.r.t. x. Since
Gε f (x) = 0 for dist(x,supp f ) > 1 we conclude that the Lp -convergence of Tε f for such
f . Using Banach-Steinhaus theorem we conclude (5.17) and the boundedness of the limiting
operator T .
HARMONIC ANALYSIS
37
Let ε > 0 we define
Z
Tε f (x) :=
f (y)dy
x−y
|y−x|≥ε
Cc∞ (R).
for f ∈
We can apply Theorem 5.4. Here k(x) = x−1 . According to the theorem in
question there exists an Lp extension of Tε and T - the so called Hilbert transform - can be
defined by (5.17) for any f ∈ Lp (Rd ), p ∈ (1, +∞).
5.4.2. Theorem about multipliers.
Definition 5.5. (The principal value of an integral.) Suppose that k ∈ L1loc (Td \ {0}). We
let
Z
Z
k(x)dx.
p.v. k(x)dx := lim
ε→0+ |x|≥ε
T
Definition 5.6. (Fourier coefficients.) Suppose that k ∈ L1loc (Td \ {0}). We let
Z
k̂(n) := p.v.
e−inx k(x)dx, n ∈ Zd ,
Td
provided that they exist for all n ∈ Zd
We have the following version of the Calderon-Zygmund theorem.
Theorem 5.7. (Theorem about multipliers) Suppose that
1)
k ∈ L1loc (Td \ {0}),
2)
Z
|k(y − x) − k(y)|dy ≤ B,
∀ x ∈ Td ,
|y|≥2|x|
3) k̂(n) exist for all n ∈ Zd and k̂ := supn |k̂(n)| < +∞.
Define the operator T : L2 (Td ) → L2 (Td )
Tcf (n) := k̂(n)fˆ(n), ∀ n ∈ Zd .
For each p ∈ (1, +∞) there exists a constant Ap depending only on p, B and k̂ such that
(5.20)
kT f kLp ≤ Ap kf kLp ,
∀ f ∈ L∞ (Td ).
In addition, the operator satisfies
Z
(5.21)
T f (x) = lim
ε→0+
|x−y|>ε
k(x − y)f (y)dy,
∀ f ∈ L∞ (Td ).
The limit in (5.21) is understood in the Lp sense for p ∈ (1, +∞).
Proof. Suppose first that k ∈ L1 (Td ). Then, thanks to Proposition 1.1 P 3) we have (5.21).
Using Plancherel identity we establish
kT f kL2 (Td ) ≤ sup |k̂(n)|kf kL2 (Td ) ,
∀ f ∈ L2 (Td ).
n
To obtain the result we only need to show weak L1 estimate for T f with f ∈ L1 (Td ). Decompose f = g + b as in the proof of Theorem 5.3. The estimate of T g is identical as in Step
38
T.KOMOROWSKI
R
3 of the proof of that theorem. The estimate of (Q̃i )c |T bi (x)|dx can be then taken directly
from the proof of Theorem 5.4, see (5.18). The remaining part of the estimates of T b can be
done in the same way as in the proof of Theorem 5.3.
Suppose now that k does not belong to L1 (Td ). For any ε > 0 set kε (x) := k(x)1[|x|>ε] (|x|)
and Tε f := kε ∗ f , f ∈ L1 (Td ). Note that for any trigonometric polynomial f ∈ P we have
limε→0+ kTε f − T f kLp (Td ) = 0 for p ∈ [1, +∞]. On the other hand, according to the already
proved part of the theorem, we also have kTε kp,p ≤ Ap for all p ∈ (1, +∞) and ε ∈ (0, 1). We
conclude therefore that limε→0+ kTε f − T f kLp (Td ) = 0 for f ∈ Lp (Td ) and p ∈ (1, +∞). The
formula (5.21) is then a consequence of the above and Proposition 1.1 P 3).
5.4.3. The Lp convergence of Fourier sums. Recall that the Fourier sums for any f ∈ L1 (T)
are given by
Z
Sn (f ; x) = Dn (x − y)f (y)dy,
T
where
Dn (x) :=
sin ((n + 1/2)x)
sin(x/2)
Theorem 5.8. For any f ∈ Lp (T) we have
lim kSn (f ) − f kLp (T) = 0.
n→+∞
Proof. Since for any f ∈ C 1+α (T) we know, see Theorem 1.8, that limn→+∞ kSn (f ) − f kL∞ =
0 it suffices only to show that there exists a constant Ap independent of n and such that
kSn (f )kLp (T) ≤ Ap kf kLp (T) ,
Recall that
D̂n (k) =

 0,

Let Mn f (x) :=
∀ f ∈ Lp (T).
|k| > n
1, |k| ≤ n.
einx f (x).
Lemma 5.9. Let h(x) := cot(x/2). We have ĥ(k) = −isign k (ĥ(0) := 0).
Proof. Let
HN (x) :=
X
sign (n)einx .
|n|≤N
We can easily verify that
HN (x) =
x
1 + eix
−
K
(x)
=
i
cot
− KN (x),
N
1 − eix
2
where
ei(N +1)x − e−iN x
.
1 − eix
R
Note that T KN (x)dx = 0. For |n| ≤ N we have
Z
Z
Z
−inx
−inx
sign n = e
HN (x)dx = i e
h(x)dx − e−inx KN (x)dx
KN (x) :=
T
T
T
HARMONIC ANALYSIS
39
thus,
Z
i
−inx
e
Z
h(x)dx = sign n +
T
T
Z
= sign n +
Since
(e−inx
− 1)(1 −
(e−inx − 1)KN (x)dx
eix )−1
∈
T
1
L (T)
Z
i
e−inx
− 1 i(N +1)x
(e
− e−iN x )dx.
1 − eix
we can take the limit as N → +∞ and obtain
e−inx h(x)dx = sign n.
T
d (k) :=sign k fˆ(k), f ∈ L2 (T). Show that
Exercise. Let Hf
Z
(5.22)
Hf (x) = i lim
cot[(x − y)/2]f (y)dy
ε→0+ [|x−y|≥ε]
for f ∈ C 1 (T) and a.e. x ∈ T.
Exercise. Verify that
1
1
Sn = M−n (I + H + S0 )Mn − Mn+1 (I + H + S0 )M−n−1 ,
2
2
where Mn f (x) := einx f (x).
p
p
Of course Mn -s are isometries on any L , S0 is clearly bounded on any L . Therefore, we
need to show only that the operator H is bounded on any Lp (T). This however follows easily
from Theorem 5.7 and the following exercise.
Exercise Prove that
Z
[cot[(y
−
x)/2]
−
cot(y/2)]dy
≤ B, ∀ x ∈ T
|y|≥2|x|
for some B > 0.
Remark. Relation with the Hilbert transform. Let

0, |x| > π

f˜(x) =

f (x), |x| ≤ π.
Suppose that f ∈ Cc1 (−π, π). We can write that
Z
(5.23)
Hf (x) = lim
cot[(x − y)/2]f˜(y)dy,
ε→0+ |x−y|≥ε
where the integral extends over R. Note that
1
sin(z/2) 2 (5.24)
≤ C|z|
cot(z/2) − z = sin(z/2) cos z − z/2
for some C > 0 and |z| ≤ π. Let
Z
H̃f (x) := 2 lim
ε→0+ |x−y|≥ε
f˜(y)
dy.
x−y
40
T.KOMOROWSKI
We recognize that the operator H̃ is, up to a multiplicity constant, equal to the Hilbert
transform of f˜. From (5.24) we have
kHf − H̃f kLp [−π,π] ≤ Ckf kLp (T) ,
f ∈ Cc1 (−π, π)
for some C > 0. Using the Lp estimate for the Hilbert transform we conclude therefore easily
that
kH̃f kLp [−π,π] ≤ kH̃f kLp (Rd ) ≤ Ckf˜kLp (R) = Ckf˜kLp (T) , f ∈ Cc1 (−π, π).
As a result we conclude that H extends to a bounded operator on Lp (T) and the conclusion
of Theorem 5.8 holds.
5.4.4. The Riesz transform. Let Rj : L2 (Rd ) → L2 (Rd ), j = 1, . . . , d be given by
ξj ˆ
d
R
f (ξ),
j f (ξ) :=
|ξ|
(5.25)
∀ f ∈ L2 (Rd ).
Note that, thanks to Plancherel’s identity (4.8), each Rj is a contraction in L2 (Rd ). We would
like to show that it is also a bounded operator in any Lp (Rd ) for all p ∈ (1, +∞).
Let
xj
, ∀ j = 1, . . . , d.
(5.26)
Kj (x) =
|x|d+1
We shall verify that each Kj satisfies the assumptions of Theorem 5.4. In fact, a more general
theorem holds
Theorem 5.10. Suppose that k(x) := Ω(x/|x|)|x|−d , where Ω : Sd−1 → R satisfies the
following conditions:
i) itR is Hölder continuous with exponent α ∈ (0, 1),
ii) Sd−1 Ω(ω)S(dω) = 0,
Then k(x) satisfies the assumptions
R of Theorem 5.4. In particular, the above means that
we can define the operator T f (x) := Rd k(x − y)f (y)dy on any Lp (Rd ) space using formula
(5.17).
In addition,
Z
(5.27)
k̂(ξ) :=
lim
e−ix·ξ k(x)dx
ε→0+,R→+∞ ε<|x|<R
exist for all ξ 6= 0 and is equal to
Z
πi
(5.28)
k̂(ξ) =
sign(ξ · ω) + log (|ξ · ω|) Ω(ω)S(dω).
2
Sd−1
For f ∈ L2 (Rd ) we have Tcf (ξ) = k̂(ξ)fˆ(ξ).
Proof. We verify the assumptions of Theorem
suffices to show 2). We have
Z
Z
(5.29)
|k(y − x) − k(y)|dy ≤
|y|≥2|x|
5.4. Note that 1) and 3) are obvious. It only
Ω((y − x)|x − y|−1 ) Ω(y|y|−1 ) dy.
−
|x − y|d
|y|d |y|≥2|x|
The left hand side of (5.29) can be estimated by
Z
|Ω((y − x)|y − x|−1 ) − Ω(y|y|−1 )|
(5.30)
dy + kΩk∞
|y|d
|y|≥2|x|
Z
|y|≥2|x|
1
1
|x − y|d − |y|d dy.
HARMONIC ANALYSIS
41
Exercise. Prove that there exists a constant C > 0 such that for x, y ∈ Rd such that
|y| ≥ 2|x| we have
(y − x)|y − x|−1 − y|y|−1 ≤ C|x||y|−1 .
(5.31)
Hint: Note that |y−x| ≥ 1/2|y| and use the standard estimate for the projection P (x) := x/|x|
that states |P (x) − P (y)| ≤ 2 max [|x|−1 , |y|−1 ]|x − y|.
Using the Hölder assumption and (5.31) we can estimate the first integral by
Z +∞
Z
dρ
min [|x|α , 1]
α
dy = CCd min [|x| , 1]
,
C
1+α
d+α
|y|
2|x| ρ
|y|≥2|x|
which upon the substitution ρ0 := |x|/ρ equals
α
Z
1/2
CCd min [|x| , 1]
0
dρ
≤ C.
ρ1−α
The second integral in (5.30) can be estimated thanks to the following.
Exercise. Show that there is a constant C > 0 such that
Z 1
1 d
|x − y|d − |y|d dy ≤ C, ∀ x ∈ R . |y|≥2|x|
Thus the condition 2) of Theorem 5.4 follows and we conclude the possibility of extending T
to a bounded operator on any Lp , p ∈ (1, +∞).
To prove (5.28) we let kε,R (x) := 1[ε≤|x|≤R] k(x) and ξ := ω 0 r, where r := |ξ| and ω 0 :=
ξ|ξ|−1 . We have
Z
Z R
dρ
0
k̂ε,R (ξ) =
S(dω)
e−iρrω ·ω Ω(ω)
ρ
Sd−1
ε
R
Since Sd−1 Ω(ω)S(dω) = 0 we can write that
Z
Z R
dρ
0
k̂ε,R (ξ) =
S(dω)
[e−iρrω ·ω − cos(rρ)]Ω(ω)
ρ
d−1
S
ε
Z
=
Iε,R (ξ, ω)S(dω),
Sd−1
where
Z
R
dρ
.
ρ
ε
Exercise. The mean value theorem for integrals, see [2], [306]. Suppose that f : [a, b] → R
is a monotone decreasing, nonnegative function and g ∈ L1 [a, b]. Then, there exists c ∈ [a, b]
such that
Z b
Z c
(5.32)
f (t)g(t)dt = f (a)
g(t)dt. Iε,R (ξ, ω) :=
a
0
[e−iρrω ·ω − cos(rρ)]
a
Exercise. Conclude from (5.32) that there exists a constant C > 0 such that for all 0 < a < b,
c>0
Z b
sin(ct) (5.33)
dt ≤ C. t
a
42
T.KOMOROWSKI
As a direct consequence of (5.33) we obtain
Lemma 5.11. There is a constant C > 0 such that
|Im Iε,R (ξ, ω)| ≤ C,
∀ ε, R > 0.
In addition
(5.34)
lim
ε→0+,R→+∞
Im Iε,R (ξ, ω) =
π
sign(ξ · ω).
2
Proof. The first estimate follows from the fact that
Z R
sin(ξ · ωρ)
Im Iε,R (ξ, ω) =
dρ
ρ
ε
and (5.33). As for (5.34) it is a consequence of
Z +∞
sin t
π
dt = ,
t
2
0
o
see e.g. [2], [492], 3 .
Exercise. Froullani integrals, see e.g. [2], paragraph [495]. Suppose that h : [0, +∞) → R
is continuous, bounded and there exists
Z +∞
h(t)
dt.
t
1
Show that for any λ, µ > 0 we have
Z R
h(λr) − h(µr)
dr = h(0) log(µ/λ) lim
ε→0+,R→+∞ ε
r
As a direct consequence of the above exercise we obtain
Lemma 5.12. We have
Z
Re Iε,R (ξ, ω) =
ε
R
cos(ξ · ωρ) − cos(rρ)
dρ
ρ
and
|Re Iε,R (ξ, ω)| ≤ C[log(|ω 0 · ω|−1 ) + 1], ∀ ε, R > 0
lim
Re Iε,R (ξ, ω) = log(|ω 0 · ω|−1 ).
ε→0+,R→+∞
Finally, let kε (x) := 1[ε≤|x|] k(x). We know, by virtue of Theorem 5.4, that for any f ∈
T f (x) = limε→0+ Tε f (x) (in the L2 –sense) and
L2 (Rd )
d
ˆ
T
ε f (ξ) = k̂ε (ξ)f (ξ).
d
ˆ
The result is then a consequence of the pointwise convergence of T
ε f (ξ) to k̂(ξ)f (ξ) and the
2
d
c
fact Tε f (ξ) converges to T f (ξ) in the L sense. This ends the proof of the theorem.
Let us describe more closely the operators corresponding to the kernels Kj . Note that by
virtue of (5.27) K̂j (ξ) corresponding to Kj (x) exist for all ξ 6= 0 and satisfy
Z
(ε,R)
K̂j (λξ) =
lim
e−ix·(λξ) Kj
(x)dx,
ε→0+,R→+∞ Rd
where
(ε,R)
Kj
(x) := 1[ε<|x|<R]
xj
.
|x|d+1
HARMONIC ANALYSIS
Changing the variables x0 := λx we obtain
Z
K̂j (λξ) =
lim
ε→0+,R→+∞ λε<|x|<λR
e−ix·ξ
43
xj
dx = K̂j (ξ).
|x|d+1
Hence, if K̂(ξ) := (K̂1 (ξ), . . . , K̂d (ξ)) we have
(5.35)
∀ ξ 6= 0.
K̂(λξ) = K̂(ξ),
For any rotation U ∈ O(Rd ) we have
Z
K̂(U ξ) =
lim
ε→0+,R→+∞ λε<|x|<λR
Z
=
lim
ε→0+,R→+∞ λε<|x|<λR
e−iU
e−ix·U ξ
−1 x·ξ
x
dx
|x|d+1
x
dx.
|x|d+1
Substituting x0 := U −1 x we conclude
(5.36)
K̂(U ξ) = U K̂(ξ),
∀ ξ 6= 0.
Proposition 5.13. Suppose that K̂ : Rd \ {0} → Rd \ {0} satisfies both (5.35) and (5.36).
Then, there exists a constant C such that
K̂(ξ) = C
ξ
,
|ξ|
ξ ∈ Rd \ {0}
and
Z
(5.37)
C :=
Sd−1
πi
x1
sign(e1 · x) + log (|e1 · x|)
S(dx).
2
|x|d+1
Proof. Of course it suffices only assume that ξ ∈ Sd−1 . Let us choose any orthonormal base
e1 , . . . , ed . Let C1 := K̂1 (e1 ). Observe that for any U = [Uij ] such that U e1 = e1 we must
have U1,j = Uj,1 = 0 for j = 2, . . . , d and therefore
K̂j (e1 ) = K̂j (U e1 ) = (U K̂(e1 ))j =
d
X
Ujk K̂k (e1 ).
k=2
But the only vector in Rd−1 that is invariant under all U 0 ∈ O(d − 1) is the null vector hence
K̂j (e1 ) = 0 for all j 6= 1. In the same fashion we prove that K̂j (ek ) = 0 for j 6= k. But the
above proves that K̂(ξ) = C(ξ)ξ, ξ ∈ Sd−1 for all j = 1, . . . , d. Note however that for any
two vectors ξ, ξ 0 ∈ Sd−1 one can find U ∈ O(d) such that ξ 0 = U ξ. Hence,
(5.36)
C(ξ 0 )ξ 0 = K̂(ξ 0 ) = K̂(U ξ) = U K̂(ξ) = C(ξ)U ξ = C(ξ)ξ 0 ,
hence C(ξ 0 ) = C(ξ) = C. Let
xj
.
|x|d+1
Theorem 5.10 (see also the proof of Theorem 5.4) allows us to conclude that the operator Rj,0
corresponding to the kernel Kj satisfies
(ε)
Kj (x) := 1[ε<|x|]
d
(ε)
[
K̂j (ξ)fˆ(ξ) = lim Kj (ξ)fˆ(ξ) = R
j,0 f (ξ),
ε→0+
44
T.KOMOROWSKI
a.e. in ξ for any f ∈ L2 (Rd ). In addition, from Proposition 5.13 we obtain
ξ
K̂(ξ) = C , ξ ∈ Rd \ {0}
|ξ|
and C is given by (5.37).
Remark. Note that for d = 1 the Riesz transform agrees, up to a constant, with the
Hilbert transform introduced in Section 5.4.1.
6. The Sobolev spaces. Elliptic partial differential equations.
6.1. The definition. Let u P
∈ L1loc (Ω) and α = (α1 , . . . , αd ) be any multi-index of nonnegative integers. Let |α| :=
αi Denote Dα := ∂xα11 . . . ∂xαdd . Then v ∈ L1loc (Ω) is called the
α-the weak derivative of u if
Z
Z
|α|
uDα ϕdx, ∀ϕ ∈ Cc|α| (Ω).
vϕdx = (−1)
(6.1)
Ω
Ω
Exercise. Show that the above definition determines v, i.e. if v, v 0 are both α-th derivatives
of u we have v = v 0 . We write Dα u := v. The space of all functions u ∈ Lp (Ω) that have all
weak derivatives of order |α| ≤ k that belong to Lp shall be denoted by W k,p (Ω), p ∈ [1, +∞].
We introduce the norm
X
kukpk,p :=
kDα ukpLp .
|α|≤k
D0 f
Here we use the convention that
= f , 0 := (0, . . . , 0).
Exercise. Show that W k,p (Ω) is a complete, normed (thus Banach) space
k,p
k,p
∞
Define W0 (Ω) the subspace of W0 (Ω) defined as the closure of Cc (Ω) in the k·kk,p -norm.
Theorem 6.1. Suppose that p ∈ [1, +∞). Then W k,p (Rd ) = W0k,p (Rd ).
k,p
k,p
d
d
∞
d
Proof. Of course
R only the inclusion W (R ) ⊂ W0 (R ) requires a proof. Let φ ∈ Cc (R )
be such that φdx = 1. Such elements have been introduced in Section 4.1 and been called
mollifiers. Let φδ (x) := δ −d φ(x/δ). Let fδ := f ∗ φδ .
Exercise. Show that Dα fδ = f ∗ Dα φδ .
Exercise. Prove that for any f ∈ W k,p (Rd ) and |α| ≤ k we have
lim kDα f − Dα fδ kLp (Rd ) = 0.
δ→0+
Hint: Use the definition of α-th weak derivative and (4.1).
The desired result is a conclusion from the above exercise.
6.2. Bessel potentials. Let a > 0 and Aa : L2 (Rd ) → L2 (Rd ) be given by
d
ˆ
A
a f (ξ) := k̂a (ξ)f (ξ),
(6.2)
where
k̂a (ξ) :=
Define
ka(0) (x) := cd
Z
0
(0)
+∞
1
.
(1 + |ξ|2 )a/2
e−t
t(d−a)/2+1
e−|x|
2 /(4t)
dt.
(0)
Lemma 6.2. We have ka ∈ L1 (Rd ) and k̂a is the Fourier transform of ka .
HARMONIC ANALYSIS
Proof. We have
Z
Z
(0)
ka (x)dx = cd
0
e−t
+∞
Z
−|x|2 /(4t)
e
t(d−a)/2+1
45
Z
+∞
dx dt = cd
ta/2−1 e−t dt < +∞.
0
Exercise. For any function ϕ ∈ S(Rd ) and f ∈ L1 (Rd ) we have
Z
Z
ˆ
f (x)ϕ(x)dx = f (x)ϕ̂(x)dx. Rd
Rd
Using the above exercise we get
Z
Z
k0 (x)ϕ̂(x)dx = cd
(6.3)
+∞
ta/2−1 e−t dt
+∞
Z
= cd
ta/2−1 e−t dt
2 /(4t)
ϕ̂(x)dx
Z
2
e−t|ξ| ϕ̂(ξ)dξ.
Rd
0
Z
t−d/2 e−|x|
Rd
0
Rd
Z
Z
+∞
= cd
Rd
2
ta/2−1 e−t(1+|ξ| ) dt ϕ̂(ξ)dξ.
0
Exercise. Show that for any b > 0, β > 0
Z +∞
tβ−1 e−bt dt = cb−β ,
0
where
Z
+∞
c :=
tβ−1 e−t dt
0
Using the above exercise we conclude that the utmost left hand side of (6.3) equals
Z
(1 + |ξ|2 )−a/2 ϕ̂(ξ)dξ
Rd
(0)
hence k̂a (ξ) = (1 + |ξ|2 )−a/2 .
Exercise. Find value cd in the statement of the above lemma.
(0)
Since now on we should denote ka by ka . It is called a Bessel potential. As a direct
consequence of the above lemma and (4.6), Aa f = ka ∗ f for f ∈ L1 (Rd ) ∩ L2 (Rd ).
After changing variable t := t/|x|2 we can write
Z +∞ −t|x|2 −1/(4t)
cd
e
e
ka (x) =
dt.
|x|d−a 0
t(d−a)/2+1
This kernel has a singularity ka (x) ∼ |x|a−d for d ≥ 2 as |x| 1.
Exercise. Prove that the Bessel potential decays rapidly as |x| 1. By the above we
mean that
lim |x|n ka (x) = 0
|x|→+∞
for any n ≥ 1.
Hint: note that supa>0 an e−a < +∞ for an arbitrary n.
q
d
Therefore ka ∈ L (R ) for any 1/q > 1 − a/d. Young’s inequality, see (3.17), implies that
(6.4)
kAa f kLr (Rd ) ≤ kka kLq (Rd ) kf kLp (Rd ) ,
46
T.KOMOROWSKI
provided that 1/p + 1/q = 1 + 1/r. Hence Aa maps Lp into Lr for any 1/r > 1/p − a/d. Note
also that for a > d we can estimate L∞ norm of Af and obtain
kAa f kL∞ (Rd ) ≤ kkkLq (Rd ) kf kLp (Rd ) ≤ Ckf kLp (Rd ) ,
(6.5)
for any p ≥ 1.
As for the ”border case” 1/r = 1/p − a/d we prove a stronger statement.
∗
Theorem 6.3. Suppose that 0 < a < d and p ∈ (1, d/a). Then, A maps Lp into Lp for
1
1 a
(6.6)
= − .
p∗
p d
Proof. Suppose that f ∈ Lp (Rd ). Let k(x) be a kernel that satisfies
lim |x|n k(x) = 0
|x|→+∞
for any n ≥ 1 and there exists C > 0 such that
|k(x)| ≤
We have
C
.
|x|d−a
Z
Z
|Af (x)| ≤
|k(y)||f (x − y)|dy +
[|y|≤ρ]
|k(y)||f (x − y)|dy.
[|y|>ρ]
(ρ)
Denote the first and the second integrals by I1 (x) and I2 (x) respectively. Let kl (x) :=
(ρ)
1[|x|≤ρ] k(x) and ku (x) := 1[|x|>ρ] k(x). We have, from (6.4)
kI1 kLp ≤ Cρa kf kLp .
On the other hand we have
kI2 kL∞ ≤ kku(ρ) kLp0 (Rd ) kf kLp (Rd )
with 1/p0 = 1 − 1/p and
0
kI2 kL∞ ≤ Cρa−d+d/p kf kLp (Rd ) .
∗ /d
Choose ρ := C1 λ−p
kf kpL1p where 1/p1 := d/p − a. We obtain then
∗
CkI1 kpLp
Ckf kLp p
md (I1 ≥ λ/2) ≤
≤
λp
λ
and 1/p∗ = 1/p − a/d.
On the other hand
kI2 kL∞ ≤ λ/2,
with an appropriate choice of C1 . In conclusion, we have shown that
∗
Ckf kLp p
, ∀ λ > 0.
md (|Af | ≥ λ) ≤
λ
for any 1 < p < d/a, p∗ given by (6.6).
Definition 6.4. We say that a subadditive operator (see (3.2)) T : Lp (µ) → Lq (ν) is of weak
type (p, q) if there exists a constant C > 0 such that for all λ > 0 we have
Ckf kLp (µ) q
ν(|T f | ≥ λ) ≤
.
λ
HARMONIC ANALYSIS
47
We have shown therefore that the operator A is of weak type (p, p∗ ) for any p ∈ [1, d/a]
with p∗ as given in (6.6). To finish the argument we shall need the following generalization
of the Marcinkiewicz interpolation theorem whose proof we present in the following section.
Theorem 6.5. Suppose that 1 ≤ p1 < p2 ≤ +∞ and pi ≤ qi , i = 1, 2, q1 6= q2 , θ ∈ (0, 1).
Assume that T : Lp1 (µ) + Lp2 (µ) → Lq1 (ν) + Lq2 (ν), is an operator that is of both weak types
(pi , qi ), i = 1, 2. Then, the operator is of strong type (p, q), where
1
θ
1−θ
+ ,
=
p
p1
p2
1
θ
1−θ
+
=
q
q1
q2
6.3. Some useful inequalities.
6.3.1. Hardy inequality.
Lemma 6.6. Suppose that p > 0 and 0 < q1 ≤ q2 < +∞. Then, there exists a constant
A > 0 such that for any h : (0, +∞) → (0, +∞) non-increasing we have
Z +∞ Z +∞ q2 dt 1/q2
q1 dt 1/q1
(6.7)
t1/p h(t)
≤A
t1/p h(t)
.
t
t
0
0
Proof. Suppose first that
Z
+∞ t
1/q1
q1 dt 1/q1
h(t)
= 1.
t
"Z
t
0
Then,
c(p, q1 )t1/p h(t) ≤
t/2
q1 ds
s1/p h(s)
s
#1/q1
≤ 1.
We can write therefore
Z +∞ Z +∞ q2 dt
q1 dt
t1/p h(t)
≤ (sup t1/p h(t))q2 −q1
t1/p h(t)
.
t
t
t
0
0
The case of a general h can be done thanks to homogeneity of both side of (6.7) with respect
to multiplication by a constant.
6.3.2. Minkowski’s inequality. For any two σ-finite measures µ, ν, p ∈ [1, +∞) and a nonnegative function f (x, y) we have
Z Z
p
1/p Z Z
1/p
p
f (x, y)µ(dx) ν(dy)
≤
f (x, y)ν(dy)
µ(dx).
(6.8)
To show (6.8) assume first that µ(X) = ν(Y ) = 1. Then, since the norm k · kLp (ν) is convex
we obtain
Z
Z
kf (x, ·)kLp (ν) µ(dx) ≥ f (x, ·)µ(dx)
Lp (ν)
Note that the inequality holds for any finite measures, since it is homogeneous w.r.t. multiplications by constants we obtain (6.8) for arbitrary finite measures. Finally, we can extend
it by approximation to arbitrary σ-finite measures.
48
T.KOMOROWSKI
6.3.3. More Hardy’s inequalities.
Lemma 6.7. Suppose that p ≥ 1 and r > 0. Then, for any h : (0, +∞) → (0, +∞) we have
(6.9)

Z

 t
p
1/p
Z +∞
1/p
Z
p
p
−r−1
−r−1
 h(s)ds t
(th(t)) t
dt
dt ≤
r 0
+∞
0
0
and
(6.10)

Z

+∞
0
p
1/p
 +∞
Z +∞
1/p
Z
p

h(s)ds tr−1 dt ≤
(th(t))p tr−1 dt
.
r 0
t
Exercise. Prove (6.10)
Proof. Let
K(x, y) :=
1[0,x] (y) y (r+1)/p
×
y
x
and
+∞
+∞
Z
Z
T f (x) :=
K(x, y)f (y)dy =
K(1, y)f (xy)dy.
0
0
Note that by Minkowski’s inequality we have
Z
+∞

1/p
Z
p

(T f (x)) dx
=
0
0
 +∞
1/p
Z
K(1, y)dy 
f p (xy)dx .
+∞
≤
0
Substituting
 +∞
p 1/p
Z

K(1, y)f (xy)dy  dx
0
Z
x0
+∞
0
:= xy we obtain that
Z
+∞
kT f kLp ≤
0
1/p
 +∞
Z
K(1, y) 
dy
f p (x)dx .
y 1/p
0
The inequality (6.9) is concluded upon the substitution f (x) := g(x)x1−(r+1)/p . The proof of
(6.10) is analogous, we choose
K(x, y) :=
1[x,+∞] (y) y (−r+1)/p
×
y
x
HARMONIC ANALYSIS
49
6.3.4. The proof of Marcinkiewicz inequality. Suppose that f ∈ Lp (µ). Let F (λ; f ) := µ(|f | >
λ). Define the non-increasing rearrangement of f as a non-increasing f ∗ : (0, +∞) → (0, +∞)
satisfying
m1 (t : f ∗ (t) > λ) = F (λ; f ).
One can define f ∗ (t) := inf[λ : F (λ; f ) ≤ t]. Note that
kf ∗ kpLp (R+ )
+∞
Z
Z
∗
p
=
(f (t)) dt = p
=p
0
λp−1 m1 (f ∗ > λ)dλ
0
0
Z
+∞
+∞
λp−1 µ(|f | > λ)dλ = kf kpLp (µ) .
Exercise. i) F (·; f ) is a non-increasing, right continuous function.
ii) f ∗ (·) is non-increasing,
iii) f ∗ (t) is the unique value that satisfies
F (f ∗ (t)+; f ) ≤ t ≤ F (f ∗ (t)−; f ),
iv) show that limt→+∞ f ∗ (t) = 0,
v) (f1 + f2 )∗ (t) ≤ f1∗ (t/2) + f2∗ (t/2).
Let
1/q − 1/q1
1/q0 − 1/q
=
.
σ :=
1/p0 − 1/p
1/p − 1/p1
For a fixed t > 0 we define
f t (x) := f (x)1[|f (x)|>f ∗ (tσ )]
and ft := f − f t . Since f t ≤ f we have F (·; f t ) ≤ F (·; f ), hence (f t )∗ ≤ f ∗ .
Note that for λ > f ∗ (tσ ) one has F (λ; f ) = F (λ; f t ), while for λ < f ∗ (tσ )
F (λ; f t ) = F (f ∗ (tσ ); f ).
Thus, for u > tσ we have (f t )∗ (u) = 0. It is elementary to see that (ft )∗ ≤ f ∗ , and (ft )∗ (u) ≤
f ∗ (tσ ).
Since T is of weak type (pi , qi ), i = 0, 1 we have
Ci kf kLpi qi
µ(|T f | ≥ λ) ≤
λ
thus
t≤
Ci kf kLpi
(T f )∗ (t)
q i
and
(T f )∗ (t) ≤ Ci t−1/qi kf kLpi .
Letting f = ft + f t we obtain
(6.11)
(T f )∗ (t) ≤ C0 t−1/q0 kf t kLp0 + C1 t−1/q1 kft kLp1
Note that
(6.12)
∗
Z
kT f kLq = k(T f ) kLq =
0
+∞ t
1/q
q dt 1/q
(T f ) (t)
t
∗
50
T.KOMOROWSKI
Coming back to (6.12), since p ≤ q we can estimate using Lemma 6.6
Z +∞ p dt 1/p
1/q
∗
q
kT f kL ≤
t (T f ) (t)
t
0
the utmost side of the right hand side of (6.12) can be bounded using (6.11). We obtain then
(6.13)
Z +∞ Z +∞ p dt 1/p
p dt 1/p
1/q−1/q
t
1/q−1/q
0
1
kT f kLq ≤ C
t
kf kLp0
+C
.
t
kft kLp1
t
t
0
0
Note however that, since p0 ≥ 1
p0 ds 1/p0
kf kLp0 =
s
(f ) (s)
s
0
Z tσ
Z +∞
Lemma 6.6
ds
ds
s1/p0 f ∗ (s) .
≤
C
s1/p0 (f t )∗ (s) ≤ C
s
s
0
0
We estimate now the first term on the right hand side of (6.13) by
"Z
p #1/p
Z tσ
+∞ ds
dt
C
t1/q−1/q0
s1/p0 f ∗ (s)
.
s
t
0
0
Z
t
+∞ 1/p0
t ∗
Substituting t0 := tσ we can write that this expression equals
Z +∞
Z t
p 1/p
ds
−p/p0
1/p0 ∗
dt
.
(6.14)
C
t
s
f (s)
s
0
0
Using (6.9) to bound (6.14) we obtain that the expression there can be estimated by
Z +∞ 1/p
ds p
C
t−p/p0 (t1/p0 f ∗ (t))p
dt
= Ckf ∗ kLp .
s
0
The estimate of the second term on the right hand side of (6.13) can be done likewise with
the help of (6.10).
Exercise. Estimate the second term on the right hand side of (6.13).
6.4. The Sobolev embeddings.
Theorem 6.8. Suppose k is a positive integer and 1 ≤ p < q < +∞. If
1
1 k
≥ − .
q
p d
then W k,p (Rd ) ⊂ Lq (Rd ) and the natural inclusion map is continuous,
Proof. Consider first the case when k = 1 and let f ∈ Cc∞ (Rd ). We have f = A2 (I − ∆)f ,
where A2 is given by (6.2). Note that
A2 (I − ∆)f = A2 f −
d
X
Kj Dj f,
j=1
where
Z
Kj g(x) := Dj A2 g(x) =
∂j k2 (x − y)g(y)dy,
g ∈ Lp (Rd ).
HARMONIC ANALYSIS
51
The singularity of ∂j k2 close to 0 is |x|1−d hence we can use (6.4) and Theorem 6.3 with a = 2.
As a result we obtain
kKj gkLr (Rd ) ≤ CkgkLp (Rd ) ,
for any 1/r ≥ 1/p − 1/d. Since A2 is a bounded operator on any Lp , p ∈ [1, +∞] we conclude
that
kf kLr (Rd ) ≤ Ckf k1,p
and the conclusion of the theorem follows for k = 1. We conduct the rest of the argument
by the induction on k. Suppose we have shown the theorem for a certain k. This implies
that W k+1,p ⊂ W 1,q for 1/q ≥ 1/p − k/d. Yet we already know that W 1,q ⊂ Lr for 1/r ≥
1/q − 1/d ≥ 1/p − (k + 1)/d and the induction assertion follows.
T
Theorem 6.9. a) If p = d/k then W k,p (Rd ) ⊂ r>1 Lr (Rd ),
b) if p > d/k then for every f ∈ W k,p (Rd ) there exists its representant such that it is a
continuous function.
Proof. We only deal with the case k = 1, the general case can be argued by induction as in
the proof of the previous theorem.
The case a) follows from the previous theorem since for any r1 > 1 we have 1/r1 ≥
1/p − 1/d = 0.
In case b) we have p > d. We can use Hölder inequality and obtain
(6.15)
|Kj g(x)| ≤ k∂j k2 kLp0 (Rd ) kgkLp (Rd )
for 1/p0 = 1 − 1/p. Note however that 1/p0 > 1 − 1/d thus k∂j k2 kLp0 (Rd ) < +∞ and Kj g ∈
L∞ (Rd ). To prove continuity let xn → x0 , as n → +∞. It suffices only to show the following
exercise.
Exercise. For any ε > 0 we can find a sufficiently large R > 0 so that
Z Z
|∂j k2 (xn − y)g(y)|dy < ε
(6.16)
sup
n≥1
[|y|≥R]
and there exists δ > 0 such that for any A ⊂ [|x| ≤ R] with md (A) < δ we have
Z
|∂j k2 (xn − y)g(y)|dy < +∞.
(6.17)
sup
n≥1
A
Hint: Both estimates follow from Hölder inequality applied as in (6.16).
Exercise. Prove the case of general k.
We also have the following.
Corollary 6.10. Suppose that k − d/p ∈ (m, m + 1] for a certain non-negative integer m.
Then for f ∈ W k,p (Rd ) there exists a representant that belongs to C m (Rd ).
Proof. Note that Dm f ∈ W k−m,p (Rd ). Using part b) of Theorem 6.9 we can find gi1 ,...,im ∈
Cb (Rd ) such that ∂im1 ,...,im f = gi1 ,...,im and i1 , . . . , im ∈ {1, . . . , d}. One can define then by
integrating out g = [gi1 ,...,im ] a function f˜ that is of class C m and such that Dm f˜ = g.
d
∞
Remark 1. Suppose that Ω is a subregion of R . Note that any function f ∈ Cc (Ω) can
be extended to f ∈ Cc∞ (Rd ). Using the embedding theorems proved so far we conclude the
following.
52
T.KOMOROWSKI
Theorem 6.11. a) If p < d/k then W0k,p (Ω) ⊂ Lq (Ω), where 1/q ≥ 1/p − k/d,
T
b) if p = d/k then W0k,p (Ω) ⊂ q≥1 Lq (Ω),
c) if k − d/p ∈ (m, m + 1] for some non-negative m then for every W0k,p (Ω) ⊂ C m (Ω).
Remark 2. An analogous result can be proved on a torus Td .
7. The Dirichlet problem for a second order elliptic equation
Lemma 7.1. Suppose that p ∈ (1, +∞), f ∈ L2 (Rd ) and u ∈ W p,2 (Rd ) are such that
∆u = f.
Then,
2
kDij
ukLp ≤ Akf kLp , ∀ i, j = 1, . . . , d
for a certain constant A > 0 depending only on p, d
Proof. We know that for any u ∈ Cc∞ (Rd )
(7.1)
2
Ri Rj ∆u = Dij
u,
i, j = 1, . . . , d,
where Ri denotes the Riesz transform. We can extend equality in (7.1) to the entire W 2,p (Rd )
2 u = R R f and the conclusion of the lemma follows from
using density argument. Thus Dij
i j
Theorem 5.10 and Proposition 5.13.
Corollary 7.2. Suppose that Q = [qij ] is a strictly positive, symmetric matrix. Then, there
exists a constant A > 0 depending only on d, p and the smallest and largest eigenvalues of Q
and such that for any u ∈ W p,2 (Rd ) and f ∈ Lp (Rd ) such that
d
X
2
qij Dij
u = f.
i,j=1
we have
2
kDij
ukLp ≤ Akf kLp ,
∀ i, j = 1, . . . , d.
Exercise. Prove the above corollary.
Assume that:
i) aij (x), i, j = 1, . . . , d are uniformly continuous functions that satisfy aij (x) = aji (x) and
there exists λ > 0 such that
1 2
|ξ| ≤ a(x)ξ · ξ ≤ λ|ξ|2 , ∀ ξ, x ∈ Rd ,
λ
ii) bi (x), i = 1, . . . , d, c(x) are measurable and satisfy
X
|bi (x)| + |c(x)| ≤ λ, ∀ x ∈ Rd .
i
Lemma 7.3. Suppose that the assumptions of Corollary 7.2 hold. Then, there exists 0 > 0,
A(p, d, 0 ) such that for any aij (x) satisfying kaij − qij k∞ ≤ 0 and f ∈ Lp (Rd ) and u ∈
W p,2 (Rd ) such that
d
X
2
L[u] :=
ai,j (x)Dij
u(x) = f (x)
i.j=1
we have
2
kDij
ukLp ≤ Akf kLp ,
∀ i, j = 1, . . . , d.
HARMONIC ANALYSIS
53
Proof. We have
d
X
LQ [u] :=
2
qij Dij
u
= L[u] −
i,j=1
d
X
2
[ai,j (x) − qij ]Dij
u
i,j=1
d
X
=f−
2
[ai,j (x) − qij ]Dij
u.
i,j=1
Hence, using Corollary 7.2 we obtain
2
kDij
ukL2
≤ Akf k
Lp
d
X
+A
2
kai,j − qij k∞ kDij
ukLp .
i,j=1
d2 A
Choosing 0
< 1/2 we conclude the lemma.
Pd
P
2u+
Let L[u] := ij=1 aij Dij
i bi Di u + cu.
Theorem 7.4. Suppose that assumptions i) and ii) hold. Then, there exists a constant C > 0
depending only on d and λ such that for any u ∈ W p,2 (Rd ), f ∈ Lp (Rd ) that satisfy Lu = f
we have
2
kDij
ukLp (Rd ) ≤ C(kf kLp (Rd ) + kukLp (Rd ) ).
(7.2)
Proof. Suppose that δ < 1 is sufficiently small so that on any ball Bδ (x0 ) we have |aij (x) −
aij (x0 )| < 0 . Here 0 is as in Lemma 7.3.
Exercise. Given ρ ∈ (0, 1) construct a function φ(x) such that φ(x) ≡ 1 for x ∈ Bρδ (x0 ),
φ(x) ≡ 0 for x 6∈ Bδ (x0 ) and
kDφkL∞ ≤
C
δ(1 − ρ)
and kD2 φkL∞ ≤
C
− ρ)2
δ 2 (1
for some C > 0.
Let v := uφ. We have
Lv = φf − φ
X
bi Di u + 2
X
aij Di φDj u + u
i
X
2
aij Dij
φ =: g.
ij
Using Lemma 7.3 we obtain
d
X
2
kDij
ukLp (Bρδ (x0 )) ≤
ij=1
d
X
2
kDij
vkLp (Rd )
ij=1
≤ A[kgkLp (Rd ) + kvkLp (Rd ) ].
But
|g(x)| ≤ |f (x)| +
C
C
|Du(x)| + 2
|D2 u(x)|.
δ(1 − ρ)
δ (1 − ρ)2
We conclude therefore that
d
X
(7.3)
2
kDij
ukLp (Bρδ (x0 )) ≤
ij=1
A kf kLp (Bδ (x0 )) +
1
1
kDukLp (Bδ (x0 )) + 2
kukLp (Bδ (x0 )) .
δ(1 − ρ)
δ (1 − ρ)2
54
T.KOMOROWSKI
We can find a covering of Rd with the balls Bδ (x0 ), x0 ∈ Z such that
X
X
0<a<
1Bρδ (x0 ) ≤
1Bδ (x0 ) ≤ a−1
x0 ∈Z
x0 ∈Z
for some a > 0. In consequence we obtain
d
X
2
kDij
ukLp (Rd ) ≤ C(kf kLp (Rd ) + kukLp (Rd ) + kDukLp (Rd ) ).
i,j=1
To obtain (7.2) we need the following interpolation result.
Lemma 7.5. There exists a constant C > 0 such that for any ε > 0 t
d
X
kDukLp (Rd ) ≤ ε
2
kDij
ukLp (Rd ) + Cε−1 kukLp (Rd ) .
i,j=1
Proof. Suppose that g ∈ C0∞ (R). Note that for x, z ∈ [0, 1], x > z
Z x
0
0
g (x) − g (z) =
g 00 (y)dy
z
hence
Z
0
x
g (x) =
Z
00
g (y)dy + g(x) − g(x − 1).
dz
x−1
x
z
Raising to p-th power both sides of the above equality, using some elementary inequalities we
obtain
kg 0 kLp (R) ≤ C(kg 00 kLp (R) + kgkLp (R) ).
For any u ∈ C0∞ (Rd ) we get
kDi u(x1 , . . . , ·, . . . , xd ) kpLp (R) ≤ C(k u(x1 , . . . , ·, . . . , xd ) kpLp (R) +kDi2 u(x1 , . . . , ·, . . . , xd ) kpLp (R) ).
i−th component
i−th component
i−th component
Integrating over the remaining variables we get
X
(7.4)
kDi ukpLp (Rd ) ≤ C(kukpLp (Rd ) +
i
d
X
2
kDij
ukpLp (R) ).
ij=1
Let v(x) = u(λx) for λ > 0. We obtain then that
kDi vkpLp (Rd ) = λp−d kDi ukpLp (Rd ) ,
kvkpLp (Rd )
= λ−d kukpLp (Rd ) ,
2
2
kDij
vkpLp (Rd ) = λ2p−d kDij
ukpLp (Rd ) .
In consequence, using (7.4) for v we obtain
X
kDi ukpLp (Rd ) ≤ C(λ−p kukpLp (Rd ) + λp
i
d
X
2
kDij
ukpLp (R) ).
ij=1
λp .
The result of the lemma follows upon the choice ε :=
Suppose that c(x) ≤ 0 and p ∈ (1, +∞). We can use the above a’priori estimate in order
to show that for any f ∈ Lp (Rd ) and ζ > 0 the equation
ζu − L[u] = f
(7.5)
has a unique solution u ∈
W 2,p (Rd ).
HARMONIC ANALYSIS
55
We use the method of continuation w.r.t. the parameter. Let
Lt [u] = (1 − t)∆u + tL[u].
Let Σ be the set of those t for which
ζu − Lt [u] = f
(7.6)
has a unique solution u ∈ W 2,p (Rd ). Note that 0 ∈ Σ for u = k2 ∗ f if ζ = 1 (k2 is the Bessel
kernel. If ζ 6= 1 the result can be concluded by a simple rescaling. Suppose now that t0 ∈ Σ.
We show that there is ε > 0 such that for (t0 − ε, t0 + ε) ⊂ Σ. Let v ∈ W 2,p (Rd ). Observe
that we have the following result.
Lemma 7.6. Suppose that t0 ∈ Σ. Then there exists C > 0 such that for any u ∈ W p,2 (Rd )
and f ∈ Lp (Rd ) we have
kukW 2,p (Rd ) ≤ Ckf kLp (Rd ) .
(7.7)
Proof. Define u := T f , where T : Lp (Rd ) → Lp (Rd ). The operator is well defined. It is closed.
Indeed, if fn → f and un := Tfn → u then, by virtue of (7.2) we have
kun kW 2,p (Rd ) ≤ C,
∀ n ≥ 1.
Hence, easily one can see that u satisfies
u − Lt0 [u] = f.
Hence u = T f . As a consequence T must be bounded, so there exists C > 0 such that
kukLp (Rd ) ≤ Ckf kLp (Rd ) .
(7.8)
Combining the above with (7.2) we conclude the result.
Denote by N v the unique solution of
N v − Lt0 [N v] = f − (Lt0 − Lt )v.
Let v1 , v2 ∈ W 2,p (Rd ). We have
N v1 − N v2 − Lt0 [N v1 − N v2 ] = (Lt − Lt0 )(v1 − v2 ).
Using (7.10) we obtain
kN v1 − N v2 kW 2,p (Rd ) ≤ C|t − t0 |kv1 − v2 kW 2,p (Rd ) .
We can choose ε > 0 so small that C|t − t0 | < 1. From Banach contraction principle there is
a unique u such that N [u] = u. It is the desired solution.
To show that Σ is closed suppose that we have tn ∈ Σ and tn → t0 . Let un be the unique
solution of
(7.9)
un − Ltn [un ] = f.
We shall need the following refinement of Theorem 7.4, which we quote from [3].
Theorem 7.7. Suppose that the assumptions i) and ii) hold. Then, there exists a constant
C > 0 depending only on d and λ such that for any u ∈ W p,2 (Rd ), f ∈ Lp (Rd ) such that
u − Lu = f we have
(7.10)
kukLp (Rd ) ≤ Ckf kLp (Rd ) .
56
T.KOMOROWSKI
We have kun kLp (Rd ) ≤ Ckf kLp (Rd ) . Using (7.2) and (7.10) we obtain that
kun kW 2,p (Rd ) ≤ C
for all n ≥ 1. For p ∈ (1, +∞) the space W 2,p (Rd ) is reflexive. One can choose a subsequence,
which we denote by the same symbol, that is weakly convergent to a certain u ∈ W 2,p (Rd ).
It is a solution to
u − Lt0 [u] = f.
(7.11)
Note that by virtue of (7.10) it is unique. Hence t0 ∈ Σ. The set Σ is therefore simultaneously
closed and open in [0, 1]. It is also nonempty, thus Σ = [0, 1]. In consequece the equation
u − L[u] = f.
(7.12)
has a unique solution u ∈ W 2,p (Rd ) for any f ∈ Lp (Rd ).
8. More on the Fourier transform. Paley-Wiener theorems.
Observe that if we know that |f (x)| ≤ Ce−K|x| , x ∈ Rd then fˆ extends onto the product
of half-planes Imζ > −K. Suppose that f ∈ L1 (Rd ) and f (x) = 0 for |x| ≥ K, where K > 0
is given. Note that
Z
fˆ(ξ) :=
e−ix·ξ f (x)dx
can be then extended to the entire Cd . Indeed, we have for ζ := ξ + iη
Z
ˆ
f (ζ) := e−ix·ξ+x·η f (x)dx.
Note also that
|fˆ(ζ)| ≤ CeK|Im ζ| , ∀ ζ ∈ Cd ,
R
where C can be chosen equal to Rd |f (x)|dx. We shall show that the converse to (8.1) also
holds. This is the contents of the following.
(8.1)
Theorem 8.1. (Paley-Wiener) Suppose that f ∈ S(Rd ) has a support contained in BK (0)
for some K > 0. Then, for a given N ≥ 0 there exists a constant CN > 0 such that
(8.2)
|fˆ(ζ)| ≤ CN (1 + |ζ|)−N eK|Imζ| ,
∀ ζ ∈ Cd .
Conversely suppose that there exists an entire extension of fˆ such that for a given K > 0 and
N ≥ 0 one can find a constant CN > 0 such that (8.2) holds. Then supp f ⊂ BK (0).
Proof. The first implication follows from an argument similar to the one used in the foregoing.
Suppose that α is an multi-index such that |α| = N . Then
Z
iN ζ α fˆ(ζ) := e−ix·ξ+x·η Dα f (x)dx.
Hence
(8.3)
(1 + |ζ|)N |fˆ(ζ)| ≤ eK|Im ζ| kf kW 1,N (Rd ) ,
∀ ζ ∈ Cd .
and the conclusion of the first part of the theorem follows.
To show the converse we start with the following one dimensional result.
HARMONIC ANALYSIS
57
Proposition 8.2. Suppose that f ∈ L1 (R) and fˆ extends to an entire function such that
(8.1) holds. Then
Z
fˆ(ξ + iη) exp{ix(ξ + iη)}dξ
R
does not depend on η for all x ∈ R.
Proof. One can consider
Z
fˆ(ζ) exp{ix · ζ}dζ,
Γ
where the contour Γ consists of Γ+ := [−A, A], Γ− := [−A, A] + iη and two vertical pieces
ΓA , Γ−A . One can easily show that
Z
Z
lim
fˆ(ζ) exp{ix · ζ}dζ = lim
fˆ(ζ) exp{ix · ζ}dζ = 0
A→+∞
ΓA
A→+∞
Γ−A
and the conclusion follows.
Exercise. Prove that for f ∈ L1 (Rd ) and fˆ such that (8.1) holds one has
Z
fˆ(ξ + iη) exp{ix · (ξ + iη)}dξ
Rd
Rd
does not depend on η ∈
for all x ∈ Rd .
d
To finish the proof of the theorem note that for x ∈ R such that |x| > K we can choose
η := λx/|x|, where λ > 0 and
Z
Z
d
ˆ
(8.4)
(2π) f (x) = f (ξ) exp{ix · ξ}dξ = fˆ(ξ + iη) exp{ix · (ξ + iη)}dξ
Rd
Rd
Z
≤ exp{λ(K − |x|)}
CN dξ
,
(1 + |ξ|)N
∀ λ > 0.
Rd
In consequence for |x| > K we obtain the conclusion, since λ > 0 can be chosen arbitrarily. Theorem 8.3. Suppose that f ∈ L1 (Rd ) possesses a Fourier transform that can be extended
to an entire function such that (8.1) holds. Then supp f ⊂ BK (0).
R
Proof. Let φ be a C ∞ (Rd ), non-negative function, supported in B1 (0) such that φdx = 1.
Let φε (x) := ε−d φ(x/ε). Let fˆε := fˆφ̂ε . Then, using the same η as in (8.4) we obtain
Z
d
ˆ
(8.5)
(2π) |fε (x)| = f (ξ)φ̂ε (ξ) exp{ix · ξ}dξ d
R
Z
Z
Prop. 8.2
ˆ
= f (ξ + iη)φ̂ε (ξ + iη) exp{ix · (ξ + iη)}dξ ≤ exp{λ(K−|x|)} |φ̂ε (ξ+iη)|dξ, ∀ λ > 0.
d
d
R
R
However
|φ̂ε (ξ + iη)| ≤
CN eε|η|
,
(1 + |ξ|)N
58
T.KOMOROWSKI
thus
Z
d
(2π) |fε (x)| ≤ CN exp{λ(K + ε − |x|)}
(8.6)
dξ
,
(1 + |ξ|)N
∀ λ > 0.
Rd
This of course implies that fε is supported in BK+ε (0). However fε → f in L1 , as ε → 0+.
In consequence f is supported in BK (0).
8.1. H p spaces on a half-plane. Denote by H := [z ∈ C :Im z > 0].
Definition 8.4. Let p > 0. Let H p (H) be the spaces of all holomorphic functions f : H → C
such that
Z
sup |f (x + iy)|p dx < +∞.
y>0
R
Suppose that p = 2 and f ∈ L2 (−∞, 0). Note that
Z 0
fˆ(ξ + iη) :=
e(−iξ+η)x f (x)dx
−∞
defines the extension of fˆ onto the half-plane H. In particular by Plancherel’s theorem
kfˆ(· +
iη)k2L2
1
=
(2π)d
Z0
e2ηx |f (x)|2 dx ≤
kf k2L2
,
(2π)d
−∞
so fˆ ∈ H 2 . We show that the converse also holds. We have the following.
Theorem 8.5. Suppose that f ∈ L1 (R) and fˆ ∈ H 1 (H). Then for x > 0 we have f (x) = 0.
Proof. Note that for any η > 0 we have
Z η
Z
dη 0
|fˆ(ξ + iη 0 )|dξ < +∞.
0
R
One can choose a sequence An → +∞, Bn → −∞ such that
Z η
Z η
0
0
ˆ
lim
|f (An + iη )|dη = lim
|fˆ(Bn + iη 0 )|dη 0 = 0.
n→+∞ 0
n→+∞ 0
From here one gets that
Zη
Zη
0
0
0
fˆ(An + iη ) exp{i(An + iη )x}dη = fˆ(Bn + iη 0 ) exp{i(Bn + iη 0 )x}dη 0 → 0,
0
0
as n → +∞. Denote by Γn the contour formed over the boundary of a rectangle An , An + iη,
Bn + iη, Bn . We have
Z
fˆ(ζ)dζ = 0.
Γn
Hence, letting n → +∞ we obtain
Z An
Z
iξx ˆ
lim f (ξ)e dξ = lim n→+∞
n→+∞
Bn
An
Bn
i(ξ+iη)x ˆ
f (ξ + iη)e
dξ HARMONIC ANALYSIS
−xη
≤e
Z
|fˆ(ξ + iη 0 )|dξ,
sup
59
∀ η > 0.
η 0 >0 R
Hence the conclusion of the Theorem follows.
Corollary 8.6. The conclusion of the above theorem holds for any f ∈ L1 (R) such that
fˆ ∈ H p (H) when p ∈ (1, +∞).
Proof. Let
ik fˆ(z)
fˆk (z) :=
.
ik + z
Exercise Prove that fk (x) = f ∗ ek (x), where ek (x) := 1(−∞,0] (x)kekx .
1
ˆ
Exercise Prove that fk ∈ H (H) for any k > 0.
Using Theorem 8.5 we conclude that fk (x) = 0 for all x > 0. Letting k → +∞ we obtain
that f (x) = 0 for all x > 0.
8.2. An application to partial differential equations. Finite speed of propagation
for solutions of strongly hiperbolic p.d.e.-s. Suppose that P (ξ, ω) is a certain polynomial of degree m. We shall assume that P is of degree m in the variable ω, i.e. it is of the
form
P (ξ, ω) := ω m + P1 (ξ)ω m−1 + . . . + Pm (ξ),
where Pi are polynomials of degree ≤ i, i = 1, . . . , m. Denote by D := (D1 , . . . , Dd ) and
τ := ∂t . For any f ∈ S(Rd+1 ) we let
Z
1
eix·ξ+iωt P (iξ, iω)fˆ(ξ, ω)dξdω.
P (D, τ )f (x, t) :=
(2π)d+1
Let us consider the following Cauchy problem.
P (D, τ )u(x, t) = 0 for t ≥ 0,
τ k u(x, 0) = fk (x) k = 0, . . . , m − 1,
(8.7)
where fk ∈ S(Rd ).
Definition 8.7. Gärding’s hyperbolicity condition. We call (8.7) hyperbolic if there exists
a real number c such that
P (iξ, iω) 6= 0
for all ξ ∈ Rd and all complex ω with Im ω ≤ −c.
Remark. This condition is of course equivalent with the fact that all m roots of
(8.8)
P (iξ, iω) = 0
lie in the half-plane Im ω > −c.
Example. Wave equation in d dimension.
2
c−2
0 ∂t − ∆x u(x, t) = 0.
2
2
d
Note that P (ξ, ω) = c−2
0 ω − |ξ| . Note that for a fixed ξ ∈ R we have
2
2
0 = P (iξ, iω) = −c−2
0 ω + |ξ|
implies ω = ±|ξ|, thus Im ω > −c for any c > 0.
60
T.KOMOROWSKI
We wish to start with the equation
P (D, τ )u(x, t) = 0
for t ≥ 0,
τ k u(x, 0) = 0,
k = 0, . . . , m − 2,
τ m−1 u(x, 0) = g(x).
(8.9)
Let us perform the Fourier transform in x of (8.9). We obtain then
P (iξ, τ )û(ξ, t) = 0
for t ≥ 0,
k
τ û(ξ, 0) = 0,
k = 0, . . . , m − 2,
τ m−1 û(ξ, 0) = ĝ(ξ).
Here
Z
û(ξ, t) :=
e−iξ·x u(x, t)dx.
Let Z(ξ, t) be the solution of the ordinary equation
P (iξ, τ )Z(ξ, t) = 0 for t ≥ 0,
τ k Z(ξ, 0) = 0,
k = 0, . . . , m − 2,
τ m−1 Z(ξ, 0) = 1.
Proposition 8.8. Suppose that Γ is a closed path that runs around each root ω of (8.8)
counter-clockwise. We have
Z
eiωt dω
1
.
Z(ξ, t) =
(2π)d Γ P (iξ, iω)
Proof. Note that
P (iξ, τ )Z(ξ, t) =
1
(2π)d
Z
eiωt dω = 0.
Γ
Also
Z
1
(iω)k dω
τ Z(ξ, 0) =
.
(2π)d Γ (iω)m + P1 (iξ)(iω)m−1 + . . . + Pm (iξ)
Expanding Γ to infinity we conclude easily that for k ≤ m − 2 the corresponding value equals
0 while for k = m − 1 it equals 1. Hence the initial conditions are satisfied.
The solution of (8.9) equals therefore
Z
1
eix·ξ Z(ξ, t)ĝ(ξ)dξ.
(8.10)
u(x, t) =
(2π)d
k
Rd
We need to verify that the integral in (8.10) converges. For that it suffices to prove that
(8.11)
(1 + |ξ|)n+1 |τ k ξ α Z(ξ, t)ĝ(ξ)| ≤ C
uniformly in ξ ∈ Rd and t ∈ [0, T ] for |α| + k ≤ m. Since we have assumed that g ∈ S(Rd ) it
suffices to prove that For that it suffices to prove that
(8.12)
|τ k Z(ξ, t)| ≤ C(1 + |ξ|)k
for some constant C > 0.
Note that |Pk (iξ)| ≤ C(1 + |ξ|k ). Hence
0 = (iω)m + P1 (iξ)(iω)m−1 + . . . + Pm (iξ)
HARMONIC ANALYSIS
61
implies
|ω|m ≤ C
m
X
(1 + |ξ|)k |ω|m−k .
k=1
Let θ := |ω|(1 +
|ξ|)−1 .
We get
θm ≤ C
m−1
X
θk .
k=1
This implies that either θ < 1, or
(8.13)
θm
≤
mM θm−1 .
As a result we conclude that
|ω| ≤ C(1 + |ξ|).
We can write
P (iξ, iω) = im
m
Y
(ω − ωk (ξ)).
k=1
Let Γ be at the distance at least 1 from all the roots ωk (ξ) and |Γ| ≤ 2πm (one can choose as
Γ a union of circles of radius 1 surrounding the roots). Then
|P (iξ, iω)| ≥ 1,
λ| ≤ C(1 + |ξ|) and we can assume that Im ω ≥ −c − 1. We have therefore
Z
k
iωt
1 (iω) e dω k
|τ Z(ξ, t)| =
2π P (iξ, iω) Γ
≤ C(1 + |ξ|)k ec(t+1)
and (8.12) follows.
A finite speed of propagation. Suppose that g is supported in BK (0). Then, for a given
N ≥ 0 there exists a constant CN > 0 such that
(8.14)
|ĝ(ζ)| ≤ CN (1 + |ζ|)−N eK|Imζ| ,
∀ ζ ∈ Cd .
Let ζ := ξ + iη and
(8.15)
1
Z(ζ, t) =
(2π)d
Z
Γ
eiωt dω
,
P (iζ, iω)
where Γ is a contour that runs counter-clockwise once around each root of
(8.16)
P (iζ, iω) = 0
and dist (Γ, the set of roots) ≥ 1. Let c∗ := − mink Im ωk , where ω1 , . . . , ωm are all the roots
of (8.15). One can easily see that
(8.17)
|Z(ζ, t)| ≤ Ce(c∗ +1)t
for some C > 0.
To estimate better the right hand side of (8.15)
P we shall need an additional assumption. To
formulate it we write the polynomial P (ξ, ω) = m
k=0 pk (ξ, ω), where pk (·, ·) is a homogeneous
polynomial of degree k.
62
T.KOMOROWSKI
Definition 8.9. We say that equation is strictly hyperbolic if for any ξ 6= 0, ξ ∈ Rd all ω-s
that are the solutions of
pm (ξ, ω) = 0
are distinct and real.
The question arises about the relation between strictly hyperbolic and hyperbolic equations.
Proposition 8.10. Any strictly hyperbolic equation is hyperbolic.
Proof. We start with the following.
Theorem 8.11. Implicit function theorem for analytic functions, cf. [2], [450], p. 430.
Suppose that F is analytic in a neighborhood W (z0 , w0 ) of point (z0 , w0 ), such that F (z0 , w0 ) =
0 and Fw (z0 , w0 ) 6= 0. Then, there exists a neighborhood U (z0 ) and a holomorphic function
f : U (z0 ) → C such that f (z0 ) = w0 , F (z, f (z)) = 0 for any z ∈ U (z0 ). Moreover, if
F (z, w) = 0 and (z, w) ∈ W (z0 , w0 ) then w = f (z).
P
m−k is a certain
Exercise. Puiseux theorem. Suppose that f (µ) = µm + m
k=1 ck (β)µ
(0)
(0)
polynomial that has exactly m distinct roots µ1 , . . . , µm for the coefficients corresponding
to β0 . Assume that c1 , . . . , cm are analytic functions in a neighborhood of β0 . Show that then
there exist a neighborhood U (β0 ) and m analytic functions µ1 (β), . . . , µm (β), β ∈ U (β0 ) such
(0)
that µi (β0 ) = µi , f (µi (β)) = 0, i = 1, . . . , m.
Hint: Show that:
m
X
m
F (µ, β) = µ +
ck (β)µm−k
k=1
satisfies the assumptions of the implicit function theorem 8.11.
Exercise. Puiseux theorem cont’d. Find the formulas for the coefficients of the power
series expansion of µk (β)
Let ξˆ := ξ/|ξ|, ρ := |ξ| > 0. Note that ωk = ρµk , where µk satisfy equation
(8.18)
ˆ µ) +
pm (ξ,
m
X
k=1
1
ˆ µ) = 0.
pm−k (ξ,
(iρ)k
ˆ . . . , µm (ξ)
ˆ be the distinct real roots of pm (ξ,
ˆ µ) = 0. Using Puiseux theorem and
Let µ1 (ξ),
d−1
the fact that S
is compact we can conclude that there exists C1 , ρ0 > 0 such that for ρ > ρ0
we have
C1
(8.19)
Im µk ≥ − , ∀ k = 1, . . . , m.
ρ
From (8.19) we conclude that Im ωk ≥ −C1 for |ξ| > ρ0 . On the other hand we have a general
bound (8.13) that shows ωk ≥ −C1 for |ξ| ≤ ρ0 .
Proposition 8.12. There exists constants C1 , C2 > 0 independent of t and such that
(8.20)
|Z(ξ + iη, t)| ≤ C1 eC2 (1+|η|)t ,
∀ ζ := ξ + iη ∈ Cd .
Proof. We need to estimate c∗ in (8.17). Let ζ̂ := ζ/|ζ|, ρ := |ζ| > 0 and µ := ω/ρ. For a
sufficiently large ρ > 0 the coefficients of equation (8.18) differ from those of
(8.21)
pm (ζ̂, µ) = 0
HARMONIC ANALYSIS
63
by terms of order at most C/ρ for some C > 0. the coefficients of equation (8.21) differ from
those of
(8.22)
pm (ξ/ρ, µ) = 0
by terms of order at most C|η|/ρ for some C > 0. Using now the argument contained in
Proposition 8.10 we conclude that
|Im ωk | ≤ C1 |η|.
Exercise. Complete the argument in the proof of the above proposition.
As a result of (8.20) and (8.14) we obtain
|Z(ζ, t)ĝ(ξ)| ≤ C1 exp C2 (1 + | Imζ|)t,
∀ ζ ∈ Cd .
Paley-Wiener theorem 8.1 implies that u(·, t) given by (8.10) has the support contained in
BC2 t (0).
We shall show now the reduction of the general problem (8.7) to considered above (8.9).
We use the Duhamel’s principle. Find any function u0 (x, t) such that τ k u(x, 0) = fk (x),
k = 0, . . . , m − 1. Let F (x, t) := P (D, τ )u0 (x, t) and v(x, t) := u(x, t) − u0 (x, t). The latter
function satisfies
P (D, τ )v(x, t) = F (x, t) for t ≥ 0,
(8.23)
τ k v(x, 0) = 0
k = 0, . . . , m − 1,
Let U (x, t; s) be the solution of
P (D, τ )U (x, t; s) = 0
for t ≥ s,
τ k U (x, s; s) = 0,
k = 0, . . . , m − 2,
τ m−1 U (x, s; s) = F (x, s).
Rt
Rt
Let v(x, t) := 0 U (x, t; s)ds. Note that τ k v(x, t) = 0 τ k U (x, t; s)ds for k = 0, . . . , m − 1.
Rt
Also τ m v(x, t) = 0 τ m U (x, t; s)ds + F (x, t). Of course τ k v(x, 0) = 0 for k = 0, . . . , m − 1.
Rt
Since Dα v(x, t) = 0 Dα U (x, t; s)ds we obtain
Z t
P (D, τ )v(x, t) =
P (D, τ )U (x, t; s)ds + F (x, t) = F (x, t).
0
9. The spaces of harmonic and analytic functions
References
[1] Pleśniak, W., (1990) Teoria aproksymacji.
[2] Fichtenholz, G. M., (1980) Rachunek r”o”rniczkowy i ca”lkowy t. II.
[3] Varadhan, S. R. S., (2007) Lecture notes in harmonic analysis Available at the web site of the author
http:www.cims.nyu.edu
[4] Zygmund, A., (1968) Fourier Series.