MATH 202B - Problem Set 3
Walid Krichene (23265217)
February 13, 2013
(3.1)
The parallelogram law for L2 norms states that for all f, g ∈ L2 ,
kf + gk22 + kf − gk22 = 2kf k22 + 2kgk22
which can be viewed as a refinement of the triangle inequality kf + gk2 ≤ kf k2 + kgk2 by writing
kf + gk22 ≤ 2kf k22 + 2kgk22 − kf − gk22
= (kf k2 + kgk2 )2 + (kf k2 − kgk2 )2 − kf − gk22
thus if kf k2 = kgk2 and kf + gk2 = kf k2 + kgk2 , then f = g a.e.
(3.1.a) show that more generally, if kf + gk2 = kf k2 + kgk2 , then there exists c ≥ 0 such one function is c
times the other almost everywhere.
answer First, observe that the case kgk2 = 0 is trivial since it implies g = 0 almost everywhere and the
result is true using the constant c = 0.
Now assume kgk2 > 0. First, we have
Z
kf + gk22 =
(f + g)2 dµ
ZX
Z
Z
=
f2 +
g2 + 2
f gdµ
X
X
Z X
= kf k22 + kgk22 + 2
f gdµ
X
Z
2
= (kf k2 + kgk2 ) + 2
f gdµ − kf k2 kgk2
X
therefore we have kf + gk2 = kf k2 + kgk2 if and only if
Now consider the function
φ:R→R
R
X
f gdµ = kf k2 kgk2 .
t 7→ φ(t) = kf + tgk22
. We have similarly
φ(t) = kf k22 + t2 kgk22 + 2t
Z
f gdµ
X
which is a polynomial of degree 2 in t. The discriminant is
∆=
Z
2
2
f gdµ − 4kgk22 kf k22 = 0
X
1
therefore φ has a real root, i.e. there exists t0 ∈ R such that φ(t) = kf + t0 gk22 = 0, i.e.
Z
(f (x) + t0 g(x))2 dµ(x) = 0
X
therefore we have for almost every x, f (x) = cg(x) where c = −t0 . Finally, we can rewrite the condition
kf + gk2 = kf k2 + kgk2 as
k(1 + c)gk2 = (|c| + 1)kgk2
i.e.
|1 + c| = |c| + 1
(since kgk2 > 0). This requires in particular that c ≥ 0.
(3.1.b) Counter example: for p ∈ {1, ∞}, find functions f, g such that kf + gkp = kf k1 + kgkp , yet neither
is a constant multiple of the other almost everywhere.
answer
Consider the measure space (R, B, µ).
• Counter example for p = 1: consider the functions
f = 1[0,1]
g = 1[1,2]
We have
kf k1 = kgk1 = 1
kf + gk1 = 2
therefore we have kf + gk1 = kf k1 + kgk1 yet the functions are not proportional almost everywhere.
• Counter example for p = ∞: consider the functions
f = 1[0,1]
g = 1[0,2]
We have
kf k∞ = kgk∞ = 1
kf + gk∞ = 2
therefore we have kf + gk∞ = kf k∞ + kgk∞ yet the functions are not proportional almost everywhere.
2
(3.2) All functions are assumed real valued. Clarkson’s inequalities extend the previous discussion to other
exponents, in a slightly weaker form.
(3.2.a)
Let p ∈ [2, ∞). For any f, g ∈ Lp , we have
kf + gkpp + kf − gkpp ≤ 2p−1 (kf kpp + kgkpp )
proof
we first show that for reals a ≥ b ≥ 0, we have
(a + b)p + (a − b)p ≤ 2p−1 (ap + bp )
(1)
consider the function
φ : [0, 1] → R
t 7→ φ(t) = (a + tb)p + (a − tb)p − 2p−1 (ap + (tb)p )
showing the desired inequality is equivalent to showing that φ(1) ≤ 0. And since we have
φ(0) = (2 − 2p−1 )ap ≤ 0
it suffices to show that for all t ∈ (0, 1], φ0 (t) ≤ 0. We have
φ0 (t) = pb(a + tb)p−1 − pb(a − tb)p−1 − 2p−1 pb(tb)p−1
a
a
= pb(tb)p−1 ( + 1)p−1 − ( − 1)p−1 − 2p−1
tb
tb
= pb(tb)p−1 ψ(t)
a
a
where ψ(t) = ( tb
+ 1)p−1 − ( tb
− 1)p−1 − 2p−1 . We have for t ∈ (0, 1]
a
a
1
p−2
p−2
(p
−
1)
(
+
1)
−
(
−
1)
bt2
tb
tb
≥ 1 and ab ≥ 1, thus
a
−1≥0
tb
ψ 0 (t) = −
but since t ≤ 1 and a ≥ b, we have
and it follows that
and using the fact that u 7→ up−2
1
t
a
a
+1≥
−1≥0
tb
tb
is increasing on [0, ∞) (since p − 2 ≥ 0), we have for all t ∈ (0, 1]
ψ 0 (t) ≤ 0
and since limt→0 ψ(t) = 0, we have for all t ∈ (0, 1], ψ(t) ≤ 0. Finally, from the expression of φ0 (t), we
conclude that for all t ∈ (0, 1], φ0 (t) ≤ 0, therefore φ is non-increasing on (0, 1] and we obtain the desired
result.
Next, it follows from inequality (1) that for all real numbers a, b,
|a + b|p + |a − b|p ≤ 2p−1 (|a|p + |b|p )
(it suffices to apply the previous inequaity)
Now consider the functions f, g. We have
Z
kf + gkpp + kf − gkpp =
|f (x) + g(x)|p + |f (x) − g(x)|p dµ(x)
ZX
≤
2p−1 (|f (x)|p + |g(x)|p )dµ(x)
X
p−1
=2
(kf kpp + kgkpp )
which proves the desired result.
3
(2)
using inequality (2)
Show that as a consequence, if kf kp = kgkp = 1, then kf + gkp ≤ 2(1 − 2−p kf − gkpp )1/p
(3.2.b)
proof
If kf kp = kgkp = 1, then inequality (2) becomes
kf + gkpp + kf − gkpp ≤ 2p
i.e.
kf + gkp ≤ (2p − kf − gkpp )1/p
= 2(1 − 2−p kf − gkpp )1/p
(3.2.c) Show that if f, g ∈ Lp satisfy kf + gkp = kf kp + kgkp , then there exists c such that one function
is c times the other almost everywhere.
proof First, if kf kp = 0, then f is zero almost everywhere, and the result holds with the constant c = 0.
Similarly, the result holds wen kgkp = 0.
Now assume without loss of generality, that kf kp ≥ kgkp > 0. Let c = kgkp /kf kp . Then we have
kcf kp = kgkp
Now observe that if kf + gkp = kf kp + kgkp , then we have1
kcf + gkp = ckf kp + kgkp = 2kgkp
now consider the functions f˜ = f /kf kp and g̃ = −g/kgkp . We have
kf˜ + g̃kp = kcf /kgkp + g/kgkp kp
= kcf + gkp /kgkp
=2
And since f˜ and g̃ have unit norm, we can apply the result of (2.b), and we obtain
kf˜ − g̃kp ≤ 2(1 − 2− pkf˜ + g̃kpp )1/p
= 2(1 − 2− p2p )1/p
=0
therefore kf˜ − g̃kp = 0, and f˜ = g̃ almost everywhere. In other words, g =
1 indeed,
kgkp
kf kp f
almost everywhere.
we have
kcf + gkp = kf + g + (1 − c)f kp
≥ kf + gkp − k(1 − c)f kp
by the triangle inequality
= kf kp + kgkp − (1 − c)kf kp
since c ∈ [0, 1]
= ckf kp + kgkp
the reverse inequality kcf + gkp ≤ ckf kp + kgkp follows simply from the triangle inequality.
4
(3.3) Let p, q ∈ (1, ∞) be a pair of conjugate exponents. Let f ∈ Lp and g ∈ Lq be C-valued functions
with strictly positive norms. Suppose that
Z
|
f ḡdµ| = kf kp kgkq
X
(3.3.a) Suppose that f, g are real-valued, non-negative, then there exists c > 0 such that g = cf p/q almost
everywhere.
proof
First, we can rewrite the equality as
Z
X
i.e.
g
f
dµ = 1
kf kp kgkq
Z
f˜g̃dµ = 1
X
where
f
kf kp
g
g̃ =
kgkq
f˜ =
Now define the function
φ:X→C
1
1
x 7→ φ(x) = f˜(x)g̃(x) − f˜(x)p − g̃(x)q
p
q
we have by Holder’s inequality, for all x ∈ X, φ(x) ≤ 0. We also have
Z
Z
1 1
φdµ =
f˜g̃dµ − −
p q
X
X
1 1
=1− −
p q
=0
therefore we must have φ = 0 almost everywhere, i.e. for almost every x,
1
1
f˜(x)g̃(x) = f˜(x)p + g̃(x)q
p
q
this corresponds to having equality in Young’s inequality for the exponential function, applied between the
points p ln f˜(x) and q ln g̃(x), with coefficients 1/p and 1/q: indeed, the above equation can be rewritten
1
˜
1
e p p ln(f (x))+ q ln g̃(x) =
1 p ln f˜(x) 1 q ln g̃(x)
e
+ e
p
q
by strict convexity of the exponential, equality only occurs if 1/p = 0, or 1/p = 1, or p ln f˜(x) = q ln g̃(x).
Since the first two cases cannot occur, we have for almost every x
p ln f˜(x) = q ln g̃(x)
i.e.
g̃(x) = f˜(x)p/q
or
g(x) =
kgkq
p/q
kf kp
5
f (x)p/q
(3.3.b) In the general case, show that there exists c ∈ C such that g = cf p/q , i.e. there exist r ≥ 0 and
θ ∈ [0, 2π] such that for almost every x,
|g(x)| = r|f (x)|p/q
arg(g(x)) = θ + arg(f (x))
mod (2π)
First, we observe that
Z
Z
|
f ḡdµ| ≤
|f ||g|dµ
X
X
≤ kf kp kgkq dµ
But since we have |
R
X
by Holder’s inequality
f ḡdµ| = kf kp kgkq , we must have equality in the previous inequalities. In particular,
Z
|f ||g|dµ =≤ kf kp kgkq dµ
X
applying (3.b) to the real-valued functions x 7→ |g(x)| and x 7→ |f (x)|, there exists r ≥ 0 such that |g| =
r|f |p/q .
R
we need to use the fact that equality holds in | X f ḡ| ≤
R Now to prove the result
R for the arguments,
|f ||g|dµ. Let us write X f ḡdµ = αeiφ , where α ≥ 0, and γ ∈ [0, 2π]. Then we have
X
Z
|
f ḡdµ| = α = <(α)
X
Z
= <( f ḡdµe−iγ )
Z X
=
<(f ḡe−iγ )dµ
X
now let
φ:X→R
x 7→ φ(x) = <(f (x)ḡ(x)e−iγ ) − |f (x)||g(x)|
we have for all x, φ(x) ≤ 0 (the real part of a complex number is no greater than its modulus). But we also
have
Z
Z
Z
φdµ =
<(f ḡe−iγ )dµ −
|f ||g|
X
X
ZX
Z
=|
f ḡdµ| −
|f ||g|
X
X
=0
therefore we have φ(x) = 0 almost everywhere, i.e. for almost every x
<(f (x)ḡ(x)e−iγ ) = |f (x)||g(x)|
i.e. f (x)ḡ(x)e−iγ is real non-negative (since equal to its modulus). Therefore for almost every x,
arg(f (x)ḡ(x)e−iγ ) = 0 mod (2π)
i.e.
arg(f (x)) − arg(g(x)) − γ = 0
this concludes the proof.
6
mod (2π)
(3.4)
Consider the measure space (R, A, µ), where
• R = ×ni=1 [ai , bi ]
• A is the set of all Borel subsets of R
• µ is the n dimensional Lebesgue measure.
A function f : R → C is said to be infinitely differentiable if f is continuous, f is infinitely differentiable
on the int(R), and each derivative extends continuously to the boundary of R2 .
Show that for all p ∈ [1, ∞), C ∞ (R) is dense in Lp (R) (in the k.kp sense).
proof Let f ∈ Lp (R), and let > 0. By definition, we can approach f from below by a non-decreasing
sequence of simple functions (sn ). Then we have sequence |sn −f |p converges pointwise to 0, and is dominated
by 2|f |p , which is integrable, therefore by the DCT
Z
lim
|sn − f |p dµ = 0
n
X
i.e. (ksn − f kp )n converges to 0. Therefore there exists a simple function s such that kf − skp ≤ /3. Next,
we cn approximate the simple function s by a continuous function s̄: let the canonical expansion of s be
given by
n
X
s=
ci 1Ei
i=1
where for all i, ci ∈ R and Ei is a Borel set of R (the Ei s are disjoint). Then we can approximate, for all
i, ci 1Ei by a continuous function s̄i (using for P
example Urysohn’s Lemma). Let s̄i be a continuous function
n
such that kci 1Ei − s̄i kp ≤ /(2n), and let s̄ = i=1 s̄i . Then we have
ks − s̄kp = k
≤
n
X
ci 1Ei − s̄i kp
i=1
n
X
kci 1Ei − s̄i kp
i=1
≤ /3
Finally, using the Stone Weirstrass theorem on the compact space R, we have that the Algebra of
complex valued polynomial functions P (R, C) is dense in the space of continuous functions (wrt the uniform
convergence metric kgk∞ = supx∈R |g(x)|).
Indeed, this algebra contains constants, separates points, and is closed under multiplication, addition,
scalar multiplication and conjugation.
Therefore there exists a polynomial p ∈ P (R) such that supx∈R |p(x) − s̄(x)| ≤ 3µ(R)
1/p , and we have
1/p
|p − s̄| dµ
Z
p
kp − s̄kp =
X
3
=
3
Z
≤
X
1/p
1
dµ
µ(R)
combining the three bounds, we obtain
kf − pkp ≤ kf − skp + ks − s̄kp + ks̄ − pkp ≤ which proves that P (R, C) is dense in Lp (R). And since the space of polynomials on R is a subset of C ∞ (R),
this proves the desired result.
7
(3.5)
Let p ∈ [1, ∞). Suppose that
• fn ∈ Lp
• fn → f almost everywhere
• there exists M < ∞ such that for all n, kfn kp ≤ M < ∞
Show that if (kfn kp )n → kf kp , then (kfn − f kp )n → 0.
proof
First, we show that for all A ∈ A,
Z
|fn |p dµ →
A
Z
|f |p dµ
A
using Fatou’s lemma on A and X \ A, we have
Z
Z
Z
p
p
|f | =
lim inf |fn | ≤ lim inf
|fn |p
n
n
A
A
A
Z
Z
Z
|fn |p
|f |p =
lim inf |fn |p ≤ lim inf
X\A
X\A
n
n
X\A
next, we rewrite the second inequality (using the fact that lim inf n −un = − lim supn un )
Z
Z
−
|f |p ≥ lim sup −
|fn |p
n
X\A
X\A
|f |p = limn X |fn |p = lim supn X |fn |p and is finite, we have
Z
Z
p
p
|f | dµ =
|f | dµ −
|f |p dµ
A
X
X\A
Z
Z
≥ lim sup
|fn |p + lim sup −
|fn |p
n
n
X
X\A
Z
Z
≥ lim sup( |fn |p −
|fn |p )
using sup un = sup vn ≥ sup(un + vn )
n
X
X\A
Z
= lim sup
|fn |p
and since
Z
R
R
R
X
n
A
Therefore we have
Z
n
Z
|fn |p ≤
lim sup
A
|f |p dµ ≤ lim inf
n
A
Z
|fn |p
A
but the last term is a lower bound on the first term, (the lim inf is always no greater than the lim sup),
therefore we have equality, and the claim is proved.
Proof of main result: let > 0 be given. Let A = {x ∈ X : |f (x)| > 0}. A is measurable. First, we
decompose the integral into
Z
Z
Z
|fn − f |p dµ =
|fn − f |p dµ +
|fn |p dµ
X
A
X\A
8
using the fact that for all x ∈ X \ A, f (x) = 0. Then by the previous claim,
R
|f |p = 0. Therefore there exists N1 such that
X\A
Z
∀n ≥ N1 ,
|fn |p ≤ /4
R
X\A
|fn |p dµ converges to
(3)
X\A
now define a new measure ν on A ∩ A as follows:
Z
|f |p dµ
∀E ∈ A ∩ A, ν(E) =
E
R
we have ν(A) = A |f |p dµ ≤ M < ∞. Therefore (A, A ∩ A, ν) is a finite measure space and we can apply
Egoroff’s theorem: we have (fn /f ) converges to 1 almost everywhere on A (note that we had to restrict this
discussion to the set A so that fn /f be defined). Thus by Egoroff’s theorem, there exists a measurable set
F ∈ A ∩ A such that
• ν(F ) ≤ /8 (we assume ν(A) > 0, otherwise f is 0 almost everywhere, in which case the result is
trivial)
• (fn /f ) converges uniformly to 1 on A \ F .
R
R
Now we seek to bound the terms F |fn − f |p dµ and A\F |fn − f |p dµ. First, we have
Z
Z
Z
p
p
|fn − f | dµ ≤
|fn | dµ +
|f |p dµ
F
and since
p
R
F
|fn | dµ converges to
F
F
p
R
|f | dµ, there exists N2 such that
Z
Z
p
∀n ≥ N2 ,
|fn − f | dµ ≤ 2
|f |p dµ + /4
F
F
F
= 2ν(F ) + /4
≤ /4 + /4
(4)
By uniform convergence, there exists N2 such that
∀n ≥ N3 , ∀x ∈ A \ E, |fn (x)/f (x) − 1| ≤
4ν(A)
(we can assume ν(A) > 0 because otherwise f is 0 a.e., in which case the result is trivial) thus
Z
Z
∀n ≥ N3 ,
|fn − f |p dµ =
|fn /f − 1|p |f |p dµ
A\F
A\F
Z
≤
dν
A\F 4ν(A)
≤
ν(A \ F )
4ν(A)
≤ /4
Finally, combining the above bounds (3) (4) (5), we have for all n ≥ max(N1 , N2 , N3 )
Z
p
Z
|fn − f | dµ =
X
Z
p
|fn − f | dµ +
F
this completes the proof.
9
Z
|fn − f | dµ +
A\F
≤
p
|fn |dµ
X\A
(5)
Let 1 ≤ p ≤ r ≤ q ≤ ∞. Let θ ∈ [0, 1] be the unique number such that
(3.6)
r−1 = θp−1 + (1 − θ)q −1
Show that for any measurable function f ,
kf kr ≤ kf kθp kf k1−θ
q
conclude that Lp ∩ Lq ⊂ Lr .
proof
We consider the following cases:
1. p = r = q = ∞. Then the result becomes kf k∞ ≤ kf kθ∞ kf k1−θ
∞ and this trivially holds (with equality)
for all θ ∈ [0, 1]
2. p < r = q = ∞. Then we have 0 = θp−1 + (1 − θ)0, thus we must have θ = 0. The result becomes
kf k∞ ≤ θkf k∞ which trivially holds with equality.
3. p ≤ r < q = ∞. Then we have r−1 = θp−1 , i.e. θ = p/r. Let M = kf k∞ be the essential supremum
(finite since f ∈ L∞ ). Then the set B = {x : |f (x)| > M } has measure 0. Then we have for all
x ∈ X \ B, |f (x)|/M ≤ 1, thus (|f (x)|/M )r−p ≤ 1 (since r − p ≥ 0). Integrating, we have
Z
Z
|f |r dµ ≤
|f |p dµM r−p
X
X
Z
Z
r
1/r
i.e. ( |f | dµ)
≤ (( |f |p dµ)1/p )p/r M (r−p)/r
X
X
(r−p)/r
i.e. kf kr ≤ (kf kp )p/r kf k∞
and we conclude observing that p/r = θ and (r − p)/r = 1 − θ.
4. p ≤ r ≤ q < ∞. Then let
a=
b=
p
rθ
q
r(1 − θ)
a and b are conjugate, since 1/a + a/b = r(θp−1 + (1 − θ)q −1 ) = 1. Then apply Holder’s inequality:
Z
Z
|f |r dµ =
|f |r dµ
X
X
Z
=
|f |p/a |f |q/b dµ
using p/a + q/b = rθ + r(1 − θ) = r
X
≤ k|f |p/a ka k|f |q/b kb
Z
Z
p/a a
1/a
= ( (|f | ) dµ) ( (|f |q/b )b dµ)1/b
X
ZX
Z
p
rθ/p
= ( |f | dµ)
( |f |q dµ)(1−θ)r/q
X
using Holder’s inequality
X
raising to the power 1/r, we obtain the desired result.
Conclusion: using the bound above, if f ∈ Lp ∩ Lq , then kf kr is finite since it is bounded by the product
of two bounded terms. Thus f ∈ Lr .
10