Math 209C Homework 2
Edward Burkard
April 20, 2010
6.
Lp Spaces
6.2. The Dual of Lp .
Exercise 17. With notation as in Theorem 6.14, if µ is semifinite, q < ∞, and Mq (g) < ∞, then {x | |g(x)| > ε}
has finite measure for all ε > 0 and hence Sg is σ-finite.
R Proof. First off we are assuming that g ∈ Lq . Now Mq (g) = sup f g | f ∈ Σ and kf kp = 1 and we are assuming
that M
Sq∞(g) < ∞. Define Gε = {x | |g(x)| > ε}. Clearly Sg is σ-finite if we can show each Gε has finite measure since
Sg = n=1 G n1 .
Assume there is an ε > 0 such that Gε has infinite measure. Since µ is semifinite we get that there is a subset
F ⊂ Gε with 0 < µ(F ) < ∞. Assume that Mq (g) = η.
1
Set c = µ(F )− p and define f ∗ = c sgn(g) χF . Notice:
∗
kf kp =
Z
∗ p
p1
|f |
Z
p
p
|c| |χF |
=
p1
1
= cµ(F ) p = 1
Thus f ∈ Σ since f |F C = 0.
q
By Exercise 14 in Chapter 1 we get for any C > 0 there is a K ⊂ Gε such that C < µ(K) < ∞. Let C = ηε
and rechoose F so that C < µ(F ) < ∞. Then
Z
Z
Z
Z
Z
h i
since F ⊂Gε ∗ ∗ = cεµ(F ) = εµ(F ) q1 > ε η
cε
>
∞ > η ≥ f g ≥ f g = csgn(g)χF sgn(g)|g| = c|g|χF =η
ε
F
F
F
F
which is obviously a contradiction!!!
Therefore Gε has finite measure, and we are done!
n
Exercise 19. Define φn ∈ (l∞ )∗ by φn (f ) =
1X
f (j). Then the sequence {φn } has a weak∗ cluster point φ, and
n j=1
φ is an element of (l∞ )∗ that does not arise from an element of l1 .
Proof. First we notice that
n
φn (f ) =
1X
f (j) ≤ kf kl∞
n j=1
since this is just an average of the values of f on [1, n] ∩ N. But this says that
φn (f )
≤ 1.
kf k∞
But this says that the operator norm of φn is always smaller than one (since the above holds for any f ∈ l∞ ). That
means that φn ∈ B ∗ = {η ∈ (l∞ )∗ | kηk ≤ 1}. Thus since l∞ is a normed vector space we can apply Anaoglu’s
Theorem to say that B ∗ is compact in (l∞ )∗ in the weak∗ topology.
But by Theorem 4.29 we have that since B ∗ is compact {φn } has a cluster point, call it φ.
1
2
P∞
Now suppose that φ did arise from a function g ∈ L1 . Then φ(f ) = j=1 f (j)g(j). Suppose that f (j) = δij where
δij is the Kronecker delta. Then
∞
X
φ(f ) =
f (j)g(j) = g(j).
j=1
n
1X
1
However, φn (f ) =
f (j) = where we are supposing that n ≥ i. Then as n → ∞ we have φn (f ) → φ(f ) = 0.
n j=1
n
This gives that
0 = φ(δij ) =
∞
X
f (j)g(j) = g(i)
j=1
so that g(i) = 0. But this holds for all i. Thus we g ≡ 0 so that φ ≡ 0.
This is a contradiction since if we let h(x) := 1 then
φn (h) = 1
for all n which means that as n → ∞ φn (h) → φ(h) = 1 which contradicts the fact that φ ≡ 0.
∴ φ did not arise from an element of l1 .
Exercise 20. Suppose supn kfn kp < ∞ and fn → f a.e.
a. If 1 < p < ∞, then Rfn → f weakly in Lp . (Given g ∈ Lq , where q is conjugate to p, andRε > 0, there exist
(i) δ > 0 such that E |g|q < ε whenever µ(E) < δ, (ii) A ⊂ X such that µ(A) < ∞ and X\A |g|q < ε, and
(iii) B ⊂ A such that µ(A \ B) < δ and fn → f uniformly on B.)
b. The result of (a) if false in general for p = 1. (Find counterexamples in L1 (R, m) and l1 .) It is, however,
true for p = ∞ if µ is σ-finite and weak convergence is replaced by weak∗ convergence.
Proof.
a. (i) This is just Corollary 3.6 where f = |g|q
q
Convergence Theorem,
Rfor qη ≥ 0 the sets Gη = {x | |g(x)| ≥ η}. 0 Then by the
R Monotone
R(ii) Define,
q
q
|g|
→
|g|
as
η
&
0.
So
we
can
find
an
η,
call
it
η
such
that
|g|
<
ε.
Let A = Gη0 . Then
X\Gη0
Gη
µ(A) < ∞ since:
Z
Z
∞>
|g|q ≥ η 0 µ(A).
|g|q ≥
A
(iii) This is just Egoroff’s Theorem.
Since supn kfn kp < ∞, suppose supn kfn kp = C. Then using Fatou’s Lemma on {|fn |p } we get:
Z
Z
Z
p
p
|f | = lim inf |fn | ≤ lim inf |fn |p ≤ C p
so that kf kp ≤ C.
Now using (i), (ii), and (iii) above we write (for g ∈ Lq ):
Z
Z
Z
Z
(fn − f )g =
(fn − f )g +
(fn − f )g +
B
A\B
(fn − f )g
X\A
For the first integral, since fn → f uniformly on B (by (iii)):
Z
q1
Z
p1 Z
Hölders
q
p
(fn − f )g ≤
|g|
≤ µ(B) sup |fn (x) − f (x)| kgkq → 0
|fn − f |
x∈B
B
B
B
as n → ∞ since fn → f a.e.
Now for the second integral, since µ(A \ B) < δ (by (i) and (iii)):
Z
! Z
!
H¨’older’s Z
Minkowski’s
p
q
(fn − f )g ≤
|fn − f |
|g|
≤
(kfn kp + kf kp ) ε < 2Cε.
A\B
A\B
A\B
3
Similarly for the third integral, we get:
Z
(fn − f )g < 2Cε
X\A
Therefore we have:
Z
lim (fn − f )g < 4Cε.
n→∞
Thus since ε was arbitraty we have that
Z
(fn − f )g → 0.
So since g was arbitrary we have that fn → f weakly.
b.
R
R
Ctex for L1 (R, m)) Let fn = χ[n,n+1] . We know that fn → f = 0. Suppose g = 1. Then fn g = 1 for all n, but f g = 0.
Thus fn 6→ f weakly.
∞
Ctex for l1 ) Let {aP
countable subset. Define fn = χ{an } . Then fn → f = 0 a.e. Suppose g = 1.
n }n=1 N be someP
Then
fn g = 1 while
f g = 0. Thus fn 6→ f weakly.
Now to prove the part that needs proving:
Since µ is σ-finite, Theorem 6.15 shows that L1 is dual to L∞ . So we want to prove:
R
Claim. If supn kfn k∞ < ∞ and fn → f a.e., then (fn − f )g → 0 for all g ∈ L1 (this is weak∗ convergence).
proof . The proof is actually exactly the same as in part a!!!
Exercise 21. If 1 < p < ∞, fn → f weakly in lp (A) iff supn kfn kp < ∞ and fn → f pointwise.
Proof.
(=⇒) Suppose fn → f weakly in lp (A). Then:
X
(fn (a) − f (a))g(a) → 0
a∈A
for all g ∈ lq (A) = (lp (A))∗ (since ν (the counting measure) is clearly semifinite) where q is the conjugate
exponent to p.
Let g = χ{a} for a ∈ A. Then
X
(fn (a) − f (a))g(a) = (fn (a) − f (a)) → 0
a∈A
so that fn (a) → f (a), hence we have pointwise convergence.
R
R
Now suppose
g ∈ lq (A). Since we have weak convergence, we know that fn g → f g < ∞. This gives
R
that supn | fn g| < ∞. Since lq (A) is a Banach space and the fact that lq (A) is isometrically isomorphic to
(lq (A))∗ (by Theorem 6.15) we get that:
sup kfn kp < ∞
n
which finishes the proof.
(⇐=) This is precisely part (a) of the previous exercise!!!
Exercise 22. Let X = [0, 1], with Lebesgue measure.
a. Let fn (x) = cos 2πnx. Then fn → 0 weakly in L2 (see Exercise 63 in §5.5), but fn 6→ 0 a.e. or in measure.
b. Let fn (x) = nχ(0, 1 ) . Then fn → 0 a.e and in measure, but fn 6→ 0 weakly in Lp for any p.
n
Proof.
4
a. Since L2 is an infinite-dimensional Hilbert space by Exercise 63 in §5.5 any orthonormal sequence in L2
converges weakly to 0. Since we have that:
Z 1 √
√
( 2 cos(2πmx))( 2 cos(2πnx))dx = δmn
0
the sequence {fn } is orthonormal in L2 it converges weakly to 0.
i
Claim. Given any ε ∈ 0, √12 and n ∈ N there is a set Gn,ε on which |fn − 0| = |fn | ≥ ε with µ(Gn,ε ) ≥ 1.
proof . Let ε = √12 .
Staying in the interval 0, n1 (one period of fn ) we get that:
1 3 5 7
1
|fn (x)| = | cos(2πnx)| = √ ⇐⇒ x =
,
,
,
8n
8n 8n 8n
2
Thus:
1 [ 3 5 [ 7 1
1
If x ∈ 0,
,
,
, then |fn (x)| ≥ √ .
8n
8n 8n
8n n
2
1 S 3 5 S 7 1
1
Define Sn, √ = 0, 8n
8n , 8n
8n , n , so that:
2
[
1
j
√
1
|fn (x)| ≥
⇐⇒ x ∈
Sn, √ +
.
2
n
2
j∈Z
T
Sn−1 1
Now notice that µ(Sn, √1 ) = 2n
. Define Gn, √1 = j=0 Sn, √1 + nj . Then since Sn, √1 + ni
Sn, √1 + nj =
2
2
2
2
2
∅ iff i 6= j we have:


n−1
n−1
n−1
[
X X 1
j
j
1
=
Sn, √1 +
µ Sn, √1 +
= .
µ(Gn, √1 ) = µ 
=
2
2
2
n
n
2n
2
j=0
j=0
j=0
Since we clearly have that for any ε <
√1 ,
2
Sn, √1 ⊂ Sn,ε the claim is proven.
2
This claim clearly implies that fn 6→ f in measure.
Now define gn (x) = e2πinx on X. Then fn (x) = Re(gn (x)). Now notice that for any p ∈ I ∩ X (=
1
(R \ Q) ∩ X) we have that the image of p in S 1 is dense (i.e. the set {gn (p)}∞
n=1 is dense in § ). Thus the
∞
set {fn (p)}n=1 is dense in [−1, 1]. Therefore it is impossible for fn (p) to converge to 0, hence fn does not
coverge pointwise to 0.
b. First we prove pointwise convergence:
Let ε > 0 and x ∈ X be given. Then |fn (x)| =
6 0 iff x ∈ 0, n1 . Let N = x1 (the smallest integer such that
N ≥ x1 ). Then for all n > N we have |fn (x)| < ε.
∴ fn → f pointwise. (Hence a.e.)
Next we show convergence in measure:
Given ε > 0 consider the set {x | |fn (x) − 0| ≥ ε} = {x | |fn (x)| ≥ ε}. Notice that |fn (x)| ≥ ε iff n ≥ ε
and x ∈ 0, n1 . So suppose that n ≥ ε. Then
µ({x | |fn (x)| ≥ ε}) =
1
.
n
But as n → ∞ we have µ({x | |fn (x)| ≥ ε}) → 0.
∴ fn → 0 in measure.
Lastly we show that fn 6→ 0 weakly in any Lp :
Cleary the function g = 1 is in Lp (X) for all p. However using the definition of weak convergence, since:
Z 1
Z 1
Z 1
fn g =
fn =
nχ(0, 1 ) = 1
0
0
0
n
5
we see that fn does not converge weakly to 0 in any Lp .
Kelliher Problems
Exercise 1. In the context of Proposition 6.13 of Folland, suppose that X is not semifinite. Show that there exists
g ∈ L∞ (X) such that kφg k < kgk∞ .
Proof. Let (X, M , µ) be our non semifinite measure space. Then by the definition of semifinite there is a subset A
of X with µ(A) = ∞ that does not have a measureable subset with nonzero finite measure.
Consider the function g = χA . Clearly g ∈ L∞ as kgk∞ = 1 (which is also clear). Now let f be any L1 on X. Then:
Z
Z
Z
f g ≤ |f g| =
|f |
A
Now on A we can approximate |f | from below by simple functions:
fn =
kn
X
an,i χEn,i
i=1
where an,i ∈ R and En,i ⊂ A measurable. Then since |fn | ≤ |f | for all n, we can dominate these bitches (i.e. use the
DCT) to get:
!
Z
Z
kn
X
|an,i |µ(En,i ) .
|fn | = lim
|f | = lim
A
n→∞
n→∞
A
i=1
Now since each En,i ⊂ A and measurable
we Rget that µ(En,i ) = 0 or ∞ by our assumption on A. However if any
R
En,i has infinite measure we get that A |f | ≥ En,i |an,i | = ∞. But this contradicts f ∈ L1 . Thus we must have that
µ(En,i ) = 0 for all n and i. Then:
Z
|f | = 0
A
R R ∴ f g = A f ≤ 0 ∀ f ∈ L1 (X)
R
∴ f g = 0 ∀ f ∈ L1 (X)
In particular we get that:
Z
kφg k = sup f g | kf k1 = 1 = 0
∴ kφg k < kgk∞ since 0 < 1.