Download - Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

Transcript
Page 1: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

Real Analysis, 2nd Edition, G.B.Folland

Chapter 6 Lp Spaces

Yung-Hsiang Huang∗

2018/04/11

6.1 Basic Theory of Lp Spaces

1. When does equality hold in Minkowski’s inequality?

Proof. (a) For p = 1, since we know the triangle inequality |(f + g)(x)| ≤ |f(x)| + |g(x)|

is always true, to characteriz the case when the equality holds in Minkowski’s inequality is

equivalent to seeking the condition for the equality |(f + g)(x)| = |f(x)|+ |g(x)| for all x, that

is f(x)g(x) ≥ 0 for a.e. x. (We consider the complex-valued functions here.)

(b) For 1 < p < ∞, in its proof, we know the equality holds in Minkowski inequality if and

only if the ones in triangle and Holder’s inequalities hold, that is, if and only if f(x)g(x) ≥ 0

for a.e. x, and there are constants α, β ≥ 0 such that |f(x)|p = α|f(x) + g(x)|p and |g(x)|p =

β|f(x) + g(x)|p for a.e. x. Combining them, we know the equality holds in Minkowski if and

only if f(x) = γg(x) for a.e. x, for some γ ≥ 0.

(c) For p =∞. Define the set Af = {{xn}} where {xn} is a sequence in X with |f(xn)| → ‖f‖∞.

We claim that ‖f + g‖∞ = ‖f‖∞ + ‖g‖∞ iff either one of the functions is 0 a.e., or Af ∩ Agis nonempty and for some {xn} ∈ Af ∩ Ag, where f(xn) → a and g(xn) → b, there is a λ > 0

such that b = λa.

(⇐) If one of them is 0 a.e., then it’s done. If not, then

‖f‖∞ + ‖g‖∞ = limn|f(xn)|+ lim

n|g(xn)| = |a|+ |b| = |a+ b| = lim

n|f(xn) + g(xn)| ≤ ‖f + g‖∞.

Combining with the Minkowski inequality, we have the desired equality.

∗Department of Math., National Taiwan University. Email: [email protected]

1

Page 2: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

(⇒) We assume both functions are not zero, so ‖f‖∞ and ‖g‖∞ > 0. Since there exists xn

such that |(f + g)(xn)| → ‖f + g‖∞ = ‖f‖∞ + ‖g‖∞. By a contradiction argument, we see

|f(xn)| → ‖f‖∞ and |g(xn)| → ‖g‖∞, that is, {xn} ∈ Af ∩ Ag. If f(xn) → ‖f‖ > 0 and

g(xn) → −‖g‖ < 0, then |f(xn) + g(xn)| → |‖f‖∞ − ‖g‖∞| < ‖f‖∞ + ‖g‖∞ = ‖f + g‖∞, a

contradiction. Similarly, it’s impossible that f(xn) → −‖f‖ and g(xn) → ‖g‖. Hence we get

the desired result with λ = ‖g‖∞‖f‖∞ .

2. Prove Theorem 6.8.

Proof. (a),(b) and (c)(⇐) are obvious, we also omit (d) since it’s proved in many textbooks.

(c)(⇒): Given m ∈ N there exists Mm ∈ N such that for all n ≥ Mm, ‖fn − f‖∞ < 1m

. Then

for such (m,n), there is Xmn ∈M such that µ(Xm

n ) = 0 and |fn − f | < 1m

on X \Xmn . Take

Ec = ∪∞m=1 ∪∞n=MmXmn

which has zero measure and fn → f uniformly on E.

(e) is proved by partition the given range [−‖f‖∞, ‖f‖∞) equally into 2k disjoint closed-open

subintervals, and for each k ∈ N, consider the proper simple function fk corresponding to the

sets {{x : ‖f‖∞

j

2k≤ |f(x)| < ‖f‖∞

j + 1

2k}, {x : ‖f‖∞ = |f(x)|}

}2k−1

j=−2k.

3. If 1 ≤ p < r ≤ ∞, Lp ∩ Lr is a Banach space with norm ‖f‖ = ‖f‖p + ‖f‖r, and if

p < q < r, the inclusion map Lp ∩ Lr → Lq is continuous.

Proof. ‖f‖q ≤ ‖f‖θp‖f‖1−θr ≤ ‖f‖θ‖f‖1−θ, where θ is defined by q−1 = θp−1 + (1− θ)r−1.

4. If 1 ≤ p < r ≤ ∞, Lp+Lr is a Banach space with norm ‖f‖ = inf{‖g‖p+‖h‖r : f = g+h},

and if p < q < r, the inclusion map Lq → Lp + Lr is continuous.

Remark 1. (2018.04.10) See Sergey Astashkin and Lech Maligranda’s paper (arXiv:1804.03469),

”Lp+Lq and Lp∩Lq are not isomorphic for all 1 ≤ p, q ≤ ∞, p 6= q” and the references therein.

Proof. Since p < q < r,

‖f‖ ≤ ‖fχ{|f |>1}‖p + ‖fχ{|f |≤1}‖r ≤ ‖fχ{|f |>1}‖q + ‖fχ{|f |≤1}‖q = ‖f‖q.

2

Page 3: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

There is a related set equality due to Alvarez [1]:

Theorem 2. Let Lpq := Lp + Lq. Then (a) Lpq + Lrs = Lmin(p,r)max(q,s) (b) Lpq · Lrs = Luv , where

u−1 = p−1 + r−1; v−1 = q−1 + s−1 (c) Lpq(Lp0q0

+ Lp1q1 ) = Lpq · Lp0q0 + Lpq · Lp1q1 .

We prove this theorem in additional exercise 6.1.3.

Remark 3. Lpq is a special case of Orlicz spaces. Its duality result to Lp′∩Lq′ and generalization

are given in Stein-Shakarchi [18, Exercise 1.24, 1.26 and Problem 1.5].

5. Suppose 0 < p < q < ∞. Then Lp 6⊂ Lq iff X contains sets of arbitrarily small positive

measure, and Lq 6⊂ Lp iff X contains sets of arbitrarily large finite measure. What about the

case q =∞

Proof. Since the first and second assertions are dual to each other, we only prove the first one.

(⇐) By assumption, there is a sequence of sets {Fn} with 0 < µ(Fn+1) < 3−1µ(Fn) for all

n ∈ N and µ(F1) < 1/3. Let En = Fn \ ∪∞n+1Fj, then we obtain a sequence of disjoint sets

En with 0 < 2−1µ(Fn) < µ(En) < 3−n. Finally we consider the function f =∑anχEn with

an = µ(En)−1/q which is in Lp but not in Lq.

(⇒) Assume there is c > 0 such that any set with positive measure has measure ≥ c. Take

f ∈ Lp and consider the sets An = {x : n ≤ |f(x)| < n + 1}. By Chebyshev’s inequality, for

n ≥ 1, either µ(An) ≥ c or µ(An) = 0. Since f ∈ Lp, only finitely many n such that µ(An) ≥ c.

So f ∈ Lq and thus Lp ⊂ Lq, which is a contradiction.

For the case q =∞, L∞ 6⊂ Lp iff µ(X) =∞. This is easy to prove.

On the other hand, Lp 6⊂ L∞ iff X contains sets of arbitrarily small positive measure. The

proof for (⇐) is the same as above except taking an = µ(En)−1/(p+1) now. The proof for (⇒)

is the same.

Remark 4. A. Villani [19] simplified Romero’s previous work [14] and show that the followings

are equivalent: Let 0 < p < q ≤ ∞. (i) Lp(µ) ⊆ Lq(µ) for some p < q; (ii) Lp(µ) ⊆ Lq(µ) for

all p < q; (iii) inf{µ(E) : µ(E) > 0} > 0. In the case q < p, condition (iii) is replaced by (iii)’

sup{µ(E) : µ(E) <∞} <∞.

This is then generalized by Miamee [12] to the more geneal relation Lp(µ) ⊆ Lq(ν).

We prove these results in additional exercise 6.1.4.

6. See Exercise 2.64.

3

Page 4: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

Proof. (i) If p1 < ∞, then consider f(x) = x−1/p1χ(0, 12

)(x) + x−1/p0χ(2,∞)(x). If p1 = ∞, then

consider f(x) = | log x|χ(0, 12

)(x) + x−1/p0χ(2,∞)(x).

(ii) If p1 <∞, then f(x) = x−1/p1| log x|−2/p1χ(0, 12

)(x) + x−1/p0| log x|−2/p0χ(2,∞)(x). If p1 =∞,

then consider f(x) = x−1/p0| log x|−2/p0χ(2,∞)(x).

(iii) Set p1 = p0 <∞ in (ii).

7. Proof. We may assume f 6≡ 0. For each q ∈ (p,∞), we know ‖f‖q ≤ ‖f‖θqp ‖f‖1−θq∞ where

θq = q−1p → 0 as q → ∞. Therefore lim supq→∞ ‖f‖q ≤ ‖f‖∞. On the other hand, given

ε ∈ (0, ‖f‖∞), there exist a measurable set E = E(ε) with µ(E) > 0 such that ‖f‖∞ − ε < |f |

on E. Since f ∈ Lp, µ(E) <∞. Integrating both sides on E, we see (‖f‖∞−ε)µ(E) <∫E|f | ≤

‖f‖qµ(E)1− 1q and hence ‖f‖∞ − ε < ‖f‖qµ(E)−

1q . Therefore, ‖f‖∞ − ε ≤ lim infq→∞ ‖f‖q.

Since ε > 0 is arbitrary, ‖f‖∞ ≤ lim infq→∞ ‖f‖q ≤ lim supq→∞ ‖f‖q ≤ ‖f‖∞.

8. Suppose µ(X) = 1 and f ∈ Lp for some p > 0, so that f ∈ Lq for q ∈ (0, p).

Proof. (a) If − log |f |q ∈ L1, then this is Jensen’s inequality. See Exercise 3.42.

If − log |f |q 6∈ L1, then∫| log |f |q| =∞ and∫

| log |f |q| =∫

log |f |qχ{|f |>1} −∫

log |f |qχ{|f |≤1} ≤∫|f |qχ{|f |>1} −

∫log |f |qχ{|f |≤1}

Since f ∈ Lq,∫

log |f |qχ{|f |≤1} = −∞. Therefore∫

log |f | = −∞ which implies the inequality.

(b) Since log(x) ≤ x− 1 for all x > 0. Let x = ‖f‖qq, we yields the first inequality. The second

assertion is a consequence of monotone convergence theorem.

(c) By (a), ‖f‖q ≥ exp(∫

log |f |) for all q ∈ (0, p), and then lim infq→0 ‖f‖q ≥ exp(∫

log |f |). On

the other hand, from (b) we know lim supq→0 log ‖f‖q ≤ lim supq→0(∫|f |q − 1)/q =

∫log |f |.

Taking the exponential of both sides, we have lim supq→0 ‖f‖q ≤ exp(∫

log |f |).

9. Proof. The first assertion is Chebyshev’s inequality. To prove the second one, we notices the

convergence in measure implies convergence a.e., up to a subsequence, and then the limit

function |f | ≤ g. This implies the desired result by LDCT and the fact |fn − f | ≤ 2g.

Remark 5. The first assertion is trivial true for p =∞, but the second one is false.

10. Suppose fn, f ∈ Lp, fn → f a.e. for some p ∈ [1,∞). Then ‖fn‖p → ‖f‖p ⇐⇒ ‖fn − f‖p → 0.

Proof. ”⇐=” part is easy (we don’t need to assume pointwise convergence here). ”=⇒” part

is an application of Fatou’s lemma to the functions 2p|fn|p + 2p|f |p − |fn − f |p ≥ 0. (Another

method is through Egoroff’s theorem and Fatou’s lemma, see Rudin [15, Exercise 3.17(b)].)

4

Page 5: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

Remark 6. Rudin [15] mentions that the proof seems to be discovered by Novinger [13], where

an application to F.Riesz’s theorem on mean convergence of Hp functions to their boundary

function is mentioned. See Duren [7, Theorem 2.6 and 2.2] and Rudin [15, Theorem 17.11].

If fn → f a.e., by Fatou’s lemma, we always have ‖f‖p ≤ lim inf ‖fn‖p. What can be said

about ‖f‖p. There is a more general and quantitative result:

Theorem 7 (Brezis-Lieb lemma[4]). Suppose fn → f a.e. and ‖fn‖p ≤ C < ∞ for all n and

for some 0 < p <∞. Then

limn→∞

‖fn‖pp − ‖fn − f‖pp = ‖f‖pp.

Another related result is the Radon-Riesz Theorem:

Theorem 8. Let p ∈ (1,∞). Suppose fk ⇀ f in Lp and ‖fn‖p → ‖f‖p, then ‖fn − f‖p → 0.

11. Proof. (a)Given zn → z in C with all zn ∈ Rf . Then given ε > 0 , since |f(x) − z| ≤

|f(x)− zn|+ |zn − z|, by choosing n large such that |zn − z| < ε, we have

{x : |f(x)− zn| < ε} ⊂ {|f(x)− z| < 2ε}.

Therefore 0 < µ({x : |f(x)− zn| < ε}) ≤ µ({|f(x)− z| < 2ε}) and hence z ∈ Rf .

(b) By definition and proof by contradiction, ‖f‖∞ ≥ sup{|w| : w ∈ Rf}. The compactness

follows from (a) and Bolzano-Weierstrauss. So there is z ∈ Rf , such that |z| = sup{|w| : w ∈

Rf}. If |z| < ‖f‖∞, then |z| < 12(|z|+ ‖f‖∞).

Hence the set {x : |f(x) − 12(|z| + ‖f‖∞)| < ε0} has zero measure for some ε0 > 0, but this

implies {x : |f(x) − 12(|z| + ‖f‖∞)| < ε} has zero measure for all 0 < ε < ε0. In particular,

choosing 1 > δ > 0 such that δ2(|z|+ ‖f‖∞) < ε0. Since

{x : |f(x)| > 1− δ2

(|z|+ ‖f‖∞)} ⊂ {x : |f(x)− 1

2(|z|+ ‖f‖∞)| < δ

2(|z|+ ‖f‖∞)},

0 = µ({x : |f(x)− 12(|z|+ ‖f‖∞)| < δ

2(|z|+ ‖f‖∞)}) ≥ µ({x : |f(x)| > 1−δ

2(|z|+ ‖f‖∞)}) = 0,

which contradicts the definition of ‖f‖∞ since 0 < 1−δ2

(|z| + ‖f‖∞) < ‖f‖∞. Therefore,

max{|w| : w ∈ Rf} = |z| = ‖f‖∞.

Remark 9. See Rudin [15, Exercise 3.19] and additional exercise 6.1.7 for an extended discus-

sion.

12. Proof. Let (X,Σ, µ) be the measure space. We assume 1 ≤ p <∞ first. If (i) for every E ∈ Σ,

µ(E) = 0 or ∞, then Lp has zero dimension, where there is nothing to prove.

5

Page 6: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

Now we assume there exists A ∈ Σ such that 0 < µ(A) < ∞. If (ii) for every E ∈ Σ,

µ(E∆A) = 0, then dim(Lp) = 1, that is, every function equals a.e. to a multiple of χA. The

inner product is defined by (αχA, βχA) = αβµ(A)2.

If (iii) there is a set C with 0 < µ(C) <∞ and µ(C∆A) > 0. We may assume B := C \A has

positive measure. Note ‖χA‖2p = µ(A)2/p, ‖χB‖2

p = µ(B)2/p, ‖χA + χB‖2p = (µ(A) + µ(B))2/p

and ‖χA−χB‖2p = (µ(A) +µ(B))2/p, so 2(‖χA‖2

p + ‖χB‖2p) = 2(µ(A)2/p +µ(B)2/p) which is not

equal to 2(µ(A) + µ(B))2/p = ‖χA + χB‖2p + ‖χA − χB‖2

p except p = 2.

For p = ∞, in case (iii), ‖χA + χB‖2∞ + ‖χA − χB‖2

∞ = 2 6= 4 = 2(‖χA‖2∞ + ‖χB‖2

∞). In

case (ii), the situation is the same as above, so it possesses inner product structure. In case

(i), if µ(X) = 0, then it’s done. For µ(X) = ∞, if there are proper subset A of X such that

µ(A) = µ(X\A) =∞, then we see there is no inner product structure as above with B = X\A;

if for every set A with µ(A) =∞, we always have µ(X \A) = 0, then Lp has dimension 1, since

for each f ∈ L∞, if f takes two distinct values on disjoint sets A,B, then either µ(A) = 0 or

µ(B) = 0. That is, f = αχX , for some scalar α.

13. Proof. The case 1 ≤ p <∞, it’s easy to show the set of simple functions on dyadic cubes with

rational coefficients is countably many, dense in Lp(Rd,m).

Next, consider F = {fh(x) = χB1(0)(x + h), h ∈ Rd}, then given f 6= g ∈ F , ‖f − g‖∞ = 1.

This shows the dense subset of L∞(Rd,m) must be uncountable.

Remark 10. See Additional Exercise 6.1.1 to characterize when is Lp(X,M , µ) separable.

14. Proof. We may assume g 6≡ 0. ‖T‖ ≤ ‖g‖∞ since ‖Tf‖p = ‖fg‖p ≤ ‖f‖p‖g‖∞. For the second

part, given ε ∈ (0, ‖g‖∞), since µ is semifinite, there is a set E with 0 < µ(E) < ∞ such that

‖g‖∞ − ε < |g| on E. Then for f = χE, we see ‖T (χE)‖p = ‖gχE‖p > (‖g‖∞ − ε)‖χE‖p, which

means ‖T‖ > ‖g‖∞ − ε. Since ε > 0 is arbitrary, ‖g‖∞ ≤ ‖T‖.

15. This is Vitali Convergence Theorem. Necessity of the hypotheses and further discussions are

given in Rudin [15, Exercise 6.10-11]. For example, in finite measure space, Vitali implies

LDCT, and there is some case Vitali works but LDCT can’t be applied. The details are in

additional exercise 6.1.8.

Proof. (Necessity) Use Chebyshev’s inequality to prove (i). Completeness of Lp implies the

existence of limit function, which controls the integrability of fn for the large n, which is the

central idea to prove (ii) and (iii).

6

Page 7: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

(Sufficiency) Given ε > 0, let E be the set in condition (iii). Take

Anm := {|fm(x)− fn(x)| ≥ ε

µ(E)}.

By (i), we know if n,m large enough, then µ(Anm) < δ, where δ is the modulus of uniform

integrability. Hence, for all large n,m∫Anm

|fn − fm|p dµ < ε.

Therefore, for all large enough n,m∫X

|fn − fm|p dµ =

∫Ec|fn − fm|p dµ+

∫Anm

|fn − fm|p dµ+

∫E\Anm

|fn − fm|p dµ

≤ 2ε+ ε+ε

µ(E)µ(E),

which implies the desired result.

16. Proof. The completeness is a revision of the proof of Theorem 6.6, we omit it. The triangle

inequality is a consequence of the inequality (a + b)p ≤ ap + bp, a, b ≥ 0, whose proof is

straightforward.

Remark 11. There are several properties for 0 < p < 1; the first is in finite measure space,

the behavior of ‖ · ‖q as q → 0, answered in exercise 8(c); the second is that the dual space of

Lp([0, 1],L ,m) is {0}, that is, it’s NOT a locally convex space; the last is the reversed Holder

and Minkowski inequalities, see Jones [10, p.252-253] and Rudin [16, p.37].

Theorem 12. Let (X,M , µ) be a measure space, 0 < p < 1 and f, g be two nonnegative and

measurable functions. Then ∫X

fg dµ ≥(fp dµ

)1/p(gp′dµ)1/p′

,

and

‖f + g‖p ≥ ‖f‖p + ‖g‖p.

6.2 The Dual of Lp

7

Page 8: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

17. Proof. First, we note that Mq(|g|) ≤ 8Mq(g) since for each f =∑n

k=1(ak + ibk)χEk ∈ Σ, the

set of simple functions that vanish outside a set of finite measure

|∫f |g||2 = |

n∑k=1

ak

∫Ek

|g|+ ibk

∫Ek

|g||2 =( n∑k=1

ak

∫Ek

|g|)2

+( n∑k=1

bk

∫Ek

|g|)2

≤( n∑k=1

ak

∫Ek

|Re(g)|+ |Im(g)|)2

+( n∑k=1

bk

∫Ek

|Re(g)|+ |Im(g)|)2

=( n∑k=1

ak( ∫

E1k

Re(g)−∫E2k

Re(g) +

∫E3k

Im(g)−∫E4k

Im(g)))2

+(· · ·)2

≤ 2( n∑k=1

ak

∫E1k

Re(g)− ak∫E2k

Re(g))2

+ 2( n∑k=1

ak

∫E3k

Im(g)− ak∫E4k

Im(g))2

+ 2(· · ·)2

+ 2(· · ·)2

≤ 2Mq(g) + 2Mq(g) + 2Mq(g) + 2Mq(g),

where Ek = E1k ∪E2

k = E3k ∪E4

k are separated by the sign of Re(g) and Im(g) respectively and

the last inequality follows from the definition of Mq(g) (by taking fr =∑n

k=1 akχE1k− akχE2

k

and fi =∑n

k=1 akχE3k− akχE4

k).

Given ε > 0. Let E := {x : |g(x)| > ε}. If µ(E) = ∞, then for each n ∈ N there is En ⊂ E

such that n < µ(En) <∞ (Exercise 1.14). However this is impossible for large n since

µ(En) ≤ 1

ε

∫|g(x)|χEn dµ =

(µ(En))1/p

ε

∫|g(x)| 1

(µ(En))1/pχEn dµ ≤

(µ(En))1/p

εMq(|g|),

that is, µ(En) is uniformly bounded (since q <∞, p > 1). So µ(E) <∞.

Remark 13. I learn from Joonyong Choi that there is another way to show the uniform

boundedness as follows:

µ(En) ≤ 1

ε

∫|g(x)|χEn dµ =

(µ(En))1/p

ε

∫g(x)

g(x)

|g(x)|(µ(En))1/pχEn dµ ≤

(µ(En))1/p

εMq(g),

where the last inequality is proved in the first paragraph of the proof of Theorem 6.14. (Note

that f(x) = g(x)

|g(x)|(µ(En))1/pχEn is bounded and vanishes outside the set En of finite measure)

18. (Von Neumann’s proof for Lebesgue-Radon-Nikodym Theorem)

Proof. (a) It’s a consequence of Holder’s inequality since λ(X) <∞.

(b) Note that for all E ∈M ,∫ERe(g) dλ + i

∫EIm(g) dλ =

∫χEg dλ =

∫χE dµ = µ(E) ≥ 0

So g is real-valued and λ({g ≤ −n−1}) = 0 for all n ∈ N. Similarly, since∫χE(1 − g) dλ =

λ(E) −∫g dλ = λ(E) −

∫χE dµ = λ(E) − µ(E) ≥ 0, λ({1 − g ≤ −n−1}) = 0 for all n ∈ N.

Combining these two results, we have 0 ≤ g ≤ 1 λ-a.e.

(c) Clearly, for each P ⊂ B, µ(P ) =∫Bg dµ =

∫χB(1 − 1) dν = 0 and for each N ⊂ A,

νs(N) = ν(∅) = 0. So νs ⊥ µ.

8

Page 9: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

Since (1 − g)−1χA ≥ 0 λ-a.e., there exists an increasing sequence of simple functions φn ≥ 0

converging to χA(1 − g)−1 λ-a.e. Since χA(1 − g)−1 < ∞ λ-a.e., φn ∈ L2(λ) for all n. For all

E ∈M , by monotone convergence theorem, we have

νa(E) = ν(A∩E) =

∫E

χA1− g

(1−g) dν = limn→∞

∫E

φn(1−g) dν = limn→∞

∫E

φng dµ =

∫E

χA1− g

g dµ,

which is the desired result.

19. Proof. Note that |φn(f)| ≤ ‖f‖∞, which means ‖φn‖(l∞)∗ ≤ 1 for all n. Then Banach-Alaoglu’s

theorem implies the sequence {φn} has a weak∗ cluster point φ. If φ(f) =∑

j f(j)g(j) for some

g ∈ l1, then by taking fi ∈ l∞ with fi(j) = δij, the Kronecker delta, we see g ≡ 0 since

g(i) = φ(fi) = limn→∞ φn(fi) = 0 for each i. But this leads a contradiction that 0 = φ(1) =

limn→∞ φn(1) = 1, where 1(j) = 1 for all j ∈ N.

20. Suppose supn ‖fn‖p <∞ and fn → f a.e.

(a) If 1 < p <∞, then fn → f weakly in Lp. (Given g ∈ Lq, here q is conjugate to p,

and ε > 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) is false in general for p = 1

Proof. (a) We omit it since every step is given in the hint and should be verified easily. A more

precise result about ‖fn‖ and ‖f‖ is Brezis-Lieb lemma, which is proved in additional exercise

6.1.5.

Note that Banach-Alaoglu theorem implies the following similar and important result: (for Lp

case, we can prove it by Cantor diagonal process without Banach-Alaoglu, see my solution file

for Stein-Shakarchi [18, Exercise 1.12(c)])

Theorem 14. Let X be a reflexive Banach space. If ‖fn‖X ≤ M <∞ for all n, then there is

a subsequence converge weakly.

Remark 15. The converse is known as Eberlein-Smulian theorem whose proof is put in addi-

tional exercise 6.2.5. Also see Brezis [3, Theorem 3.19 and its remarks]

(b) For L1(R,m), consider fn = χ(n,n+1) with 1 ≡ g ∈ L∞(R,m).

For l1, consider fn(j) = 1n

if j ≤ n, and 0 otherwise. If there is weak convergence for fn, then

by considering gi(j) = δij ∈ l∞ the Kronecker delta, we see f ≡ 0. On the other hand, for

1 ≡ g ∈ l∞, 0 =∫fg 8

∫fng = 1, which is a contradiction.

9

Page 10: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

21. If 1 < p <∞, fn → f weakly in lp(A) iff supn ‖fn‖p <∞ and fn → f pointwise.

Proof. (⇒) By weak convergence, for any g ∈ lq, there exists Mg <∞ such that supn |∫fng| <

Mg . Then uniform boundedness principle implies supn ‖fn‖p is finite. The pointwise conver-

gence is proved by taking g = χ{a} for each a ∈ A.

(⇐) This is exercise 20 (a).

Remark 16. (⇒) is also true for p = 1 and p = ∞ due to Proposition 6.13 works in both

case (counting measure is semifinite.) An interesting theorem for l1(N) is the Schur’s property

which asserts the equivalence between weak convergence and norm convergence. (See additional

exercise 6.2.3.)

(⇐) is discussed in exercise 20 (b).

22. Let X = [0, 1] with Lebesgue measure.

(a) Let fn(x) = cos(2πnx). Then fn → 0 weakly in L2 (See Exercise 63 in Section 5.5)

but fn 9 0 a.e. or in measure.

(b) Let fn(x) = nχ(0,1/n). Then fn → 0 a.e. and in measure, but fn 9 0 weakly in Lp

for any p.

(a) This is Riemann-Lebesgue lemma, which holds not only for L2 but also for Lp, (p ∈ [1,∞)),

and weak-* convergence in L∞; I put my comment on various proofs below:

(1) I think the easy way to understand is through Bessel’s inequality (cf: Exercise 5.63). But

this method does not work on Lp, except L2.

(2) The second way is through integration by parts, and one needs the density theorem (simple

functions or C∞c functions). This method is adapted to our claim, except for weak convergence

in L1. But for L1 case, it’s a simple consequence of squeeze theorem in freshman’s Calculus.

It’s easy to see fn 6→ 0 a.e. or in measure.

(b) The first assertion is easy to prove. To see it’s not weak convergent, take g ≡ 1 in Lp′.

Remark 17. From this problem, we know for finite measure space (E,M , µ), we have the

following figure from Dibenedetto [6, p.294](1st edition: p.268) with some modifications (the

red and green words.)

10

Page 11: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

23. See Exercise 1.16 for the notion of locally measurable sets.

Proof. (a) If 0 < µ(E) < ∞, then 0 = µ(E ∩ E) > 0, a contradiction. If µ is semifinite, then

for every locally null set E with µ(E) = ∞, there exist F ⊂ E with 0 < µ(F ) < ∞, but

0 = µ(E ∩ F ) = µ(F ) > 0, a contradiction.

(b) If ‖f‖∗ = 0, then for each 1n

there exist an ∈ (0, 1n) such that {|f | > an} is locally null.

Since {|f | > an} ⊇ {|f | > 1n}, {|f | > 1

n} is locally null. Given F ∈ M with µ(F ) < ∞, we

have 0 ≤ µ(F ∩ {f 6= 0}) = µ(F ∩ ∪n{|f | > 1n}) ≤

∑n µ(F ∩ {|f | > 1

n}) = 0, that is, {f 6= 0}

is locally null.

Given f, g ∈ L∞ and then given ε > 0. Note that

{|f+g| > ‖f‖∗+‖g‖∗+ε} ⊆ {|f |+|g| > ‖f‖∗+‖g‖∗+ε} ⊆ {|f | > ‖f‖∗+ε/2}∪{|g| > ‖g‖∗+ε/2}.

11

Page 12: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

Since the latter set is locally null, {|f + g| > ‖f‖∗ + ‖g‖∗ + ε} is locally null. Then ‖f + g‖∗ ≤

‖f‖∗ + ‖g‖∗ + ε for all ε > 0 and therefore ‖f + g‖∗ ≤ ‖f‖∗ + ‖g‖∗.

Given 0 6= λ ∈ C and f ∈ L∞, since {|λf | > |λ|‖f‖∗ + ε} = {|f | > ‖f‖∗ + ε|λ|} is locally null

for each ε > 0 and {|λf | > |λ|‖f‖∗ − ε} = {|f | > ‖f‖∗ − ε|λ|} is not locally null for each ε > 0,

we see ‖λf‖∗ = |λ|‖f‖∗. Consequently, ‖ · ‖∗ is a norm on L∞.

Given {fn} ⊂ L∞ with∑

n ‖fn‖∗ < ∞. We are going to show∑

n fn ∈ L∞. More precisely,

‖∑

n fn‖∗ ≤∑

n ‖fn‖∗ <∞. This is true since for each ε > 0{|∑n

fn| >∑n

‖fn‖∗ + ε}⊆{∑

n

|fn| >∑n

‖fn‖∗ + ε}⊆ ∪n

{|fn| > ‖fn‖∗ +

ε

2n

},

where each indexed-set in the last set is locally null. Therefore, (L∞, ‖ · ‖∗) is complete.

The last assertion is due to ‖ · ‖∗ = ‖ · ‖∞ here, which is a consequence of (a).

24. In contrast to Proposition 6.13 and Theorem 6.14, we do NOT need to assume µ is semifinite

or Sg = {g 6= 0} is σ-finite here. Also see additional exercise 6.2.2.

Proof. (i) Given f ∈ L1 with ‖f‖1 = 1. Since, for each n ∈ N, {|f | > 1n} has finite measure

and the set {|g| > ‖g‖∗} is locally null, the intersection of these two sets are null. Therefore

|f(x)g(x)| ≤ |f(x)|‖g‖∗ a.e.. Hence sup{|∫fg| : ‖f‖1 = 1} ≤ ‖g‖∗.

Given ε > 0, since {|g| > ‖g‖∗ − ε} is not locally null, there is a measurable set A with

finite measure such that B = A ∩ {|g| > ‖g‖∗ − ε} has nonzero finite measure. Consider

h = sgn(g)χB(µ(B))−1, then ‖h‖1 = 1 and

sup{|∫fg| : ‖f‖1 = 1} ≥ |

∫hg| = (µ(B))−1

∫B

|g| > ‖g‖∗ − ε.

Therefore sup{|∫fg| : ‖f‖1 = 1} ≥ ‖g‖∗.

(ii) Now we consider the counterpart of Theorem 6.14: Given ε > 0. If {|g| > M∞(g) + ε} is

not locally null, then there is a set A with finite measure, such that B = A∩{|g| > M∞(g)+ ε}

has nonzero finite measure. But for h = sgn(g)χB(µ(B))−1, ‖h‖1 = 1 and hence

M∞(g) ≥ |∫hg| = (µ(B))−1

∫B

|g| > M∞(g) + ε,

which is a contradiction. So {|g| > M∞(g) + ε} is locally null for each ε > 0. Therefore

‖g‖∗ ≤M∞(g) <∞. Now by the first paragraph in (i), M∞(g) ≤ ‖g‖∗.

25. Proof. If µ is finite, then the proof is essentially the same as the original proof except the role

of Theorem 6.14 is replaced by Exercise 24.

12

Page 13: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

Now for decomposable µ, let {Fa}a∈A be the decomposition of µ. Let ga ∈ L∞(Fa) be so

φ�L1(Fa)= φga and define g = ga on each Fa. Then g ∈ L∞ since for each α ∈ R, g−1((α,∞)) =

∪ag−1a (α,∞)) is measurable by property (iv) in the definition of decomposable measur and

since ‖g‖∗ = supa ‖ga‖∗ ≤ supa ‖φ�L1(Fa)‖ ≤ ‖φ‖.

Since for each f ∈ L1, the set En = {|f | > 1n} has finite measure and µ(En) =

∑a µ(En ∩ Fa).

Therefore for each k ∈ N only finite many index a such that µ(En∩Fa) > 1k, and only countably

many indices a such that µ(En ∩ Fa) > 0. Furthermore, only countably many indices a such

that µ({|f | > 0} ∩ Fa) > 0. We call this index set I and thus,

φ(f) =∑i∈I

φ(fχFi) =∑i∈I

∫Fi

fgi =

∫fg,

where we apply LDCT in the last equality.

Remark 18. The definition of decomposable measure we used here should be precisely called

almost decomposable due to (iii) µ(E) =∑µ(E∩F ) is required to be true for finite measure

set E only. (cf: Bogachev [2, 1.12.131-132] and Fremlin [8, Vol. 2].)

A subtle difference is the following facts:

µ is decomposable ⇒ µ is semifinite : µ is almost decomposable.

(⇒) is straightforward. The counterexample for (:) is recorded in additional exercise 6.2.6.

From Bogachev [2, 1.12.134], we know µ is semifinite ; µ is decomposable.

Example 19. I learned these examples from Professor Nico Spronk’s Notes

(i) The first is an application of Exercise 25 which is not a consequence of Theorem 6.15.

Let I be an uncountable set, and M be the σ-algebra of sets of countable or co-countable. µ is

the counting measure. Then the collection M of all locally measurable sets is P (I), the power

set of I and we can define the saturation µ of µ on M (cf: Exercise 1.16).

Note that L1(I,M , µ) = L1(I, M , µ) and {{i}}i∈I is a decomposition of for µ. The theorem

we just proved implies that (L1)∗ ∼= L∞.

(ii) The second example sits outside Theorem 6.15 and Exercise 25!

Let X = {f, i} and consider µ : P (X) → [0,∞] be the unique measure satisfying µ({f}) = 1

and µ({i}) = ∞. Note that {i} is locally null. Then L∞ is the span of χ{f}. On the other

hand L1 is also the span of χ{f}. Then by definition, (L1)∗ ∼= R ∼= L∞.

But the existence of infinite atom {i} means that we cannot partition X into subsets of finite

measure, that is, µ is neither semifinite nor decomposable!

13

Page 14: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

Remark 20. Further results related to Exercise 23-25 may also be confirmed with Cohn [5,

Chapter 3].

6.3 Some Useful Inequalities

26. The case p = 1 is a consequence of Fubini-Tonelli’s Theorem. The case p =∞ is apparent.

27. (Hilbert’s inequality, a special case of Schur’s lemma, Theorem 6.20, our proof here is essential

the same as the one given there.)

Proof. By Minkowski’s inequality,(∫ ∞0

|∫ ∞

0

f(y)

x+ ydy|p dx

)1/p

=(∫ ∞

0

|∫ ∞

0

f(xz)

1 + zdz|p dx

)1/p

≤∫ ∞

0

(∫ ∞0

|f(xz)|p dx)1/p 1

1 + zdz

= ‖f‖p∫ ∞

0

z−1/p(1 + z)−1 dz = ‖f‖pπ csc(π/p).

To compute the last integral, one use the residue theorem in one complex variable.

28. This exercise extends Corollary 6.20 with T = J1. Further generalizations are given in the

exercise 29 and 44.

Proof. For p = ∞, it is straightforward. For 1 < p < ∞, using change of variables and

Minkowski’s inequality, we have(∫ ∞0

∣∣∣ 1

∫ x

0

f(t)(x− t)α−1 dt∣∣∣p dx)1/p

≤(∫ ∞

0

(∫ 1

0

|f(xw)|(1− w)α−1 dw)pdx)1/p

≤∫ 1

0

(∫ ∞0

|f(xw)|p(1− w)αp−p dx)1/p

dw = ‖f‖p∫ 1

0

(1− w)α−1w−1/p dw = ‖f‖pΓ(α)Γ(1− p−1)

Γ(α + 1− p−1).

A counterexample for p = 1 is χ(0,1)(t). Note that∫ ∞0

x−α∫ x

0

(x− t)α−1χ(0,1)(t) dt dx

=

∫ ∞1

x−α∫ 1

0

(x− t)α−1 dt dx+

∫ 1

0

x−α∫ x

0

(x− t)α−1 dt dx

=

∫ ∞1

x−α(x∗)α−1 dx+ α−1,

where x∗ ∈ (x − 1, x) for all x ∈ (1,∞) by mean value theorem. We consider α > 1 and

0 < α ≤ 1 separately. For α > 1, the integrand is larger than x−α(x−1)α−1 = (1− 1x)α(x−1)−1

which is not integrable on (2,∞). For α ≤ 1, the integrand is larger than x−αxα−1 = x−1 which

is not integrable on (1,∞). So Jα(χ(0,1)(t)) 6∈ L1(0,∞).

14

Page 15: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

Remark 21. Is the constant Γ(1−p−1)Γ(α+1−p−1)

sharpest ? When does the equality hold? See Rudin

[15, Exercise 3.14] for the Hardy’s inequality, that is, the case α = 1.

Remark 22. A similar operator to fractional integral operator is Riesz and Bessel potential

operators, which is strongly related to Sobolev embedding theorem. See Stein [17, Chapter V]

and Ziemer [20, Chapter 2]. Also see exercise 44-45.

29. (Hardy’s inequality)

Proof. Again, we prove this without using Theorem 6.20. By Minkowski’s inequality,(∫ ∞0

x−r−1[ ∫ x

0

h(y) dy]pdx)1/p

=(∫ ∞

0

xp−r−1[ ∫ 1

0

h(xz) dz]pdx)1/p

≤∫ 1

0

(∫ ∞0

xp−r−1(h(xz))p dx)1/p

dz =

∫ 1

0

(∫ ∞0

yp−r−1h(y)p dy)1/p

zrp−1 dz

=p

r

(∫ ∞0

yp−r−1h(y)p dy)1/p

.

(∫ ∞0

xr−1[ ∫ ∞

x

h(y) dy]pdx)1/p

=(∫ ∞

0

xp+r−1[ ∫ ∞

1

h(xz) dz]pdx)1/p

≤∫ ∞

1

(∫ ∞0

xp+r−1(h(xz))p dx)1/p

dz =

∫ ∞1

(∫ ∞0

yp+r−1h(y)p dy)1/p

z−rp−1 dz

=p

r

(∫ ∞0

yp+r−1h(y)p dy)1/p

.

30. Proof. (a)Since all terms in the integrand is nonnegative, by Fubini-Tonelli, we can interchange

the order of integrations freely.∫ ∫K(xy)f(x)g(y) dx dy =

∫ ∫K(z)f(

z

y)g(y)y−1 dz dy =

∫ ∫f(z

y)g(y)y−1 dy K(z) dz

≤∫

(

∫[f(

z

y)]py−p dy)1/p(

∫g(y)q dy)1/qK(z) dz

=

∫(

∫[f(w)]p(w/z)pz/w2 dw)1/pK(z) dz(

∫g(y)q dy)1/q

=

∫z−1+ 1

pK(z) dz(

∫[f(w)]pwp−2 dw)1/p(

∫g(y)q dy)1/q.

(b) is proved by (a) with p = q = 2 and g(y) =∫k(zy)f(z) dz.

31. Proof. (Sketch) Mathematical induction and apply the original Holder’s inequality to ‖·‖rr.

32. (Hilbert-Schmidt integral operator.)

15

Page 16: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

Proof. By Tonelli-Fubini’s theorem and Holder’s inequality, we know for µ-a.e. x, the integral∫Y

|K(x, y)||f(y)| dν(y) ≤(∫

Y

|K(x, y)|2 dν(y))1/2

‖f‖L2(ν) <∞.

Then Tf(x) is well-defined a.e. and |Tf(x)| ≤( ∫

Y|K(x, y)|2 dν(y)

)1/2

‖f‖L2(ν). Integrating

Tf with respect to dµ(x), we see ‖Tf‖2L2 ≤ ‖K‖2

L2‖f‖2L2 .

Theorem 23. The above T : L2(ν)→ L2(µ) is compact provided these L2 space are separable.

Proof. Let {φi(x)} and {ϕj(y)} be the orthonormal basis of L2(µ) and L2(ν) respectively. By

Fubini’s theorem and standard characterizations of complete set in Hilbert space, {φi(x)ϕj(y) =:

Φij(x, y)}i,j is an orthonormal basis for L2(µ × ν). Hence we can write K =∑aijΦij. For

N = 1, 2, · · · , let

KN(x, y) =∑i+j≤N

aijΦij

and TKN is the corresponding operator with kernel KN . Then TKN has finite rank. On the

other hand, ‖K − KN‖L2 → 0 as N → ∞. So by the previous exercise ‖T − TKN‖ → 0 and

hence T is compact. (This standard result is proved by Cantor diagonal process.)

33. Proof. Since x−1/p and∫ x

0f(t) dt are continuous, Tf(x) is continuous on (0,∞). Given ε > 0.

Since for any x ∈ (0,∞)

|Tf(x)− T (fχ(0,n))(x)| ≤ x−1/p

∫ x

0

|f(t)− fχ(0,n)(t)| dt ≤ ‖f − fχ(0,n)‖q,

we can choose a large N such that ‖Tf − T (fχ(0,N))‖∞ < ε. Since for x > N

|T (fχ(0,N))(x)| ≤ x−1/p

∫ N

0

|f(t)| dt,

the left hand side is less than ε if x is sufficient large. Combine these inequalities and since ε is

arbitrary, Tf ∈ C0(0,∞). The boundedness of T is a consequence of Holder’s inequality.

34. Proof. Since f is absolutely continuous on any [ε, 1], we know for any [x, y] ⊂ (0, 1),

|f(x)− f(y)| = |∫ y

x

f ′(s)ds| ≤(∫ y

x

s−1p

pp−1

) p−1p(∫ y

x

s|f ′(s)|p)1/p

(i) If p > 2, then by the above calculation, for any sequence xn → 0, {f(xn)} forms a Cauchy

sequence. Therefore limx→0 f(x) exists.

(ii) If p = 2, given ε > 0 and pick y = y(ε) ∈ (0, 12) such that

∫ y0t|f ′(t)|2 dt < ε2, then for any

x ∈ (0, y),

0 ≤ |f(x)|| log x|1/2

=|f(y)−

∫ yxf ′(t)dt|

| log x|1/2≤ |f(y)|| log x|1/2

+| log y − log x|1/2ε| log x|1/2

.

16

Page 17: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

Hence there exist δ = δ(ε) ∈ (0, y) such that for any x ∈ (0, δ), the last term is less than 2ε.

(iii) If p ∈ (1, 2), given ε > 0 and pick y = y(ε) ∈ (0, 12) such that

∫ y0t|f ′(t)|p dt < εp, then for

any x ∈ (0, y),

0 ≤ |f(x)|x1−(2/p)

=|f(y)−

∫ yxf ′(t)dt|

x1−(2/p)≤ |f(y)|x1−(2/p)

+ Cp|y

p−2p−1 − x

p−2p−1 |

p−1p ε

x1−(2/p).

Hence there exist δ = δ(ε) ∈ (0, y) such that for any x ∈ (0, δ), the last term is less than 2ε.

(iv)If p = 1, given ε > 0 and pick y = y(ε) ∈ (0, 12) such that

∫ y0t|f ′(t)| dt < ε, then for any

x ∈ (0, y),

0 ≤ |f(x)|x−1

=|f(y)−

∫ yxf ′(t)dt|

x−1≤ |f(y)|

x−1+x−1ε

x−1.

Hence there exist δ = δ(ε) ∈ (0, y) such that for any x ∈ (0, δ), the last term is less than 2ε.

From(ii)-(iv), we see the second and third assertions hold.

6.4 Distribution Functions and Weak Lp

35. For any measurable f and g we have [cf ]p = |c|[f ]p and [f + g]p ≤ 2([f ]pp + [g]pp)1p ; hence

weak Lp is a vector space. Moreover, the ”balls” {g : [g − f ]p < r} (r > 0, f ∈ weak

Lp) generate a topology on weak Lp that makes weak Lp into a topological vector

space.

Proof. The first assertion follows from definition; the second assertion is through the fact

{|f + g| > λ} ⊂ {|f |+ |g| > λ} ⊂ {|f | > λ

2} ∪ {|g| > λ

2}.

The addition (x,y) 7→ x+y is continuous under the given topology by definition and the second

assertion.

The scalar multiplication (a, x) 7→ ax is continuous by definition and the above two assertions.

Remark 24. Normability of weak Lp, see Grafakos [9, p.14-15] and additional exercise 6.4.1.

36. If f ∈ weak Lp and µ({x : f(x) 6= 0}) < ∞, then f ∈ Lq for all q < p. On the other

hand, if f ∈(weak Lp) ∩ L∞, then f ∈ Lq for all q > p.

Proof. (i) By Proposition 6.24 and assumption, we see q − p < 0 and

1

q

∫|f |q dµ =

∫ 1

0

+

∫ ∞1

αq−1µ({x : |f | > α}) dα ≤M

∫ 1

0

αq−1 dα + C

∫ ∞1

αq−1−p dα <∞,

17

Page 18: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

for some positive constants M,C.

(ii) Similarly, we see q − p > 0 and

1

q

∫|f |q dµ =

∫ ‖f‖∞0

+

∫ ∞‖f‖∞

αq−1µ({x : |f | > α}) dα ≤ C

∫ ‖f‖∞0

αq−p−1 dα <∞,

for some positive constant C.

37. Draw a picture to prove Proposition 6.25 !!

38. Suppose (X,M , µ) be a measure space, f ≥ 0, measurable and finite a.e.

Let p ∈ (0,∞), E2k = {x : f(x) > 2k} and Fk = {x : 2k < f(x) ≤ 2k+1}. Prove that

f ∈ Lp(µ)⇐⇒∞∑

k=−∞

2kpµ(Fk) <∞⇐⇒∞∑

k=−∞

2kpµ(E2k) <∞.

Proof. The first ⇐⇒ and second ⇐= are trivial. The second =⇒ is by applying Tonelli’s

theorem (Fubini) to counting measure on N, which is σ-finite: (cf: Exercise 2.46)

∞∑k=−∞

2kpµ(E2k) =∞∑

k=−∞

2kp∞∑i=k

µ(Fi) =∞∑

i=−∞

µ(Fi)i∑

k=−∞

2kp =∞∑

i=−∞

µ(Fi)2ip2p <∞.

Remark 25. We can prove the second =⇒ without using Fubini-Tonelli as follows:

Accoridng to Proposition 6.24, it suffices to show∫∞

0αp−1λf (α) dα is finite. This is true since

2k(p−1)+kλf (2k+1) =

∫ 2k+1

2k2k(p−1)λf (2

k+1) dα ≤∫ 2k+1

2kαp−1λf (α) dα

≤∫ 2k+1

2k2(k+1)(p−1)λf (2

k) dα = 2(k+1)(p−1)+kλf (2k).

The desired result follows by summing k from −∞ to ∞.

39. If f ∈ Lp, then limα→0 αpλf (α) = limα→∞ α

pλf (α) = 0.

Proof. Since f ∈ Lp, 0 ≤ αpλf (α) ≤∫{|f |>α} |f |

p → 0 as α →∞. On the other hand, for each

ε > 0, there is δ > 0 such that∫{|f |≤δ} |f |

p < ε. Thus αp(λf (α) − λf (δ)) ≤∫{α<|f |≤δ} |f |

p < ε

for all 0 < α < δ. We are done by taking lim supα→0 on both sides.

Remark 26. From the above proof, we see that, instead of assuming f ∈ Lp, we can assume

either∫∞

0αp dλf (α) or

∫∞0αp−1λf (α) dα is fintie, although according to Proposition 6.23 and

6.24, they are equivalent.

18

Page 19: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

40. If f is a measurable function on X, its decreasing rearrangement is the function

f ∗(0,∞)→ [0,∞] defined by

f ∗(t) = inf{α : λf (α) ≤ t} (where inf =∞).

(a) f ∗ is decreasing. If f ∗(t) <∞ then λf (f∗(t)) ≤ t, and if λf (α) <∞ then f ∗(λf (α)) ≤

α.

(b) λf = λf∗ where λf∗ is defined with respect to Lebesgue measure on (0,∞).

(c) If λf (α) <∞ for all α > 0 and limα→∞ λf (α) = 0 ( so that f ∗(t) <∞ for all t > 0),

and φ is a nonnegative Borel measurable function on (0,∞), then∫Xφ ◦ |f | dµ =∫∞

0φ ◦ f ∗(t) dt. In particular, ‖f‖p = ‖f ∗‖p for 0 < p <∞.

(d) If 0 < p <∞, [f ]p = supt>0 t1pf ∗(t).

(e) The name ”rearrangement” for f ∗ comes from the case where f is a nonnegative

function on (0,∞) assuming four or five different values and draw the graphs of f

and f ∗.

Proof. (a) It’s easy to see f ∗ is nonincreasing. If f ∗(t) < ∞, then {α : λf (α) ≤ t} 6= ∅ and

hence by definition for each n, there is εn ∈ (0, 1n) such that λf (f

∗(t) + εn) ≤ t. Therefore

λf (f∗(t)) = lim

n→∞λf (f

∗(t) + εn) ≤ t.

If λf (α) <∞, then for each β ≥ α, λf (β) ≤ λf (α) and hence f ∗(λf (α)) ≤ α.

(b) For t, y ∈ (0,∞). Since f ∗(t) > y iff λf (y) > t, {t : f ∗(t) > y} = (0, λf (y)). Therefore,

λf∗(y) = λf (y) for all y ∈ (0,∞).

(c) Apply Proposition 6.23.

(d) Given t > 0, If f ∗(t) = 0, then t1/pf ∗(t) ≤ [f ]p. If f ∗(t) > 0 then by definition, for all

0 < ε < f ∗(t), we have

t(f ∗(t)− ε)p < λf (f∗(t)− ε)(f ∗(t)− ε)p ≤ [f ]pp.

Taking p-th root and letting ε→ 0, we have t1/pf ∗(t) ≤ [f ]p. Since t > 0 is arbitrary,

supt>0

t1/pf ∗(t) ≤ [f ]p.

On the other hand, given α > 0 and assume λf (α) > 0 without loss of generality. Since for any

0 < ε < λf (α), we have f ∗(λf (α)− ε) > α. Therefore

supt>0

t1/pf ∗(t) ≥ (λf (α)− ε)1/pf ∗(λf (α)− ε) > α(λf (α)− ε)1/p.

Letting ε→ 0, we have supt>0 t1/pf ∗(t) ≥ [f ]p.

19

Page 20: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

Remark 27. See Lieb-Loss [11, Chapter 3-4] and Dibenedetto [6, Chapte 9] for further dis-

cussions and applications on higher-dimensional rearrangement and Riesz’s rearrangement in-

equality.

6.5 Interpolation of Lp Spaces

41. Suppose 1 < p ≤ ∞ and 1p

+ 1q

= 1. If T is a bounded linear operator on Lp such

that∫

(Tf)g =∫f(Tg) for all f, g ∈ Lp ∩ Lq, then T extends uniquely to a bounded

linear operator on Lr for all r ∈ [p, q] (if p < q) or [q, p] if q < p). We should assume µ

is semifinite when p =∞, see the errata sheet provided by Prof. Folland.

Proof. (i) If q < p < ∞ or p < q < ∞, then we define T on Lp ∩ Lq the same as the original

bounded linear map on Lp. Given f ∈ Lp∩Lq and g ∈ Lp∩Lq, we have, by Holder’s inequality,

|∫

(Tf)g| = |∫f(Tg)| ≤ ‖f‖q‖T‖p→p‖g‖p, (1)

where ‖T‖p→p is the operator norm of T on Lp. Since Lp ∩ Lq is dense in Lp, for any g ∈ Lp,

we choose gn ∈ Lp ∩ Lq such that gn → g in Lp, then the sequence {∫

(Tf)gn} is Cauchy in C

and we define ∫(Tf)g = lim

n→∞

∫(Tf)gn.

This definition is independent of the choice of gn since the union of two approximation se-

quences is still an approximation sequence and the corresponding sequence is still Cauchy in

C. Therefore we know (1) holds for any g ∈ Lp with ‖g‖p = 1, and hence

‖Tf‖q ≤ ‖f‖q‖T‖p→p

for any f ∈ Lp ∩Lq. Since Lp ∩Lq is dense in Lq, this inequality holds for any f ∈ Lq. So T is

bounded on Lq. The desired boundedness of T on Lr follows from Riesz-Thorin theorem.

(ii) If p = ∞, then q = 1. First, we show T maps L∞ ∩ L1 → L1 continuously by Theorem

6.14: Let Σ be the set of simple functions that vanish outside a set of finite measure. Given

f ∈ L∞ ∩ L1 and g ∈ Σ ⊂ L∞ ∩ L1, then fg ∈ L1 and we have, by Holder’s inequality, for any

‖g‖∞ = 1

|∫

(Tf)g| = |∫f(Tg)| ≤ ‖f‖1‖T‖, (2)

where ‖T‖ is the operator norm of T on L∞. Therefore

M1(Tf) := sup{|∫

(Tf)g| : ‖g‖∞ = 1} ≤ ‖f‖1‖T‖.

20

Page 21: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

By Theorem 6.14 and Exercise 17, we have Tf ∈ L1 and ‖Tf‖ ≤ ‖T‖‖f‖1. Since L1 ∩ L∞

is dense in L1, we can extend T to L1 continuously with norm less than ‖T‖. The desired

boundedness of T on Lr follows from Riesz-Thorin again.

The uniqueness part is due to the fact Lp ∩ Lq is dense in Lr which is proved by considering

cut-off in domain and range simultaneous.

Remark 28. Here I emphasis that case (i) can also be proved by the method given in (ii), but

the density argument in (i) may not be suitable for case (ii).

42. Prove the Marcinkiewicz theorem in the case p0 = p1.

Proof. ∫X

|Tf |q = q

∫ ∞0

sq−1µ({|Tf | > s}) ds

= q

∫ ‖f‖p0

sq−1µ({|Tf | > s}) ds+ q

∫ ∞‖f‖p

sq−1µ({|Tf | > s}) ds

≤ q

∫ ‖f‖p0

sq−1(C0‖f‖p

s)q0 ds+ q

∫ ∞‖f‖p

sq−1(C1‖f‖p

s)q1 ds

= q( Cq0

0

q − q0

+Cq1

1

q1 − q

)‖f‖qp

43. Let H be the Hardy-Littlewood maximal operator on R. Compute Hχ(0,1) explicitly.

Show that it is in Lp for all p > 1 and in weak L1 but not in L1, and that its Lp

norm tends to ∞ like 1p−1

as p→ 1+, although ‖χ(0,1)‖p = 1 for all p.

Proof. By direct computations,

Hχ(0,1)(x) =

1

2(1−x)for x ≤ 0,

1 for x ∈ (0, 1),

12x

for x ≥ 1.

(3)

For t > 0,

m({Hχ(0,1)(x) > t}) =

0 for t > 1,

1 for t ∈ [12, 1],

1t− 1 for t ∈ (0, 1

2),

(4)

which is easy to see the weak (1,1) estimate with constant 1 and strong (p,p) estimate with

constant 1(p−1)2p−1 + 1.

21

Page 22: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

44. Let Iα be the fractional integration operator of Exercise 61 in §2.6. If 0 < α < 1,

1 < p < 1α, and 1

r= 1

p− α, then Iα is weak type (1, 1

1−α) and strong type (p, r) with

respect to Lebesgue measure on (0,∞). Also see the last paragraph in section 6.6.

Proof. This exercise is essentially the same as the next one. We change our α in this problem

to β and put the following parameters in the next exercise: n = 1, α = 1− β, f = gχ(0,∞) and

K(x) := |x|−αχ(0,∞). Using the results there, we have the desired weak type estimate.

45. If 0 < α < n, define an operator Tα on functions on Rn by

Tαf(x) =

∫|x− y|−αf(y) dy.

Then Tα os weak type (1, nα

) and strong type (p, r) with respect to Lebesgue measure

on Rn, where 1 < p < nn−α and 1

r= 1

p− n−α

n.

Proof. Given f ∈ Lp, we may assume ‖f‖p = 1.

Fix µ > 0 to be determined later, we decompose

K(x) := |x|−α = K(x)χ{|x|≤µ} +K(x)χ{|x|>µ}

=: K1(x) +K∞(x).

Then Tαf = K1 ∗ f +K∞ ∗ f . The first term is an Lp function since it’s the convolution of an

L1 function K1 and Lp function f and similarly, the second term is an L∞ function.

Since for each λ > 0

m({|K ∗ f | > 2λ}) ≤ m({|K1 ∗ f | > λ}) +m({|K∞ ∗ f | > λ})

Note that m({|K1 ∗ f | > λ}) ≤ ‖K1∗f‖ppλp

≤ (‖K1‖p1λ

)p = c1(µn−α

λ)p and ‖K∞‖p′ = c2µ

−n/r. Choose

µ such that c2µ−n/r = λ and therefore ‖K∞ ∗ f‖∞ ≤ λ, that is, m({|K∞ ∗ f | > λ}) = 0. Then

m({|K ∗ f | > 2λ}) ≤ c1(µn−α

λ)p = c3λ

−r

The special case for p = 1 gives the weak (1, nα−1) estimate. The strong (p,r) estimate follows

from Marcinkiewicz interpolation theorem.

Remark 29. There are examples for the non-strong type at end point p = 1 and non-embedding

phenomena at p = nn−α given in Stein [17, Section V.1]. For simplicity, we construct one-

dimensional case only, but they can extend to higher-dimensional case easily.

22

Page 23: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

(i) Take f = χ(−12, 12

) . Then ‖f‖1 = 1 and we see (Tαf)(x) ≥(|x| + 1

2

)−α≥ 0 for all |x| ≥ 1.

Hence

∞ =

∫|x|≥1

(2|x|)−1 ≤∫|x|≥1

(|x|+ 1

2

)−α 1α ≤

∫ ∣∣∣(Tαf)(x)∣∣∣ 1α .

(ii) Let f(x) = |x|α−1(log 1/|x|)(α−1)(1+ε) for |x| ≤ 12

and 0 for |x| > 12, where 0 < ε� 1. Then

f ∈ L1

1−α (R). But

(Tαf)(0) =

∫|x|≤ 1

2

|x|−1(log1

|x|)(α−1)(1+ε) dx =∞,

as long as (α− 1)(1 + ε) ≥ −1. Therefore Tαf 6∈ L∞.

Acknowledgement

The author would like to thank Joonyong Choi for his reminder that the proof for problem 17 has

to consider En (not E) and to consider complex-valued g (not only real-valued). (2017.08.20)

References

[1] Sergio A Alvarez. Lp arithmetic. The American mathematical monthly, 99(7):656–662, 1992.

[2] Vladimir I Bogachev. Measure theory, volume 1. Springer Science & Business Media, 2007.

[3] Haim Brezis. Functional Analysis, Sobolev Spaces and Partial Differential Equations. Springer

Science & Business Media, 2010.

[4] Haım Brezis and Elliott Lieb. A relation between pointwise convergence of functions and

convergence of functionals. Proceedings of the American Mathematical Society, 88(3):486–

490, 1983.

[5] Donald L Cohn. Measure Theory. Springer Science & Business Media, 2nd edition, 2013.

[6] Emmanuele DiBenedetto. Real Analysis. Springer Science & Business Media, 2nd edition,

2016.

[7] Peter L Duren. Theory of Hp spaces, volume 38. IMA, 1970.

[8] David Heaver Fremlin. Measure theory. Torres Fremlin, 2000.

[9] Loukas Grafakos. Classical Fourier Analysis, volume 249. Springer, 3rd edition, 2014.

23

Page 24: Real Analysis, 2nd Edition, G.B.Folland Chapter 6 L …homepage.ntu.edu.tw/~d04221001/Notes/Problems and Solutions Secti… · Real Analysis, 2nd Edition, G.B.Folland Chapter 6 LpSpaces

[10] Frank Jones. Lebesgue Integration on Euclidean Space. Jones & Bartlett Learning, revised

edition, 2001.

[11] Elliott H Lieb and Michael Loss. Analysis, volume 14. American Mathematical Society,

Providence, RI, 2nd edition, 2001.

[12] AG Miamee. The inclusion Lp(µ) ⊂ Lq(ν). The American Mathematical Monthly, 98(4):342–

345, 1991.

[13] WP Novinger. Mean convergence in ˆ{} spaces. Proceedings of the American Mathematical

Society, 34(2):627–628, 1972.

[14] Juan L. Romero. When is Lp(µ) contained in Lq(µ)? 90(3):203–206, 1983.

[15] Walter Rudin. Real and Complex Analysis. Tata McGraw-Hill Education, third edition, 1987.

[16] Walter Rudin. Functional analysis. McGraw-Hill, Inc., New York, 2nd edition, 1991.

[17] Elias M Stein. Singular Integrals and Differentiability Properties of Functions, volume 2.

Princeton University Press, 1970.

[18] Elias M Stein and Rami Shakarchi. Functional analysis: Introduction to Further Topics in

Analysis, volume 4. Princeton University Press, 2011.

[19] Alfonso Villani. Another note on the inclusion Lp(µ) ⊂ Lq(µ). The American Mathematical

Monthly, 92(7):485–487, 1985.

[20] William P Ziemer. Weakly Differentiable Functions: Sobolev Spaces and Functions of Bounded

Variation, volume 120. Springer-Verlag, 1989.

24