PARTIAL SOLUTIONS TO REAL ANALYSIS, FOLLAND

ZHENGJUN LIANG

Abstract. This following are partial solutions to exercises on Real Analysis, Folland, writtenconcurrently as I took graduate real analysis at the University of California, Los Angeles. LastUpdated: November 18, 2019

Contents

1. Chapter 1-Measures 22. Chapter 2-Integration 23. Chapter 3-Signed Measures and Differentiation 114. Chapter 4-Point Set Topology 235. Chapter 5-Elements of Functional Analysis 316. Chapter 6-Lp Spaces 427. Chapter 7-Radon Measures 528. Chapter 8-Elements of Fourier Analysis 529. Chapter 9-Elements of Distribution Theory 60

1

1. Chapter 1-Measures

Folland 1.10

Given a measure space (X,M, µ) and E ∈ M, define µE(A) = µ(A ∪ E) for A ∈ M. ThenµE is a measure.

Proof. First of all, µE(∅) = µ(∅ ∩ E) = µ(∅) = 0. Also, suppose A1, A2, ... ∈ M, µE(⋃∞i=1) =

µ(E∩(⋃∞i=1Ai)) = µ(

⋃∞i=1(Ai∩E)) =

∑∞i=1 µ(Ai∩E) =

∑∞i=1 µE(Ai). Then µE is a measure. �

2. Chapter 2-Integration

Folland 2.6

The supremum of an uncountable family of measurable R-valued functions on X can failto be measurable (unless the σ-algebra is really special).

Proof. Let X = R, M = L. We know that there is A ⊂ R such that A 6∈ L. Then we definefx : R → R by fx = χ{x} for each x ∈ R. Notice that f := supx∈A fx = supx∈A χ{x} = χA, and f

is not measurable since f−1([1,∞)) = A is non-measurable while [1,∞) is Borel. �

Folland 2.7

Suppose that for each α ∈ R we are given a set Eα ⊂ Eβ whenever α < β,⋃α∈REα = X

and⋂α∈REα = ∅. Then there is a measurable function f : X → R such that f(x) ≤ α on

Eα and f(x) ≥ α on Ecα for every α.

Proof Sketch. Show that f(x) := inf{α ∈ R : x ∈ Eα} works. �

Folland 2.8

If f : R→ R is monotone, then f is Borel measurable.

Proof. We shall show the conclusion by showing that for every a ∈ R, f−1([a,∞)) is an interval,which is Borel measurable. Given x, y ∈ f−1([a,∞)), for any z ∈ [x, y], since f is monotone,f(z) ∈ [f(x), f(y)] and thus z ∈ f−1([a,∞)). Thus f−1([a,∞)) is an interval and we finish theproof. �

Folland 2.9

Let f : [0, 1]→ [0, 1] be the Cantor function, and let g(x) = f(x) + x.(a) g is a bijection from [0, 1] to [0, 2], and h = g−1 is continuous from [0, 2] to [0, 1].(b) If C is the Cantor set, m(g(C)) = 1.(c) By Exercise 29 Chapter 1, g(C) contains a Lebesgue non-measurable set A. Let

B = g−1(A). Then B is Lebesgue measurable but not Borel.(d) There exists a Lebesgue measurable function F and a continuous function G on R

such that F ◦G is not Lebesgue measurable.

2

Proof. (a) Since f is monotone increasing, if y 6= x, without loss of generality we assume y >x and f(y) ≥ f(x). Then g(y) = f(y) + y > f(x) + x = g(x), so g is injective andmonotone increasing. g is continuous as a sum of two continuous functions, and g(0) =0 and g(1) = f(1) + 1 = 2, by intermediate value theorem the whole [0, 2] is mappedby g and g is surjective. To show that h is continuous, it suffices to show that g is openmap, and to that end it suffices to show that g maps open intervals to open intervals sinceevery open set is a disjoint union of them. But that is apparent since g is monotone andthus g(a, b) = (g(a), g(b)). Thus h is continuous on [0, 1].

(b) Since g is surjective, [0, 2] = g([0, 1] \ C) t g(C), so it suffices to show that m(g([0, 1] \ C)) =1. [0, 1] \ C is open since [0, 1] and C are both closed, [0, 1] \ C is just a disjoint union ofopen intervals on which f is constant. Therefore, g([0, 1] \ C) is a disjoint union of openintervals of the form (f(a) + a, f(b) + b) = (f(a) + a, f(a) + b) and thus m(a, b) =m(f(a)+a, f(b)+b). By countable additivity of m we get 1 = m([0, 1] \ C) = g([0, 1] \ C)and thus m(g(C)) = 1.

(c) Notice that B = g−1(A) ⊂ g−1(g(C)) = C since g is a bijection. Also m(B) ≤ m(C) = 0.By completeness of Lebesgue measure B is Lebesgue measurable. If B is Borel, h−1(B) =g(B) = A is Borel by continuity of h, a contradiction. Hence B is not Borel.

(d) Let F = χB and G = h. Then F ◦ G : [0, 2] → [0, 1] such that F is Lebesgue measurable(since B ∈ L ) and h is continuous. Notice that

(F ◦G)−1([1,∞)) = G−1 ◦ F−1([1,∞)) = G−1(B) = g(B) = A 6∈ Lso F ◦G is not Lebesgue measurable. This finishes the proof.

�

Folland 2.11

Suppose that f is a function on R×Rk such that f(x, ·) is Borel measurable for each x ∈ Rand f(·, y) is continuous for each y ∈ Rk. For n ∈ N, define fn as follows. For i ∈ Z letai = i/n, and for ai ≤ x ≤ ai+1 let

fn(x, y) =f(ai+1, y)(x− ai)− f(ai, y)(x− ai+1)

ai+1 − aiThen fn is Borel measurable on R × Rk and fn → f pointwise; hence f is Borel measur-able on R × Rk. Conclude by induction that every function Rn that is continuous in eachvariable separately is Borel measurable.

Proof. Observe that

|fn(x, y)− f(x, y)| = f(ai+1, y)(x− ai)− f(ai, y)(x− ai+1)− f(x, y)(x− ai+1) + f(x, y)(x− ai)ai+1 − ai

�

Folland 2.19

Suppose {fn} ⊂ L1(µ) and fn → f uniformly.(a) If µ(X) <∞, then f ∈ L1(µ) and

∫fn →

∫f

(b) If µ(X) =∞, the conclusions of (a) can fail.

Proof.3

(a) First of all, since fn → f uniformly, and fn is measurable for each n, f is measurable. Letε > 0, since fn → f uniformly, there is some N ∈ N such that when n ≥ N , |fn(x) −f(x)| < ε for all x. Then |f | < |fN |+ ε, and∫

|f | <∫|fN |+ εµ(X) <∞

which implies that f ∈ L1. Fix ε and N above, let g = max{|f1|, ..., |fN |, |f | + ε}, wecan see that g is clearly measurable and |fn| ≤ g for all n. Then we apply the dominatedconvergence theorem and get that

∫fn →

∫f .

(b) Let fn =1

nχ[−n,n]. First we show that fn → 0 uniformly. Let ε > 0 and N = d1ε e. When

n > N , |fn − 0| = |fn| < ε, so fn → 0 uniformly. However,∫fn = 2 6= 0 for all n, showing

that conclusions of (a) can fail.

�

Folland 2.21

Suppose fn, f ∈ L1 and fn → f a.e., then∫|fn − f | → 0 iff

∫|fn| →

∫|f |.

Proof. Suppose∫|fn − f | → 0, |

∫|fn − f | | → 0. Since |

∫|fn| −

∫|f | | ≤ |

∫|fn − f | | → 0,∫

|fn| →∫|f |.

Conversely, suppose∫|fn| →

∫|f |, then

∫|fn|+ |f | →

∫2|f |. Notice that fn, f ∈ L1, fn− f ∈ L1,

and thus |fn|, |f |, |fn|+ |f |, and |fn − f | ∈ L1. Also, fn → f a.e., it is clear that |fn − f | → 0 and|fn| + |f | → 2f a.e.. Since ||fn| − |f || ≤ |fn| + |f | and

∫|fn| + |f | →

∫2f , by previous problem∫

|fn − f | → 0. Then the proof is complete. �

Folland 2.25

Let f(x) = x−1/2 if 0 < x < 1, f(x) = 0 otherwise. Let {rn}∞1 be an enumeration of therationals, and set g(x) =

∑∞1 2−nf(x− rn).

(a) g ∈ L1(m), and in particular g <∞ a.e.(b) g is discontinuous at every point and unbounded on every interval, and it remains

so after any modification on a Lebesgue null set.(c) g2 <∞ a.e. but g2 is not integrable on any interval.

Proof. (a) For each n, we denote fn = 2−nf(x− rn) and have∫|fn(x)|dx =

∫2−n|f(x− rn)|dx = 2−n

∫ 1

0x−1/2dx = 2−n · 2 = 2−(n−1)

and thus∞∑1

∫|fn(x)|dx = 1 + 2−1 + 2−2 + ... = 2 <∞

Then ∫|g| =

∫ ∞∑1

|2−nf(x− rn)| =∞∑1

∫|fn(x)| <∞

and thus g ∈ L1(m). In particular g < ∞ a.e. since otherwise suppose g = ∞ on U suchthat m(U) > 0, we have

∫|g| ≥ ∞ ·m(U) =∞, a contradiction.

4

(b) We directly prove the result for slightly modifyed g, and the previous statement easilyfollows.

Now suppose we modify g on a m−null set F , and let h be the modified function suchthat h = g on R \ F . Pick x0 ∈ R, for all δ > 0, Bδ(x0) contains some rn. Since Bδ(x0) is

open, we choose k so large that Ik := (rn−1

k, rn+

1

k) ⊂ Bδ(x0). Notice that Ik \ Ik+1 con-

tains some x1 ∈ R \ F since it has positive measure. Then we choose such xi inductivelysuch that xi ∈ Ik+i−1 \ Ik+i. Then xi → rn. Then observe that 2−nf(xi − rn) → ∞by our knowledge about the function x−1/2. Then since we pick xi such that xi /∈ F ,h(xi) = g(xi) → ∞ as i → ∞. Then h is unbounded on Bδ(x0) and cannot be contin-uous at x0 (unbounded ocscillation). Also for an arbitrary interval I, pick x ∈ I and δ > 0small enough such that Bδ(x) ∈ I. Then f is unbounded on Bδ(x) and thus unboundedon I, and we show that the conclusions for g remains true after modifying g on a null set.

(c) Since g < ∞ a.e., g2 < ∞ a.e. Now we show that g is not integrable on any interval.Given an interval, we can assume it to be [a, b] since removing a null set won’t change theresult of integration. Then since g is non-negative,∫ b

a|g2| ≥

∫ b

a

∞∑1

2−2nf2(x− rn)

≥∫ b

a2−2nf2(x− rn) where rn ∈ [a, b]

≥ 2−2n∫ b

rn

f2(x− rn)dx

= 2−2n∫ b−rn

0f2(x)dx = 2−2n

∫ b−rn

0x−1dx

which fails to converge and thus tends to infinity. Then g2 is not integrable on I and ourproof is complete.

�

Folland 2.32

Suppose µ(X) <∞. If f and g are complex-valued measurable functions on X, define

ρ(f, g) =

∫|f − g|

1 + |f − g|dµ

Then ρ is a metric on the space of measurable functions if we identify functions that areequal a.e. and fn → f with respect to this metric iff fn → f in measure.

Proof. We first verify that ρ is a well-defined metric.

(a) ρ(f, g) =∫ |f − g|

1 + |f − g|dµ ≥

∫0 dµ = 0

(b) If f = g a.e.,

ρ(f, g) =

∫|f − g|

1 + |f − g|dµ =

∫0 dµ = 0

since a measure zero set doesn’t change the result of integration.

(c) ρ(f, g) =∫ |f − g|

1 + |f − g|=∫ |g − f |

1 + |g − f |= ρ(g, f)

5

(d) Notice that

|f − g|1 + |f − g|

+|g − h|

1 + |g − h|− |f − h|

1 + |f − h|

=|f − g|+ |g − h| − |f − h|+ |f − g||g − h|+ |f − g||g − h||f − h|+ |g − h||f − g|

(1 + |f − g)(1 + |g − h|)(1 + |f − h|)

≥ |f − g|+ |g − h| − |f − h|(1 + |f − g)(1 + |g − h|)(1 + |f − h|)≥ 0

so ρ(f, g) + ρ(g, h) ≥ ρ(f, h) and thus ρ(f, g) + ρ(g, h) ≥ ρ(f, h)

Then ρ is a metric.We then show that fn → f in this metric iff fn → f in measure. Suppose fn → f in measure. Let

ε > 0, and En := {x : |fn(x)− f(x)| ≥ ε

µ(X)}, which is possible since µ(X) <∞.

ρ(fn, f) =

∫|fn − f |

1 + |fn − f |dµ

=

∫|fn − f |

1 + |fn − f |χEndµ+

∫|fn − f |

1 + |fn − f |χ(En)cdµ

= µ(En) +ε

µ(X)µ(Ecn)

Then limn→∞ ρ(fn, f) ≤ 0+ε = ε. Since ε is arbitrary, this shows that actually limn→∞ ρ(fn, f) =0 and this measn that fn → f in this metric.Conversely, suppose fn → f with respect this metric. If fn 6→ f in measure, there are some δ, εsuch that there are infinitely many n such that µ{x : |fn(x) − f(x)| ≥ ε} ≥ δ, and we denote thisset En. Thus

ρ(fn, f) ≥∫En

1− 1

1 + |f − fn|≥∫En

1− 1

1 + εdµ =

ε

1 + εµ(En) ≥ εδ

1 + ε

for infinitely many n, contradiction to convergence in this metric and we finish our proof. �

Folland 2.34

Suppose |fn| ≤ g ∈ L1 and fn → f in measure.(a)

∫f → lim

∫fn

(b) fn → f in L1

Proof. I will slightly reverse the order of the two parts of the problem. Observe that if fn → f inL1,

∫|fn − f | → 0 and thus |

∫fn −

∫f | = |

∫(fn − f)| ≤

∫|fn − f | → 0 and

∫f → lim

∫fn. So

at this point it suffices to prove (b) to show both (a) and (b).fn → f in measure, so we can find a subsequence {fnj} that converges to f a.e. Since fnj → f

a.e. and |fnj | ≤ g ∈ L1, by dominated convergence theorem (also paragraph 2 sentence 1 of

page 61) fnj → f in L1. We now claim that {fn} actually converges to f in L1. Suppose in the

contrary that there exists ε > 0 such that there are infinitely many nk with∫|fnk − f | ≥ ε, then

we arrange them into a subsequence fnk . Since fn → f in measure, fnk → f in measure, andthere is a subsequence of fnk which we call {gn} for convenience that converges to f a.e. Then|gn − f | → 0 a.e. Since |fn| ≤ g, |gn − g| ≤ 2g ∈ L1. By dominated convergence theorem gn → g

6

in L1. However,∫|gn − f | ≥ ε, contradiction. This contradiction shows our claim that fn → f in

L1 and we are done. �

Folland 2.38

Suppose fn → f in measure and gn → g in measure.(a) fn + gn → f + g in measure(b) fngn → fg in measure if µ(X) <∞, but not necessarily if µ(X) =∞

Proof.

(a) Let ε > 0. Since fn → f in measure and gn → g in measure, as n→∞,

µ{|fn − f | ≥ε

2} → 0

µ{|gn − g| ≥ε

2} → 0

Notice that

{|fn + gn − fn − gn| ≥ ε} ⊂ {|fn − f |+ |gn − g| ≥ ε}

⊂ {|fn − f | ≥ε

2} ∪ {|gn − g| ≥

ε

2}

and thus

µ{|fn + gn − fn − gn| ≥ ε} ≤ µ{|fn − f | ≥ε

2}+ µ{|gn − g| ≥

ε

2}

Which tends to 0 as n→∞. Then fn + gn → f + g in measure.(b) We first prove a technical lemma:

Lemma 1. If fn → f in measure and µ(X) <∞, f2n → f in measure.

Proof of lemma. Since fn → f in measure, some subsequence {fnj} → f a.e. Then

{f2nj} → f2 a.e. Since µ(X) < ∞, f2nj → f2 in measure by Egoroff’s theorem1. We now

show that the whole sequence actually converge to f2 in measure by contradiction. Letε > 0. Suppose that there is some δ > 0 such that there are infinitely many n such that

µ{x : |f2n − f2| ≥ ε} ≥ δthen we arrange such f2n into a sequence f2ni . Notice that we still have fni → f in measure

and thus fn′i → f a.e for a further subsequence {fn′i}. Thus f2n′i→ f2 a.e. Since µ(X) <

∞, we should have f2n′i→ f2 in measure by our argument above, but by our construction

this cannot happen, a contradition. Then f2n → f2 in measure. �

By our lemma and part (a), (fn + gn)2 → (f + g)2 in measure (i.e. f2n + 2fngn + g2n →f2 + 2fg + g2 in measure) and −f2n → −f and −g2n → −g in measure. Then by part (a)again we have fngn → fg in measure.

If µ(X) = ∞, we give a counterexample of functions R → R and the measure is

Lebesgue measure. let f(x) = g(x) = x and fn(x) = gn(x) = x +1

nχ[n,n+1). Then

clearly fn → f and gn → g in measure. However,

fn(x)gn(x) = (x+1

nχ[n,n+1))

2 = x2 +2x

nχ[n,n+1) +

1

n2χ[n,n+1)

1This is an easy corollary of Egoroff’s theorem and is also mentioned as a side remark in Folland P62.

7

and however large n is, for x ∈ [n, n+ 1), we have f(x)g(x) ≥ x2 + 2, and µ[n, n+ 1) = 1.This shows that fngn doesn’t converge to fg in measure.

�

Folland 2.40

In Egoroff’s theorem, the hypothesis ”µ(X) < ∞” can be replaced by ”|fn| ≤ g for all n,where g ∈ L1(µ).”

Proof. Let E1(k) :=⋃∞m=1{x : |fm(x) − f(x)| ≥ 1

k}, as defined in the proof of Egoroff’s theorem.

The condition µ(X) < ∞ is used to justify µ(E1(k)) < ∞ for any fixed k, so if we can show thatµ(E1(k)) <∞, we are done. Observe that since |fn − f | ≤ 2g2,

E1(k) ⊂∞⋃m=1

{x : g(x) ≥ 1

2k} = {x : g(x) ≥ 1

2k}

Then if µ(E1) =∞ for some k,∫|g| ≥ 1

2k· ∞ =∞, a contradiction. �

Folland 2.41

If µ is σ-finite and fn → f a.e. there exists measurable E1, E2, ... ⊂ X such thatµ((⋃∞

1 Ej)c) = 0 and fn → f uniformly on each Ej .

Proof. We first suppose that µ(X) is actually finite. By Egoroff’s theorem, for each k there issome Ek such that µ(Ek)

c < 2−k and fn → f uniformly on Ek. Set Fn =⋃n

1 Ei, then {Fn} is

an increasing sequence and {(Fn)c} is a decreasing sequence. Also µ(F cn) ≤ µ(Ecn) < 2−k. Sinceµ(F c1 ) ≤ µ(X) <∞, by continuity from above,

µ(

∞⋃1

Ej)c = µ(

∞⋃1

Fj)c = µ(

∞⋂1

F cj ) = limj→∞

µ(F cj ) = 0

and fn → f uniformly on each Ej .If X is σ-finite, then X = X1 t X2 t ... such that µ(Xi) < ∞ for each i. On each i there are{Eik}∞k=1 such that µ(Xi − (

⋃∞k=1E

ik)) = 0 and fn → f uniformly on each Eik. Then consider

{Eik}∞k,i=1, fn → f uniformly on each Eik, and

µ(⋃i,k

Eik)c = µ[

∞⋃i=1

(Xi −∞⋃k=1

Eik)] =∞∑i=1

µ(Xi −∞⋃k=1

Eik) = 0

�

Folland 2.43

Suppose that µ(X) <∞ and f : X× [0, 1]→ C is a function such that f(·, y) is measurablefor each y ∈ [0, 1] and f(x, ·) is continuous for each x ∈ X.

(a) If 0 < ε, δ < 1 then Eε,δ = {x : |f(x, y)− f(x, 0)| ≤ ε for all y < δ} is measurable.(b) For any ε > 0 there is a set E ⊂ X such that µ(E) < ε and f(·, y) → f(·, 0)

uniformly on Ec as y → 0.

2We adopt the assumption in Folland that fn → f everywhere.

8

Proof. (a) Let Q ∩ [0, δ) be enumerated as {yn}. Then we define

Eε,n := {x : |f(x, yn)− f(x, 0)| ≤ ε}Notice that f(x, yn) and f(x, 0) are both measurable, |f(x, yn) − f(x, 0)| is measurable.Thus Eε,n is measurable. Consider

E :=∞⋂n=1

Eε,n =∞⋂n=1

{x : |f(x, yn)− f(x, 0)| ≤ ε}

Then E is measurable and clearly Eε,δ ⊂ E. Conversely, suppose x ∈ E. Then |f(x, yn) −f(x, 0)| ≤ ε for all yn. Let y < δ, then we can find {ynk} that converges to y. Since|f(x, ynk) − f(x, 0)| ≤ ε for all ynk , send nk to infinity we get |f(x, y) − f(x, 0)| ≤ εand thus x ∈ Eε,δ. Then E = Eε,δ and thus Eε,δ is measurable.

(b) Let ε > 0. Choose a monotone decreasing sequence {δn} ∈ [0, 1) such that δn → 0 fromabove, notice that Eε,δ1 ⊂ Eε,δ2 ⊂ Eε,δ3 .... Consider Fε,i = (Eε,δi)

c, then Fε,1 ⊃ Fε,2 ⊃Fε,3 ⊃ Fε,4... and µ(Fε,1) ≤ µ(X) <∞. Thus by continuity from above we have

µ(

∞⋂i=1

Fε,i) = limi→∞

µ(Fε,i) = 0

Therefore, for any γ > 0, there is some N ∈ N such that when n > N , µ(Fε,n) < γ,|f(x, y) − f(x, 0)| < ε for x ∈ (Fε,n)c and y ≤ δn. Therefore given γ > 0 and m ∈ Nwe can choose N(δ,m) such that µ(F 1

N(δ,m),m) < γ and |f(x, y) − f(x, 0)| < 1

N(δ,m) for

x ∈ F 1N(δ,m)

,m and y ≤ δm. Let E =⋂∞m=1 F 1

N(δ,m),m, then µ(E) <

1

mfor all m and for all

m, |f(x, y) − f(x, 0)| < 1

mfor all x provided y ≤ δm, which means that f(·, y) → f(·, 0)

uniformly as y → 0.�

Folland 2.46

Let X = Y = [0, 1], M = N = B[0,1]. µ = Lebesgue meaure, and ν = counting measure.

If D = {(x, x) : x ∈ [0, 1]} is the diagonal in X × Y , then∫∫

χD dµdν,∫∫

χD dνdµ, and∫χD d(µ× ν) are all unequal.

Proof. Notice that∫∫χD dµdν =

∫ [ ∫χD dµ(x)

]dν(y) =

∫ [ ∫(χD)y dµ(x)

]dν(y)

=

∫ [ ∫χDy dµ(x)

]dν(y) =

∫ [ ∫χ{(y,y)} dµ(x)

]dν(y) =

∫0 dν(y) = 0

and∫∫χD dνdµ =

∫ [ ∫χD dν(y)

]dµ(x) =

∫ [ ∫(χD)x dν(y)

]dµ(x) =

∫ [ ∫χDx dν(y)

]dµ(x)

=

∫ [ ∫χ{(x,x)}dν(y)

]dµ(x) =

∫dµ(x) = 1

9

Now notice that∫χD d(µ× ν) = µ× ν(D), and that

µ× ν(D) = inf{∞∑1

µ(Ai)ν(Bi) : Ai, Bi ∈ B[0,1], D ⊂∞⋃i=1

(Ai ×Bi)}

We want to show that there is some i such that µ(Ai) > 0 and ν(Bi) =∞, since then∑∞

1 µ(Ai)ν(Bi) >µ(Ai)ν(Bi) = ∞. And taking infimum we get µ × ν(D) = ∞. By definition of counting measure,ν(Bi) < ∞ if and only if |Bi| < ∞, and thus µ(Bi) = 0. Similaly, if µ(Ai) > 0, ν(Ai) = ∞since if |Ai| < ∞ there should be µ(Ai) = 0. Observe that

⋃iAi = [0, 1] and

⋃iBi = [0, 1], so⋃

i(Ai ∩ Bi) = [0, 1] and thus µ(⋃i(Ai ∩ Bi)) =

∑∞i=1 µ(Ai ∩ Bi) = 1. Then there is some i such

that µ(Ai ∩Bi) > 0, and this means that µ(Ai) ≥ µ(Ai ∩Bi) > 0 and ν(Bi) ≥ ν(Ai ∩Bi) =∞, sowe show what we want to show and finishes the proof. �

Folland 2.55

Let E = [0, 1]× [0, 1]. Investigate the existence and equality of∫E fdm

2,∫ 10

∫ 10 f(x, y)dxdy,

and∫ 10

∫ 10 f(x, y)dydx for f(x, y) = (x− 1

2)−3 if 0 < y < |x− 1

2 | and f(x, y) = 0 otherwise.

Proof. Notice that∫E f dm

2 =∫fχE dm2, and that f ≤ 0 on E1 = [0,

1

2] and f ≥ 0 on E2 =

[1

2, 1]. Now∫

Ef dm2 =

∫fχE dm2 =

∫(fχE)+ dm2 −

∫(fχE)− dm2 =

∫fχE2 dm

2 −∫fχE1 dm

2

Clearly fχE2 ∈ L+(m2). Then by Tonelli,∫fχE2 dm

2 =

∫ 1

12

[ ∫ 1

0f(x, y) dy

]dx =

∫ 1

12

[ ∫ x− 12

0(x− 1

2)−3 dy

]dx

=

∫ 1

12

(x− 1

2)−2 dx = lim

n→∞

∫ 1

12+ 1n

(x− 1

2)−2 dx (Monotone Convergence)

= limn→∞

n− 2 =∞

And therefore∫E f dm

2 doesn’t exist.

Then we examine the existence and equality of∫ 10

∫ 10 f(x, y) dydx. We have∫ 1

0

∫ 1

0f(x, y) dydx =

∫ 1

0

[ ∫ 1

0f(x, y) dy

]dx =

∫ 1

0

[ ∫fxχ[0,1] dy

]dx

=

∫ 1

0

[ ∫(fxχ[0,1])

+ dy −∫

(fxχ[0,1])− dy

]dx (1)

Notice that if x ≥ 1

2, (fxχ[0,1])

+ = (x− 1

2)−3χ[0,x− 1

2] and (fxχ[0,1])

− = 0; if x <1

2, (fxχ[0,1])

+ = 0

and (fxχ[0,1])− = (

1

2− x)−3χ[0, 1

2−x]. Also∫

(x− 1

2)−3χ[0,x− 1

2] dy =

∫ x− 12

0(x− 1

2)−3 dy = (x− 1

2)−2

10

and

−∫

(1

2− x)−3χ[0, 1

2−x] dy = −

∫ 12−x

0(1

2− x)−3 dy = (x− 1

2)−2

Therefore,

(1) =

∫ 1

0(x− 1

2)−2 dx =

∫((x− 1

2)−2χ[0,1])

+ dx−∫

((x− 1

2)−2χ[0,1])

− dx

=

∫ 1

12

(x− 1

2)−2 dx−

∫ 12

0(1

2− x)−2 dx

However, by our work of part one,∫ 1

12(x − 1

2)−2 dx = ∞ and therefore

∫ 10

∫ 10 f(x, y) dydx doesn’t

exist as well.We eventually investigate the existence and equality of

∫ 10

∫ 10 f(x, y) dxdy. Notice that f(x, y) =

0 for y ≥ 1

2. Then∫ 1

0

∫ 1

0f(x, y) dxdy =

∫ 1

0

[ ∫ 1

0f(x, y) dx

]dy =

∫ 12

0

[ ∫ 1

0f+ dx−

∫ 1

0f− dx

]dy

=

∫ 12

0

[ ∫ 1

12+yf+ dx−

∫ 12−y

0f− dx

]dy

=

∫ 12

0

[ ∫ 1

12+y

(x− 1

2)−3 dx−

∫ 12−y

0(1

2− x)−3 dx

]dy

=

∫ 12

0−2 · (4− y−2)− 2(y−2 − 4) dy = 0

and both∫ 1

20 −2 · (4− y−2) dy and

∫ 120 2(y−2 − 4) dy <∞. Then the integral exists and equals 0.

Now we complete the proof. �

3. Chapter 3-Signed Measures and Differentiation

Folland 3.4

If ν is a signed measure and λ, µ are positive measures such that µ = λ − ν, then λ ≥ v+

and µ ≥ ν−.

Proof. Observe that ν = λ − ν = ν+ − ν−, so λ − ν+ = µ − ν−. Let ρ := λ − ν+ = µ − ν−, thenclearly ρ is a signed measure. Notice that we finish our proof if we can show that ρ is a positivemeasure. Since v+ ⊥ v−, there are E and F such that X = E tF and ν+(F ) = 0 and ν−(E) = 0.Then for any measurable A ⊂ E, ρ(A) = µ(A) − ν−(A) = µ(A) ≥ 0, so E is positive; also forany measurable B ⊂ F , ρ(B) = λ(B) − ν+(B) = λ(B) ≥ 0, so F is also positive. Thus take anyC ∈M, ρ(C) = ρ(C ∩ E) + ρ(C ∩ F ) ≥ 0, so ρ is positive and we finish our proof. �

Folland 3.7

Suppose that ν is a signed measure on (X,M) and E ∈M.(a) ν+(E) = sup{ν(F ) : F ∈M, F ⊂ E} and ν−(E) = − inf{ν(F ) : F ∈M, F ⊂ E}.(b) |ν|(E) = sup{

∑n1 |ν(Ej)| : n ∈ N, E1, ..., En are disjoint, and ∪n1 Ej = E}

11

Proof. (a) Let µ = sup{ν(F ) : F ∈ M, F ⊂ E} and λ = − inf{ν(F ) : F ∈ M, F ⊂ E}.The idea is to show that µ, λ are well-defined (positive) measures such that µ ⊥ λ andν = µ − λ. Then by uniqueness of Jordan decomposition we get the result. We first showthat µ and λ are positive. This is true since for every measurable E, µ(E) ≥ ν(∅) forevery E since ∅ ∈ M and ∅ ⊂ E. Similarly we can show λ(E) ≥ 0. We then show µ isa well-defined measure. µ(∅) = ν(∅) = 0. For countable additivity, suppose E1, E2, . . .disjoint, we denote E =

⋃iEi for convenience. Then

µ(E) = sup{ν(F ) : F ∈M, F ⊂ E} = sup{ν(F ∩ E) : F ∈M, F ⊂ E}

= sup{ν(∞⋃i=1

(F ∩ Ei)) : F ∈M, F ⊂ E} = sup{∞∑i=1

ν(F ∩ Ei) : F ∈M, F ⊂ E}

= sup{∞∑i=1

ν(Fi) : Fi ∈M, Fi ⊂ Ei} =

∞∑i=1

sup{ν(Fi) : Fi ∈M, Fi ⊂ Ei}

if we notice F ⊂ E iff Fi = F ∩ Ei ⊂ Ei for each i. Similarly we can show that λ is a well-defined positive measure. We then show that µ ⊥ λ. By Hahn decomposition, we haveX = P t N where P is positive and N is negative. Then for any E ∈ M, we write it asE = (E ∩ P ) t (E ∩ N). Notice that F ⊂ E iff F = F1 t F2, where F1 ⊂ E ∩ P andF2 ⊂ E ∩N , so

µ(E) = sup{ν(F ) : F ∈M, F ⊂ E} = sup{ν(F1 t F2) : F1, F2 ∈M, F1 ⊂ E ∩ P, F2 ⊂ E ∩N}= sup{ν(F1) + ν(F2) : F1, F2 ∈M, F1 ⊂ E ∩ P, F2 ⊂ E ∩N}= sup{ν(F1) : F1 ∈M, F1 ⊂ E ∩ P}+ sup{ν(F2) : F2 ∈M, F2 ⊂ E ∩N}= sup{ν(F1) : F1 ∈M, F1 ⊂ E ∩ P} = sup{ν(F ∩ P ) : F ∈M, F ⊂ E}= ν(E ∩ P )

and similarly we have λ(E) = ν(E ∩ N) and µ(E) + λ(E) = ν(E). The result follows byuniqueness of Jordan decomposition.

(b) First observe that for E ∈M,

|ν(E)| = |ν+(E)− ν−(E)| ≤ |ν+(E)|+ |ν−(E)| = |ν|(E)

Then by countable additivity,

|ν|(E) = sup{n∑1

|ν|(Ej) : n ∈ N, E1, ..., En are disjoint, and ∪n1 Ej = E}

≥ sup{n∑1

|ν(Ej)| : n ∈ N, E1, ..., En are disjoint, and ∪n1 Ej = E}

Conversely, if we apply the Hahn decomposition used above, we get |ν|(E) = |ν|(E ∩ P ) +|ν|(E ∩N) where P is positive and N is negative. Notice that

0 ≤ |ν|(E ∩ P ) = ν+(E ∪ P ) + ν−(E ∪ P ) = ν+(E ∪ P ) = ν+(E ∪ P )− ν−(E ∪ P ) = v(E ∪ P )

So |ν|(E ∩ P ) = |ν(E ∪ P )|. Similarly |ν|(E ∩N) = |ν(E ∪N)|. Therefore

|ν|(E) = |ν(E ∩ P )|+ |ν(E ∩N)| (where (E ∩ P ) ∪ (E ∩N) = E)

≤ sup{n∑1

|ν(Ej)| : n ∈ N, E1, ..., En are disjoint, and ∪n1 Ej = E}

Two directions combined, we prove the equality and finish the proof.�

12

Folland 3.12

For j = 1, 2, let µj , νj be σ-finite measures on (Xj ,Mj) such that vj � µj . Then ν1×ν2 �µ1 × µ2 and

d(ν1 × ν2)(µ1 × µ2)

(x1, x2) =dν1dµ1

(x1)dν2dµ2

(x2)

Proof. First we show that ν1 × ν2 � µ1 × µ2. Suppose µ1 × µ2(E) = 0 for some E ∈ M1 ⊗M2,then χE ∈ L+. By Tonelli,

0 = µ1 × µ2(E) =

∫∫χE dµ1dµ2 =

∫µ1(E

y) dµ2(y)

Then µ1(Ey) = 0 for µ1-a.e. y and since ν1 � µ1, ν1(E

y) = 0 for µ2-a.e. y. Since ν2 � µ2,ν1(E

y) = 0 for ν2-a.e. y. By Tonelli again,

(ν1 × ν2)(E) =

∫χE d(ν1 × ν2) =

∫ (∫χE dν1(x)

)dν2(y)

=

∫ν1(E

y) dν2(y) = 0

Thus ν1 × ν2 � µ1 × µ2.Then we show that

d(ν1 × ν2)d(µ1 × µ2)

(x1, x2) =dν1dµ1

(x1)dν2dµ2

(x2)

We claim thatdνidµi≥ 0 a.e. Suppose in the contrary that

dνidµi

< 0 on some E with µi(E) > 0.

Then

νi(E) =

∫Ef dµi < 0

contradicting to our assumption that νi is positive, so the claim is true. Thendν1dµ1

(x1)dν2dµ2

(x2) ∈

L+(M1 ×M2), and by Tonelli, we get for any E ∈M1 ⊗M2,∫χE

d(ν1 × ν2)d(µ1 × µ2)

d(µ1 × µ2) =

∫χE d(ν1 × ν2) =

∫∫χE dν1dν2 =

∫∫χE (

dν1dµ1

dµ1)(dν2dµ2

dµ2)

=

∫χE

dν1dµ1

(x1)dν2dµ2

(x2) d(µ1 × µ2)

Then we have shown thatd(ν1 × ν2)d(µ1 × µ2)

=dν1dµ1

(x1)dν2dµ2

(x2)

a.e. and thus prove the result since we identify the derivative functions with their equivalenceclasses. �

Folland 3.16

µ, ν are measures on (X,M) with ν � µ, and let λ = µ+ ν. If f =dν

dλ, show that 0 ≤ f <

1 µ-a.e. anddν

dµ=

f

1− f.

13

Proof. We first show that 0 ≤ f < 1 µ-a.e. Suppose in the contrary that f ≥ 1 on some E withµ(E) > 0,

ν(E) =

∫Ef dλ ≥ λ(E) = µ(E) + ν(E)

But this implies that µ(E) ≤ 0, a contradiction.

We then show thatdν

dµ=

f

1− f. We claim that µ � λ and λ � ν. If λ(E) = µ(E) + ν(E) = 0,

µ(E). Conversely, if µ(E) = 0, since ν � µ, ν(E) = 0 and thus λ(E) = ν(E) + µ(E) = 0. So our

claim is true and(dµdλ

)(dλdµ

)= 1. By additivity of derivatives, we have

dµ

dλ+dν

dλ=dλ

dλ= 1 and

thereforedµ

dλ= 1− dν

dλ. Then

f

1− f=

dν/dλ

1− dν/dλ=dν/dλ

dµ/dλ=dν

dλ· dλdµ

Since we have ν � λ and λ� µ,f

1− f=dν

dλ· dλdµ

=dν

dµand we finish our proof. �

Folland 3.18

Let ν be a complex measure on (X,M). L1(ν) = L1(|ν|), and if f ∈ L1(ν), then |∫fdν| ≤∫

|f |d|ν|.

Proof. We first show that L1(ν) = L1(|ν|). By Lebesgue-Radon-Nikodym, we have dν = fdµwhere µ is a positive measure, and thus d|ν| = |g|dµ. If f ∈ L1(|ν|),

∞ >

∫|f |d|ν| =

∫|f ||g|dµ =

∫|fg|dµ ≥

∣∣∣∣ ∫ |f |gdµ∣∣∣∣ =

∫|f |dν (2)

showing that f ∈ L1(ν) and that L1(|ν|) ⊂ L1(ν). Conversely if f ∈ L1(ν), f ∈ L1(|νr| + |νi|).Then since |ν| ≤ |vr|+ |vi| are all positive measures we have

∞ >

∫|f |d(|vr|+ |vi|) ≥

∫|f |d|ν|

showing that f ∈ L(|ν|). Thus L1(ν) = L1(|ν|). Moreover by (1) we have∣∣∣∣ ∫ fdν

∣∣∣∣ ≤ ∫ |f |dν ≤ ∫ |f |d|ν|which finishes the proof. �

Remark. This is basically a formal check using definitions.

Folland 3.20

If ν is a complex measure on (X,M) and ν(X) = |ν|(X), then ν = |ν|.

Proof. By Lebesgue-Randon-Nikodym we have dν = fdµ for some positive measure µ, and thusd|ν| = |f |dµ. Then ∫

fdµ =

∫|f |dµ =⇒

∫(|f | − f)dµ = 0 (3)

14

Since |f | − f ≥ 0, (2) implies that |f | − f = 0 µ-a.e., and thus |f | = f since we identify f asequivalence class in L1. Thus d|ν| = dν and thus |ν| = ν, as desired. �

Folland 3.21

Let ν be a complex measure on (X,M). If E ∈M, define

µ1(E) = sup

{ n∑1

|ν(Ej)| : n ∈ N, E1, ..., En disjoint and E =n⋃j=1

Ej

}

µ3(E) = sup

{∣∣∣∣ ∫Efdν

∣∣∣∣ : |f | ≤ 1

}

Proof. We first show that µ1 ≤ µ3. Define f :=∑n

j=1 sgn(ν(Ei))χEi , and since |sgn(ν(Ei))| ≤ 1,

|f | ≤ 1. Thus we have∣∣∣∣ ∫Efdν

∣∣∣∣ =

∣∣∣∣ n∑j=1

∫Ei

sgn(ν(Ei))dν

∣∣∣∣ =

∣∣∣∣ n∑j=1

sgn(ν(Ei))ν(Ei)

∣∣∣∣ =n∑j=1

|ν(Ei)|

And taking supremum over {En}n satisfying the conditions of µ1 we get µ1(E) ≤ µ3(E) and

thus µ1 ≤ µ3. We then show that µ3 = |ν|. Define f = dν/d|ν| and by proposition 3.13b |f | =|f | = 1 ≤ 1. Moreover, Lebesgue-Radon-Nikodym gives dν = gdµ for some positive µ, and thusdν = fdν = fdν and d|ν| = |g|dν. Then

dν/d|ν|dν =fdµ · fdµ|f |dµ

=|f |2(dµ)2

|f |dµ= |f |dµ = d|ν|

and thus∫E fdν =

∫E d|ν| = |ν|(E), showing that µ3 ≥ |ν|. On the other hand µ3(E) ≤

sup{∫E |f |d|ν| : |f | ≤ 1} ≤

∫E d|ν| = |ν|(E) and thus µ3 = |ν|. Eventually we show that ν3 ≤ ν1.

Let φ :=∑n

1 ckχEk where |ck| ≤ 1 for all k, Eis are disjoint and⋃ni=1Ei = E. We have∣∣∣∣ ∫

Eφdν

∣∣∣∣ ≤ n∑k=1

∣∣∣∣ck ∫Ek

χEkdν

∣∣∣∣ =n∑k=1

|ck||ν(Ek)| ≤n∑k=1

|ν(Ek)| ≤ µ1(E)

Let |f | ≤ 1, choose 〈φn〉n simple functions that approximate f from below (meaning that |φn| ≤ 1for all n) in L1 since simple functions are dense in L1 and apply dominated convergence theorem(since |φn| ≤ χE ∈ L1 for all n) we have |

∫E fdν| ≤ µ1(E). Taking supremum over f we have

µ3(E) ≤ µ1(E). Eventually we have µ1 = µ3 = ν, as desired.

�

Folland 3.23

A useful variant of the Hardy-Littlewood maximal function is

H∗f(x) = sup

{1

m(B)

∫B|f(y)| dy : B is a ball and x ∈ B

}Show that Hf ≤ H∗f ≤ 2nHf .

Proof. First inequality: We first observe that x ∈ B(r, x) for any r > 0. Thus

{B(r, x) : r > 0} ⊂ {B : B is a ball and x ∈ B}15

and {1

m(B(r, x))

∫B(r,x)

|f(y)| dy : r > 0

}⊂{

1

m(B)

∫B|f(y)| dy : B is a ball and x ∈ B

}Therefore

sup

{1

m(B(r, x))

∫B(r,x)

|f(y)| dy : r > 0

}≤ sup

{1

m(B)

∫B|f(y)| dy : B is a ball and x ∈ B

}which means Hf ≤ H∗f .Second inequality: We observe that

H∗f(x) = supr>0

{1

m(Br)

∫Br

|f(y)| dy : Br is a ball of radius r such that x ∈ Br}

= supr>0

{1

m(B(r, x))

∫Br

|f(y)| dy : Br is a ball of radius r such that x ∈ Br}

≤ supr>0

1

m(B(r, x))

∫B(2r,x)

|f(y)| dy (since B(2r, x) contains all Br 3 x)

= supr>0

2n

m(B(2r, x))

∫B(2r,x)

|f(y)| dy

= 2nHf(x)

so we finish the proof. �

Folland 3.24

If f ∈ L1loc and f is continuous at x, then x is in the Lebesgue set of f .

Proof. Let ε > 0. Since f is continuous at x, there is δ > 0 such that whenever |y − x| < δ,|f(y)− f(x)| < ε. Then when r < δ,

1

m(B(r, x))

∫B(r,x)

|f(y)− f(x)| dx < ε m(B(r, x))

m(B(r, x))= ε

and thus

limr→0

1

m(B(r, x))

∫B(r,x)

|f(y)− f(x)| dy = 0

which means that x ∈ Lf and we finish the proof. �

Folland 3.25

If E is a Borel set in Rn, the density DE(x) of E at x is defined as

DE(x) = limr→0

m(E ∩B(r, x))

m(B(r, x))

whenever the limit exists.(a) Show that DE(x) = 1 for a.e. x ∈ E and DE(x) = 0 for a.e. x ∈ Ec.(b) Find examples of E and x such that DE(x) is a given number α ∈ (0, 1), or such

that DE(x) does not exist.

16

Proof. (a) We define f := χE . Then clearly f ∈ L1loc and thus m((Lf )c) = 0. Therefore, for

a.e. x ∈ E, x ∈ Lf and thus

limr→0

1

m(B(r, x))

∫B(r,x)

|f(y)− f(x)| dy = limr→0

1

m(B(r, x))

∫B(r,x)

|f(y)− 1| dy = 0

This implies

0 = limr→0

1

m(B(r, x))

∫B(r,x)

(f(y)− 1) dy

= limr→0

1

m(B(r, x))

∫B(r,x)

f(y)− limr→0

1

m(B(r, x))

∫B(r,x)

1 dy

= limr→0

1

m(B(r, x))

∫χE∩B(r,x) − lim

r→0

1

m(B(r, x))m(B(r, x))

= limr→0

m(E ∩B(r, x))

m(B(r, x))− 1

for a.e x ∈ E and thusm(E ∩B(r, x))

m(B(r, x))= 1 for a.e. x ∈ E.

For a.e. x ∈ Ec, x ∈ Lf and thus

limr→0

1

m(B(r, x))

∫B(r,x)

|f(y)− f(x)| dy = limr→0

1

m(B(r, x))

∫B(r,x)

|f(y)− 0| dy = 0

Using an argument similar to the one above we can show thatm(E ∩B(r, x))

m(B(r, x))= 0 for a.e.

x ∈ Ec.(b) Let E = [0, 1] and x = 0, then

limr→0

m(E ∩B(r, x))

B(r, x)= lim

r→0

m[0, r)

m(−r, r)=

1

2∈ (0, 1)

For the other example,

E = {0} t∞⊔n=1

(1

22n+1,

1

22n

)We want to find two subsequences converging to different limit, and thus show that the

limit doesn’t exist. First we let {rk} := { 1

22k}, x = 0, then rk → 0, and m(E ∩B(rk, x)) =

1/22k+1

1− 1/4=

4

3· 1

22k+1=

1

3· 1

22k−1. So

limk→∞

m(E ∩B(rk, x))

m(B(rk, x))= lim

k→∞

13 ·

122k−1

122k−1

=1

3

Then we let {r′k} be1

22k+1, and we have m(E∩B(r′k, x)) =

1/22k+2

1− 1/4=

4

3· 1

22k+2=

1

6· 1

22k−1.

So

limk→∞

m(E ∩B(r′k, x))

m(B(r′k, x))= lim

k→∞

16 ·

122k−1

122k−1

=1

6

Then {rk} and {r′k} both converge to 0, but

limk→∞

m(E ∩B(rk, x))

m(B(rk, x))6= lim

k→∞

m(E ∩B(r′k, x))

m(B(r′k, x))

17

showing that DE(0) doesn’t exist and we finish the proof.�

Folland 3.28

If F ∈ NBV , let G(x) = |µF |((−∞, x]). Prove that |µF | = µTF by showing that G = TFvia the following steps.

(a) From the definition of TF , TF ≤ G(b) |µF (E)| ≤ µTF (E) when E is an interval, and hence when E is a Borel set.(c) |µF | ≤ µTF , and hence G ≤ TF .

Proof. (a) By definition,

TF = sup{n∑j=1

|F (xj)− F (xj−1)| : n ∈ N,−∞ < x0 < ... < xn = x}

= sup{n∑j=1

|µF (−∞, xj ]− µF (−∞, xj−1]| : n ∈ N,−∞ < x0 < ... < xn = x}

= sup{n∑j=1

|µF (xj , xj−1]| : n ∈ N,−∞ < x0 < ... < xn = x} (1)

Since (xj−1, xj ] (j = 1, ..., n) are disjoint, and⋃nj=1(xj−1, xj ] = (x0, x], (1) ≤ |µF |(x0, x] ≤

|µF |(−∞, x] and thus TF ≤ G.(b) We first suppose E = (a, b] is an h-interval, then

|µF (E)| = |µF (a, b]| = |µF (−∞, b]− µF (−∞, a]| = |F (b)− F (a)|

≤ sup{n∑j=1

|F (xj)− F (xj−1)| : n ∈ N, a = x0 < ... < xn = b}

= TF (b)− TF (a) = µTF (−∞, b]− µTF (−∞, a]

= µTF (a, b] = µTF (E)

In general, for E ∈ BR, by regularity

|µF (E)| = | inf{∞∑j=1

µF (aj , bj ] : E ⊂∞⋃i=1

(aj , bj ]}| ≤ inf{∞∑j=1

∣∣µF (aj , bj ]∣∣ : E ⊂

∞⋃i=1

(aj , bj ]} (2)

But |µF (aj , bj)| ≤ µTF (aj , bj ] for every j, so

(2) ≤ inf{∞∑j=1

µTF (aj , bj ] : E ⊂∞⋃i=1

(aj , bj ]} = µTF (E)

and we prove the result.(c) By exercise 21, for E ∈ BR,

|µF (E)| ≤ sup{n∑i=1

|µF (Ei)| : E =

n⊔i=1

Ei} ≤ sup{n∑i=1

µTF (Ei) : E =

n⊔i=1

Ei}

= sup{µTF (

n⊔i=1

Ei) : E =

n⊔i=1

Ei} = µTF (E)

18

In particular,

G(x) = |µF |(−∞, x] ≤ µTF (−∞, x] = TF (x)− TF (−∞) = TF (x)

Since G(x) = |µF |(−∞, x] for a complex Borel measure |µF | , G ∈ NBV and |µF | = µG. ThenTF = G ∈ NBV and µTF = µG = |µF |. �

Folland 3.31

Let F (x) = x2 sin(x−1) and G(x) = x2 sin(x−2) for x 6= 0, and F (0) = G(0) = 0.(a) F and G are differentiable everywhere (including x = 0)(b) F ∈ BV ([−1, 1]), but G 6∈ BV ([−1, 1]).

Proof. (a) When x 6= 0, both F and G are compositions of elementary functions and are thusdifferentiable, so it suffices to verify for x = 0 in both cases.

limh→0

F (x+ h)− F (0)

h= lim

h→0

h2 sin(1/h)

h= lim

h→0h sin(1/h) = 0

since | sin(1/h)| ≤ 1. Also,

limh→0

G(x+ h)−G(0)

h= lim

h→0

h2 sin(1/h2)

h= lim

h→0h sin(1/h2) = 0

since | sin(1/h2)| ≤ 1. Thus F and G are differentiable everywhere.(b) F ′(x) = 2x sin(x−1) − cos(x−1) when x 6= 0 and F ′(0) = 0. Then |F ′| ≤ 3 in [−1, 1]

and thus F ∈ BV [−1, 1]. For G, choose xj =

√2

(n− j + 1)π(j = 1, 2, ..., n), notice that

x0 > 0 and xn < 1. Then

TF (1)− TF (−1) ≥n∑j=0

|G(xj)−G(xj−1)|

=

n∑j=0

∣∣∣∣ 2

(n− j + 1)πsin

n− j2

π − 2

(n− j + 2)πsin

n− j + 1

2π

∣∣∣∣ ≥ n+2∑j=2

2

jπ

This partition gets finer as n increses, but TF (1) − TF (−1) ≥∑n+2

j=22jπ → ∞ as n → ∞

since harmonic series diverge. Thus G 6∈ BV ([−1, 1]).�

Folland 3.32

If F1, F2, ..., F ∈ NBV and Fj → F pointwise, then TF ≤ lim inf TFj .

Proof. Fix x ∈ R, let N ∈ N and (x0, x1, ..., xN ) be a partition such that −∞ < x0 < ... < xN =x. Thenn∑i=1

|F (xi)− F (xi−1)| = lim infj→∞

N∑i=1

|Fj(xi)− Fj(xi−1)|

≤ lim infj→∞

{sup{

n∑i=1

|Fj(xi)− Fj(xi−1)| : n ∈ N,−∞ < x0 < ... < xn = x}}

≤ lim infj→∞

TFj

19

Since N and the partition are arbitrary, taking supremum over them we get TF ≤ lim inf TFj . �

Folland 3.36

Let G be a continuous increasing function on [a, b] and let G(a) = c, G(b) = d.(a) If E ⊂ [c, d] is a Borel set, then m(E) = µG(G−1(E)).

(b) If f is a Borel measure and integrable function on [c, d], then∫ dc f(y)dy =∫ b

a f(G(x))dG(x). In particular,∫ dc f(y)dy =

∫ ba f(G(x))G′(x)dx if G is absolutely

continuous.(c) The validity of (b) may fail if G is merely right continuous rather than continuous.

Proof. (a) We first consider the case where E = (c1, d1] is an h-interval in [c, d]. Since G iscontinuous increasing, we can conclude using intermediate value theorem that [c, d] =G[a, b].Claim 1. For any interval I, G−1(I) is also an interval.

Proof of Claim. Suppose I is an interval. For x < y in G−1(I), if z ∈ [x, y], G(z) ∈[f(x), f(y)] ⊂ I since G is continuous increasing. Then z ∈ G−1(I) and I is an inter-val. �

Claim 2. G−1(c1, d1] = (a1, b1] where G(a1) = c1 and G(b1) = d1.

Proof of Claim. [a, b] = G−1(−∞, c1) tG−1(c1, d1] tG−1(d1,+∞). Since G is continuous,G pulls back open (closed) sets to open (closed) sets. Combined with claim 1, we concludethat G−1(−∞, c1] = [a, a1] and G−1(d1,+∞) = (b1, b]. This forces G−1(c1, d1] = (a1, b1].Observe that G−1(c1) 6= ∅, and G is continuous increasing, c1 = supG[a, a1] = G(a1).Similarly, d1 = supG(a1, b1] = G(b1). This establishes the claim. �

Then µG(G−1(E)) = µG(a1, b1] = G(b1)−G(a1) = d1 − c1 = m(E).For E ∈ BR, by regularity,

m(E) = inf{∞∑j=1

(cj , dj ] : E ⊂∞⋃j=1

(cj , dj ]} = inf{∞∑j=1

µG(G−1(cj , dj ]) : E ⊂∞⋃j=1

(cj , dj ]} (1)

Shrinking the intervals if necessary, we may assume (cj , dj ] ⊂ [c, d]. By above claim,

(1) = inf{∞∑j=1

µG(aj , bj ] : E ⊂∞⋃i=1

(G(aj), G(bj)]} (2)

Claim 3. E ⊂⋃∞j=1(G(aj), G(bj)] if and only if G−1(E) ⊂

⋃∞j=1(aj , bj ].

Proof of Claim. Suppose G−1(E) ⊂⋃∞j=1(aj , bj ],

E ⊂ G(∞⋃j=1

(aj , bj ]) ⊂∞⋃i=1

G(aj , bj ] =∞⋃j=1

(cj , dj ]∞⋃j=1

(G(aj), G(bj)]

by claim 2. Conversely, suppose E ⊂⋃∞j=1(G(aj), G(bj)],

G−1(E) ⊂ G−1(∞⋃j=1

(cj , dj ]) =

∞⋃j=1

G−1(cj , dj ] =

∞⋃j=1

(aj , bj ]

Then we prove the claim. �20

Then by claim 2 and regularity of µG,

(2) = inf{∞∑j=1

µG(aj , bj ] : G−1(E) ⊂∞⋃j=1

(aj , bj ]} = µG(G−1(E))

and we prove the result.(b) Dealing with f+ and f− separately, we may assume f ∈ L+. Choose a sequence of simple

functions {φn} ↑ f . We may assume φn =∑m

i=1 aiχEi vanishes outside [c, d], i.e. Ei ⊂[c, d] for all i. Then∫ d

cφn(y)dy =

m∑i=1

aim(Ei) =

m∑i=1

aiµG(G−1(Ei))

=

m∑i=1

ai

∫ b

aχG−1(Ei)dG (since G−1(Ei) ⊂ [a, b] ∀ i)

=

∫ b

a

m∑i=1

aiχG−1(Ei)dG (3)

Observation 4. x ∈ G−1(Ei) if and only if G(x) ∈ Ei.Then χG−1(Ei) = χEi(G(x)) and (3) =

∫φn(G(x))dG(x). Sending n → ∞, by mono-

tone convergence theorem, we get∫ dc f(y)dy =

∫ ba f(G(x))dG(x). In particular, if G is

absolutely continuous, G is differentiable a.e., so dG(x) = G′(x)dx a.e. and∫ dc f(y)dy =∫ b

a f(G(x))G′(x)dx.(c) We define G : [0, 3] → [1, 3] such that G(x) = 0 on [0, 1), 1 on [1, 2), and 2 on [2, 3].

By extending G to be 0 at (−∞, 0) and 2 at (3,+∞) we may assume G ∈ NBV . Thensuppose f ≡ 1,

∫[0,3] f(x)dx = 3. However,∫

[0,3]f(G(x))dG(x) =

∫(−∞,3]

f(G(x))dG(x)

=

∫(−∞,1]

f(G(x))dG(x) +

∫(1,2]

f(G(x))dG(x) +

∫(2,3]

f(G(x))dG(x)

= µG(−∞, 1] + µG(1, 2] + µG(2, 3]

= G(1) + (G(2)−G(1)) + (G(3)−G(2))

= 1 + 1 + 0 = 2 6= 3

showing that the conclusion of (b) may fail.�

Folland 3.39

If {Fj} is a sequence of nonnegative functions on [a, b] such that F (x) =∑∞

1 Fj(x) < ∞for all x ∈ [a, b], then F ′(x) =

∑∞1 F ′j(x) for a.e. x ∈ [a, b].

Proof. Without loss of generality we may assume that Fj ∈ NBV for all j. This is because wecan extend Fj to R such that F (x) = F (b) for all x ≥ b and F (x) = F (a) for all x ≤ a, and theresulted function is still bounded increasing and thus in BV . Consider G(x) = F (x+) − F (−∞),then G ∈ NBV and G′ = F ′ m-a.e. Then there is a Borel measure µFj such that Fj(x) =

21

µFj (−∞, x]. Consider the Leabesgue-Radon-Nikodym representation of µFj

dµFj = dλj + gjdm ⇐⇒ µFj = λj +

∫gjdm (1)

Here λj is a positive measure since µ and∫gj are both positive, and gj ∈ L1(m). We set µ =∑∞

j=1 µFj , λ =∑∞

j=1 λj , and g =∑∞

j=0 gj . Now

dµF = dλ+ g dm ⇐⇒ µF = λ+

∫g dm (2)

Observe that λ ⊥ m since m(E) = 0 implies λj(E) = 0 (since λj ⊥ m) and thus λ(E) =∑∞j=1 λj(E) = 0. Also g ∈ L1(m) since∫

|g| dm =

∫g dm ≤ µ(X) =

∞∑j=1

µFj (X) =∞∑j=1

Fj(b) = F (b) <∞

Thus (2) is the a.e. unique LRN representation of µ. On the other hand,

F (x) =

∞∑j=1

µj(−∞, x] = (

∞∑j=1

µj)(−∞, x] = µ(−∞, x] <∞

for some finite Borel measure µ. Then F ∈ NBV and (2) is the a.e. unique LRN representation

of µ. Then F ′j(x) = limr→0µFj (Er)

m(Er)= gj(x) a.e. for Er = (x, x+r] or (x−r, x] which shrink nicely

to x. Also we have F ′(x) = limr→0µF (Er)

m(Er)= g(x) a.e. This means that F ′ = g =

∑∞j=1 gj =∑∞

j=1 Fj a.e. and this finishes the proof. �

Folland 3.41

Let A ⊂ [0, 1] be a Borel set such that 0 < m(A∩I) < m(I) for every subinterval I of [0, 1].(a) Let F (x) = m([0, x] ∩ A). Then F is absolutely continuous and strictly increasing

on [0, 1], but F ′ = 0 on a set of positive measure.(b) Let G(x) = m([0, x] ∩ A)−m([0, x] \ A). Then G is absolutely continuous on [0, 1],

but G is not monotone on any subinterval of [0, 1].

Proof. (a) As we did in the previous problem, we may assume that F ∈ NBV . Let ε > 0.

Take δ = ε. For finite disjoint (a1, b1), ..., (aN , bN ), if∑N

1 (bj − aj) < δ,

N∑1

|F (bj)− F (aj)| =N∑1

m((aj , bj ] ∩A) <N∑1

m((aj , bj ]) =N∑1

(bj − aj) < ε

so F is absolutely continuous. For x1 < x2 in [0, 1],

F (x1)− F (x1) = m([0, x2] ∩A)−m([0, x1] ∩A) = m((x1, x2] ∩A) > 0

so F is strictly increasing. To establish the last part, note first that since F is absolutely

continuous, F (1) = F (1)− F (0) =∫ 10 F

′(t)dt. Also

F (1) = m([0, 1] ∩A) =

∫ 1

0χA dm

Therefore F ′ = χA a.e. on [0, 1]. Since m([0, 1] \ A) = m[0, 1] −m([0, 1] ∩ A) > 0, χA = 0on a set of positive measure and thus F ′ = 0 on a set of positive measure.

22

(b) Let ε > 0 and δ = ε, then for finite disjoint (a1, b1), ..., (aN , bN ) in [0, 1], if∑N

1 (bj−aj) < δ,

|G(bj)−G(aj)| = |m([0, bj ] ∩A)−m([0, aj ] ∩A)−m([0, bj ] \ A) +m([0, aj ] \ A)|= |m((aj , bj ] ∩A)−m((aj , bj ] \ A)| ≤ |m((aj , bj ] ∩A) +m((aj , bj ] \ A)|= m(aj , bj) = (bj − aj)

and thus∑N

1 |G(bj) − G(aj)| ≤∑N

1 (bj − aj) < δ = ε, showing that G is absolutelycontinuous on [0, 1]. Suppose G is mototone on some interval I which can be assumed to(a, b), either G′ ≥ 0 or G′ ≤ 0. However, since G is absolutely continuous,∫ b

aG′(x)dm = G(b)−G(a) = m((a, b] ∩A)−m((a, b] \ A) =

∫ b

aχA − χAc

Then G′(x) = χA − χAc m-a.e. However since m(A ∩ I) < m(I), I contains a positivemeasure of points in both A and Ac and G′ must assume both −1 and 1, contradiction.So G is not monotone in any interval.

�

4. Chapter 4-Point Set Topology

Folland 4.8

If X is an infinite set with the cofinite topology and {xj} is a sequence of distinct points inX, then xj → x for every x ∈ X.

Proof. Let x ∈ X. We take a neighborhood U of x. Since U◦ is open, (U◦)c contains only finitelymany points in X. Therefore, since {xj} is a sequence of distinct points, starting at some suf-ficiently large N we have xn ∈ U◦ ⊂ U . Thus xj → x. Since x is arbitrary, this finishes theproof. �

Folland 4.13

If X is a topological space, U is open in X and A is dense in X, then U = U ∩A.

Proof. Clearly U ∩A ⊂ U . Conversely, suppose x ∈ U , then for any neighborhood N of x, sinceN◦ is also a neighborhood of x, N◦ ∩ U 6= ∅. Since U is open, N◦ ∩ U is open in X. Since A isdense in X, N◦ ∩ U contains some element of A. Then ∅ 6= N◦ ∩ (U ∩ A) ⊂ N ∩ (U ∩ A). SinceN is arbitrary, this means that x ∈ U ∩A. Thus U ⊂ U ∩A. Two directions combined, we provethat U = U ∩A. �

Folland 4.15

If X is a topological space, A ⊂ X is closed, and g ∈ C(A) satisfies g = 0 on ∂A, then theextension of g to X defined by g(x) = 0 for x ∈ Ac is continuous.

Proof. Let’s call the extension f . By considering real and imaginary parts seperately, we mayassume that g and f are R-valued. Since {(a, b)} generate the usual topology on R, it suffices toverify that f−1[(a, b)] is open in X for each (a, b). We split up to two cases:

(a) If 0 6∈ (a, b), clearly f−1[(a, b)] = g−1[(a, b)] ⊂ A◦. Since g is continuous, g−1[(a, b)] is openin A. Then g−1[(a, b)] = U ∩ A = (U ∩ A◦) ∪ (U ∩ ∂A) for some U open in X. Notice that

23

∂A ∩ g−1[(a, b)] = ∅ since 0 6∈ (a, b), (U ∩ ∂A) must be empty and g−1[(a, b)] = U ∩ A◦ isopen in X. Then f−1[(a, b)] is open in X.

(b) Suppose 0 ∈ (a, b), then

f−1(a, b) = f−1(a, 0) ∪ f−1({0}) ∪ f−1(0, b)= g−1(a, 0) ∪ f−1({0}) ∪ g−1(0, b)= g−1(a, 0) ∪ ∂A ∪Ac ∪ g−1(0, b)= g−1(a, 0) ∪ g−1({0}) ∪ g−1(0, b) ∪Ac

= g−1(a, b) ∪Ac (∗)

Since g is continuous, g−1(a, b) is open in A, meaning that g−1(a, b) = U ∩ A for some Uopen in X. Then

(∗) = (U ∩A) ∪Ac = U ∪Ac

which is open in X. Thus f−1(a, b) is open in X.

In both cases we have f−1(a, b) open in X, so f is continuous. This finishes the proof. �

Folland 4.20

If A is a countable set and Xα is a first (resp. second) countable space for each α ∈ A,then

∏α∈AXα is first (resp. second) countable.

Proof. (a) Xα is first countable for all α ∈ A. Suppose x = 〈xα〉α∈A ∈ X :=∏α∈AXα,

then for each xα there is a countable neighborhood base Nα. We claim that finite inter-sections of sets in π−1α (Nα), where α ∈ A, form a countable neighborhood base of x. Wedenote this countable neighborhood base N and first show that it’s countable. First ofall π−1α (Nα) is countable, and A is countable, so C :=

⋃α∈A π

−1α (Nα) is countable, and

we enumerate them as C1, C2, .... We use Cn to denote the collection of finite intersectionsof sets in {C1, ..., Cn}, so each Cn is finite. N ⊂

⋃n∈N Cn, and the latter is a countable

union of finite elements, and is therefore countable. Thus N is countable. We then showthat N is indeed a neighborhood base of x. First of all, since Nα is a neighborhood baseof xα, for every Nα ∈ Nα, xα ∈ Nα and thus x = π−1α (Nα). Any finite intersections ofsets like π−1α (Nα) must still contain x. Thus every element of N contains x. Suppose U isopen the product topology on

∏α∈AXα, and x ∈ U . U must take form

∏α∈A Uα, where

Uα = Xα for all but finitely many α. Thus we suppose Uα 6= Xα for α1, ..., αn. Since Nαiis a neighborhood base of xαi , there is some Vαi ∈ Nαi such that x ∈ Vαi ⊂ Uαi . Then wehave

⋂n1 π−1αi (Vαi) ∈ N such that x ∈ π−1αi (Vαi) ⊂ U . Then our claim is true. Since N is

countable and x is arbitrary, we show that X :=∏α∈AXα is first countable.

(b) Xα is second countable for all α ∈ A. Then for each α there is a countable base Nα ofXα. We claim that finite intersections of sets in π−1α (Nα) is a countable base of X :=∏α∈AXα. First of all, using exactly the same technique as above can we prove that the

collection, which we name N , is countable. We then show that N is actually a base. Firstof all, let x = 〈xα〉α∈A ∈ X, each xα ∈ Vα ∈ Nα for some Vα. Then we just randomlypick an α ∈ A and we have x ∈ π−1α (Vα). Next, suppose we have U =

⋂n1 π−1αi (Uαi) and

V =⋂m

1 π−1βi

(Vβi) in N and x = 〈xα〉α∈A ∈ U ∩ V . For our convenience, we denote

U =∏α∈A Uα and V =

∏α∈A Vα, where Uα 6= Xα only for {αi} and Vα 6= Xα only for

{βi}. We construct a family {Wα} the following way:1. If Uα = Vα = Xα, let Wα = Xα.

24

2. If Uα = Xα 6= Vα, we know that there is Wα ∈ Nα such that xα ∈ Wα ⊂ Vα = Vα ∩ Uα.The case where Uα 6= Xα = Vα is similar.

3. If Uα 6= Xα and Vα 6= Xα, since Nα is a base, there is some Wα ∈ Nα such that x ∈Wα ⊂ Uα ∩ Vα.

Then W :=∏α∈AWα ⊂

∏α∈A Uα ∩

∏α∈A Vα = U ∩ V , and W ∈ N since Wα 6= Xα

only for finitely many α, and for every such α Wα ∈ Nα. Thus N is a base. Since N iscountable, X is second countable.

�

Folland 4.22

Let X be a topological space, (Y, ρ) a complete metric space, and {fn} a sequence in Y X

such that supx∈X ρ(fn(x), fm(x)) → 0 as m,n → ∞. There is a unique f ∈ Y X such thatsupx∈X ρ(fn(x), f(x))→ 0 as n→∞. If each fn is continuous, so is f .

Proof. (a) Define f such that f(x) = limn→∞ fn(x), and we claim that supx∈X ρ(fn(x), f(x))→ 0 as n→∞. Let ε > 0. Since supx∈X ρ(fn(x), fm(x))→ 0 as m,n→∞, we can pick Nlarge enough such that supx∈X ρ(fn(x), fm(x)) < ε/2 if m,n > N . Also, by our definitionof f(x), for every x we can pick large enough mx > N such that ρ(fmx(x), f(x)) < ε/2.Then for every x, if n > N ,

ρ(fn(x), f(x)) ≤ ρ(fn(x), fmx(x)) + ρ(fmx(x), f(x)) < ε

and thus sup(fn(x), f(x)) < ε. This proves that sup ρ(fn(x), f(x)) → 0. Moreover, sup-pose g has the property that supx∈X ρ(fn(x), g(x))→ 0, for every x we have ρ(fn(x), g(x))→0 ⇐⇒ fn(x) → g(x). Since (Y, ρ) is a metric space, the limit is unique and g(x) = f(x).Then g = f and f is unique.

(b) Suppose each fn is continuous. Let x ∈ X, then fn is continuous at x for all n. Let ε > 0.Then there is some N > 0 such that supx∈X ρ(fN (x), f(x)) < ε/3. By continuity of fN ,A := f−1n [B(ε/3, fn(x))] is an open neighborhood of x. Let y ∈ A, then

ρ(f(y), f(x)) ≤ ρ(f(y), fn(y)) + ρ(fn(y), fn(x)) + ρ(fn(x), f(x)) < ε

and thus y ∈ A ⊂ (f−1(B(ε, f(x))))◦ since A is open. To show that f is continuous at x,since open balls generate the topology on Y , it suffices to show that f−1(B) is a neighbor-hood of x for every open ball B containing f(x). Since B 3 f(x) is open, we can take asmall enough ε > 0 such that B(ε, f(x)) ⊂ B. Then by what we did above f−1(B(ε, f(x)))is a neighborhood containing x. Then f−1(B) ⊃ f−1(B(ε, f(x))) is also a neighborhoodcontaining x. Thus f is continuous at x. Since x is arbitrary, this shows that f is continu-ous.

�

Folland 4.24

A Hausdorff space X is normal iff X satisfies the conclusion of Urysohn’s lemma iff X sat-isfies the conclusion of the Tietze extension theorem.

Proof. We use (1), (2), (3) to denote these three statements respectively.We first show (1) ⇐⇒ (2). The fact that (1) =⇒ (2) is trivial. Conversely, first we notice thatX is T1 since X is Hausdorff. Given disjoint closed sets A and B, by Urysohn’s lemma there is acontinuous f : X → [0, 1] such that f ≡ 0 on A and f ≡ 1 on B. Then take U = f−1[0, 1/2)and V = f−1(1/2, 1]. Since f is continuous, U and V are open, and clearly U and V are disjoint

25

since [0, 1/2) and (1/2, 1] are disjoint. Then U is an open set containing A and V is an open setcontaining B such that U ∩ V = ∅. Then X is normal.We then show that (2) ⇐⇒ (3). Suppose we have (2). Since X is Hausdorff, by the previousparagraph we have X is normal. Then (3) holds automatically. Conversely, suppose we have (3),given disjoint closed sets A and B, define f : A ∪ B → [0, 1] to be f |A ≡ 0 and f |B ≡ 1, thenf ∈ C(A ∪ B, [0, 1]). By (3) we have F ∈ C(X, [0, 1]) such that F = f on A ∪ B, i.e. F ≡ 0 on Aand F ≡ 1 on B. Then (2) holds. �

Folland 4.38

Suppose that (X, T ) is a compact Hausdorff space and T ′ is another topology on X. IfT ′ is strictly stronger than T , then (X, T ′) is Hausdorff but not compact. If T ′ is strictlyweaker than T , then (X, T ′) is compact but not Hausdorff.

Proof. (a) Suppose T ′ is stricly stronger than T . For x 6= y ∈ X, since (X, T ) is Hausdorff,there are disjoint closed A,B such that x ∈ A and y ∈ B. But (X, T ′) ) (X, T ), soA,B ∈ T ′. Thus (X, T ′) is Hausdorff. Suppose in the contrary that (X, T ′) is compact.Consider mapping f : (X, T ′) → (X, T ) defined by x 7→ x, which is clearly bijective.For U open in (X, T ), f−1(U) = U open in (X, T ′) since T ′ is strictly stronger than T ,so f is continuous. If (X, T ′) is compact, then f is a continuous bijection mapping from acompact space to a Hausdorff space and is thus a homeomorphism. But T ′ ) T , a contra-diction. Thus (X, T ′) cannot be compact.

(b) Suppose T ′ is strictly weaker than T . Take an open cover U of X in T ′, U is also an opencover in T and thus has a finite subcover. Thus (X, T ′) is compact. Suppose (X, T ′) isHausdorff, consider g : (X, T ) → (X, T ′) defined by x 7→ x, then g is bijective. For Uopen in (X, T ′), f−1(U) = U ∈ T ′ ⊂ T and is therefore open. Then g is continuous bijec-tion between a compact space and a Hausdorff space and is therefore a homeomorphism.which is not possible since T ′ stricly weaker than T . Thus (X, T ′) is not compact.

�

Folland 4.43

For x ∈ [0, 1), let∑∞

1 an(x)2−n (an(x) = 0 or 1) be the base-2 decimal expansion of x.(If x is a dyadic rational, choose the expansion such that an(x) = 0 for n large.) Then the

sequence 〈an〉 in {0, 1}[0,1) has no pointwise convergent subsequence.

Proof. Take any subsequence 〈ank〉 of 〈an〉, consider x =∑∞

k=1 2−n2k . It is clear that x is not adyadic rational (since there are no consecutive 1s or 0s), so the expression is uniquely determinedhere. Now we have an2k

= 1 and other ank = 0. Since we have infinitely many alternating terms,〈ank(x)〉 fails to converge. Thus 〈an〉 has no pointwise convergent subsequence. �

26

Folland 4.56

Define φ : [0,∞]→ [0, 1] by φ(t) = t/(t+ 1) for t ∈ [0,∞] and φ(∞) = 1.(a) φ is strictly increasing and φ(t+ s) ≤ φ(t) + φ(s).(b) If (Y, ρ) is a metric space, then φ◦ρ is a bounded metric on Y that defines the same

topology as ρ.(c) If X is a topological space, the function ρ(f, g) = φ(supx∈X ||f(x) − g(x)|) is a

metric on CX whose associated topology is the topology of uniform convergence.(d) If X = Rn and Un = B(n, 0) for all n, the function

ρ(f, g) =

∞∑1

2−nφ( supx∈Un

|f(x)− g(x)|)

is a metric on CX whose associated topology is the topology of locally uniform con-vergence.

Proof. (a) We first show that φ is strictly increasing. Suppose t1 < t2 <∞, then

φ(t2)− φ(t1) =t2

t2 + 1− t1t1 + 1

=t2(t1 + 1)− t1(t2 + 1)

(t2 + 1)(t1 + 1)=

t2 − t1(t2 + 1)(t1 + 1)

> 0

Suppose t1 < t2 =∞, then since t1 <∞, φ(t1) < 1 = φ(t2). We next show that φ(t+ s) ≤φ(t) + φ(s). If one of t, s is ∞, which we may assume to be t, then

φ(t+ s) = φ(∞) ≤ φ(∞) + 1 = φ(t) + φ(s)

If t, s <∞, then

φ(t+ s)− (φ(t) + φ(s)) =t+ s

t+ s+ 1− t

t+ 1− s

s+ 1

=1− 1

t+ s+ 1− 1 +

1

t+ 1− 1 +

1

s+ 1=

t+ 1 + s+ 1

(t+ 1)(s+ 1)− t+ s+ 2

t+ s+ 1

≤ t+ 1 + s+ 1

(t+ 1)(s+ 1)− t+ s+ 2

t+ s+ ts+ 1=

t+ 1 + s+ 1

(t+ 1)(s+ 1)− t+ s+ 2

(t+ 1)(s+ 1)= 0

showing the result.

(b) For convenience, define ρ′ := φ ◦ ρ =ρ

ρ+ 1. It is easy to see ρ′ is bounded by 1. We

denote the topology generated by ρ and ρ′ using T and T ′, respectively, and the collectionof open balls in T and T ′ are named E and E ′, respectively. We know that T (E) = T , andT (E ′) = T ′. Then it suffices to show E ⊂ T ′ and E ′ ⊂ T . Let Bρ(x, r) ∈ E , and we claimthat Bρ(x, r) = Bρ′(x,

rr+1). To show the claim, suppose we have y ∈ Y , then ρ(x, y) < r

if and only if ρ′(x, y) = φ ◦ ρ(x, y) < φ(r) = r/(r + 1) since φ is strictly increasing. Inother words, y ∈ Bρ(x, r) if and only if y ∈ Bρ′(x, r

r+1) and thus the claim is true. Thus

Bρ(x, r) = Bρ′(x,rr+1) ∈ E ′ ⊂ T ′ as desired. Similarly we can show that Bρ′(x, r) =

Bρ(x,r

1−r ) since φ(r/1 − r) = r. Thus E ′ ⊂ E ⊂ T . Now we have T = T (E) ⊂ T ′ and

T ′ = T (E ′) ⊂ T , showing that ρ and ρ′ generate the same topology on Y .(c) We first verify that ρ is a metric on CX .

Non-Negativity: ρ(f, g) ≥ 0 since φ ≥ 0Identity of Indiscernibles: ρ(f, g) = 0 iff φ(supx∈X |f(x)−g(x)|) = 0 iff supx∈X |f(x)−g(x)| = 0 iff |f(x)− g(x)| for every x iff f = g.

Symmetry: ρ(f, g) = φ(supx∈X |f(x)− g(x)|) = φ(supx∈X |g(x)− f(x)|) = ρ(g, f)27

Triangular Inequality: For f, g, h ∈ CX , ρ(f, g) + ρ(g, h) = φ(supx∈X |f(x) −g(x)|) + φ(supx∈X |g(x) − h(x)|) ≥ φ(supx∈X |f(x) − g(x)| + supx∈X |g(x) − h(x)|) ≥φ(supx∈X |f(x)− h(x)| = ρ(f, h).

We know that ρu(f, g) = supx∈X ||f(x) − g(x)||, and ρu is a metric that generates thetopology of uniform convergence, and by (b) we know that ρu and ρ = φ ◦ ρu generatethe same topology, so the associated topology of ρ is also the topology of uniform conver-gence.

(d) Suppose 〈fn〉n converges locally uniformly to f . By definition

ρ(fn, f) =∞∑i=1

2−iφ

(supx∈U i

|fn(x)− f(x)|)

Let ε > 0. Choose N large enough such that 2−N+1 < ε. Since UN−1 is compact, bylocally uniform convergence fn|UN−1 → f |UN−1 uniformly (and automatically convergesuniformly on U i for all i < N). Then

ρ(fn, f) =N−1∑i=1

2−iφ

(supx∈U i

|fn(x)− f(x)|)

+

∞∑i=N

2−iφ

(supx∈U i

|fn(x)− f(x)|)

≤N−1∑i=1

2−iφ

(supx∈U i

|fn(x)− f(x)|)

+∞∑i=N

2−i

≤N−1∑i=1

2−iφ

(supx∈U i

|fn(x)− f(x)|)

+ ε

Sending n to infinity, we get limn→∞ ρ(fn, f) < ε. Since ε is arbitrary, limn→∞ ρ(fn, f) =0, implying convergence in ρ. Conversely, suppose limn→∞ ρ(fn, f) = 0 and K ⊂ Rncompact. We can take N large enough such that UN contains K, so it suffices to showthat fn|UN → f |UN . Actually, we prove a stronger claim: for all n ∈ N, fk|Un → f |Ununiformly. Suppose not, there is some N ∈ N such that fk|UN 6→ f |UN uniformly. Thenthere is some δ > 0 such that for infinitely many k we have supx∈Un |fk(x) − f(x)| ≥ δ.

Notice that this means that for infinitely many k, we have supx∈Un |fk(x) − f(x)| ≥ δ forn ≥ N . Then

ρ(f, fk) ≥∞∑n=N

2−nφ(δ) =2−N

1− 1/2· δ

δ + 1= 2−N+1 δ

δ + 1

for infinitely many k so fk doesn’t converge to f in ρ, a contradiction. This finishes theproof.

�

Folland 4.61

Theorem 4.43 remains valid for maps from a compact Hausdorff space X into a completemetric space Y provided the hypothesis of pointwise boundedness is replaced by pointwisetotal boundedness.

Proof. Let ε > 0. Since F is equicontinuous, for each x ∈ X there is open Ux of x such that

ρ(f(x), f(y)) <ε

4for all y ∈ Ux and f ∈ F . Since X is compact, we can choose x1, ..., xn ∈ X

such that⋃n

1 Uxj = X. By pointwise total boundedness, {f(xj) : f ∈ F, 1 ≤ j ≤ n} is atotally bounded subset of Y being a finite union of totally bounded subsets of Y . Then we can

28

pick finite {z1, ..., zm} that isε

4-dense in it. Let A = {x1, ..., xn} and B = {z1, ..., zm}, then BA is

finite. For each φ ∈ BA let

Fφ = {f ∈ F : ρ(f(xj), φ(xj)) <ε

4for 1 ≤ j ≤ n}

clearly ∪φ∈BA = F , and we claim that each Fφ has diameter ≤ ε, so we obtain a finite ε-densesubset of F by picking one f from each Fφ that is non-empty. To prove this claim, suppose we

have f, g ∈ Fφ. Then ρ(f, φ) <ε

4and ρ(g, φ) <

ε

4on A and we have ρ(f, g) <

ε

2on A. If x ∈ X,

we have x ∈ Uxj for some j, and then

ρ(f(x), g(x)) ≤ ρ(f(x), f(xj)) + ρ(f(xj), g(xj)) + ρ(g(xj), g(x)) < ε

This shows that F is totally bounded. Then F is totally bounded. Since Y is complete, by ex-ercise 4.22 C(X,Y ) is complete. Being a closed and totally bounded subset of complete metricspace C(X,Y ), F is compact. �

Folland 4.63

Let K ∈ C([0, 1] × [0, 1]). For f ∈ C([0, 1]), let Tf(x) =∫ 10 K(x, y)f(y)dy. Then Tf ∈

C([0, 1]), and {Tf : ||f ||u ≤ 1} is precompact in C([0, 1]).

Proof. We first show that Tf ∈ C([0, 1]). Let ε > 0. Since K ∈ C([0, 1] × [0, 1]) and [0, 1] × [0, 1]is compact, K is uniformly continuous on [0, 1] × [0, 1]. Thus there is a δ > 0 such that when|x− y| < δ, |K(x)−K(y)| < ε. Then for x1, x2 ∈ [0, 1] such that |x1 − x2| < δ,

|Tf(x1)− Tf(x2)| ≤∫ 1

0|K(x1, y)−K(x2, y)|f(y)dy < ε

∫ 1

0f(y)dy

Observe that f ∈ C([0, 1]), so by extreme value theorem |f | ≤ M for some M > 0. Then|Tf(x1) − Tf(x2)| < Mε. Since ε is arbitrary, Tf is uniformly continuous on [0, 1] and thusTf ∈ C([0, 1]).We then show that {Tf : ||f ||u ≤ 1} is precompact in C([0, 1]). For convenience, we use F todenote the family {Tf : ||f ||u ≤ 1}. Let ε > 0. Since K ∈ C([0, 1] × [0, 1]) and [0, 1] × [0, 1]is compact, K is uniformly continuous on [0, 1] × [0, 1]. By uniform continuity, there is a δ > 0such that |K(x) − K(y)| < ε whenever x,y ∈ [0, 1] × [0, 1] satisfy |x − y| < δ. Let x1 ∈ [0, 1].Consider Ux1 := (x1−δ, x1+δ) and by shrinking δ if necessary we may assume Ux1 ⊂ [0, 1]. Whenx2 ∈ Ux1 ,

|Tf(x2)− Tf(x1)| ≤∫ 1

0|K(x2, y)−K(x1, y)|f(y)dy < ε

∫ 1

0f(y)dy ≤ ε · 1 = ε

The last inequality is validated since |f | ≤ 1. Then F is equicontinuous at x1 and thus equicon-tinuous. Since K is continuous on a compact set, |K| ≤ M for some M > 0 by extreme value

theorem. Then for x ∈ [0, 1], |Tf(x)| ≤∫ 10 |K(x, y)|f(y)dy ≤ M for all f such that ||f ||u ≤ 1 and

is thus pointwise bounded. Since [0, 1] is compact Hausdorff, by Arzela-Ascoli, F is precompactas desired. �

29

Folland 4.64

Let (X, ρ) be a metric space. A function f ∈ C(X) is called Holder continuous of exponentα if the quantity

Nα(f) = supx 6=y

|f(x)− f(y)|ρ(x, y)α

is finite. If X is compact, {f ∈ C(X) : ||f ||u ≤ 1 and Nα ≤ 1} is compact in C(X).

Proof. We first notice that X is a compact metric space and is therefore compact Hausdorff. Forour convenience, we define F := {f ∈ C(X) : ||f ||u ≤ 1 and Nα ≤ 1}. Let ε > 0, x ∈ X consider

Ux = B(ε1/α, x).3 Thus for any f ∈ F , y ∈ Ux,

|f(y)− f(x)| ≤ Nα(f)ρ(x, y)α ≤ ρ(x, y)α < ε

Thus F is equicontinuous at x and is thus equicontinuous. Furthermore F is clearly pointwisebounded since it is uniformly bounded by 1. By Arzela-Ascoli, F is precompact, so it suffices toshow that F is closed. Suppose f ∈ F , then for every ε > 0, there is f ′ ∈ F such that ||f ′−f ||u <ε and Nα(f ′) ≤ 1. Then ||f ||u < ||f ′||u + ε ≤ 1 + ε. Since ε is arbitrary, ||f ||u ≤ 1. Also we have

Nα(f ′) = supx 6=y|f ′(x)− f ′(y)|

ρ(x, y)α≤ 1. For x 6= y ∈ X, |f(x) − f(y)| ≤ |f(x) − f ′(x)| + |f ′(x) −

f ′(y)| + |f ′(y) − f(y)| < 2ε + ρ(x, y)α. But ε is arbitrary, so |f(x) − f(y)| ≤ ρ(x, y)α and thus

Nα(f) = supx 6=y|f(x)− f(y)|ρ(x, y)α

≤ 1. Thus f ∈ F and thus F is closed. This means that F = F is

compact and this finishes the proof. �

Folland 4.68

Let X and Y be compact Hausdorff spaces. The algebra generated by functions of theform f(x, y) = g(x)h(y), where g ∈ C(X) and h ∈ C(Y ), is dense in C(X × Y ).

Proof. We denote the algebra using A. For f ∈ A, f = gh for g ∈ C(X) and f ∈ C(Y ). Thenf = gh = gh. Continuity is componentwise and therefore preserved under conjugation, so g ∈C(X) and h ∈ C(Y ) and f ∈ A. Thus A is closed under conjugation. Suppose (x1, y1) 6= (x2, y2).Then x1 6= x2 or y1 6= y2 and we may assume x1 6= x2. Remember that X is compact Hausdorffand therefore normal, and {x1} and {x2} are disjoint closed sets in X since singletons are closedin Hausdorff spaces. By Urysohn’s lemma, there is a g ∈ C(X) such that g(x1) = 0 and g(x2) =1. Let h ≡ 1 in C(Y ). Then f := gh ∈ A satisfies

f(x1, y1) = g(x1)h(y1) = 0 6= 1 = g(x2)h(y2) = f(x2, y2)

and thus A separate points. Notice that f ≡ 1 · 1 = 1 ∈ A so f doesn’t vanish at any x0. ByStone-Weierstrass, A is dense in C(X × Y ). �

Folland 4.69

Let A be a non-empty set, and let X = [0, 1]A. The algebra generated by the coordinatemaps πα : X → [0, 1] (α ∈ A) and the constant function 1 is dense in C(X).

3Here 1/α is defined since α > 0.

30

Proof. We denote the algebra using A. For πα ∈ A, πα = πα since it is a function to [0, 1] ⊂ R.Thus A is closed under conjugation. Suppose we have distinct x = 〈xα〉α∈A and y = 〈yα〉α∈A in[0, 1]A, then xα 6= yα at some α0. Then πα0(x) = xα0 6= yα0 = πα0(y). Thus A separates points.Since 1 ∈ A, A doesn’t vanish at any x ∈ [0, 1]A. By Stone-Weierstrass, A is dense in C(X). �

Folland 4.76

If X is normal and second countable, there is a countable family F ⊂ C(X, I) that sepa-rates points and closed sets.

Proof. Since X is second countable, let B be a countable base of X. For each pair (U, V ) ∈ B × Bsuch that U ⊂ V , by Urysohn’s lemma, there is some function f : X → I such that f(U) = 0and f(V c) = 1 since U ∩ V c ⊂ V ∩ V c = ∅ and U, V c are closed. For each such pair (U, V ),we pick one particular function f that satisfies the conditions above, and let F be defined as thecollection of such functions. |F| ≤ |B × B|, and the latter set is countable since B is countable,so F is countable. We proceed to show that F separates points and closed sets. Given E ⊂ Xclosed and x ∈ Ec, since Ec is open, there is some V ∈ B such that x ∈ V ⊂ Ec. Then E =(Ec)c ⊂ V c. We claim that there is some U ′ open such that x ∈ U ′ ⊂ U ′ ⊂ V . To show theclaim, we notice that {x} and V c are disjoint closed sets, so by normality there are disjoint openV ′ ⊃ V c and U ′ 3 x. Given U ′ ∩ V ′ = ∅, U ′ ⊂ (V ′)c and thus U ′ ⊂ (V ′)c since (V ′)c is closed.Then x ∈ U ′ ⊂ U ′ ⊂ (V ′)c ⊂ (V c)c = V , as desired. Since U ′ is open, we have U ∈ B such thatx ∈ U ⊂ U ′. Then x ∈ U ⊂ U ′ ⊂ V for U, V ∈ B. By definition there is some f ∈ F such thatf(U) = 0 and f(V c) = 1. Remember that x ∈ U and E ⊂ V c, so f(x) = 0 6∈ {1} = f(E), asdesired. �

5. Chapter 5-Elements of Functional Analysis

Folland 5.3

If Y is complete, so is L(X ,Y).

Proof. Pick a Cauchy sequence {Tn}n in L(X ,Y) under the norm metric. Then {Tnx}n is Cauchyfor each x since ‖Tnx − Tmx‖ ≤ ‖Tn − Tm‖‖x‖ → 0 as n,m → ∞ since ‖Tn − Tm‖ → 0 asn,m → ∞. Thus {Tnx} converges. Define Tx := limn→∞ Tnx. We first show that T ∈ L(X ,Y).Let ε > 0. Since {Tn} is Cauchy, there is some N > 0 such that when m,n ≥ N , ||Tn − Tm|| < ε.Then ||Tnx−Tmx||/||x|| < ε for all x 6= 0. Sending m to infinity, we have ||Tn(x)−T (x)||/||x|| < εfor all x 6= 0. In particular, ||TN (x) − T (x)|| < ε||x|| for all x. Since TN is bounded, suppose||TNx|| ≤ CN ||x|| for all x. Then

||T (x)|| ≤ ||TN (x)||+ ||T (x)− TN (x)|| ≤ (CN + ε)||x||for all x and thus T ∈ L(X ,Y). Furthermore, for the ε and N given above, when n > N , ||Tn(x)−T (x)||/||x|| < ε for all x 6= 0 and thus ||Tn − T || = sup{||Tn(x)− T (x)||/||x|| : x 6= 0} < ε, showingthat ||Tn − T || → 0, as desired. �

Folland 5.8

Let (X,M) be a measurable space, and let M(X) be the space of finite signed measureson (X,M). Then ||µ|| = |µ|(X) is a norm on M(X) that makes M(X) into a Banachspace.

31

Proof. We first verify that ||µ|| := |µ|(X) is a norm.

• ||µ1 + µ2|| = |µ1 + µ2|(X) ≤ |µ1|(X) + |µ2|(X) = ||µ1||+ ||µ2||• By Lebesgue-Randon-Nikodym, there is some positive measure ν such that dµ = fdν.

Then d(λµ) = λdµ = λfdν Then ||λµ|| = |λµ|(X) =∫|λf |dν = |λ|

∫|f |dν = |λ||µ|(X) =

|λ| · ||µ||• If ||µ|| = 0, then µ(X) = 0 and µ = 0 since any subset of a measure zero set (in this case,

every measurable set) has measure zero.

We then show that M(X) is a Banach space under the given norm by showing that every ab-solutely convergent series in M(X) converges under this norm. Suppose we have µ1, µ2, ... ∈M(X) such that

∑∞n=1 ||µn|| < ∞. Then

∑∞n=1 |µn|(X) < ∞. Define ν :=

∑∞n=1 µn. Notice

that∑m

n=1 µn(X) ≤∑m

n=1 |µ|(X) for all m and thus sending m → ∞ gives∑∞

n=1 µn(X) ≤∑mn=1 |µ|(X) < ∞, showing that ν is a finite signed measure. Thus limm→∞ ‖ν −

∑mn=1 µn‖ =

‖v −∑∞

n=1 νn‖ = 0, showing that∑∞

n=1 converges to ν ∈ M(X), as desired. Thus M(X) is aBanach space. �

Folland 5.9

Let Ck([0, 1]) be the space of functions on [0, 1] possessing continuous derivatives up toorder k on [0, 1], including one-sided derivatives at endpoints.

(a) If f ∈ C([0, 1]), then f ∈ Ck([0, 1]) iff f is k times continuously differentiable on

(0, 1) and limx↓0 f(j)(x) and limx↑1 f

(j)(x) exist for j ≤ k.

(b) ‖f‖ =∑k

0 ‖f (j)‖u is a norm on Ck([0, 1]) that makes Ck([0, 1]) into a Banach space.

Proof. (a) If f ∈ Ck([0, 1]), then clearly f ∈ Ck(0, 1) and limx↓0 f(j)(x) = f (j)(0) and

limx↑1 f(j)(x) = f (j)(1) for all j ≤ k. Conversely, suppose f is k times continuously

differentiable on (0, 1) and limx↓0 f(j)(x) and limx↑1 f

(j)(x) exist for j ≤ k, we want to

show that f ∈ Ck[0, 1]. Since f ∈ Ck(0, 1), it suffices to show that f is continuouslydifferentiable at 0 and 1. So we proceed by induction. The base case where j = 0 istrue since f ∈ C[0, 1]. Suppose f is j times continuously differentiable at 0. Notice that

limx↓0f (j)(x)− f (j)(0)

x= limx↓0 f

(j)(c) for some c ∈ (0, x] by mean value theorem, and

thus c ↓ 0 as x ↓ 0. Thus limx↓0f (j)(x)− f (j)(0)

x= limc↓0 f

(j)(c) which is assumed to

exist. Thus f (j+1)(0) = limc↓0 f(j)(c), showing that f is j + 1 times continuously differen-

tiable at 0. Then it follows by induction that f is Ck at 0. Similarly we can show f is Ck

at 1. Then f ∈ Ck([0, 1]), as desired.

(b) We first show that ‖f‖ =∑k

0 ‖f (j)‖u is a norm. ‖f + g‖ =∑k

0 ‖(f + g)(j)‖u =∑k

0 ‖f (j) +

g(j)‖u ≤∑k

0(‖f (j)‖u + ‖g(j)‖u) =∑k

0 ‖f (j)‖u +∑k

0 ‖g(j)‖u = ‖f‖ + ‖g‖. Also ‖λf‖ =∑k0 ‖(λf)(j)‖u =

∑k0 ‖λf (j)‖u = |λ|

∑k0 ‖f (j)‖u = |λ|‖f‖. ‖f‖ = 0 implies ‖f‖u ≤∑k

0 ‖f (j)‖u = 0 and thus f ≡ 0 since f is continuous. We then show that this norm makes

Ck([0, 1]) into a Banach space. Pick a Cauchy sequence {fn} in Ck([0, 1]) and let ε > 0.

Then there is N > 0 such that m,n > N implies ‖fn − fm‖ =∑k

j=0 ‖f(j)n − f (j)m ‖u < ε.

In particular, ‖fn − fm‖ < ε for n,m > N and thus {fn} is uniformly Cauchy. Since fnis continuous and C([0, 1]) is complete, fn → f for some f ∈ C([0, 1]). We now claim

that f ∈ Ck([0, 1]) and f(j)n → f (j) uniformly for all j ≤ k. We prove the claim by in-

duction. For k = 0, f ∈ C([0, 1]) and fn → f uniformly. Suppose f ∈ C l([0, 1]) and32

f(j)n → f (i) uniformly for all j ≤ l, we try to show the result for l + 1. Fix the ε andm,n above, ‖f l+1

n − f l+1m ‖ < ε for all m,n > N . Thus {f l+1

n } is uniformly Cauchy. Sincefn, fm ∈ Ck([0, 1]), f l+1

n and f l+1m are continuous and thus f l+1

n → g for some g ∈ C([0, 1]).Notice that f ln(x)−f ln(0) =

∫ x0 f

l+1n (t)dt. Since f l+1

n → g uniformly and f l+1n is continuous,

by undergraduate analysis, sending n → ∞ we get f l(x) − f l(0) =∫ x0 g(t)dt. By funda-

mental theorem of calculus, f l+1(x) = g(x), showing the result, and the claim follows by

induction. By the claim, ‖fn− f‖ =∑k

j=0 ‖f(j)n − f (j)m ‖u → 0 as n→∞ and f ∈ Ck([0, 1]),

showing that C([0, 1]) is complete under this norm and is thus a Banach space, as desired.�

Folland 5.15

Suppose that X and Y are normed vector spaces and T ∈ L(X ,Y). Let N (T ) = {x ∈ X :Tx = 0}.

(a) N (T ) is a closed subspace of X .(b) There is a unique S ∈ L(X/N (T ),Y) such that T = S ◦ π where π : X → X/M is

the projection. Moreover, ||S|| = ||T ||.

Proof. (a) Since T ∈ L(X ,Y), T is continuous. Since {0} is closed in Y, N (T ) = T−1({0}) isclosed in X .

(b) We first show that such S exists. We define S : X/N (T ) → Y by S(x + N (T )) := T (x),and claim that S ∈ L(X/N (T ),Y). To show the claim, we first show that S is well-defined. If x+N (T ) = x′ +N (T ), x′ = x+ y for some y ∈ N (T ). Thus

S(x′ +N (T )) = S(x+ y +N (T )) = T (x+ y) = Tx+ Ty = Tx = S(x+N (T ))

showing that S is well-defined. We then show that S is linear. This is true since

S[(x+N (T )) + (y +N (T ))] = S(x+ y +N (T ))

=T (x+ y) = T (x) + T (y) = S(x+N (T )) + S(y +N (T ))

Now we show that S is bounded. We claim a stronger result that any constant that boundsT also bounds S. To show this claim, we suppose ||Tx||Y ≤ C||x||X for all x ∈ X . Thenfor any y ∈ N (T ), ||S(x+N (T ))||Y = ||Tx||Y = ||Tx+ Ty||Y = ||T (x+ y)||Y ≤ C||x+ y||Xand thus ||S(x + N (T ))||Y ≤ C · inf{||x + y|| : y ∈ N (T )} = C||x + N (T )||, prov-ing the claim. Next we show that such S is unique. Suppose S1 ◦ π = S2 ◦ π = T ,then for any x + N (T ) ∈ X/N (T ), there is some x such that π(x) = x + N (T ). ThenS1(x + N (T )) = S1 ◦ π(x) = S2 ◦ π(x) = S2(x + N (T )), showing that S1 ≡ S2 and thatS is unique. Eventually we show that ||S|| = ||T ||. By our claim above, ||S|| = inf{C :||S(x +N (T ))||Y ≤ C||x +N (T )|| for all x} ≤ inf{C : ||Tx||Y ≤ C||x||X for all x} = ||T ||.Conversely we have ||T || = ||S ◦ π|| ≤ ||S|| · ||π|| = ||S|| and thus ||T || = ||S||, finishing theproof.

�

Folland 5.20

If M is a finite-dimensional subspace of a normed vector space X , there is a closed sub-space N such that M∩N = {0} and M+N = X .

Proof. First consider the case where M is one dimensional. Then we may write M = Kx1, whereK is C or R and x1 is a non-zero vector in M. We may assume ‖x1‖ = 1, otherwise we just do

33

some scaling to make this happen. Then we define f : M → K by f(λx1) = λ. Since ‖f‖ =sup{||f(x)|| : ‖x‖ = 1} = sup{‖f(λx1)‖ : ‖λx1‖ = 1} = sup{|λ| : |λ|‖x1‖ = 1} = sup{|λ| : |λ| =1} = 1, f is a bounded linear functional. By Hahn-Banach theorem we extend f to an F ∈ X ∗,and claim that F−1({0}) is a desired subspace N . To prove the claim, we first notice that F isbounded linear by assumption and is thus continuous. Since {0} is closed, N = F−1({0}) is alsoclosed. Moreover, for any λx1 in M, λx1 ∈ N iff F (λx1) = f(λx1) = λ = 0 iff λx1 = 0, soM∩N = {0}. Eventually, for x ∈ X , we write x = F (x)x1 + (x− F (x)x1). Clearly F (x)x1 ∈M,and F (x−F (x)x1) = F (x)−F (x)F (x1) = F (x)−F (x) = 0, so x−F (x)x1 ∈ N . Then x ∈M+Nand thus M+N = X . Then our claim is true.We now finish the proof by induction. Suppose the results holds for dimension ≤ n, we showthat it holds for dimension n + 1. Suppose we have M′ with dimension n + 1, we may chooseM ⊂ M′ an n-dimensional subspace of M′. By induction hypothesis we can choose a closedN ⊂ X such that M ∩ N = {0} and M + N = X . Let y ∈ M′ \ M, we have y = m + xfor m ∈ M and x ∈ N . Clearly x 6∈ M, and thus dim(M + Kx) = dim(M) + dim(Kx) =n + 1 = dim(M′). Since M + Kx ⊂ M′, M + Kx = M′. Since Kx is a one-dimensionalsubspace of N ′, we use the same technique as above to choose some closed N ′ ⊂ N such thatKx ∩ N ′ = {0} and Kx + N ′ = N . Since N ′ is closed in N and N is closed, N ′ is closed inX . Now we claim that N ′ is a closed subspace such that M′ ∩ N ′ = {0} and M′ + N ′ = X .M′ ∩ N ′ = (M + Kx) ∩ N ′. If m + kx ∈ N ′ ⊂ N for m ∈ M, since kx ∈ N , m ∈ N and thusm = 0. Then kx ∈ N ′ and kx = 0. Thus m + kx = 0 and M′ ∩ N ′ = (M + Kx) ∩ N ′ = {0}.Moreover, M′ + N ′ = M + Kx + N ′ = M + (Kx + N ′) = M + N = X . The result follows byinduction. �

Folland 5.21

If X and Y are normed vector spaces, define α : X ∗ × Y∗ → (X × Y)∗ by α(f, g)(x, y) =f(x) + g(y). Then α is an isomorphism which is isometric if we use the norm ||(x, y)|| =max(||x||, ||y||) on X × Y, the corresponding operator norm on (X × Y)∗, and the norm||(f, g)|| = ||f ||+ ||g|| on X ∗ × Y∗.

Proof. We first show that α is an isomorphism. Let h ∈ (X × Y)∗, then we claim that β : h 7→(f, g), where f and g are in X ∗ × Y∗ such that f(x) = h(x, 0) and g(y) = h(0, y), is the inverse ofα. First of all, β◦α(f, g) = β(h), where h(x, y) = f(x)+g(y). Suppose β◦α(f, g) = (β◦α(f, g)1, β◦α(f, g)2), then (β ◦ α(f, g)1(x), β ◦ α(f, g)2(y)) = (β(h)1(x), β(h)2(y)) = (h(x, 0), h(0, y)) =(f(x), g(y)) for any x ∈ X and y ∈ Y. Thus β ◦ α(f, g) = (f, g). On the other hand, for h ∈(X × Y)∗, α ◦ β(h)(x, y) = α(f, g)(x, y) = f(x) + g(y) = h(x, 0) + h(0, y) = h(x, y) wheref and g are defined at the very beginning. Then α ◦ β(h) = h. Thus β = α−1 is the two-sidedinverse. The inverse of a linear map is linear, so it remains to verify that β is bounded. Observethat sup{||β(h)|| : ||h|| = 1} = sup{||(f, g)|| : ||h|| = 1} = sup{||f || + ||g|| : ||h|| = 1} < ∞ since||f || and ||g|| are bounded. Then α is an isomorphism.

34

We then show that α is an isometry. First of all,

‖α(f, g)‖ = sup{‖α(f, g)(x, y)‖ : ‖(x, y)‖ = 1} = sup{‖α(f, g)(x, y)‖ : ‖(x, y)‖ = 1}= sup{‖f(x) + g(y)‖ : ‖(x, y)‖ = 1} ≥ sup{‖f(sgnf · x) + g(sgng · y)‖ : ‖(x, y)‖ = 1}= sup{‖(sgnf)f(x) + (sgng)g(y)‖ : max(‖x‖, ‖y‖) = 1}= sup{‖(sgnf)f(x) + (sgng)g(y)‖ : max(‖x||, ‖y‖) = 1}= sup{(sgnf)f(x) + (sgng)g(y) : max(‖x‖, ‖y‖) = 1}= sup{‖f(x)‖+ ‖g(y)‖ : max(‖x‖, ‖y‖) = 1} ≥ sup{‖f(x)‖+ ‖g(y)‖ : ‖x‖ = 1, ‖y‖ = 1}= sup{‖f(x)‖ : ‖x‖ = 1}+ sup{‖g(y)‖ : ‖y‖ = 1} = ‖f‖+ ‖g‖ = ‖(f, g)‖

Conversely, notice that if ||x|| ≤ 1, then there is some |λ| ≥ 1 such that ||λx|| = 1 and thus||f(x)|| ≤ |λ| · ||f(x)|| = ||λf(x)|| = ||f(λx)||. Therefore, for any ||x|| ≤ 1, we can find a corre-sponding x′ such that ||x′|| = 1 and ||f(x)|| ≤ ||f(x′)||. This means that sup{||f(x)|| + ||g(y)|| :max(||x||, ||y||) = 1} ≤ sup{||f(x)|| + ||g(y)|| : ||x|| = 1, ||y|| = 1} = sup{||f(x)|| : ||x|| =1}+ sup{||g(y)|| : ||y|| = 1} = ||f ||+ ||g|| = ||(f, g)||. And sup{||f(x)||+ ||g(y)|| : max(||x||, ||y||) =1} = sup{||α(f, g)(x, y)|| : ||(x, y)|| = 1} = ||α(f, g)||. This shows that ||α(f, g)|| = ||(f, g)|| andthus α is an isometry. �

A Better Proof Idea. We first show that α is isometric and then show that it is bijective. If α isisometric, α−1 is also isometric and automatically bounded and thus α is an isomorphism. Thisavoids constructing an explicit inverse. �

Folland 5.22

Suppose X and Y are normed vector spaces and T ∈ L(X ,Y).(a) Define T+ : Y∗ → X ∗ by T+f = f ◦ T . Then T+ ∈ L(Y∗,X ∗) and ‖T+‖ = ‖T‖. T+

is called the adjoint or transpose of T .(b) Applying the construction in (a) twice, one obtains T++ ∈ L(X ∗∗,Y∗∗). If X and Y

are identified with their natural images ˆcX and Y in X ∗∗ and Y∗∗, then T++|X =T .

(c) T+ is injective iff the range of T is dense in Y.(d) If the range of T+ is dense in X ∗, then T is injective; the converse is true if X is

reflexive.

Proof. (a) ‖T+f‖ ≤ ‖T‖‖f‖ and thus ‖T+‖ = sup{‖T+f‖ : ‖f‖ = 1} ≤ ‖T‖ < ∞, meaningthat T+ is bounded. It remains to show that ‖T+‖ ≥ ‖T‖. Let ε > 0, choose x ∈ Xsuch that ‖x‖ = 1 and ‖Tx‖ > ‖T‖ − ε. If ‖Tx‖ = 0, then automatically ‖T+‖ ≥ ‖T‖.Otherwise we can choose f ∈ Y∗ such that ‖f‖ = 1 and f(Tx) = ‖Tx‖. Then

‖T+‖ ≥ ‖T+f‖ ≥ ‖T+fx‖ = ‖f(Tx)‖ = ‖Tx‖ ≥ ‖T‖ − εand ‖T+‖ ≥ ‖T‖ since ε is arbitrary.

(b) T++f = f ◦ T+. Given x ∈ X , (T++x)(f) = x ◦ T+(f) = x(T+f) = T+f(x) = f(Tx).

Then T++x = ˆ(Tx) and thus T++x = Tx if we identify Tx with its canonical image ˆ(Tx).(c) Suppose T (X ) is dense in Y and T+(f) = T+(g). Then f ◦ T = g ◦ T . Let y ∈ Y,

by denseness choose a sequence {yn}n ∈ T (X ) that converges to y. Since yn ∈ T (X ),f(yn) = g(yn) for all n. Notice that f and g are both bounded linear and thus continuous.Sending n → ∞, we get f(y) = g(y). Thus f = g and we show that T+ is injective.

Conversely, suppose T+ is injective. If T (X ) is not dense in Y, pick y 6∈ T (X ). Then thereis some f ∈ Y∗ such that f(y) := inf

z∈T (X )‖y − z‖. Clearly f 6≡ 0 since f(y) 6= 0, and

35

f |T (X ) ≡ 0. However,

T+f = f ◦ T ≡ 0 = 0 ◦ T = T+0

contradicting injectivity. This shows the conclusion.(d) Suppose T+(Y∗) is dense in X ∗. By (c), T++ is injective and since T++|X = T , T is in-

jective. Conversely, suppose T is injective, since X = X ∗∗, T++ = T , and thus T++ isinjective. By (c) again the range of T+ is dense in X ∗.

�

Remark. This problem is quite standard. You just follow your inituition and everything is clear.Recognizing the relationship between (c) and (d) will save a lot of work. Nevertheless, the resultis important being an analog of the one in finite-dimensional linear algebra.

Folland 5.25

If X is a Banach space and X ∗ is separable, then X is separable.

Proof. Let {fn}∞1 be a countable dense subset of X ∗. For each n choose xn ∈ X with ||xn|| =1 and |fn(xn)| ≥ 1

2 ||fn||. We claim that the linear combinations of {xn}∞1 are dense in X . Toprove the claim, we first define M to be the closure of linear combinations of {xn}∞1 . Then Mis a closed subsapce of X . Suppose there is x ∈ X \ M, then there is some f ∈ X ∗ such thatf(x) := infy∈M ||x− y||. Let ε > 0. By denseness of {fn}∞1 we have some fn such that ||f − fn|| <ε. In particular |fn(xn)− f(xn)| < ε for the corresponding xn and thus |fn(xn)| < ε since xn ∈Mand thus f(xn) = 0. Since |fn(xn)| ≥ 1

2 ||fn||, ||fn|| < 2ε and ||f || < ||fn|| + ε < 3ε. Since ε isarbitrary, ||f || = 0 and thus f ≡ 0. But this means that 0 = f(x) = infy∈M ||x − y|| and thus

x ∈ M = M, a contradiction. Thus M = X and linear combinations of {xn}∞1 is dense in X .We know that linear combinations of {xn}∞1 with coefficients whose real and imaginary parts areboth rational, which we call N , is dense in linear combinations of {xn}∞1 and is thus dense in X .N is countable if we identify it as a countable union (union over n) of countable sets (the set ofcoefficients). Then N is a countable dense subset of X and thus X is separable. �

Remark. The key to solving this problem is finding the correct definition of denseness to be used.I started off tring to use the neighborhood definition of denseness, but I didn’t find a way to use“linear combination” as suggested by the hint of the book. I then realized that linear combina-tion endows a space structure, so I should consider the whole space spanned by {xn}∞1 . The solu-tion naturally follows.

Folland 5.27

There exists meager subsets of R whose complements have Lebesgue measure zero.

Proof. For our convenience, we define Im = [m,m + 1] for all m ∈ Z. We first show that thereis a meager subset of Im for all m whose complement in Im has Lebesgue measure zero. To showthis, we first claim that for every n > 1, there is a generalized Cantor set Kn on Im such thatm(Im \ Kn) = 1/n. To show the claim we define Kn this way: Kn,0 = Im, and suppose we havedefined Kn,j , define Kn,j+1 by removing the middle αjth content from every interval that makes

36

up Kn,j , where αj = 1/(n+ j − 1)2 for j ≥ 1. Thus

m(Kn) =

∞∏i=1

(1− αj) =

(1− 1

n2

)(1− 1

(n+ 1)2

). . .

=1− 1/n

1− 1/(n+ 1)· 1− 1/(n+ 1)

1− 1/(n+ 2). . .

= limk→∞

1− 1/n

1− 1/(n+ k)= 1− 1

n

and thus m(Im \ Kn) = m(Im) −m(Kn) = 1 − (1 − 1/n) = 1/n, showing that the claim is true.We know that Kn is nowhere dense, so K :=

⋃∞n=1Kn is meager. Meanwhile 1 ≥ m(K) ≥ 1− 1/n

for all n and thus m(K) = 1. Thus m(Im \ K) = 0. Therefore, on every Im we have Mm ⊂ Immeager such that m(Im \ Mm) = 0. Let M =

⋃m∈ZMm. Then M is meager and

m(R \ M) = m

[ ⋃m∈Z

(Im \ Mm)

]≤⋃m∈Z

m(Im \ Mm) = 0

and thus m(R \ M) = 0, as desired. �

Another Proof. Equivalently we prove that there is a residual set of measure 0. Let {qn}n be anenumeration of Q. Let ε > 0, define Bn,ε := B(ε2−n, qn). It is clear that Bε :=

⋃nBn,ε is an open

dense subset of R. Then (Bε)c is nowhere dense and thus Bε is residual. Define B :=

⋂nB1/n,

then B is residual as a countable intersection of residual sets. Moreover m(B) ≤ m(B1/n) = 2/nfor every n and thus m(B) = 0, as desired. �

Folland 5.29

Let Y ∈ L1(µ), where µ is counting measure on N, and let X = {f ∈ Y :∑∞

1 n|f(n)| <∞}, equipped with the L1 norm.

(a) X is proper dense subspace of Y, hence X is not complete.(b) Define T : X → Y by Tf(n) = nf(n). Then T is closed but not bounded.(c) Let S = T−1. Then S : Y → X is bounded and surjective but not open.

Proof. (a) Let ε > 0, g ∈ Y. We want to find some f ∈ X such that ||g − f ||1 < ε.Recall that simple functions on N are dense in L1(µ), so we can pick some simple func-tion f :=

∑ni=1 ckχEk , where Ek is a measurable subset of N and ck < ∞, such that

||f − g||1 < ε. We claim that each Ek is finite. Suppose not, there is some Ek that hasinfinite cardinality. Then

∫|f |dµ ≥ |ck|µ(Ek) = ∞, contradicting f ∈ L1(µ) and showing

the claim. Thus f(n) 6= 0 for only finitely many n ∈ N, and clearly∑∞

1 n|f(n)| < ∞.Then f ∈ X and ‖g − f‖1 < ε, as desired. This shows that X is a dense subspace of Y.Now consider f on N such that f(n) = 1/n2, we know that

∫|f |dµ =

∑∞1 1/n2 < ∞

and thus f ∈ Y. However,∑∞

1 n|f(n)| =∑∞

1 1/n = ∞, so f 6∈ X . Thus X is a proper

dense subspace of Y. Then X = Y 6= X , so X is not closed. Since a complete subspace ofa metric space must be closed, X is not complete.

(b) We first show that T is closed, i.e. Γ(T ) is closed in X ×Y, i.e. Γ(T ) = Γ(T ). Let (f, g) bea limit point in Γ(T ), we have {(fn, gn)} ∈ γ(T ) such that (fn, gn) → (f, g) in the productnorm. We want to show g = Tf . Let ε > 0, by convergence there is some large N suchthat when n > N , ‖(fn, gn) − (f, g)‖ < ε. This means that max(‖fn − f‖1, ‖gn − g‖1) < ε

37

for large enough n. Then

‖g − Tf‖1 ≤ ‖g − gn‖1 + ‖gn − Tfn‖1 + ‖Tfn − Tf‖1

≤ ε+ 0 +

∫m|fn(m)− f(m)|dµ

We define hn(m) := m(|fn(m)|+|f(m)|), and∫|hn|dµ =

∫m|fn(m)|dµ+

∫m|f(m)|dµ <∞

since fn, f ∈ X . Thus hn ∈ L1 and m|fn(m)−f(m)| ≤ hn(m). By dominated convergence,sending n → ∞ we get ‖g − Tf‖ < ε. Since ε is arbitrary, ‖g − Tf‖ = 0. Then

∫|g −

Tf |dµ =∑∞

1 |g(n)− Tf(n)| = 0 and thus g(n) = Tf(n) for all n, from which we concludeg = Tf , as desired. Then (f, g) ∈ Γ(T ), showing that it is closed.

We then show that it is not bounded. Notice that fn := χ{n} satisfied ‖fn‖1 = |f(n)| =1 for all n. Then sup{‖Tf‖ : ‖f‖1 = 1} ≥ supn ‖Tfn‖ = supn nf(n) → ∞ and thus T isunbounded.

(c) To make S well-defined, we need to show that T is bijective. For g ∈ Y, define f(n) :=g(n)/n for all n. (Here we assume 0 6∈ N) Then

∑∞1 n|f(n)| =

∑∞1 |g(n)| =

∫|g|dµ < ∞

and thus f ∈ X . Also the most importantly Tf(n) = nf(n) = g(n) and thus g = Tf ,showing that T is surjective. Suppose f1 6= f2, f1(n) 6= f2(n) for some n. Then Tf1(n) =nf1(n) 6= nf2(n) = Tf2(n) and thus Tf1 6= Tf2, showing that T is injective. Then Tis bijective as desired and S is well-defined. We now claim that S is defined such thatSg(n) = g(n)/n. To show the claim, observe that TSg(n) = ng(n)/n = g(n) and thusTS is the identity. Similarly we can show that ST is also the identity. Then the claim istrue. We need to show that S is bounded. This is true since

sup{‖Sg‖ : ‖g‖ = 1} = sup

{∫ ∣∣∣∣g(n)

n

∣∣∣∣dµ :

∫|g|dµ = 1

}≤ sup

{∫|g|dµ :

∫|g|dµ = 1

}= 1

Also S is surjective since T is bijective. Eventually, if S = T−1 is open, T is continuousand thus bounded, a contradiction, so S is not open, as desired. This finishes the proof.

�

Folland 5.37

Let X and Y be Banach spaces. If T : X → Y is a linear map such that f ◦ T ∈ X ∗ forevery f ∈ Y∗, then T is bounded.

Proof. Let f ∈ Y∗. f ◦T is bounded and thus continuous. Then f ◦T is closed and thus Γ(f ◦T ) ={(x, f ◦ T (x)) : x ∈ X} is closed. We define hf (x, y) := (x, f(y)) and it is continuous since

continuity is component-wise. (In particular f is bounded and thus continuous) Then h−1f (Γ(f ◦T )) = {(x, y) : (x, f(y)) = (x, f ◦ T (x))} is closed. We claim that

⋂f∈Y∗ h

−1f (Γ(f ◦ T )) = Γ(T ).

To show this claim, we first observe that Γ(T ) ⊂ h−1f (Γ(f ◦ T )) for every f and thus Γ(T ) ⊂⋂f∈Y∗ h

−1f (Γ(f ◦ T )). Conversely, suppose (x, y) ∈

⋂f∈Y∗ h

−1f (Γ(f ◦ T )), then f(y) = f ◦ T (x) for

all f ∈ Y∗. If y 6= x, since bounded linear functionals on Y separate points, there is some g suchthat g(y) 6= g◦T (x), a contradiction. Then (x, y) ⊂ Γ(T ), showing our claim.

⋂f∈Y∗ h

−1f (Γ(f ◦T ))

is closed since each h−1f (Γ(f ◦ T )) is closed, so Γ(T ) is closed by the claim. Then T is closed and

thus bounded by closed graph theorem. �

Remark. The condition X and Y are Banach spaces hints using the closed graph theorem. There-fore the goal is reduced to showing that Γ(T ) is closed. The key step here is expressing Γ(T ) as

38

a (potentially) huge intersection of closed sets. A immature observation: sometimes there mightnot be a single object that satisfies the desired property, but considering a (huge) arbitrary union(mostly for open sets) or intersection (mostly for open sets) may work. In many other situations,if a union is still not clear enough, further expressing a union as a huge cartesian product andapply nice theorems like Tychonoff gives desired results.

Folland 5.38

Let X and Y be Banach spaces, and let {Tn} be a sequence in L(X ,Y) such that limTnxexists for every x ∈ X. Let Tx = limTnx; then T ∈ L(X ,Y).

Proof. Since addition and multiplication respect limits, T is linear. {Tnx} converges and in par-ticular bounded for each x. Since X is a Banach space, by uniform boundedness supm ‖Tm‖ < Mfor some M > 0. Then

‖Tx‖ = limn→∞

‖Tmx‖ ≤ supm‖Tm‖‖x‖ ≤M‖x‖

and thus T is bounded, as desired. �

Remark. This is a direct application of uniform boundedness principle. Uniform boundednessprinciple is proved cleverly, but its applications seem to be straightforward at most times.

Folland 5.42

Let En be the set of all f ∈ C([0, 1]) for which there exists x0 ∈ [0, 1] such that |f(x) −f(x0)| ≤ n|x− x0| for all x ∈ [0, 1].

(a) En is nowhere dense in [0, 1].(b) The set of nowhere differentiable functions is residual in C([0, 1]).

Proof. (a) We claim that given En, every f ∈ En can be uniformly approximated by a piece-wise linear function, whose linear pieces, finite in number, have slope ≥ 2n or ≤ −2n.To show the claim, we first observe that f is continouous on [0, 1] compact and there-fore uniformly continuous on [0, 1]. Let ε > 0, we know that there is some δ > 0 suchthat when |x − y| < δ, |f(x) − f(y)| < ε

2 . We define δ′ := min( ε4n , δ). Then con-

sider a partition 0 = x0 ≤ x1 ≤ · · · ≤ xn = 1 such that xj − xj−1 < δ′ for allj. We construct a piecewise linear function g by connecting f(xj−1) and f(xj) for ev-ery j, and we may assume that every such linear segment has slope whose absolute value≥ 2n, since if any linear piece has slope whose absolute value < 2n, we can replace itwith a bell-shaped graph, i.e. a wedge such that the ascending piece has slope 2n and de-scending piece has slope −2n. The refined partition still satisfies all the assumptions men-tioned above. Now notice that for the g we just constructed, in every [xj−1, xj ], for anyx, y ∈ [xj−1, xj ], |g(x) − g(y)| ≤ min( ε2 , 2n|x − y|) ≤ min( ε2 , 2nδ

′) = ε2 . It remains to show

that supx∈[0,1] |f(x)− g(x)| < ε. For any x ∈ [0, 1], we can find a j such that x ∈ [xj−1, xj ],then

|g(x)− f(x)| ≤ |g(x)− g(xj)|+ |g(xj)− f(xj)|+ |f(x)− f(xj)| <ε

2+ε

2= ε

showing our claim. Notice that |g(x) − g(x0)| ≥ 2n|x − x0| > n|x − x0| (since n ≥ 1)for some x 6= x0 lying in the same line segment as x0 and thus g 6∈ En Therefore, by ourclaim, for any ε > 0 and f ∈ En, we can find g 6∈ En such that supx∈[0,1] |f(x) − g(x)| < ε

and thus En is nowhere dense in [0, 1], as desired.39

(b) We denote the set of nowhere differentiable functions using C, and want to show that Cc ismeager. Suppose f ∈ Cc, then f is differentiable at some x0 ∈ [0, 1]. We define φ(x) :=f(x)− f(x0)

x− x0where φ(x0) := f ′(x0). and claim that it is continuous on [0, 1]. It is clear

that φ is continuous at x ∈ [0, 1] \ {x0}, so it suffices to show continuity at x0. This istrue since

limx→x0

φ(x) = limx→x0

f(x)− f(x0)

x− x0= f ′(x0) = φ(x0)

Since φ is continuous, it is bounded on [0, 1] by extreme value theorem. Then f ∈ Em for

some m. Then Cc ⊂⋃nEn and Cc =

⋃n(En∩Cc). Since each En is nowhere dense, E

◦n = ∅

and thus En ∩ Cc◦ ⊂ E

◦n = ∅. Then En ∩ Cc is nowhere dense and Cc is meager. Thus C is

residual, finishing the proof.�

Folland 5.45

The space C∞(R) of all infinitely differentiable functions on R has a Frechet space topol-

ogy with respect to which fn → f iff f(k)n → f (k) uniformly on compact sets for all k ≥ 0.

Proof. We define pk,l(f) = sup|x|≤k |f (l)(x)| and verify that it is a semi-norm. This is true be-

cause pk,l(f + g) = sup|x|≤k |(f + g)(l)(x)| = sup|x|≤k |f (l)(x) + g(l)(x)| ≤ sup|x|≤k |f (l)(x)| +sup|x|≤k |g(l)(x)| = pk,l(f) + pk,l(g), and pk,l(rf) = sup|x|≤k |(rf)(l)(x)| = sup|x|≤k |rf (l)(x)| =

|r| sup|x|≤k |f (l)(x)| = |r|pk,l(f). Also {pk,l}k,l∈N is clearly countable since the index set N × N is

countable. By theorem 5.14(b), fn → f in the topology generated by these seminorms iff pk,l(fn−f) = sup|x|≤k |f

(l)n (x)−f (l)(x)| → 0 iff f

(l)n → f (l) uniformly on all compact subsets since any com-

pact subset of R is closed and bounded and is eventually contained in some larger enough [−k, k].Eventually we verify that {pk,l}k,l∈N makes makes C∞ a Frechet space, i.e. a complete Hausdorfftopological vector space. Let f 6= 0, there is some x0 ∈ R such that f(x0) 6= 0. Then considera large enough k such that x0 ∈ [−k, k] and thus pk,0(f) = sup|x|≤k |f(x)| ≥ f(x0) > 0. Thus

by 5.16 (a) C∞ is Hausdorff. It remains to show that it is complete. Suppose we have a Cauchy

sequence 〈fn〉n in C∞(R), then pk,l(fn − fm) = sup|x|≤k |f ln(x)− f (l)m (x)| → 0 as m,n→∞. Then

〈f (l)n |[−k,k]〉 is uniformly Cauchy for all l and k. Since each f(l)n |[−k,k] is continuous and C[k, k] is

complete, f(l)n |[−k,k] uniformly converges to some gk,l ∈ C[−k, k]. Now we define gl on R such that

gl(x) = gk,l(x) where x ∈ [−k, k]. This function is well-defined since gk,l(x) = limn→∞ f(l)n (x)

for all k which means that the definition of gk,l(x) is independent of k. Eventually we observe

that f(l)n → gl locally uniformly by our definition of gl, and we claim that gl = g

(l)0 . To show

this claim, we shall do an induction on l as in problem 5.9. The base case is trivial since g0 = g0.

Suppose we have gl = g(l)0 ,

gl(x) = limn→∞

f (l)n (x) = limn→∞

∫ x

0f (l+1)n (t)dt

Notice that f(l+1)n is continuous and therefore bounded on [0, x], so by dominated convergence

theorem,

gl(x) = limn→∞

∫ x

0f (l+1)n (t)dt =

∫ x

0limn→∞

f (l+1)n (t)dt =

∫ x

0gl+1(t)dt

40

and by fundamental theorem of calculus we get g′l(x) = gl+10 (x) = gl+1, and the claim follows by

induction. Thus f(k)n → g

(k)0 locally uniformly for every k and thus fn → g0 by the previous part

of the problem. g(l)0 = gl is continuous for all l, so g0 ∈ C∞(R) and thus the space is complete.

This finishes the proof. �

Folland 5.48

Suppose that X is a Banach space.(a) The norm-closed unit ball B = {x ∈ X : ‖x‖ ≤ 1} is also weakly closed.(b) If E ⊂ X is bounded with respect to the norm, so is its weak closure.(c) If F ⊂ X ∗ is bounded with respect to the norm, so is its weak* closure.(d) Every weak*-Cauchy sequence in X ∗ converges.

Proof. (a) We show the result by showing that the complement of B is weakly open. Supposey ∈ Bc. Since B is norm closed, by Hahn-Banach there is a linear functional f such thatf(y) = δ where δ := infx∈B ‖x − y‖. Then f−1[B(y, δ)] is a weakly open ball excluding B.Hence Bc is weakly open and B is weakly closed.

(b) Since E is bounded with respect to norm, E ⊂ B = {x ∈ X : ‖x‖ ≤ M} for someM ≥ 0. By appropriate scaling if necessary, we may assume M = 1. Then by a) B isweakly closed and hence the weak closure of E is contained in B. It follows that the weakclosure of E is bounded in norm as well.

(c) Suppose F ⊂ X ∗I s bounded in norm. Similar to above, without loss of generality we mayassume F ⊂ B, where B is the norm closed unit ball as defined in a). By Alaoglu, B iscompact in the weak* topology on X and hence closed. Then the weak* closure of F isalso contained in B and it follows that it is weak* bounded.

(d) Let 〈fn〉n be a weak* Cauchy sequence in X ∗, then fn − fm → 0 as n,m → ∞ and thus(fn − fm)(x) → 0 as n,m → ∞ and eventually ‖fn(x) − f(x)‖ → 0 as n,m → ∞. Thus〈fn(x)〉n is Cauchy for each x and thus converges since K = {R,C} is complete. We setf(x) := limn fn(x), then f ∈ X∗ and fn(x) → f(x) for all x ∈ X , showing that fn → f inweak* topology. This finishes the proof.

�

Folland 5.51

A vector subspace of a normed vector space X is norm-closed iff it is weakly closed.

Proof. Suppose V ⊂ X is norm closed. We show that X is weakly closed by showing that thecomplement of V is open. Suppose x ∈ V c, then since V is norm closed, it follows by Hahn-Banach that there is a linear functional f such that f(x) = δ where δ = infy∈V ‖x − y‖. Thenf−1[B(δ, f(x))] is a weakly open neighborhood of x excluding V [weakly open because f as a lin-ear functional is assumed to be continuous in weak topology and B(δ, f(x)) is open]. Thus V c isweakly open and V is weakly closed. Conversely, suppose V ⊂ X is weak closed. Let x ∈ V inthe norm topology, then there is 〈xn〉n such that xn → x in the norm topology, i.e. ‖xn − x‖ → 0as n → ∞. For any f ∈ X ∗, |f(xn − x)| ≤ ‖f‖ · ‖xn − x‖ → 0 as n → ∞, so xn → x in weaktopology and thus x ∈ V since V is weak-closed. �

41

Folland 5.53

Suppose that X is a Banach space and {Tn}, {Sn} are sequences in L(X ,X ) such thatTn → T strongly and Sn → S strongly.

(a) If {xn} ⊂ X and ‖xn − x‖ → 0, then ‖Tnxn − Tx‖ → 0.(b) TnSn → TS strongly.

Proof. (a) Notice that

‖Tnxn − Tx‖ ≤ ‖Tnxn − Tnx‖+ ‖Tnx− Tx‖= ‖Tn(xn − x)‖+ ‖Tnx− Tx‖≤ ‖Tn‖‖xn − x‖+ ‖Tnx− Tx‖ (∗)

For any x ∈ X , since {Tn} converges strongly, {Tnx} converges and in particular boundedin the norm metric, i.e. supn ‖Tx‖ < ∞ for every x. Since X is a Banach space, by uni-form boundedness we have supn ‖Tn‖ < M for some large M > 0. Then (∗) ≤ M‖xn −x‖ + ‖Tnx − Tx‖. Sending n to infinity, ‖xn − x‖ → 0 and ‖Tnx − Tx‖ → 0 (strongconvergence). Thus ‖Tnxn − Tx‖ → 0, as desired.

(b) For every x ∈ X , Tnx → Tx and Sn → Sx since Tn → T and Sn → S strongly. ThenTnxSnx → TxSx or equivalently TnSnx → TSx. Since x is arbitrary, TnSn → TSstrongly. This finishes the proof.

�

6. Chapter 6-Lp Spaces

Folland 6.5

Suppose 0 < p < q < ∞. Then Lp 6⊂ Lq iff X contains sets of arbitrarily small positivemeasure, and Lq 6⊂ Lp iff X contains sets of arbitrarily large finite measure. What aboutthe case q =∞?

Proof. Lp 6⊂ Lq: Suppose Lp 6⊂ Lq, then there is some f ∈ Lp \ Lq. Consider En := {x :|f(x)| > n} for n ∈ N, we have

∞ > ‖f‖pp ≥ ‖fχEn‖pp =

∫|fχEn |pdµ > npµ(En)

and thus µ(En) < ‖f‖pp/np. Therefore µ(En)→ 0 as n→∞ since f ∈ Lp and thus ‖f‖pp <∞. Now it suffices to show that µ(En) > 0 for each n. If in the contrary µ(En) = 0, letFn := Ecn, we have

‖f‖pp =

∫|f |pdµ =

∫|f |pχFndµ <∞

notice that also we have

‖f‖qq =

∫|f |p|f |q−pdµ =

∫|f |p|f |q−pχFndµ ≤ nq−p

∫|f |pχFndµ = nq−p‖f‖pp <∞

contradicting f 6∈ Lq. Conversely, suppose X contains sets of arbitrarily small positivemeasure, then we can pick a disjoint family of sets {En}n such that 0 < µ(En) < 2−n.This is because we can take a family {Fn}n of sets such that µ(Fn) < 2−n, and definingEn := Fn \

⋃∞n+1Ei creates the desired subsets {En}n. Now we define f :=

∑∞1 anχEn ,

42

where an = µ(En)−1/p. Notice that

‖f‖pp =

∫|f |pdµ =

∫ ∣∣∣∣ ∞∑n=1

µ(En)−1/pχEn

∣∣∣∣pdµ =

∫ ( ∞∑n=1

µ(En)−1/pχEn

)pdµ

= limm→∞

∫ ( m∑n=1

µ(En)−1/pχEn

)pdµ = lim

m→∞

∫ m∑n=1

χEndµ = limm→∞

m∑n=1

µ(En) (1)

since Ens are disjoint. Then by our assumption 0 < (1) < limm→∞∑m

n=1 2−n < ∞,showing that f ∈ Lp. The same thing won’t happen on ‖f‖q, since

‖f‖qq =

∫|f |qdµ =

∫ ∣∣∣∣ ∞∑n=1

µ(En)−1/pχEn

∣∣∣∣qdµ =

∫ ∞∑n=1

µ(En)−q/pχEndµ

≥ limm→∞

m∑n=1

(1/µ(En))q/p−1 ≥ limm→∞

m∑n=1

(2q/p−1)n ≥ limm→∞

m =∞

which fails to converge. Thus ‖f‖q =∞, showing that f 6∈ Lq.Lq 6⊂ Lp: Suppose X has finite measure, then by proposition 6.12 Lp ⊂ Lq, a contradiction.

Conversely, suppose X contains sets of arbitrarily large measure, then we can find disjoint{En}n such that 1 ≤ µ(En) < ∞ for all n: First we chose F1 with ∞ > µ(F1) ≥ 1, andthen suppose we have chosen Fn, choose Fn+1 such that µ(Fn+1) ≥ 2

∑n1 µ(Fn). Thus by

considering En := Fn \⋃n−1

1 Ei we get the desired subsets. Consider f :=∑∞

1 anχEn ,

where an = µ(En)−1/q. The rest of the proof is similar to above.q =∞: In the case q = ∞, we claim that Lp 6⊂ Lq iff X contains subsets of arbitrarily

small measure, and that Lq 6⊂ Lp only if X has infinite measure. For the first part ofthe statement, we use the same proof as above, except setting f =

∑∞1 anχEn , where

an = µ(En)−1/(p+1). For the second part of the statement, if X has infinite measure, wecan consider the same f as in the proof of Lq 6⊂ Lp above and that f is clearly in L∞.This finishes the proof. Note that for the second part, the if direction fails. Consider thefollowing silly example: let X = {0, 1} such that 0 has infinite measure and 1 has zeromeasure. For every f ∈ L∞, f must be actually bounded, and thus ‖f‖p =

∫|f |p =

|f(1)|p <∞, showing that f ∈ Lp.�

Folland 6.9

Suppose 1 ≤ p <∞. If ‖fn−f‖p → 0, then fn → f in measure, and hence some subsquenceconverges to f a.e. On the other hand, if fn → f in measure and |fn| ≤ g ∈ Lp for all n,then ‖fn − f‖p → 0.

Proof. First of all suppose ‖fn → f‖p → 0 for some 1 ≤ p <∞. Let ε > 0. We define En,ε := {x :|fn(x)− f(x)| ≥ ε}. Then

‖fn − f‖p =

(∫|fn − f |p

)1/p

≥(∫

En,ε

|fn − f |p)1/p

≥ (εpµ(En,ε))1/p = εµ(En,ε)

1/p

and thus µ(En,ε) ≤ ε−p(‖fn − f‖p)p → 0 as n → ∞ since ‖fn → f‖p → 0 as n → ∞. Thismeans that fn → f in measure, as desired. In particular, since fn → f in measure, there is a sub-sequence of 〈fn〉n that converges to f pointwise a.e. On the other hand, suppose fn → f in mea-

sure and |fn| ≤ g ∈ Lp for all n. We first make the observation that g ∈ Lp implies (∫|g|p)1/p <

43

∞ and thus∫|g|p < ∞, meaning that gp ∈ L1. Now since fn → f in measure, there is some

subsequence 〈fnk〉k that converges to f a.e. Since 〈fnk〉k is a subsequence, we also have |fnk | < gfor all k. Sending k →∞ we also obtain |f | ≤ g a.e. Therefore, |fnk − f |p ≤ (|fnk |+ |f |)p ≤ (2g)p.Remember our observation that gp ∈ L1, we have |fnk − f |p ≤ (2g)p = 2pgp ∈ L1. By dominatedconvergence theorem, we have

limn→∞

∫|fnk − f |

p =

∫limn→∞

|fnk − f |p = 0

and hence taking 1/pth power on each side we get ‖fnk − f‖p → 0. We now claim that actually‖fn − f‖p → 0. Suppose not, there is some δ > 0 such that ‖fn − f‖p ≥ δ for infinitely manyfn. We arrange them to a new sequence, which for our convenience we call 〈gn〉n. Since 〈gn〉n isessentially a subsequence of 〈fn〉n, we still have gn → f in measure and |gn| ≤ g ∈ Lp for all n.By exactly the same reasoning as above we can show that there is a further subsequence 〈gnk〉nsuch that ‖gnk − f‖p → 0, meaning that for large enough k we have ‖gnk − f‖p < δ, contradictingour assumption. Thus our claim is true, meaning that ‖fn − f‖p → 0, as desired. �

Folland 6.13

Lp(Rn,m) is separable for 1 ≤ p <∞. However, L∞(Rn,m) is not separable.

Proof. We first show that Lp(R,m) is separable for 1 ≤ p < ∞. Let F be the family of simplefunctions of form

∑nj=1 ajχFj where aj ∈ Q and Fj is a finite union of measurable rectangles

with rational-length sides. We claim that F is a countable dense subset of Lp(R,m). First of allwe need to show that F is countable. Notice that there are countably many such sets Fj sincethere are countably many such measurable rectangles, and countably many such aj , so there arecountably many characteristic functions of the form ajχRj . Identifying each sum

∑nj=1 ajχFj

as (a1χF1 , ...) we know that there are countably many of them. Thus F is countable. We pro-ceed to show that F is dense. Let f ∈ Lp(Rn,m) and ε > 0. Since simple functions are densein Lp(Rn,m), we may assume that f is simple and can be written as

∑n1 bjχEj . Fix the ε. By

denseness of rationals, we have some aj such that |aj − bj | < [ε/m(Ej)]1/p for each j. Moreover,

by regularity of Lebesgue measure, for each Ej there is a finite union of measurable rectangles,which may be taken to have rational coordinates by denseness of rationals and which we call Fj ,such that m(Ej4Fj) < ε/|aj |p. Then(∫ ∣∣∣∣ n∑

j=1

(bjχEj − ajχFj )∣∣∣∣pdm)1/p

=

( n∑j=1

∫|bjχEj − ajχFj |pdm

)1/p

(1)

For each j, ∫|bjχEj − ajχFj |pdm ≤

∫|(bj − aj)χEj |pdm+

∫|aj(χEj − χFj )|pdm

≤|bj − aj |pm(Ej) + |aj |pm(Ej4Fj) < 2ε

and thus (1) < (2nε)1/p. Since ε is arbitrary, this shows that F is dense, as desired.We then show that L∞(R,m) is not separable. It suffices to give an uncountable family F ⊂ L∞

such that ‖f − g‖∞ ≥ 1 for all f, g ∈ F with f 6= g. This is because if we take an open ball ofradius 1 around each f ∈ F , we obtain an uncountable disjoint collection of open balls. Then forany countable subset of L∞(R,m), we must have some open ball in this collection not containingany of the points, meaning that this subset cannot be dense. Now we give the desired subset:consider F := {χBr}r∈R>0 , where Br is the open ball of radius r. Suppose f = χBr 6= g = χBr′ ,we must have r 6= r′ and we may assume r < r′. Then for x ∈ Br′ \ Br, a positive measure

44

set, we have |f(x) − g(x)| = 1, and thus ‖f − g‖∞ ≥ 1. The family is clearly uncountable, asdesired. �

Folland 6.19

Define φn ∈ (l∞)∗ by φn(f) = n−1∑n

1 f(j). Then the sequence {φn} has a weak* clusterpoint φ, and φ is an element of (l∞)∗ that does not arise from an element of l1.

Proof. First of all, since f ∈ l∞, f is essentially bounded by some M > 0, and in the case ofcounting measure this means that f is actually bounded by M . φn(f) takes the arithmetic meanof the first n terms, and is thus bounded by M for all n. Therefore

‖φn‖ = sup{|φn(f)| : ‖f‖∞ = 1} ≤ 1

by above reasoning. This means that φn ∈ B∗, the unit ball of (l∞)∗. By Alaoglu, B∗ is compactin the weak* topology. We define Km := {φn : n ≥ m}, and notice that 〈Km〉m is a familyof closed sets (here it means weak* closure) with finite intersection property. Since each Km ⊂B∗ = B∗, we have

⋂mKm 6= ∅. Pick φ ∈

⋂mKm, we claim that φ is a weak* cluster point of

〈φn〉n. To show this claim, take a weak* neighborhood of φ, since φ ∈ Km for each m, U ∩Km 6=∅ for each m. This means that for each m, there is some φn ∈ U with n ≥ m. Then 〈φ〉n isfrequently in U and thus φ is a weak* cluster point. This shows the claim.We then show that φ doesn’t arise from l1. Suppose in the contrary that it does, then φ(f) =ϕg(f) :=

∫fgdµ for some g ∈ l1. Consider a special family of functions {fk}k := {χ{k}}k, then

φ(fk) =∫χ{k}gdµ = g(k). Notice that φn(fk) = 0 for n < k, and φn(fk) = 1/n for n ≥ k. Since

φ is a weak* cluster point of 〈φn〉n, φ(fk) is a cluster point of 〈φn(fk)〉n. Then a subsequence of〈φn(fk)〉n converges to φ(fk) and φ(fk) = 0. Then g(k) = 0 for all k, meaning that φ ≡ 0.However, consider f ≡ 1 which lies in l∞ since it is uniformly bounded, φn(f) ≡ 1 for all n andtherefore φ(f) = 1 ≥ 0, a contrasiction. Then φ doesn’t arise from l1. �

Folland 6.20

Suppose supn ‖fn‖p <∞ and fn → f a.e.(a) If 1 < p <∞, then fn → f weakly in Lp

(b) The result of (a) is false in general for p = 1. It is, however, true for p = ∞ if µ isσ-finite and weak convergence is replaced by weak* convergence.

Proof. (a) Let 1 < p < ∞ and q the conjugated of p. By Riesz representation theorem, toprove fn → f weakly in Lp, it suffices to prove that ϕg(fn) → ϕg(f) for all g ∈ Lq, whereϕg(f) =

∫fg. Given g ∈ Lq and let ε > 0, we have the following observations, which we

shall prove at the end of this question:1. There is some δ > 0 such that

∫E |g|

q < ε whenever µ(E) < δ.2. There is an A ⊂ X such that µ(A) <∞ and

∫X \ A |g|

q < ε.

3. There is B ⊂ A such that µ(A \ B) < ε and fn → f uniformly on B.45

Based on the observations, we have

|ϕfn(g)− ϕf (g)| =∣∣∣∣ ∫ (fn − f)g

∣∣∣∣ ≤ ∣∣∣∣ ∫B

(fn − f)g

∣∣∣∣+

∣∣∣∣ ∫A \ B

(fn − f)g

∣∣∣∣+

∣∣∣∣ ∫X \ A

(fn − f)g

∣∣∣∣≤∫B|(fn − f)g|+

∫A \ B

|(fn − f)g|+∫X \ A

|(fn − f)g|

≤(∫

B|(fn − f)|p

)1/p(∫B|g|q)1/q

+

(∫A \ B

|g|q)1/q

‖f − fn‖p +

(∫X \ A

|g|q)1/q

‖fn − f‖p

<

(∫B|fn − f |p

)1/p

‖gn − g‖q + 4Mε1/q ≤ ‖fnχB − fχB‖uµ(B) + 4Mε1/q (1)

since fn → f uniformly on B, ‖fnχB − fχB‖u → 0. Also notice that µ(B) ≤ µ(A) < ∞.

Therefore, sending n→∞, we get limn→∞ |ϕfn(g)− ϕf (g)| < 4Mε1/q. Since ε is arbitrary,this shows that limn→∞ |ϕfn(g) − ϕf (g)| = 0, as desired. Thus it remains to show theobservations to finish the proof.1. Since |g|q is measurable, ν(E) :=

∫E |g|

qdµ is a well-defined measure, and it is clearthat ν is absolutely continuous to µ since integrating on a measure-zero set we get 0.Notice that g ∈ L1, so ν(E) ≤ ‖g‖q is a finite signed measure. Then the result followsby the ε− δ definition of absolute continuity.

2. Since |g|q ∈ L+ and∫|g|q <∞, K := {x : |g|q > 0} is σ-finite. Then we can write K =⋃∞

1 En where each En has finite measure, and define Kn :=⋃n

1 En for our convenience.Notice that ν(K) =

∫K |g|

qdµ = limn→∞ ν(Kn) by continuity from below, where ν isdefined as above. Since ν(K) < ∞, this means that there is some N ∈ N such that|ν(K)− ν(KN )| < ε. Take A = KN , we have

|ν(X)− ν(A)| = |ν(K)− ν(A)| < ε =⇒ |ν(X \ A)| < ε =⇒∫X \ A

|g|pdµ < ε

as desired.3. Since µ(A) <∞ and fnχA → fχA a.e., this result is immediate by Egoroff’s theorem.Now the proof is complete.

(b) First of all we show that the result of (a) is false by giving two counterexamples. First ofall, in L1(R,m) consider fn = 1

nχ(0,n).∫|fn|dm =

∫1

nχ(0,n) =

1

nm(0, n) = 1

for all n and thus supn ‖fn‖1 < ∞. Also it is clear that fn → 0 a.e. However, for g ≡ 1 ∈L∞ (since ∞ is the conjugate of 1), ϕg(fn) =

∫fndm = 1 for all n and thus ϕg(fn) 6→ 0 =

ϕg(0). Thus fn 6→ f weakly in L1. Next we consider fn = 1nχ{1,...,n} ∈ l

1, and we use µ forcounting measure. Notice that∫

fndµ =

∫1

nχ{1,...,n} =

1

n· n = 1

for all n and thus supn ‖f‖1 < ∞. Also fn → 0 a.e. However, for g ≡ 1 ∈ l∞, we haveϕg(fn) =

∫fndµ = 1 for all n. Thus ϕg(fn) 6→ 0 = ϕg(0) and hence fn 6→ 0 weakly.

We then show that the result is true for p = ∞ im the context of µ σ-finite and weak*convergence. Since µ is σ-finite, L∞ = (L1)∗. By Riesz representation, it suffices to showthat ϕfn → ϕf , and this yields to show that ϕfn(g)→ ϕf (g) for all g ∈ L1.

|ϕfn(g)− ϕf (g)| =∣∣∣∣ ∫ fng −

∫fg

∣∣∣∣ ≤ ∫ |fng − fg| (1)

46

Suppose supn ‖fn‖∞ < M , we have |fng − fg| ≤ 2M |g| ∈ L1 for all n. Also fn → f a.e.and g essentially bounded implies fng → fg a.e. Applying dominated convergence we get(1)→ 0, as desired.

�

Folland 6.22

Let X = [0, 1] with the Lebesgue measure.(a) Let fn(x) = cos 2πnx. Then fn → 0 weakly in L2, but fn 6→ 0 a.e. or in measure.(b) Let fn(x) = nχ(0,1/n). Then fn → 0 a.e. and in measure, but fn 6→ 0 weakly in Lp

for any p.

Proof. (a) We first show that fn → 0 weakly in L2. By Riesz representation, it suffices toshow that ϕg(fn) :=

∫fng converges to ϕg(0) = 0 for all g ∈ L2. To this end, let g ∈

L2 and ε > 0. By denseness of simple functions, we can choose a simple function φ :=∑n1 aiχEi such that ‖g − φ‖2 < ε. Furthermore, since each Ei is measurable and m(Ei) ≤

m([0, 1]) = 1 < ∞, by regularity we can find a finite union of intervals, which we callFi, such that m(Fi4Ei) < ε/(n|ai|) for each i. For convenience, let’s call the redefinedfunction φ′ :=

∑m1 biχUi(=

∑n1 aiχFi), where Ui are intervals with end points ai and bi.

Now

|ϕφ′(fn)| =∣∣∣∣ ∫ φ′ cos 2πnx

∣∣∣∣ =

∣∣∣∣ m∑i=1

bi

∫Ui

cos 2πnx

∣∣∣∣ ≤ m∑i=1

bi

∣∣∣∣ ∫ bi

ai

cos 2πnx

∣∣∣∣≤

m∑i=1

∣∣∣∣ bi2πnsin 2πnx

∣∣∣∣biai

∣∣∣∣ ≤ m∑i=1

|bi|πn→ 0 as n→∞

By assumption we have ‖g − φ‖2 ≤ ε, and

‖φ− φ′‖2 =

∫|φ− φ′| ≤

n∑i=1

|ai|m(Fi4Ei) < ε

so by Minkowski inequality we have ‖g − φ′‖ < 2ε. Since the assignment g 7→ ϕg is isomet-ric, we have ‖ϕφ′ − ϕg‖ < 2ε. Since ‖fn‖2 =

∫[0,1] cos 2πnx ≤ 1, it follows that

|ϕφ′(fn)− ϕg(fn)| ≤ ‖ϕφ′ − ϕg‖‖fn‖2 ≤ ‖ϕφ′ − ϕg‖ < ε

for all n and sending n → ∞ we have limn→∞ |ϕg(fn)| < ε. Since ε is arbitrary, we havelimn→∞ |ϕg(fn)| = 0. Since g is arbitrary, this shows that fn → 0 weakly. Moreover,we show that no subsequence of 〈fn〉n converges to 0 a.e. Let 〈fnk〉k be a subsequence of〈fn〉n, suppose fnk → 0 a.e., we have |f2nk | = | cos2(2πnkx)| ≤ χ[0,1] ∈ L1([0, 1],m).

Applying dominated convergence we should get∫

cos2 2πnkx→ 0 as k →∞. However,∫cos2(2πnkx) =

∫cos 4πnx+ 1

2=

1

8πnsin 4πnx

∣∣∣∣10

+1

2=

1

26= 0

a contradiction. It immediately follows that fn 6→ 0 a.e. If fn → 0 in measure, there is asubsequence fnk → 0 a.e., hence fn 6→ 0 in measure as well.

(b) First of all fn → 0 a.e. since for every x ∈ [0, 1], when n is large enough 1n < x and hence

fn(x) = 0. fn → 0 in measure since for every ε ≥ 0, µ{|fn| ≥ ε} ≤ 1n → 0 as n → ∞. We

then show that fn 6→ 0 weakly in Lp for any p. Let p ∈ (1,∞) consider its conjugate q.Clearly g ≡ 1 ∈ Lq since [0, 1] has finite measure. Then ϕg(fn) = |

∫fn| =

∫nχ(0,1/n) = 1

47

for all n, and hence ϕg(fn) doesn’t converge. Since we already showed in class that ϕg is awell-defined linear functional, this shows that fn 6→ 0 weakly.

�

Folland 6.26

Complete the proof of Theorem 6.18 for p = 1.

Proof. Suppose p = 1, and q =∞ is its Holder conjugate. Since |K(x, y)f(y)| ∈ L+, by Tonelli,∫ [ ∫|K(x, y)f(y)|dν(y)

]dµ(x) ≤

∫∫|K(x, y)f(y)|dµ(x)dν(y)

=

∫ [ ∫|K(x, y)|dµ(x)

]|f(y)|dν(y) ≤ C

∫|f(y)|dν(y) (1)

Since f ∈ L1(ν), the last integral is finite, and thus∫|K(x, y)f(y)|dν(y) is finite for a.e.-x. This

shows that Tf(x) converges absoluetely for a.e.-x and also∫|Tf(x)|dµ(x) ≤

∫ [ ∫|K(x, y)f(y)|dν(y)

]dµ(x) ≤ C

∫|f(y)|dν(y) <∞(2)

by (1), showing that Tf is defined in L1(µ). Eventually, we can read from (2) that ‖Tf‖1 ≤C‖f‖1. This finishes the proof. �

Remark. This is a analogous but simpler argument conpared to the proof of Theorem 6.18.

Folland 6.36

If f ∈ weakLp and µ({x : f(x) 6= 0}) <∞, then f ∈ Lq for all q < p. On the other hand, iff ∈ (weakLp) ∩ L∞, then f ∈ Lq for all q > p.

Proof. Part 1. First of all, since µ{x : f(x) 6= 0} < ∞, we may assume µ{x : f(x) 6= 0} ≤ C.It follows that λf (α) ≤ C for all α. Also, since f ∈ weakLp, [f ]p = supα>0 α

pλf (α) < ∞. Wemay assume αpλf (α) ≤ M for some M > 0. Given that α > 0, we have λf (α) ≤ M/αp. Since0 < p <∞, by proposition 6.24,

‖f‖qq =

∫ ∞0

αq−1λf (α)dα = q

∫ 1

0αq−1λf (α)dα+ q

∫ ∞1

αq−1λf (α)dα

= qC

∫ 1

0αq−1dα+ qM

∫ ∞1

αq−p−1dα

= Cαq∣∣∣∣10

+ qM

(1

q − pαq−p

∣∣∣∣∞1

)(1)

Since q > 0, Cαq∣∣10

= C, and for q < p, q − p < 0 and thus qM( 1q−pα

q−p∣∣∞1

) = − qMq−p . It follows

that ‖f‖qq = (1) <∞ for all q < p, showing that f ∈ Lq for all q < p.Part 2. Suppose f ∈ weakLp ∩ L∞. Similar to above, we have λf (a) ≤ M/αp for some M > 0.Since f ∈ L∞, there is some β such that λ(α) = 0 when α > β. Then

‖f‖qq = q

∫ ∞0

αq−1λf (α)dα = qM

∫ β

0αq−p−1dα = qM

1

q − pαq−p

∣∣∣∣β0

= qM1

q − pβq−p <∞

for all q ∈ (p,∞). Since we already assume f ∈ L∞, f ∈ Lq for all q > p. This finishes theproof. �

48

Remark. The key technique in this problem is splitting up an integral into different scales, whichis a standard technique when integrals behave differently at different scales. In part 1, the in-tegral is nice in large scale, so we bound it in small scale; in part 2 the integral is nice in smallscale, so we bound it in large scale.

Folland 6.45

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

Tαf(x) =

∫|x− y|−αf(y)dy

Then Tα is weak type (1, q) where q = n/α, and strong type (p, r) where r−1 + 1 = p−1 +na−1.

Proof. We define K(x, y) := |x − y|−α, which is clearly X × Y measurable, and let q denote thesame thing as in the problem. Observe that

βqλK(x,·)(β) = βqm{y : |x− y|−α > β} = βqm{y : |x− y| < β−1/α}

≤ βqmB(x, β−1/α) ≤ βqC1 + β−n/α = βqC1β−q = C1

for some constant C1 > 0. Therefore, [βqλK(x,·)(β)]1/q ≤ C1/q1 for all β and thus [K(x, ·)]q ≤

C1/q1 for some C1 > 0. By symmetry of K(x, y), we can similarly prove that [K(·, y)]q ≤ C2

for some C2 > 0 for all y. By setting C = max(C1/q1 , C2) we obtain [K(x, ·)]q ≤ C for all x

and [K(·, y)]q ≤ C for all y. Now applying theorem 6.36 to q and p as given in the problem weconclude the proof. �

Folland 6.37

If f is a measurable function and A > 0, let E(A) = {x : |f(x)| > A}, and set

hA = fχX \ E(A) +A(sgnf)χE(A), gA = f − hA = (sgnf)(|f | −A)χE(A)

Then

λgA(α) = λf (α+A), λhA(α) =

{λf (α) α < A0 α ≥ A

Proof. First of all,

λgA(α) = µ{|gA| > α} = µ{|(sgnf)(|f | −A)χE(A)| > α} = µ{|(|f | −A)χE(A)| > α} (1)

On E(A) we have |f | > A and therefore (|f | −A)χE(A) > 0. It follows that

(1) = µ{(|f | −A)χE(A) > α} = µ{x ∈ E(A) : |f(x)| −A > α} = µ[E(A) ∩ {|f | > A+ α}] (2)

Observe that |f(x)| > A + α implies |f(x)| > A and hence x ∈ E(A), so {|f | > A + α} ⊂ E(A)and therefore (2) = µ{|f | > A+ α} = λf (α+A). For the second part of the problem,

λhA(α) = µ{|hA| > α} = µ{|fχX \ E(A) +A(sgnf)χE(A)| > α} (3)

We now claim that {|fχX \ E(A) +A(sgnf)χE(A)| > α} = {|fχX \ E(A)| > α}∪{|A(sgnf)χE(A)| >α}. To prove the claim, we first prove the forward inclusion. x ∈ {|fχX \ E(A) +A(sgnf)χE(A)| >α} means that |fχX \ E(A)(x) + A(sgnf)χE(A)(x)| > α. Since X \ E(A) and E(A) are disjoint,we have either

|fχX \ E(A)(x) +A(sgnf)χE(A)(x)| = |fχX \ E(A)(x) + 0| = |fχX \ E(A)(x)| > α49

or

|fχX \ E(A)(x) +A(sgnf)χE(A)(x)| = |0 +A(sgnf)χE(A)(x)| = |A(sgnf)χE(A)(x)| > α

meaning that x ∈ {|fχX \ E(A)| > α} ∪ {|A(sgnf)χE(A)| > α}. Conversely, suppose x ∈{|fχX \ E(A)| > α} ∪ {|A(sgnf)χE(A)| > α}. If x ∈ {|fχX \ E(A)| > α}, x ∈ X \ E(A) andhence A(sgnf)χE(A)(x) = 0. It follows that

|fχX \ E(A)(x) +A(sgnf)χE(A)(x)| = |fχX \ E(A)(x) + 0| = |fχX \ E(A)(x)| > α

and x ∈ {|fχX \ E(A) + A(sgnf)χE(A)| > α}. Similarly we can show the result for the case wherex ∈ {|A(sgnf)χE(A)| > α}. Therefore we have shown the claim. By the claim and the fact that{|fχX \ E(A)| > α} and {|A(sgnf)χE(A)| > α} are disjoint by the disjointness of X \ E(A) andE(A), we obtain

(3) = µ{|fχX \ E(A)| > α}+ µ{|A(sgnf)χE(A)| > α}

If α < A, x ∈ {|fχX \ E(A)| > α} iff x ∈ X \ E(A) and |f(x)| > α iff α < |f(x)| ≤ A, and

(3) = µ{|fχX \ E(A)| > α}+ µ{|A(sgnf)χE(A)| > α} = µ{α < |f | ≤ A}+ µ{|χE(A)| > α/A}= µ{α < |f | ≤ A}+ µ(E(A)) = µ{α < |f | ≤ A}+ µ{|f | > A} = µ{|f | > α} = λf (α)

If α ≥ A, |f(x)| ≤ A ≤ α for all x ∈ X \ E(A) and hence {|fχX \ E(A)| > α} = ∅. Moreover

{|A(sgnf)χE(A)| > α} = {χE(A) > α/A ≥ 1} = ∅so (3) = µ{|fχX \ E(A)| > α}+ µ{|A(sgnf)χE(A)| > α} = 0. This finishes the proof. �

Folland 6.38

f ∈ Lp iff∑∞−∞ 2kpλf (2k) <∞.

Proof. Suppose f ∈ Lp. Then∞∑−∞

2kpλf (2k) =∞∑−∞

2k2k(p−1)λf (2k) = 2p∞∑−∞

2k−12(k−1)(p−1)λf (2k) (1)

Notice that on [2k−1, 2k), α ≥ 2k−1 and λf (α) ≥ λf (2k) (since λf is a decreasing function), so

αp−1λf (α) ≥ (2k−1)p−1λf (2k). Hence∫[2k−1,2k)

αp−1λf (α)dα ≥∫[2k−1,2k)

(2k−1)p−1λf (2k) = 2k−12(k−1)(p−1)λf (2k) (2)

and it follows by additivity that

(1) ≤ 2p∞∑−∞

∫[2k−1,2k)

αp−1λf (α)dα (3)

Notice that the collection of intervals {[2k−1, 2k)}k are pairwise disjoint, and that⋃∞−∞[2k−1, 2k) =

(0,∞). We have

(3) ≤ 2p∞∑−∞

∫αp−1λf (α)χ[2k−1,2k)dα (4)

50

Since |αp−1λf (α)χ[2k−1,2k)| ≤ αp−1λf (α) ∈ L1 by proposition 6.24, by dominated convergencetheorem,

(4) = 2p∫ ∞∑−∞

αp−1λf (α)χ[2k−1,2k)dα = 2p∫αp−1λf (α)χ⋃∞

−∞[2k−1,2k)dα

= 2p∫αp−1λf (α) ≤ 2p

p‖f‖pp <∞

and the result follows. Conversely, suppose∑∞−∞ 2kpλf (2k) < ∞,

∑∞−∞ 2kpµ{|f | > 2k} < ∞. For

our convenience, define Kn := {2n < |f | ≤ 2n+1},∫|f |p ≤

∞∑−∞

2(n+1)pµ(Kn) ≤∞∑−∞

2(n+1)pλf (2n) ≤ 2p∞∑−∞

2npλf (2n) <∞

showing that f ∈ Lp, as desired. This finishes the proof. �

Folland 6.41

Suppose 1 < p ≤ ∞ and p−1 + q−1 = 1. If T is a bounded operator on Lp such that∫(Tf)g =

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

on Lr for all r in [p, q] (if p < q) or [q, p] (if q < p).

Proof. First of all we notice that 1 ≤ q < ∞. We use Σ to denote the space of simple functionsthat vanish outside a set of finite measure. For f ∈ Lp ∩ Lq, Tf ∈ Lp and is thus measurable.Moreover, for any g ∈ Σ, clearly g ∈ Lp ∩ Lq, and thus ‖g(Tf)‖1 ≤ ‖g‖q‖Tf‖p < ∞ by Holder’sinequality, meaning that g(Tf) ∈ L1. Also, by assumption

Mq(Tf) := sup

{∣∣∣∣ ∫ g(Tf)

∣∣∣∣ : g ∈ Σ, ‖g‖p = 1

}≤ sup

{∣∣∣∣ ∫ f(Tg)

∣∣∣∣ : g ∈ Σ, ‖g‖p = 1

}≤ sup{‖f(Tg)‖1 : g ∈ Σ, ‖g‖p = 1}≤ sup{‖f‖q‖Tg‖p : g ∈ Σ, ‖g‖p = 1} (Holder)

≤ sup{‖f‖q‖T‖op‖g‖p : g ∈ Σ, ‖g‖p = 1}≤ sup{‖f‖q‖T‖op : g ∈ Σ, ‖g‖p = 1}= ‖T‖op‖f‖q <∞

We also notice that since f ∈ Lp and T is a bounded operator on Lp, Tf ∈ Lp. If p < ∞,∫|Tf |p < ∞. Hence {|Tf |p 6= 0} = {Tf 6= 0} is σ-finite. If p = ∞, µ the background mea-

sure is assumed to be semifinite. Now applying theorem 6.14 on Folland we obtain that Tf ∈ Lq.That is, we prove that Tf ∈ Lq for any f ∈ Lp ∩ Lq.Observe that Lp ∩ Lq is dense in Lq since simple functions, which are dense in Lp, are containedin Lp ∩ Lq. Therefore for g ∈ Lq we can choose {fn}n ⊂ Lp ∩ Lq such that fn → g in Lq. Nowwith a little abuse of notation we extend T to Lq by defining

Tg = limn→∞

Tfn

where the limit here refers to limit in Lq.Claim. T is well-defined. That is, T is a bounded linear operator on Lq and it is independent ofthe sequence {fn}.

51

Proof of Claim. Linearity of T easily follows from linearity of T on Lp ∩ Lq and linearity of limit.Notice that ‖Tfm−Tfn‖q ≤ Cq‖fm− fn‖q → 0 as m,n→∞, so {Tfn}n is Cauchy and convergessince Lq is complete. Since the limit is unique, it must be Tg. Hence Tg ∈ Lq and ‖Tg−Tfn‖q →0. This shows that for any fn → g in Lq, Tfn → Tg in Lq. Then this definition indeed defines alinear operator on Lq and the definition is independent of the choice of the sequence. (Since forany such sequence {fn}n the limit in Lq will be Tg. �

Hence for f + g ∈ Lp + Lq where f ∈ Lp and g ∈ Lq, T (f + g) = Tf + Tg ∈ Lp + Lq and thusT is a linear operator on Lp + Lq. Moreover, we showed above that T is of strong type (p, p) (byassumption) and strong type (q, q) (by claim). By Riesz-Thorin applied with p and q as given inthe problem, T can be extended to a bounded operator for r where r is given as in the problem.It remains to show that the extention is unique. Suppose there is another extension T ′, h ∈ Lr.Without loss of generality suppose r ∈ [p, q], h = f + g for some f ∈ Lp and g ∈ Lq. Then

T ′h = T ′(f + g) = T ′f + T ′g = Tf + T ′g

Choose gn ∈ Lp ∩ Lq such that gn → g ∈ Lq. Then T ′g = limn→∞ Tgn = Tg and henceT ′h = Tf + Tg = Th. This shows that T is unique. �

7. Chapter 7-Radon Measures

Folland 7.21

Let {fα}α∈A be a subset of C(X) where X is compact and {cα}α∈A be a family of com-plex numbers. If for each finite set B ⊂ A there is µB ∈ M(X) such that ‖µB‖ ≤ 1 and∫fαdµB = cα for α ∈ B, then there is µ ∈ M(X) such that ‖µ‖ ≤ 1 and

∫fαdµ = cα for

all α ∈ A.

Proof. 4 First of all since X is compact the uniform norm on C(X) makes sense, making C(X) anormed vector space. Then by Alaoglu’s theorem B∗ := {Iµ ∈ C(X)∗ : ‖Iµ‖ ≤ 1} is compactin the weak* topology of C(X). We define Mα to be the set of measures µ such that ‖Iµ‖ ≤ 1and

∫fαdµ = cα. Mα is non-empty since {α} is a finite subset of A. We claim that {Mα}α∈A

has finite intersection property. To show the claim, we first show that Mα is closed for each α.Let B := {α1, ..., αn} be a finite subset of A and by assumptions in the problem there is someµB ∈ B∗ such that ‖IµB‖ ≤ 1 and

∫fαdµB = cα for each α ∈ B, meaning that µB ∈

⋂α∈BMα

and thus showing that⋂α∈BMα is non-empty. Thus the claim is true. Notice that Mα ⊂ B∗

for each α, and that Mα = B∗ ∩ {µ ∈ M(X) : µ(fα) = cα} = B∗ ∩ {µ ∈ M(X) : µ(fα) =

cα} = B∗ ∩ fα−1

({cα)} where fα : µ 7→ µ(fα) is the canonical embedding. Since fα is bounded

and thus continuous, fα−1

({cα}) is closed and therefore each Mα is closed as a closed subset of acompact set. Then {Mα}α∈A is a family of closed subsets of B∗ compact with finite intersectionproperty and thus

⋂α∈AMα 6= ∅. Let µ ∈

⋂α∈AMα, ‖µ‖ ≤ 1 and

∫fαdµ = cα for all α ∈ A, as

desired. �

8. Chapter 8-Elements of Fourier Analysis

Folland 8.4

If f ∈ L∞ and ‖Tyf − f‖∞ → 0 as y → 0, then f agrees a.e. with a uniformly continuousfunction.

4In this proof I may use µ and Iµ interchangably, which is common in this situation.

52

Proof. The main goal of this proof is to show that h(x) := limn→∞A1/nf(x) (where Arf(x) =1

m(B(r,x))

∫B(r,x) f(y)dy) is well-defined, and that it is the desired function. Before establishing

this result, we prove some important claims.Claim 1. Arf is uniformly continuous for any r > 0.

Proof of Claim. Let r > 0. For our convenience we use gr to denote Arf , then

‖τygr − gr‖u = supx|gr(x)− gr(x− y)| = sup

x

1

m(B(r, x))

∣∣∣∣ ∫B(r,x)

f(z)dz −∫B(r,x−y)

f(z)dz

∣∣∣∣= sup

x

1

m(B(r, x)

∣∣∣∣ ∫B(r,x−y)

f(z − y)− f(z)dz

∣∣∣∣≤ sup

x

1

m(B(r, x))

∫B(r,x−y)

|τyf(z)− f(z)|dz

≤ supx

1

m(B(r, x))

∫B(r,x−y)

‖τyf − f‖∞dz

= sup ‖τyf − f‖∞ = ‖τyf − f‖∞which tends to 0 as y → 0 and thus gr = Arf is uniformly continuous. �

Claim 2. Arf is uniformly Cauchy as r → 0.

Proof of Claim. Let ε > 0. Since ‖τyf − f‖∞ → 0 as y → 0, there is some δ > 0 such that when|y| < δ, ‖τyf − f‖∞ < ε. Therefore, if r1, r2 < δ,∣∣∣∣ 1

m(B(r1, x))

∫B(r1,x)

f(y)dy − f(x)

∣∣∣∣ ≤ 1

m(B(r1, x))

∫B(r1,x)

|f(y)− f(x)|dy

=1

m(B(r1, x))

∫B(r1,x)

|f(x− (x− y))− f(x)|dy < ε

Since |x− y| ≤ r1 < δ. Analogously we can prove that∣∣∣∣ 1

m(B(r2, x))

∫B(r2,x)

f(y)dy − f(x)

∣∣∣∣ < ε

and therefore

‖Ar1f −Ar2f‖u = supx

∣∣∣∣ 1

m(B(r1, x))

∫B(r1,x)

f(y)dy − 1

m(B(r2, x))

∫B(r2,x)

f(y)dy

∣∣∣∣≤ sup

x

∣∣∣∣ 1

m(B(r1, x))

∫B(r1,x)

f(y)dy − f(x)

∣∣∣∣+

∣∣∣∣ 1

m(B(r2, x))

∫B(r2,x)

f(y)dy − f(x)

∣∣∣∣< sup

x2ε = 2ε

and thus Arf is uniformly Cauchy as r → 0. �

We now prove that h is uniformly continuous. Since {A1/nf}n is uniformly Cauchy by claim 2,{A1/nf(x)}n is a Cauchy sequence for each x and therefore converges since R is complete. Letε > 0, there is some N ∈ N such that when n,m > N , |A1/nf(x) − A1/mf(x)| < ε for all x.Sending m → ∞ gives us |A1/nf(x) − h(x)| < ε, and since x is arbitrary {A1/n}n must convergesuniformly to h.Claim 3. h is uniformly continuous.

Proof of Claim. Let ε > 0. Since {A1/nf}n → h uniformly, there is some N large enough suchthat ‖A1/Nf − h‖u < ε/3. By claim 1 A1/Nf is uniformly continuous, so there is some δ > 0 such

53

that when |y| < δ, |A1/Nf(x− y)−A1/Nf(x)| < ε/3 for all x. Now for all x and |y| < ε,

|h(x− y)− h(x)| ≤ |h(x− y)−A1/n(x− y)|+ |A1/n(x− y)−A1/n(x)|+ |A1/n(x)− h(x)|

<ε

3+ε

3+ε

3= ε

showing that h is uniformly continuous and finishes the claim. �

Since f ∈ L∞, f integrated on any bounded measurable set must admit finite value and thus f islocally integrable. By theorem 3.18 on Folland, h agrees with f a.e. Since h is uniformly continu-ous, this finishes the proof. �

Folland 8.8

Suppose that f ∈ Lp(R). If there exists h ∈ Lp(R) such that

limy→0‖y−1(τ−yf − f)− h‖p = 0

we call h the strong Lp derivative of f . If f ∈ Lp(Rn), Lp partial derivatives of f are de-fined similarly. Suppose that p and q are conjugate exponents, f ∈ Lp, g ∈ Lq, and the Lp

derivative ∂jf exists. Then ∂j(f ∗ g) exists (in the ordinary sense) and equals (∂jf) ∗ g.

Proof. First of all we show that the definition (∂jf) ∗ g makes sense. Since ∂jf is the Lp deriv-ative, it lies in Lp, and g ∈ Lq by assumtion. It follows from proposition 8.8 that ‖∂jf ∗ g‖u ≤‖∂jf‖p‖g‖q <∞, showing that (∂jf) ∗ g is well defined. Behold that

|(h−1(τ−h(f ∗ g)− f ∗ g)− ∂jf ∗ g)(x)| =∫ ∣∣∣∣f(x− y + hej)− f(x− y)

h− ∂jf(x− y)

∣∣∣∣∣∣∣∣g(y)

∣∣∣∣dy (∗)

where ej is the unit vector on the jth component. For our convenience, we denotef(x+hej)−f(x)

has Fh(x) and ∂jf(x) as F (x). Note that Fh ∈ Lp for each h since f ∈ Lp, and F = ∂jf ∈ Lp, soF − Fh ∈ Lp. Now

(∗) = |(Fh − F ) ∗ g(x)| ≤ ‖(Fh − F ) ∗ g‖u ≤ ‖Fh − F‖p‖g‖qand sending h → 0, our assumption limy→0 ‖y−1(τ−yf − f) − h‖p = 0 tells us exactly that‖Fh − F‖p tends to 0 and hence

0 = limh→0|(h−1(τ−h(f ∗ g)− f ∗ g)− ∂jf ∗ g)(x)| = lim

h→0

∣∣∣∣f ∗ g(x+ h)− f ∗ g(x)

h− ∂jf ∗ g

∣∣∣∣showing that ∂j(f ∗ g) exists in an ordinary sense and that it equals ∂jf ∗ g, as desired. �

Folland 8.9

If f ∈ Lp(R), the Lp derivative of f (which we call h) exists iff f is absolutely continu-ous on every bounded interval (perhaps after modification on a null set) and its pointwisederivatice f ′ is in Lp, in which case h = f ′ a.e.

Proof. Only If. Let [a, b] be a bounded interval since addition or deletion of one single pointdoesn’t affect the result of integration. It is well known that we can construct a g ∈ C∞c suchthat Suppg = [a, b] and

∫g = 1 using the bump function.

Claim 1. For each t > 0, |g(x)| ≤ C(1 + |x|)−2 for some C > 0.54

Proof of Claim. Since g ∈ Cc, by extreme value theorem we have |g| ≤ M for some M > 0.Consider C := M(1 + max(|a|, |b|))2. Then

C(1 + |x|)−2 = M

(1 + max(|a|, |b|)

1 + |x|

)2

≥M ≥ |g(x)|

showing the claim. �

Observation 2. Since f ∈ Lp and claim 1 holds, theorem 8.15 gives f ∗ gt → f a.e. as t→ 0.Observation 3. Let t > 0. Since g ∈ Cc, so is gt, and thus gt ∈ Lq where q is the conjugate of p,and by problem 8.8 we have (f ∗ gt) exists and

(f ∗ gt)′ = h ∗ gt → h a.e. as t→ 0

where the limit is attained by applying theorem 8.15 since h ∈ Lp.For convenience, we use {kn}n to denote the sequence {h ∗ g1/n}n. Since h ∈ Lp[a, b], by theorem8.14 a), h ∗ gt → h in Lp[a, b], and in particular {kn} is bounded in Lp[a, b]. Since [a, b] has finitemeasure b− a, by proposition 6.12 we have

‖knχ[a,b]‖1 ≤ ‖knχ[a,b]‖p(b− a)1−1/p

and hence {kn} is bounded in L1[a, b]. Notice that

f ∗ g1/n(x)− f ∗ g1/n(a) =

∫ x

a(f ∗ g1/n)′ =

∫ x

ah ∗ g′1/n =

∫ b

aknχ[a,x]

and sending n to infinity and apply dominated convergence theorem (since |knχ[a,x]| ≤ |kn| is

bounded in L1[a, b]) we obtain

f(x)− f(b) =

∫ x

bh(t)dt

showing that f is abosolutely continuous on [a, b] with a possible modification on a null set andthat f ′ = h a.e.If. For y > 0, notice that

f(x+ y)− f(x)

y− f ′(x) =

1

y

∫ y

0[f ′(x+ t)− f ′(x)]dt

Therefore ∥∥∥∥f(x+ y)− f(x)

y− f ′(x)

∥∥∥∥p

=

([∫ y

0

1

y[f ′(x+ t)− f ′(x)]dt

]pdx

) 1p

≤∫ y

0

(∫ ∣∣∣∣f ′(x+ t)− f ′(x)

y

∣∣∣∣pdx) 1p

dt

≤ 1

y

∫ y

0

(∫|f ′(x+ t)− f ′(x)|pdx

) 1p

dt

=1

y

∫ y

0‖τ−tf ′ − f ′‖pdt

where the first inequality is obtained by Minkowski’s inequality for integrals. Since ‖τ−tf ′‖p =‖f ′‖p, by triangular inequality we have ‖τ−tf ′ − f ′‖pχ[0,y] ≤ 2‖f ′‖pχ[0,y] ∈ L1 for all y > 0 sincef ′ ∈ Lp. Moreover, by proposition 8.5,

limt→0‖τ−tf ′ − f ′‖p = 0 (1)

55

so applying dominated convergence theorem, we get

limy→0+

∥∥∥∥f(x+ y)− f(x)

y− f ′(x)

∥∥∥∥p

= limy→0+

1

y

∫ y

0‖τ−tf ′ − f ′‖pdt

=1

y

∫ y

0limy→0+

‖τ−tf ′ − f ′‖pdt (2)

Let ε > 0. By (1) we know that there is some δ > 0 such that when |t| < δ, ‖τ−tf ′ − f ′‖p < ε.Thus when y < δ,

(2) < y

∫ y

0εdt = ε

showing that

limy→0+

∥∥∥∥f(x+ y)− f(x)

y− f ′(x)

∥∥∥∥p

= 0

The case where y < 0 is the same is we replace every [0, y] with [y, 0]. It follows that

limy→0

∥∥∥∥f(x+ y)− f(x)

y− f ′(x)

∥∥∥∥p

= 0

meaning that the Lp derivative of f exists and equals to its usual derivative. �

Folland 8.14

(Wirtinger’s Inequality) If f ∈ C1([a, b]) and f(a) = f(b) = 0, then∫ b

a|f(x)|2dx ≤

(b− aπ

)2 ∫ b

a|f ′(x)|2dx

Proof. Step 1. We show that by change of variable it suffices to assume that a = 0, b = 12 .

To see this, suppose the inequality holds for a = 0, b = 12 , i.e. for f ∈ C1([0, 12 ]) and f(0) =

f(12) = 0, ∫ 1/2

0|f(x)|2dx ≤

(1

2π

)2 ∫ 1/2

0|f ′(x)|2dx (1)

Now, given an f ∈ C1([a, b]) with f(a) = f(b) = 0, we consider g(x) := f(2(b − a)x + a). Since2(b − a)x + a ∈ [a, b], x ∈ [0, 12 ], meaning that g is defined on [0, 12 ]. Also since 2(b − a)x + a is

clearly a C1 function g is still C1 as f . Thus g ∈ C1([0, 12 ]) and g(0) = g(12) = 0. Applying ourassumption (1) we get ∫ 1/2

0|g(x)|2dx ≤

(1

2π

)2 ∫ 1/2

0|g′(x)|2dx

Notice that by change of variable we have

LHS =

∫ 1/2

0|f(2(b− a)x+ a)|2dx =

1

2(b− a)

∫ b

a|f(x)|2dx

and since g′(x) = [f(2(b− a)x+ a)]′ = 2(b− a)f ′(2(b− a)x+ a), we also have

RHS =

(1

2π

)2 ∫ 1/2

0[2(b− a)]2f ′(2(b− a)x+ a)dx

=

(1

2π

)2

[2(b− a)]21

2(b− a)

∫ b

a|f ′(x)|2dx

56

and putting them together we get∫ b

a|f(x)|2dx ≤

(b− aπ

)2 ∫ b

a|f ′(x)|2dx

as desired.Step 2. Now that we have assume without loss of generality that f is defined on [0, 12 ], we ex-

tend f to [−12 ,

12 ] by setting f(−x) = −f(x), and then extend f to be periodic on R with period

1. This is extension is well-defined at the overlapping points since f(n2 ) = 0 for n ∈ Z. We check

that f ∈ C1(T).We use f ′(12) to denote the left one-sided derivative of f at 1

2 , and use f ′(−12) to denote the right

one-sided derivatice of f at −12 . For h > 0, we have∣∣∣∣f(12)− f(12 − h)

h− f ′

(1

2

)∣∣∣∣ =

∣∣∣∣−f(−12) + f(−1

2 + h)

h− f ′

(1

2

)∣∣∣∣and thus

0 = limh→0

∣∣∣∣f(12)− f(12 − h)

h− f ′

(1

2

)∣∣∣∣ = limh→0

∣∣∣∣f(−12 + h)− f(−1

2)

h− f ′

(1

2

)∣∣∣∣showing that f ′(−1

2) = f ′(12), and we denote this common quantity by L. Since f is C1,

limx→1/2−

f ′(x) = L = limx→−1/2+

f ′(x)

and the way we extend f makes sure that this result holds for [2n−12 , 2n+12 ] for every n ∈ Z. It fol-

lows that f is C1 at the endpoint of T. f is C1 in the interior of T by assumption, so f ∈ C1(T).Step 3. We want to use Parseval’s identity to conclude the result.By proposition 7.9, C(T) is dense in L2(T) and in particular C(T) ⊂ L2(T). Hence f, f ′ ∈ L2(T).

By Parseval’s identity, ‖f‖2 = ‖f‖2 and ‖f ′‖2 = ‖f ′‖2. Thus∫ 1/2

0|f(x)|2dx =

∫T|f(x)|2dx = ‖f‖22 =

∑k∈Z|f(k)|2

and using integration by parts we have

f(k) =

∫Tf(x)e−2πikxdx =

∫ 1/2

0f(x)e−2πikx

=1

2πikf(x)e−2πikx

∣∣∣∣1/20

+

∫ 1/2

0

1

2πikf ′(x)e−2πikdx

= 0 +1

2πik

∫ 1/2

0f ′(x)e−2πikx =

1

2πikf ′(k)

implying |f(k)| = | 12πk ||f ′(k)| and thus∫ 1/2

0|f(x)|2dx = ‖f‖22 = ‖f‖22 =

∑k∈Z|f(k)|2 ≤

∑k∈Z

(1

2πk

)2

|f ′(k)| ≤∑k∈Z

(1

2π

)2

|f ′(k)|2

=

(1

2π

)2

‖f ′‖22 =

(1

2π

)2

‖f ′‖22 =

(1

2π

)2 ∫ 1/2

0|f ′(x)|2dx

showing the result as desired. This completes the proof. �

57

Folland 8.26

The aim of this exercise is to show that the inverse Fourier transform of e−2π|ξ| on Rn is

φ(x) =Γ(12(n+ 1))

π(n+1)/2(1 + |x|2)−(n+1)/2

(a) If β > 0, e−β = π−1∫∞−∞(1 + t2)−1e−iβtdt.

(b) If β ≥ 0, e−β =∫∞0 (πs)−1/2e−se−β

2/4sds.

(c) Let β = 2π|ξ| where ξ ∈ Rn; the the formula in (b) expresses e−2π|ξ| as a super-position of dilated Gauss kernels. Use proposition 8.24 again to derive the assertedformula for φ.

Proof. (a) By (8.37), we have φ(x) = 1π(1+x2)

and therefore

e−2π|ξ| = Φ(ξ) = φ(ξ) =

∫R

e−2πiξt

π(1 + t2)dt (1)

If β > 0, β = 2πξ for some ξ > 0, and then plugging in (1) we get

e−β =

∫R

e−iβt

π(1 + t2)dt = π−1

∫ ∞−∞

(1 + t2)−1e−iβtdt

as desired.(b) Notice that

e−β =1

π

∫1

1 + t2e−iβt =

1

π

∫ ∞0

e−iβt(∫ ∞

0e−(1+t

2)sds

)dt

=1

π

∫ ∞0

∫ ∞−∞

e−(1+t2)se−iβtdtds (1)

where the last step holds since e−(1+t2)se−iβt ∈ L+ and Tonelli’s theorem justifies the

interchange of order of integrals. Now substituting z = βt/2π, we obtain

(1) =1

π

∫ ∞0

e−s∫ ∞−∞

2π

βe−s 4π

2z2

β2 e−2πizdtds =2

β

∫ ∞0

e−s∫ ∞−∞

e4π2sβ2

z2e−2πizdzds

=2

β

∫ ∞0

e−sF(e− 4π2s

β2z2

)(1)ds =2

β

∫ ∞0

e−s(

4πs

β2

)−1/2e−π

β2

4πsds (Prop 8.24)

=2

β

∫ ∞0

e−s(β2

4πs

)1/2

e−β2

4s =

∫ ∞0

(πs)−1/2e−se−β2/4sds

showing the result as desired.(c) Since β = 2π|ξ|, now

e−β = e−2π|ξ| =

∫ ∞0

(πs)−1/2e−se−4π2|ξ|2

4s ds =

∫ ∞0

(πs)−1/2e−se−π2|ξ|2s ds

And computing the inverse Fourier transform we have

(e−2π|ξ|)∨(x) =

∫Rne−2π|ξ|e2πiξ·xdξ =

∫Rn

∫ ∞0

(πs)−1/2e−se−π2|ξ|2s e2πiξ·xdsdξ (2)

58

Note that |(πs)−1/2e−se−π2|ξ|2s e2πiξ·x| ∈ L+, so by Tonelli’s theorem we have∫

Rn×[0,∞)|(πs)−1/2e−se−

π2|ξ|2s e2πiξ·x|ds⊗ ξ

=

∫ ∞0

∫Rn|(πs)−1/2e−se−

π2|ξ|2s e2πiξ·x|dξds

=

∫ ∞0

∫Rn|(πs)−1/2e−se−

π2|ξ|2s |dξds

=

∫ ∞0|(πs)−1/2e−s|

(∫Rn|e−

π2|ξ|2s |dξ

)ds

=

∫ ∞0|(πs)−1/2e−s|(π · s

π2)2/nds

=

∫ ∞0|πs|−1/2|e−s|

(s

π

)n/2ds (since s > 0)

=π−1+n2

∫ ∞0

sn+12−1e−sds = π−

1+n2 Γ(

n+ 1

2) <∞

where the last step is by proposition 2.55. Then (πs)−1/2e−se−π2|ξ|2s e2πiξ·x ∈ L1, and we

can apply Fubini-Tonelli to interchange the order of integral in (2). Thus we get

(2) =

∫ ∞0

(πs)−1/2e−s(∫

Rne−

π2|ξ|2s e2πiξxdξ

)ds

=

∫ ∞0

(πs)−1/2e−s(e−π2|ξ|2s )∨(x)ds

=

∫ ∞0

(πs)−1/2e−s(s

π

)n/2e−s|x|

2ds (Prop 8.24)

=1

πn+12

∫ ∞0

sn−12 e−s(1+|x|

2)ds (3)

Substituting z = −s(1 + |x|2), we get

(3) =1

πn+12

(1

1 + |x|2

)n+12∫ ∞0

zn+12−1e−sdz =

Γ(12(n+ 1))

π(n+1)/2(1 + |x|2)−(n+1)/2

as desired. This finishes the proof.�

Folland 8.30

If f ∈ L1(Rn), f is continuous at 0, and f ≥ 0, then f ∈ L1.

59

Proof. Observe that

‖f‖1 =

∫|f(ξ)|dξ =

∫f(ξ)dξ

=

∫limt→0

f(ξ)e−π|tξ|2e2πiξ·0dξ

≤ limt→0

∫f(ξ)e−π|tξ|

2e2πiξ·0dξ (Fatou′s Lemma)

= f(0) <∞The last equality holds by Theorem 8.35 and the fact that Gauss kernel fits the theorem; the lastinequality holds since f is continuous and thus bounded at 0. �

9. Chapter 9-Elements of Distribution Theory

Folland 9.6

If f is absolutely continuous on compact subsets of an interval U ⊂ R, the distributionderivative f ′ ∈ D′(U) coincides with the pointwise (a.e.-defined) derivative of f .

Proof. Let f ′ be the distribution derivative and g pointwise a.e.-defined derivative. For any φ ∈C∞c (U), Suppφ = K for some K compact and thus f is absolutely continuous on K. It followsthat ∫

Kφf ′ = −

∫Kfφ′ = −

∫Kfdφ =

∫φdf =

∫Kφg (1)

by absolute continuity of f and properties of distribution derivative.Since U is an open interval, we may assume U = (a, b), and let Kn := [a + 1

n , b −1n ]. Taking

φ = χKn in (1) we obtain f ′ = g a.e. on Kn. We may suppose En ⊂ Kn is the set on which f ′

and g don’t agree, and hence En is a null set for each n. Since U =⋃∞

1 Kn, f ′ and g disagree onat most

⋃∞1 En which is still a null set. Thus f ′ and g agree a.e. on U , as desired. �

60