PARTIAL SOLUTIONS TO REAL ANALYSIS, FOLLAND
ZHENGJUN LIANG
Abstract. This following are partial solutions to exercises on Real Analysis, Folland, written
concurrently as I took graduate real analysis at the University of California, Los Angeles. Last
Updated: November 18, 2019
Contents
1.
2.
3.
4.
5.
6.
7.
8.
9.
Chapter 1-Measures
Chapter 2-Integration
Chapter 3-Signed Measures and Differentiation
Chapter 4-Point Set Topology
Chapter 5-Elements of Functional Analysis
Chapter 6-Lp Spaces
Chapter 7-Radon Measures
Chapter 8-Elements of Fourier Analysis
Chapter 9-Elements of Distribution Theory
1
2
2
11
23
31
42
52
52
60
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. Then
µE is a measure.
S∞
Proof. S
First of all, µES
(∅) = µ(∅ ∩ E) =
µ(∅)
=
0.
Also,
suppose
A
,
A
,
...
∈
M,
µ
(
1
2
E
i=1 ) =
P∞
P∞
∞
µ(E ∩ ( ∞
A
))
=
µ(
(A
∩
E))
=
µ(A
∩
E)
=
µ
(A
).
Then
µ
is
a
measure.
i
i
i
i
E
E
i=1
i=1
i=1
i=1
2. Chapter 2-Integration
Folland 2.6
The supremum of an uncountable family of measurable R̄-valued functions on X can fail
to 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 define
fx : R → R̄ 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
S
Suppose
that
for
each
α
∈
R
we
are
given
a
set
E
⊂
E
whenever
α
<
β,
α
β
α∈R Eα = X
T
and α∈R Eα = ∅. Then there is a measurable function f : X → R such that f (x) ≤ α on
Eα and f (x) ≥ α on Eαc 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 the
proof.
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
(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 and
monotone 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 mapped
by g and g is surjective. To show that h is continuous, it suffices to show that g is open
map, and to that end it suffices to show that g maps open intervals to open intervals since
every open set is a disjoint union of them. But that is apparent since g is monotone and
thus 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 of
open intervals on which f is constant. Therefore, g([0, 1] \ C) is a disjoint union of open
intervals 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
Proof.
(F ◦ G)−1 ([1, ∞)) = G−1 ◦ F −1 ([1, ∞)) = G−1 (B) = g(B) = A 6∈ L
so 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 ∈ R
and f (·, y) is continuous for each y ∈ Rk . For n ∈ N, define fn as follows. For i ∈ Z let
ai = 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 − ai
Then fn is Borel measurable on R × Rk and fn → f pointwise; hence f is Borel measurable on R × Rk . Conclude by induction that every function Rn that is continuous in each
variable 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.
R
R
(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. Let
> 0, since fn → f uniformly, there is some N ∈ N such that when n ≥ N , |fn (x) −
f (x)| < for all x. Then |f | < |fN | + , and
Z
Z
|f | < |fN | + µ(X) < ∞
which implies that f ∈ L1 . Fix and N above, let g = max{|f1 |, ..., |fN |, |f | + }, we
can see that g is clearly measurableR and |fRn | ≤ g for all n. Then we apply the dominated
convergence theorem and get that fn → f .
1
(b) Let fn = χ[−n,n] . First we show that fn → 0 uniformly. Let > 0 and N = d 1 e. When
n
R
n > N , |fn − 0| = |fn | < , 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
R
|fn − f | → 0 iff
R
|fn | →
R
|f |.
R
R
R
R
R
Proof.
Suppose
|fn − f | → 0, | |fn − f | | → 0. Since | |fn | − |f | | ≤ | |fn − f | | → 0,
R
R
|fn | → |f |.
R
R
R
R
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, fnR→ f a.e., it is Rclear that |fn − f | → 0 and
R|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 < xP< 1, f (x) = 0 otherwise. Let {rn }∞
1 be an enumeration of the
−n f (x − r ).
2
rationals, and set g(x) = ∞
n
1
(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) g 2 < ∞ a.e. but g 2 is not integrable on any interval.
Proof.
(a) For each n, we denote fn = 2−n f (x − rn ) and have
Z
Z
Z 1
−n
−n
|fn (x)|dx = 2 |f (x − rn )|dx = 2
x−1/2 dx = 2−n · 2 = 2−(n−1)
0
and thus
∞ Z
X
|fn (x)|dx = 1 + 2−1 + 2−2 + ... = 2 < ∞
1
Then
Z
|g| =
Z X
∞
|2
−n
f (x − rn )| =
1
∞ Z
X
|fn (x)| < ∞
1
and thus g ∈ L1 (m). In particular
g < ∞ a.e. since otherwise suppose g = ∞ on U such
R
that 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 easily
follows.
Now suppose we modify g on a m−null set F , and let h be the modified function such
that h = g on R \ F . Pick x0 ∈ R, for all δ > 0, Bδ (x0 ) contains some rn . Since Bδ (x0 ) is
1
1
open, we choose k so large that Ik := (rn − , rn + ) ⊂ Bδ (x0 ). Notice that Ik \ Ik+1 conk
k
tains some x1 ∈ R \ F since it has positive measure. Then we choose such xi inductively
such that xi ∈ Ik+i−1 \ Ik+i . Then xi → rn . Then observe that 2−n f (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 continuous at x0 (unbounded ocscillation). Also for an arbitrary interval I, pick x ∈ I and δ > 0
small enough such that Bδ (x) ∈ I. Then f is unbounded on Bδ (x) and thus unbounded
on I, and we show that the conclusions for g remains true after modifying g on a null set.
(c) Since g < ∞ a.e., g 2 < ∞ 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 the
result of integration. Then since g is non-negative,
Z b
Z bX
∞
2
|g | ≥
2−2n f 2 (x − rn )
a
a
Z b
≥
1
2−2n f 2 (x − rn ) where rn ∈ [a, b]
a
≥2
−2n
−2n
Z b
f 2 (x − rn )dx
rn
Z b−rn
−2n
2
Z b−rn
f (x)dx = 2
=2
x−1 dx
0
0
which fails to converge and thus tends to infinity. Then g 2 is not integrable on I and our
proof is complete.
Folland 2.32
Suppose µ(X) < ∞. If f and g are complex-valued measurable functions on X, define
Z
|f − g|
ρ(f, g) =
dµ
1 + |f − g|
Then ρ is a metric on the space of measurable functions if we identify functions that are
equal 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.
R |f − g|
R
(a) ρ(f, g) =
dµ ≥ 0 dµ = 0
1 + |f − g|
(b) If f = g a.e.,
Z
Z
|f − g|
ρ(f, g) =
dµ = 0 dµ = 0
1 + |f − g|
since a measure zero set doesn’t change the result of integration.
R |f − g|
R |g − f |
=
= ρ(g, f )
(c) ρ(f, g) =
1 + |f − g|
1 + |g − f |
5
(d) Notice that
|f − g|
|g − h|
|f − h|
+
−
1 + |f − g| 1 + |g − 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
> 0, and En := {x : |fn (x) − f (x)| ≥
}, which is possible since µ(X) < ∞.
µ(X)
Z
|fn − f |
dµ
ρ(fn , f ) =
1 + |fn − f |
Z
Z
|fn − f |
|fn − f |
c dµ
χEn dµ +
χ
=
1 + |fn − f |
1 + |fn − f | (En )
= µ(En ) +
µ(Enc )
µ(X)
Then limn→∞ ρ(fn , f ) ≤ 0 + = . Since 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, such that there are infinitely many n such that µ{x : |fn (x) − f (x)| ≥ } ≥ δ, and we denote this
set En . Thus
Z
Z
1
δ
1
≥
1−
dµ =
µ(En ) ≥
ρ(fn , f ) ≥
1−
1 + |f − fn |
1+
1+
1+
En
En
for infinitely many n, contradiction to convergence in this metric and we finish our proof.
Folland 2.34
1
SupposeR |fn | ≤ g ∈
R L and fn → f in measure.
(a) f → lim fn
(b) fn → f in L1
Proof.
R I will slightly reverse the
R orderR of the Rtwo parts of the
R problem. ObserveR that if fnR → f in
L1 , |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 inRL1 . Suppose in the
contrary that there exists > 0 such that there are infinitely many nk with |fnk − f | ≥ , then
we arrange them into a subsequence fnk . Since fn → f in measure, fnk → f in measure, and
there 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
R
in L1 . However, |gn − f | ≥ , 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) fn gn → f g in measure if µ(X) < ∞, but not necessarily if µ(X) = ∞
Proof.
(a) Let > 0. Since fn → f in measure and gn → g in measure, as n → ∞,
µ{|fn − f | ≥ } → 0
2
µ{|gn − g| ≥ } → 0
2
Notice that
{|fn + gn − fn − gn | ≥ } ⊂ {|fn − f | + |gn − g| ≥ }
⊂ {|fn − f | ≥ } ∪ {|gn − g| ≥ }
2
2
and thus
µ{|fn + gn − fn − gn | ≥ } ≤ µ{|fn − f | ≥ } + µ{|gn − g| ≥ }
2
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) < ∞, fn2 → f in measure.
Proof of lemma. Since fn → f in measure, some subsequence {fnj } → f a.e. Then
{fn2j } → f 2 a.e. Since µ(X) < ∞, fn2j → f 2 in measure by Egoroff’s theorem1. We now
show that the whole sequence actually converge to f 2 in measure by contradiction. Let
> 0. Suppose that there is some δ > 0 such that there are infinitely many n such that
µ{x : |fn2 − f 2 | ≥ } ≥ δ
then we arrange such fn2 into a sequence fn2i . Notice that we still have fni → f in measure
and thus fn0i → f a.e for a further subsequence {fn0i }. Thus fn20 → f 2 a.e. Since µ(X) <
i
∞, we should have fn20 → f 2 in measure by our argument above, but by our construction
i
this cannot happen, a contradition. Then fn2 → f 2 in measure.
By our lemma and part (a), (fn + gn )2 → (f + g)2 in measure (i.e. fn2 + 2fn gn + gn2 →
2
f + 2f g + g 2 in measure) and −fn2 → −f and −gn2 → −g in measure. Then by part (a)
again we have fn gn → f g in measure.
If µ(X) = ∞, we give a counterexample of functions R → R and the measure is
1
Lebesgue measure. let f (x) = g(x) = x and fn (x) = gn (x) = x + χ[n,n+1) . Then
n
clearly fn → f and gn → g in measure. However,
1
2x
1
fn (x)gn (x) = (x + χ[n,n+1) )2 = x2 +
χ
+ χ
n
n [n,n+1) n2 [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 fn gn doesn’t converge to f g 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).”
1
}, as defined in the proof of Egoroff’s theorem.
k
The condition µ(X) < ∞ is used to justify µ(E1 (k)) < ∞ for any fixed k, so if we can show that
µ(E1 (k)) < ∞, we are done. Observe that since |fn − f | ≤ 2g 2,
∞
[
1
1
E1 (k) ⊂
{x : g(x) ≥
} = {x : g(x) ≥
}
2k
2k
Proof. Let E1 (k) :=
S∞
m=1 {x : |fm (x) − f (x)| ≥
m=1
Then if µ(E1 ) = ∞ for some k,
R
|g| ≥
1
· ∞ = ∞, a contradiction.
2k
Folland 2.41
If µSis σ-finite and fn → f a.e. there exists measurable E1 , E2 , ... ⊂ X such that
c
µ(( ∞
1 Ej ) ) = 0 and fn → f uniformly on each Ej .
Proof. We first suppose that µ(X) is actually finite. By Egoroff’s theorem, S
for each k there is
some Ek such that µ(Ek )c < 2−k and fn → f uniformly on Ek . Set Fn = n1 Ei , then {Fn } is
an increasing sequence and {(Fn )c } is a decreasing sequence. Also µ(Fnc ) ≤ µ(Enc ) < 2−k . Since
µ(F1c ) ≤ µ(X) < ∞, by continuity from above,
∞
∞
∞
[
[
\
µ( Ej )c = µ( Fj )c = µ( Fjc ) = lim µ(Fjc ) = 0
1
1
j→∞
1
and fn → f uniformly on each Ej .
If X is σ-finite, then X = XS1 t X2 t ... such that µ(Xi ) < ∞ for each i. On each i there are
∞
i
i
{Eki }∞
k=1 such that µ(Xi − ( k=1 Ek )) = 0 and fn → f uniformly on each Ek . Then consider
i
∞
i
{Ek }k,i=1 , fn → f uniformly on each Ek , and
∞
∞
∞
∞
[
[
[
X
[
µ( Eki )c = µ[ (Xi −
Eki )] =
µ(Xi −
Eki ) = 0
i,k
i=1
k=1
i=1
k=1
Folland 2.43
Suppose that µ(X) < ∞ and f : X × [0, 1] → C is a function such that f (·, y) is measurable
for each y ∈ [0, 1] and f (x, ·) is continuous for each x ∈ X.
(a) If 0 < , δ < 1 then E,δ = {x : |f (x, y) − f (x, 0)| ≤ for all y < δ} is measurable.
(b) For any > 0 there is a set E ⊂ X such that µ(E) < and f (·, y) → f (·, 0)
uniformly on E c as y → 0.
2We adopt the assumption in Folland that f → f everywhere.
n
8
Proof.
(a) Let Q ∩ [0, δ) be enumerated as {yn }. Then we define
E,n := {x : |f (x, yn ) − f (x, 0)| ≤ }
Notice that f (x, yn ) and f (x, 0) are both measurable, |f (x, yn ) − f (x, 0)| is measurable.
Thus E,n is measurable. Consider
E :=
∞
\
E,n =
n=1
∞
\
{x : |f (x, yn ) − f (x, 0)| ≤ }
n=1
Then E is measurable and clearly E,δ ⊂ E. Conversely, suppose x ∈ E. Then |f (x, yn ) −
f (x, 0)| ≤ for all yn . Let y < δ, then we can find {ynk } that converges to y. Since
|f (x, ynk ) − f (x, 0)| ≤ for all ynk , send nk to infinity we get |f (x, y) − f (x, 0)| ≤ and thus x ∈ E,δ . Then E = E,δ and thus E,δ is measurable.
(b) Let > 0. Choose a monotone decreasing sequence {δn } ∈ [0, 1) such that δn → 0 from
above, notice that E,δ1 ⊂ E,δ2 ⊂ E,δ3 .... Consider F,i = (E,δi )c , then F,1 ⊃ F,2 ⊃
F,3 ⊃ F,4 ... and µ(F,1 ) ≤ µ(X) < ∞. Thus by continuity from above we have
µ(
∞
\
i=1
F,i ) = lim µ(F,i ) = 0
i→∞
Therefore, for any γ > 0, there is some N ∈ N such that when n > N , µ(F,n ) < γ,
|f (x, y) − f (x, 0)| < for x ∈ (F,n )c and y ≤ δn . Therefore given γ > 0 and m ∈ N
1
for
we can choose N (δ, m) such that µ(F 1 ,m ) < γ and |f (x, y) − f (x, 0)| < N (δ,m)
N (δ,m)
T
1
x ∈ F 1 ,m and y ≤ δm . Let E = ∞
F 1 ,m , then µ(E) <
for all m and for all
m=1
N (δ,m)
N (δ,m)
m
1
for all x provided y ≤ δm , which means that f (·, y) → f (·, 0)
m, |f (x, y) − f (x, 0)| <
m
uniformly as y → 0.
Folland 2.46
Let X = Y = [0, 1], M = N = B[0,1] . µ = Lebesgue meaure,
measure.
RR and ν = counting
RR
If
χD dµdν,
χD dνdµ, and
R D = {(x, x) : x ∈ [0, 1]} is the diagonal in X × Y , then
χD d(µ × ν) are all unequal.
Proof. Notice that
ZZ
Z Z
Z Z
y
χD dµdν =
χD dµ(x) dν(y) =
(χD ) dµ(x) dν(y)
Z Z
Z Z
Z
=
χDy dµ(x) dν(y) =
χ{(y,y)} dµ(x) dν(y) = 0 dν(y) = 0
and
ZZ
Z Z
Z Z
Z Z
χD dνdµ =
χD dν(y) dµ(x) =
(χD )x dν(y) dµ(x) =
χDx dν(y) dµ(x)
Z Z
Z
=
χ{(x,x)} dν(y) dµ(x) = dµ(x) = 1
9
Now notice that
R
χD d(µ × ν) = µ × ν(D), and that
∞
∞
X
[
µ × ν(D) = inf{
µ(Ai )ν(Bi ) : Ai , Bi ∈ B[0,1] , D ⊂
(Ai × Bi )}
1
i=1
P
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 ) >S0, ν(Ai ) = ∞
S
since if |Ai | < ∞ there should be
S
Sµ(Ai ) = 0. Observe
P∞ that i Ai = [0, 1] and i Bi = [0, 1], so
(A
∩
B
)
=
[0,
1]
and
thus
µ(
(A
∩
B
))
=
i
i
i
i
i=1 µ(Ai ∩ Bi ) = 1. Then there is some i such
i
i
that µ(Ai ∩ Bi ) > 0, and this means that µ(Ai ) ≥ µ(Ai ∩ Bi ) > 0 and ν(Bi ) ≥ ν(Ai ∩ Bi ) = ∞, so
we show what we want to show and finishes the proof.
Folland 2.55
R
R1R1
Let E = [0, 1] × [0, 1]. Investigate the existence and equality of E f dm2 , 0 0 f (x, y)dxdy,
R1R1
1
and 0 0 f (x, y)dydx for f (x, y) = (x − )−3 if 0 < y < |x − 21 | and f (x, y) = 0 otherwise.
2
R
R
1
Proof. Notice that E f dm2 = f χE dm2 , and that f ≤ 0 on E1 = [0, ] and f ≥ 0 on E2 =
2
1
[ , 1]. Now
2
Z
Z
Z
Z
Z
Z
f dm2 = f χE dm2 = (f χE )+ dm2 − (f χE )− dm2 = f χE2 dm2 − f χE1 dm2
E
Clearly f χE2 ∈ L+ (m2 ). Then by Tonelli,
Z 1 Z x− 1
Z
Z 1Z 1
2
1 −3
2
(x − ) dy dx
f χE2 dm =
f (x, y) dy dx =
1
1
2
0
0
2
2
Z 1
Z 1
1
1
=
(x − )−2 dx = lim
(x − )−2 dx (Monotone Convergence)
n→∞
1
1
1
2
2
+
2
2
n
= lim n − 2 = ∞
n→∞
R
And therefore E f dm2 doesn’t exist.
R1R1
Then we examine the existence and equality of 0 0 f (x, y) dydx. We have
Z 1Z 1
Z 1Z 1
Z 1Z
f (x, y) dydx =
f (x, y) dy dx =
fx χ[0,1] dy dx
0
0
0
0
0
Z 1Z
Z
+
−
=
(fx χ[0,1] ) dy − (fx χ[0,1] ) dy dx (1)
0
1
1
1
, (fx χ[0,1] )+ = (x − )−3 χ[0,x− 1 ] and (fx χ[0,1] )− = 0; if x < , (fx χ[0,1] )+ = 0
2
2
2
2
1
−
−3
and (fx χ[0,1] ) = ( − x) χ[0, 1 −x] . Also
2
2
Z
Z x− 1
2
1 −3
1
1
(x − ) χ[0,x− 1 ] dy =
(x − )−3 dy = (x − )−2
2
2
2
2
0
Notice that if x ≥
10
and
Z
−
1
( − x)−3 χ[0, 1 −x] dy = −
2
2
Z 1 −x
2
0
1
1
( − x)−3 dy = (x − )−2
2
2
Therefore,
Z 1
Z
Z
1
1
1
(x − )−2 dx = ((x − )−2 χ[0,1] )+ dx − ((x − )−2 χ[0,1] )− dx
2
2
2
0
Z 1
Z 1
2 1
1
(x − )−2 dx −
( − x)−2 dx
=
1
2
2
0
(1) =
2
R1
R1R1
1
However, by our work of part one, 1 (x − )−2 dx = ∞ and therefore 0 0 f (x, y) dydx doesn’t
2
2
exist as well.
R1R1
We eventually investigate the existence and equality of 0 0 f (x, y) dxdy. Notice that f (x, y) =
1
0 for y ≥ . Then
2
Z 1
Z 1Z 1
Z 1 Z 1
Z 1Z 1
2
+
−
f dx −
f dx dy
f (x, y) dxdy =
f (x, y) dx dy =
0
0
0
2
=
0
0
Z 1 Z 1
f + dx −
1
+y
2
0
1
(x − )−3 dx −
1
2
+y
2
0
2
Z 1
2
=
2
0
0
−
f dx dy
0
Z 1 Z 1
=
Z 1 −y
Z 1 −y
2
0
1
−3
( − x) dx dy
2
−2 · (4 − y −2 ) − 2(y −2 − 4) dy = 0
0
R 12
−2 · (4 − y −2 ) dy and
and both 0
Now we complete the proof.
R 12
0
2(y −2 − 4) dy < ∞. Then the integral exists and equals 0.
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 − , then
clearly ρ is a signed measure. Notice that we finish our proof if we can show that ρ is a positive
measure. Since v + ⊥ v − , there are E and F such that X = E t F 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 for
any measurable B ⊂ F , ρ(B) = λ(B) − ν + (B) = λ(B) ≥ 0, so F is also positive. Thus take any
C ∈ 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
Pn ) : F ∈ M, F ⊂ E} and ν (E) = − inf{ν(F ) :n F ∈ M, F ⊂ E}.
(b) |ν|(E) = sup{ 1 |ν(Ej )| : n ∈ N, E1 , ..., En are disjoint, and ∪1 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 and
ν = µ − λ. Then by uniqueness of Jordan decomposition we get the result. We first show
that µ and λ are positive. This is true since for every measurable E, µ(E) ≥ ν(∅) for
every E since ∅ ∈ M and ∅ ⊂ E. Similarly we can show λ(E) ≥ 0. We then show µ is
a well-defined measure. µ(∅)
S = ν(∅) = 0. For countable additivity, suppose E1 , E2 , . . .
disjoint, we denote E = i Ei for convenience. Then
µ(E) = sup{ν(F ) : F ∈ M, F ⊂ E} = sup{ν(F ∩ E) : F ∈ M, F ⊂ E}
= sup{ν(
∞
[
(F ∩ Ei )) : F ∈ M, F ⊂ E} = sup{
i=1
∞
X
ν(F ∩ Ei ) : F ∈ M, F ⊂ E}
i=1
∞
∞
X
X
= sup{
ν(Fi ) : Fi ∈ M, Fi ⊂ Ei } =
sup{ν(Fi ) : Fi ∈ M, Fi ⊂ Ei }
i=1
i=1
if we notice F ⊂ E iff Fi = F ∩ Ei ⊂ Ei for each i. Similarly we can show that λ is a welldefined positive measure. We then show that µ ⊥ λ. By Hahn decomposition, we have
X = P t N where P is positive and N is negative. Then for any E ∈ M, we write it as
E = (E ∩ P ) t (E ∩ N ). Notice that F ⊂ E iff F = F1 t F2 , where F1 ⊂ E ∩ P and
F2 ⊂ 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 by
uniqueness 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,
n
X
|ν|(E) = sup{
|ν|(Ej ) : n ∈ N, E1 , ..., En are disjoint, and ∪n1 Ej = E}
≥ sup{
1
n
X
|ν(Ej )| : n ∈ N, E1 , ..., En are disjoint, and ∪n1 Ej = E}
1
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
X
|ν(Ej )| : n ∈ N, E1 , ..., En are disjoint, and ∪n1 Ej = E}
1
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 )
dν2
dν1
(x1 )
(x2 )
(x1 , x2 ) =
(µ1 × µ2 )
dµ1
dµ2
Proof. First we show that ν1 × ν2 µ1 × µ2 . Suppose µ1 × µ2 (E) = 0 for some E ∈ M1 ⊗ M2 ,
then χE ∈ L+ . By Tonelli,
ZZ
Z
0 = µ1 × µ2 (E) =
χE dµ1 dµ2 = µ1 (E y ) dµ2 (y)
Then µ1 (E y ) = 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,
Z
Z Z
(ν1 × ν2 )(E) = χE d(ν1 × ν2 ) =
χE dν1 (x) dν2 (y)
Z
= ν1 (E y ) dν2 (y) = 0
Thus ν1 × ν2 µ1 × µ2 .
Then we show that
d(ν1 × ν2 )
dν1
dν2
(x1 , x2 ) =
(x1 )
(x2 )
d(µ1 × µ2 )
dµ1
dµ2
dνi
dνi
We claim that
≥ 0 a.e. Suppose in the contrary that
< 0 on some E with µi (E) > 0.
dµi
dµi
Then
Z
νi (E) =
f dµi < 0
E
contradicting to our assumption that νi is positive, so the claim is true. Then
dν2
dν1
(x1 )
(x2 ) ∈
dµ1
dµ2
L+ (M1 × M2 ), and by Tonelli, we get for any E ∈ M1 ⊗ M2 ,
Z
Z
ZZ
ZZ
d(ν1 × ν2 )
dν1
dν2
χE
d(µ1 × µ2 ) = χE d(ν1 × ν2 ) =
χE dν1 dν2 =
χE (
dµ1 )(
dµ2 )
d(µ1 × µ2 )
dµ1
dµ2
Z
dν1
dν2
= χE
(x1 )
(x2 ) d(µ1 × µ2 )
dµ1
dµ2
Then we have shown that
dν1
dν2
d(ν1 × ν2 )
=
(x1 )
(x2 )
d(µ1 × µ2 )
dµ1
dµ2
a.e. and thus prove the result since we identify the derivative functions with their equivalence
classes.
Folland 3.16
dν
µ, ν are measures on (X, M) with ν µ, and let λ = µ + ν. If f =
, show that 0 ≤ f <
dλ
dν
f
1 µ-a.e. and
=
.
dµ
1−f
13
Proof. We first show that 0 ≤ f < 1 µ-a.e. Suppose in the contrary that f ≥ 1 on some E with
µ(E) > 0,
Z
f dλ ≥ λ(E) = µ(E) + ν(E)
ν(E) =
E
But this implies that µ(E) ≤ 0, a contradiction.
dν
f
We then show that
=
. We claim that µ λ and λ ν. If λ(E) = µ(E) + ν(E) = 0,
dµ
1−f
µ(E). Conversely, if µ(E) = 0, since ν µ, ν(E) = 0 and thus λ(E) = ν(E) + µ(E) = 0. So our
dµ dλ dµ
dν
dλ
claim is true and
= 1. By additivity of derivatives, we have
+
=
= 1 and
dλ dµ
dλ
dλ
dλ
dν
dµ
=1−
. Then
therefore
dλ
dλ
f
dν/dλ
dν/dλ
dν dλ
=
=
=
·
1−f
1 − dν/dλ
dµ/dλ
dλ dµ
Since we have ν λ and λ µ,
dν dλ
dν
f
=
·
=
1−f
dλ dµ
dµ
and we finish our proof.
Folland 3.18
1
1
1
Let
R ν be a complex measure on (X, M). L (ν) = L (|ν|), and if f ∈ L (ν), then |
|f |d|ν|.
R
f dν| ≤
Proof. We first show that L1 (ν) = L1 (|ν|). By Lebesgue-Radon-Nikodym, we have dν = f dµ
where µ is a positive measure, and thus d|ν| = |g|dµ. If f ∈ L1 (|ν|),
Z
Z
Z
Z
Z
∞ > |f |d|ν| = |f ||g|dµ = |f g|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
Z
Z
∞ > |f |d(|vr | + |vi |) ≥ |f |d|ν|
showing that f ∈ L( |ν|). Thus L1 (ν) = L1 (|ν|). Moreover by (1) we have
Z
Z
Z
f dν ≤ |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 = f dµ for some positive measure µ, and thus
d|ν| = |f |dµ. Then
Z
Z
Z
f dµ =
|f |dµ =⇒
14
(|f | − f )dµ = 0
(3)
Since |f | − f ≥ 0, (2) implies that |f | − f = 0 µ-a.e., and thus |f | = f since we identify f as
equivalence 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
X
n
n
[
µ1 (E) = sup
|ν(Ej )| : n ∈ N, E1 , ..., En disjoint and E =
Ej
1
j=1
Z
f dν : |f | ≤ 1
µ3 (E) = sup
E
P
Proof. We first show that µ1 ≤ µ3 . Define f := nj=1 sgn(ν(Ei ))χEi , and since |sgn(ν(Ei ))| ≤ 1,
|f | ≤ 1. Thus we have
Z
n Z
n
n
X
X
X
sgn(ν(Ei ))dν =
f dν =
sgn(ν(Ei ))ν(Ei ) =
|ν(Ei )|
E
j=1
Ei
j=1
j=1
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 thus
dν = f dν = f dν and d|ν| = |g|dν. Then
dν/d|ν|dν =
f dµ · f dµ
|f |2 (dµ)2
=
= |f |dµ = d|ν|
|f |dµ
|f |dµ
R
R
f
dν
=
and thus
E
E d|ν|R = |ν|(E), showing that µ3 ≥ |ν|. On the other hand µ3 (E) ≤
R
sup{ E |fP
|d|ν| : |f | ≤ 1} ≤ E d|ν| = |ν|(E) and thus µ3 = |ν|. Eventually
we show that ν3 ≤ ν1 .
S
Let φ := n1 ck χEk where |ck | ≤ 1 for all k, Ei s are disjoint and ni=1 Ei = E. We have
Z
Z
n
n
n
X
X
X
ck
φdν ≤
χEk dν =
|ck ||ν(Ek )| ≤
|ν(Ek )| ≤ µ1 (E)
E
k=1
Ek
k=1
k=1
Let |f | ≤ 1, choose hφn in simple functions that approximate f from below (meaning that |φn | ≤ 1
for all n) in L1 since simple functions are dense
in L1 and apply dominated convergence theorem
R
(since |φn | ≤ χE ∈ L1 for all n) we have | E f dν| ≤ 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
Z
1
H ∗ f (x) = sup
|f (y)| dy : B is a ball and x ∈ B
m(B) B
∗
n
Show that Hf ≤ H f ≤ 2 Hf .
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))
Z
|f (y)| dy : r > 0
⊂
B(r,x)
1
m(B)
Z
|f (y)| dy : B is a ball and x ∈ B
B
Therefore
Z
Z
1
1
sup
|f (y)| dy : r > 0 ≤ sup
|f (y)| dy : B is a ball and x ∈ B
m(B(r, x)) B(r,x)
m(B) B
which means Hf ≤ H ∗ f .
Second inequality: We observe that
Z
1
∗
|f (y)| dy : Br is a ball of radius r such that x ∈ Br
H f (x) = sup
m(Br ) Br
r>0
Z
1
= sup
|f (y)| dy : Br is a ball of radius r such that x ∈ Br
m(B(r, x)) Br
r>0
Z
1
≤ sup
|f (y)| dy (since B(2r, x) contains all Br 3 x)
r>0 m(B(r, x)) B(2r,x)
Z
2n
= sup
|f (y)| dy
r>0 m(B(2r, x)) B(2r,x)
= 2n Hf (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 > 0. Since f is continuous at x, there is δ > 0 such that whenever |y − x| < δ,
|f (y) − f (x)| < . Then when r < δ,
Z
m(B(r, x))
1
|f (y) − f (x)| dx <
=
m(B(r, x)) B(r,x)
m(B(r, x))
and thus
Z
1
lim
|f (y) − f (x)| dy = 0
r→0 m(B(r, x)) B(r,x)
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) = lim
r→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 ∈ E c .
(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
Z
Z
1
1
|f (y) − f (x)| dy = lim
|f (y) − 1| dy = 0
lim
r→0 m(B(r, x)) B(r,x)
r→0 m(B(r, x)) B(r,x)
This implies
Z
1
0 = lim
(f (y) − 1) dy
r→0 m(B(r, x)) B(r,x)
Z
Z
1
1
= lim
f (y) − lim
1 dy
r→0 m(B(r, x)) B(r,x)
r→0 m(B(r, x)) B(r,x)
Z
1
1
= lim
χE∩B(r,x) − lim
m(B(r, x))
r→0 m(B(r, x))
r→0 m(B(r, x))
m(E ∩ B(r, x))
= lim
−1
r→0
m(B(r, x))
m(E ∩ B(r, x))
= 1 for a.e. x ∈ E.
m(B(r, x))
For a.e. x ∈ E c , x ∈ Lf and thus
Z
Z
1
1
|f (y) − f (x)| dy = lim
|f (y) − 0| dy = 0
lim
r→0 m(B(r, x)) B(r,x)
r→0 m(B(r, x)) B(r,x)
for a.e x ∈ E and thus
Using an argument similar to the one above we can show that
x ∈ Ec.
(b) Let E = [0, 1] and x = 0, then
m(E ∩ B(r, x))
= 0 for a.e.
m(B(r, x))
m(E ∩ B(r, x))
m[0, r)
1
= lim
= ∈ (0, 1)
r→0 m(−r, r)
B(r, x)
2
For the other example,
∞ G
1
1
,
E = {0} t
22n+1 22n
lim
r→0
n=1
We want to find two subsequences converging to different limit, and thus show that the
1
limit doesn’t exist. First we let {rk } := { 2k }, x = 0, then rk → 0, and m(E ∩ B(rk , x)) =
2
1
1/22k+1
4
1
1
= · 2k+1 = · 2k−1 . So
1 − 1/4
3 2
3 2
1
1
· 2k−1
m(E ∩ B(rk , x))
1
= lim 3 21
=
k→∞
k→∞
m(B(rk , x))
3
22k−1
lim
Then we let {rk0 } be
So
1
2
, and we have m(E∩B(rk0 , x)) =
2k+1
1/22k+2
4 1
1 1
= · 2k+2 = · 2k−1 .
1 − 1/4
3 2
6 2
1
1
m(E ∩ B(rk0 , x))
1
6 · 22k−1
=
lim
=
0
1
k→∞
k→∞
m(B(rk , x))
6
22k−1
lim
Then {rk } and {rk0 } both converge to 0, but
m(E ∩ B(rk0 , x))
m(E ∩ B(rk , x))
6= lim
k→∞
k→∞
m(B(rk , x))
m(B(rk0 , x))
lim
17
showing that DE (0) doesn’t exist and we finish the proof.
Folland 3.28
If F ∈ N BV , let G(x) = |µF |((−∞, x]). Prove that |µF | = µTF by showing that G = TF
via 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,
n
X
TF = sup{
|F (xj ) − F (xj−1 )| : n ∈ N, −∞ < x0 < ... < xn = x}
= sup{
= sup{
j=1
n
X
j=1
n
X
|µF (−∞, xj ] − µF (−∞, xj−1 ]| : n ∈ N, −∞ < x0 < ... < xn = x}
|µF (xj , xj−1 ]| : n ∈ N, −∞ < x0 < ... < xn = x}
(1)
j=1
S
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)|
n
X
≤ sup{
|F (xj ) − F (xj−1 )| : n ∈ N, a = x0 < ... < xn = b}
j=1
= TF (b) − TF (a) = µTF (−∞, b] − µTF (−∞, a]
= µTF (a, b] = µTF (E)
In general, for E ∈ BR , by regularity
∞
∞
∞
∞
X
[
X
[
|µF (E)| = | inf{
µF (aj , bj ] : E ⊂
(aj , bj ]}| ≤ inf{
µF (aj , bj ] : E ⊂
(aj , bj ]} (2)
j=1
i=1
j=1
i=1
But |µF (aj , bj )| ≤ µTF (aj , bj ] for every j, so
(2) ≤ inf{
∞
X
µTF (aj , bj ] : E ⊂
j=1
∞
[
(aj , bj ]} = µTF (E)
i=1
and we prove the result.
(c) By exercise 21, for E ∈ BR ,
n
n
n
n
X
G
X
G
|µF (E)| ≤ sup{
|µF (Ei )| : E =
Ei } ≤ sup{
µTF (Ei ) : E =
Ei }
i=1
i=1
= sup{µTF (
n
G
i=1
Ei ) : E =
n
G
i=1
Ei } = µTF (E)
i=1
18
i=1
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 ∈ N BV and |µF | = µG . Then
TF = G ∈ N BV 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 thus
differentiable, so it suffices to verify for x = 0 in both cases.
F (x + h) − F (0)
h2 sin(1/h)
= lim
= lim h sin(1/h) = 0
h→0
h→0
h→0
h
h
since | sin(1/h)| ≤ 1. Also,
lim
G(x + h) − G(0)
h2 sin(1/h2 )
= lim
= lim h sin(1/h2 ) = 0
h→0
h→0
h→0
h
h
2
since | sin(1/h )| ≤ 1. Thus F and G are differentiable everywhere.
0
0
(b) F 0 (x) = 2x sin(x−1 ) − cos(x−1 ) when x 6= 0 and
r F (0) = 0. Then |F | ≤ 3 in [−1, 1]
2
(j = 1, 2, ..., n), notice that
and thus F ∈ BV [−1, 1]. For G, choose xj =
(n − j + 1)π
x0 > 0 and xn < 1. Then
n
X
TF (1) − TF (−1) ≥
|G(xj ) − G(xj−1 )|
lim
j=0
n+2
X 2
2
n−j
2
n−j+1
sin
π−
sin
π ≥
(n − j + 1)π
2
(n − j + 2)π
2
jπ
j=0
j=2
P
2
This partition gets finer as n increses, but TF (1) − TF (−1) ≥ n+2
j=2 jπ → ∞ as n → ∞
since harmonic series diverge. Thus G 6∈ BV ([−1, 1]).
=
n
X
Folland 3.32
If F1 , F2 , ..., F ∈ N BV 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. Then
n
N
X
X
|F (xi ) − F (xi−1 )| = lim inf
|Fj (xi ) − Fj (xi−1 )|
i=1
j→∞
i=1
≤ lim inf
j→∞
sup{
n
X
|Fj (xi ) − Fj (xi−1 )| : n ∈ N, −∞ < x0 < ... < xn = x}
i=1
≤ lim inf TFj
j→∞
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)).
Rd
(b) If f is a Borel measure and integrable function on [c, d], then c f (y)dy =
Rb
Rd
Rb
0
a f (G(x))dG(x). In particular, c f (y)dy = a 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 is
continuous 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 interval.
Claim 2. G−1 (c1 , d1 ] = (a1 , b1 ] where G(a1 ) = c1 and G(b1 ) = d1 .
Proof of Claim. [a, b] = G−1 (−∞, c1 ) t G−1 (c1 , d1 ] t G−1 (d1 , +∞). Since G is continuous,
G pulls back open (closed) sets to open (closed) sets. Combined with claim 1, we conclude
that 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 = sup G[a, a1 ] = G(a1 ).
Similarly, d1 = sup G(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,
∞
∞
∞
∞
X
[
X
[
m(E) = inf{ (cj , dj ] : E ⊂
(cj , dj ]} = inf{
µG (G−1 (cj , dj ]) : E ⊂
(cj , dj ]}
j=1
j=1
j=1
(1)
j=1
Shrinking the intervals if necessary, we may assume (cj , dj ] ⊂ [c, d]. By above claim,
∞
∞
X
[
(1) = inf{
µG (aj , bj ] : E ⊂
(G(aj ), G(bj )]}
j=1
Claim 3. E ⊂
S∞
j=1 (G(aj ), G(bj )] if and only if G
Proof of Claim. Suppose G−1 (E) ⊂
E ⊂ G(
∞
[
(aj , bj ]) ⊂
j=1
(2)
i=1
∞
[
S∞
−1 (E) ⊂
j=1 (aj , bj ].
S∞
j=1 (aj , bj ],
∞
[
G(aj , bj ] =
i=1
by claim 2. Conversely, suppose E ⊂
(cj , dj ]
j=1
∞
[
(G(aj ), G(bj )]
j=1
S∞
j=1 (G(aj ), G(bj )],
∞
∞
∞
[
[
[
G−1 (E) ⊂ G−1 ( (cj , dj ]) =
G−1 (cj , dj ] =
(aj , bj ]
j=1
j=1
j=1
Then we prove the claim.
20
Then by claim 2 and regularity of µG ,
∞
∞
X
[
(2) = inf{
µG (aj , bj ] : G−1 (E) ⊂
(aj , bj ]} = µG (G−1 (E))
j=1
j=1
and we prove the result.
(b) Dealing with f + and f − separately, we mayP
assume f ∈ L+ . Choose a sequence of simple
m
functions {φn } ↑ f . We may assume φn =
i=1 ai χEi vanishes outside [c, d], i.e. Ei ⊂
[c, d] for all i. Then
Z d
m
m
X
X
φn (y)dy =
ai m(Ei ) =
ai µG (G−1 (Ei ))
c
=
i=1
m
X
i=1
Z b
ai
i=1
=
a
χG−1 (Ei ) dG (since G−1 (Ei ) ⊂ [a, b] ∀ i)
Z bX
m
ai χG−1 (Ei ) dG (3)
a i=1
Observation 4. x ∈ G−1 (Ei ) if and only Rif G(x) ∈ Ei .
Then χG−1 (Ei ) = χEi (G(x)) and (3) = φn (G(x))dG(x). Sending n → ∞, by monoRb
Rd
tone convergence theorem, we get c f (y)dy = a f (G(x))dG(x). In particular, if G is
Rd
absolutely continuous, G is differentiable a.e., so dG(x) = G0 (x)dx a.e. and c f (y)dy =
Rb
0
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 GRto be 0 at (−∞, 0) and 2 at (3, +∞) we may assume G ∈ N BV . Then
suppose f ≡ 1, [0,3] f (x)dx = 3. However,
Z
Z
f (G(x))dG(x) =
f (G(x))dG(x)
[0,3]
(−∞,3]
Z
Z
Z
=
f (G(x))dG(x) +
f (G(x))dG(x) +
f (G(x))dG(x)
(−∞,1]
(1,2]
(2,3]
= 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) =
P
0
for all x ∈ [a, b], then F 0 (x) = ∞
F
j (x) for a.e. x ∈ [a, b].
1
P∞
1
Fj (x) < ∞
Proof. Without loss of generality we may assume that Fj ∈ N BV for all j. This is because we
can extend Fj to R such that F (x) = F (b) for all x ≥ b and F (x) = F (a) for all x ≤ a, and the
resulted function is still bounded increasing and thus in BV . Consider G(x) = F (x+) − F (−∞),
then G ∈ N BV and G0 = F 0 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
Z
dµFj = dλj + gj dm ⇐⇒ µFj = λj + gj dm (1)
R
1
Here
P∞ λj is a positive
P∞ measure since
Pµ∞and gj are both positive, and gj ∈ L (m). We set µ =
j=1 µFj , λ =
j=1 λj , and g =
j=0 gj . Now
Z
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) =
P∞
1
j=1 λj (E) = 0. Also g ∈ L (m) since
Z
Z
∞
∞
X
X
|g| dm = g dm ≤ µ(X) =
µFj (X) =
Fj (b) = F (b) < ∞
j=1
j=1
Thus (2) is the a.e. unique LRN representation of µ. On the other hand,
∞
∞
X
X
F (x) =
µj (−∞, x] = (
µj )(−∞, x] = µ(−∞, x] < ∞
j=1
j=1
for some finite Borel measure µ. Then F ∈ N BV and (2) is the a.e. unique LRN representation
µF (Er )
of µ. Then Fj0 (x) = limr→0 j
= gj (x) a.e. for Er = (x, x+r] or (x−r, x] which shrink nicely
m(Er )
P
µF (Er )
to x. Also we have F 0 (x) = limr→0
= g(x) a.e. This means that F 0 = g = ∞
j=1 gj =
m(Er )
P∞
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 = 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 ∈ N BV . Let > 0.
P
Take δ = . For finite disjoint (a1 , b1 ), ..., (aN , bN ), if N
1 (bj − aj ) < δ,
N
X
1
|F (bj ) − F (aj )| =
N
X
m((aj , bj ] ∩ A) <
1
N
X
N
X
m((aj , bj ]) =
(bj − aj ) < 1
1
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
R 1 0the last part, note first that since F is absolutely
continuous, F (1) = F (1) − F (0) = 0 F (t)dt. Also
Z 1
F (1) = m([0, 1] ∩ A) =
χA dm
0
Therefore F 0 = χA a.e. on [0, 1]. Since m([0, 1] \ A) = m[0, 1] − m([0, 1] ∩ A) > 0, χA = 0
on a set of positive measure and thus F 0 = 0 on a set of positive measure.
22
(b) Let > 0 and δ = , then for finite disjoint (a1 , b1 ), ..., (aN , bN ) in [0, 1], if
PN
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 )
PN
and thus 1 |G(bj ) − G(aj )| ≤
1 (bj − aj ) < δ = , showing that G is absolutely
continuous on [0, 1]. Suppose G is mototone on some interval I which can be assumed to
(a, b), either G0 ≥ 0 or G0 ≤ 0. However, since G is absolutely continuous,
Z b
Z b
0
χA − χAc
G (x)dm = G(b) − G(a) = m((a, b] ∩ A) − m((a, b] \ A) =
PN
a
a
Then G0 (x) = χ
A − χAc m-a.e. However since m(A ∩ I) < m(I), I contains a positive
measure of points in both A and Ac and G0 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 in
X, 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 finitely
many points in X. Therefore, since {xj } is a sequence of distinct points, starting at some sufficiently large N we have xn ∈ U ◦ ⊂ U . Thus xj → x. Since x is arbitrary, this finishes the
proof.
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, since
N ◦ is also a neighborhood of x, N ◦ ∩ U 6= ∅. Since U is open, N ◦ ∩ U is open in X. Since A is
dense in X, N ◦ ∩ U contains some element of A. Then ∅ =
6 N ◦ ∩ (U ∩ A) ⊂ N ∩ (U ∩ A). Since
N is arbitrary, this means that x ∈ U ∩ A. Thus U ⊂ U ∩ A. Two directions combined, we prove
that 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 the
extension 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 may
assume that g and f are R-valued. Since {(a, b)} generate the usual topology on R, it suffices to
verify 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 open
in 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◦ is
open 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 U
open 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 isQa countable set and Xα is a first (resp. second) countable space for each α ∈ A,
then α∈A Xα is first (resp. second) countable.
Q
(a) Xα is first countable for all α ∈ A. Suppose x = hxα iα∈A ∈ X := α∈A Xα ,
then for each xα there is a countable neighborhood base Nα . We claim that finite intersections of sets in πα−1 (Nα ), where α ∈ A, form a countable neighborhood base of x. We
denote this countable neighborhood base N and first show
it’s countable. First of
S that
−1
−1
all πα (Nα ) is countable, and A is countable, so C := α∈A πα (Nα ) is countable, and
we enumerate them as C1 , C2 , .... We use Cn to denote
the collection of finite intersections
S
of 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 show
that N is indeed a neighborhood base of x. First of all, since Nα is a neighborhood base
of xα , for every Nα ∈ Nα , xα ∈ Nα and thus x = πα−1 (Nα ). Any finite intersections of
x. Thus every element of N containsQx. Suppose U is
sets like πα−1 (Nα ) must still contain
Q
open the product topology on α∈A Xα , 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 Nαi
is a neighborhood
base of xαi , there is some Vαi ∈ Nαi such that x ∈ Vαi ⊂ Uαi . Then we
T
have n1 πα−1
(V
)
∈
N such that x ∈ πα−1
(Vαi ) ⊂QU . Then our claim is true. Since N is
αi
i
i
countable and x is arbitrary, we show that X := α∈A Xα is first countable.
(b) Xα is second countable for all α ∈ A. Then for each α there is a countable base Nα of
Xα . We claim that finite intersections of sets in πα−1 (Nα ) is a countable base of X :=
Q
α∈A Xα . 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. First
of all, let x = hxα iα∈A ∈ X, each xα ∈ Vα ∈ Nα for some Vα . Then we
T just randomly
pick an α ∈ A and we have x ∈ πα−1 (Vα ). Next, suppose we have U = n1 πα−1
(Uαi ) and
i
T
−1
V = m
π
(V
)
in
N
and
x
=
hx
i
∈
U
∩
V
.
For
our
convenience,
we
denote
α α∈A
Q 1 βi βi
Q
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 .
Proof.
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 ∩
QVα .
Q
Q
Then W := α∈A Wα ⊂ 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 is
countable, 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 that
supx∈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 > 0. Since supx∈X ρ(fn (x), fm (x)) → 0 as m, n → ∞, we can pick N
large enough such that supx∈X ρ(fn (x), fm (x)) < /2 if m, n > N . Also, by our definition
of f (x), for every x we can pick large enough mx > N such that ρ(fmx (x), f (x)) < /2.
Then for every x, if n > N ,
ρ(fn (x), f (x)) ≤ ρ(fn (x), fmx (x)) + ρ(fmx (x), f (x)) < and thus sup(fn (x), f (x)) < . This proves that sup ρ(fn (x), f (x)) → 0. Moreover, suppose 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 > 0.
Then there is some N > 0 such that supx∈X ρ(fN (x), f (x)) < /3. By continuity of fN ,
A := fn−1 [B(/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)) < and thus y ∈ A ⊂ (f −1 (B(, 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 neighborhood of x for every open ball B containing f (x). Since B 3 f (x) is open, we can take a
small enough > 0 such that B(, f (x)) ⊂ B. Then by what we did above f −1 (B(, f (x)))
is a neighborhood containing x. Then f −1 (B) ⊃ f −1 (B(, f (x))) is also a neighborhood
containing x. Thus f is continuous at x. Since x is arbitrary, this shows that f is continuous.
Folland 4.24
A Hausdorff space X is normal iff X satisfies the conclusion of Urysohn’s lemma iff X satisfies 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 that
X is T1 since X is Hausdorff. Given disjoint closed sets A and B, by Urysohn’s lemma there is a
continuous 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 set
containing 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 previous
paragraph 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, then
f ∈ 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 A
and F ≡ 1 on B. Then (2) holds.
Folland 4.38
Suppose that (X, T ) is a compact Hausdorff space and T 0 is another topology on X. If
T 0 is strictly stronger than T , then (X, T 0 ) is Hausdorff but not compact. If T 0 is strictly
weaker than T , then (X, T 0 ) is compact but not Hausdorff.
(a) Suppose T 0 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 0 ) ) (X, T ), so
A, B ∈ T 0 . Thus (X, T 0 ) is Hausdorff. Suppose in the contrary that (X, T 0 ) is compact.
Consider mapping f : (X, T 0 ) → (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 0 ) since T 0 is strictly stronger than T ,
so f is continuous. If (X, T 0 ) is compact, then f is a continuous bijection mapping from a
compact space to a Hausdorff space and is thus a homeomorphism. But T 0 ) T , a contradiction. Thus (X, T 0 ) cannot be compact.
(b) Suppose T 0 is strictly weaker than T . Take an open cover U of X in T 0 , U is also an open
cover in T and thus has a finite subcover. Thus (X, T 0 ) is compact. Suppose (X, T 0 ) is
Hausdorff, consider g : (X, T ) → (X, T 0 ) defined by x 7→ x, then g is bijective. For U
open in (X, T 0 ), f −1 (U ) = U ∈ T 0 ⊂ T and is therefore open. Then g is continuous bijection between a compact space and a Hausdorff space and is therefore a homeomorphism.
which is not possible since T 0 stricly weaker than T . Thus (X, T 0 ) is not compact.
Proof.
Folland 4.43
P
−n (a (x) = 0 or 1) be the base-2 decimal expansion of x.
For x ∈ [0, 1), let ∞
n
1 an (x)2
(If x is a dyadic rational, choose the expansion such that an (x) = 0 for n large.) Then the
sequence han i in {0, 1}[0,1) has no pointwise convergent subsequence.
P
−n2k . It is clear that x is not a
Proof. Take any subsequence hank i of han i, consider x = ∞
k=1 2
dyadic rational (since there are no consecutive 1s or 0s), so the expression is uniquely determined
here. Now we have an2k = 1 and other ank = 0. Since we have infinitely many alternating terms,
hank (x)i fails to converge. Thus han i 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
∞
X
ρ(f, g) =
2−n φ( sup |f (x) − g(x)|)
1
x∈U n
is a metric on CX whose associated topology is the topology of locally uniform convergence.
Proof.
(a) We first show that φ is strictly increasing. Suppose t1 < t2 < ∞, then
φ(t2 ) − φ(t1 ) =
t2
t1
t2 (t1 + 1) − t1 (t2 + 1)
t2 − t1
−
=
=
>0
t2 + 1 t1 + 1
(t2 + 1)(t1 + 1)
(t2 + 1)(t1 + 1)
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
t+s
t
s
−
−
t+s+1 t+1 s+1
1
1
1
t+1+s+1
t+s+2
=1 −
−1+
−1+
=
−
t+s+1
t+1
s+1
(t + 1)(s + 1) t + s + 1
t+1+s+1
t+s+2
t+1+s+1
t+s+2
≤
−
=
−
=0
(t + 1)(s + 1) t + s + ts + 1
(t + 1)(s + 1) (t + 1)(s + 1)
showing the result.
ρ
(b) For convenience, define ρ0 := φ ◦ ρ =
. It is easy to see ρ0 is bounded by 1. We
ρ+1
denote the topology generated by ρ and ρ0 using T and T 0 , respectively, and the collection
of open balls in T and T 0 are named E and E 0 , respectively. We know that T (E) = T , and
T (E 0 ) = T 0 . Then it suffices to show E ⊂ T 0 and E 0 ⊂ T . Let Bρ (x, r) ∈ E, and we claim
r
that Bρ (x, r) = Bρ0 (x, r+1
). To show the claim, suppose we have y ∈ Y , then ρ(x, y) < r
0
if and only if ρ (x, y) = φ ◦ ρ(x, y) < φ(r) = r/(r + 1) since φ is strictly increasing. In
r
other words, y ∈ Bρ (x, r) if and only if y ∈ Bρ0 (x, r+1
) and thus the claim is true. Thus
r
0
0
Bρ (x, r) = Bρ0 (x, r+1 ) ∈ E ⊂ T as desired. Similarly we can show that Bρ0 (x, r) =
r
Bρ (x, 1−r
) since φ(r/1 − r) = r. Thus E 0 ⊂ E ⊂ T . Now we have T = T (E) ⊂ T 0 and
0
T = T (E 0 ) ⊂ T , showing that ρ and ρ0 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 ≥ 0
Identity 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 )
φ(t + s) − (φ(t) + φ(s)) =
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 the
topology of uniform convergence, and by (b) we know that ρu and ρ = φ ◦ ρu generate
the same topology, so the associated topology of ρ is also the topology of uniform convergence.
(d) Suppose hfn in converges locally uniformly to f . By definition
∞
X
−i
ρ(fn , f ) =
2 φ sup |fn (x) − f (x)|
i=1
x∈U i
Let > 0. Choose N large enough such that 2−N +1 < . Since U N −1 is compact, by
locally uniform convergence fn |U N −1 → f |U N −1 uniformly (and automatically converges
uniformly on U i for all i < N ). Then
X
N
−1
∞
X
−i
−i
ρ(fn , f ) =
2 φ sup |fn (x) − f (x)| +
2 φ sup |fn (x) − f (x)|
x∈U i
i=1
≤
N
−1
X
2−i φ
2−i φ
x∈U i
i=1
≤
N
−1
X
i=1
i=N
x∈U i
X
∞
sup |fn (x) − f (x)| +
2−i
i=N
sup |fn (x) − f (x)| + x∈U i
Sending n to infinity, we get limn→∞ ρ(fn , f ) < . Since is arbitrary, limn→∞ ρ(fn , f ) =
0, implying convergence in ρ. Conversely, suppose limn→∞ ρ(fn , f ) = 0 and K ⊂ Rn
compact. We can take N large enough such that UN contains K, so it suffices to show
that fn |U N → f |U N . Actually, we prove a stronger claim: for all n ∈ N, fk |U n → f |Un
uniformly. Suppose not, there is some N ∈ N such that fk |U N 6→ f |U N uniformly. Then
there is some δ > 0 such that for infinitely many k we have supx∈U n |fk (x) − f (x)| ≥ δ.
Notice that this means that for infinitely many k, we have supx∈U n |fk (x) − f (x)| ≥ δ for
n ≥ N . Then
∞
X
2−N
δ
δ
ρ(f, fk ) ≥
2−n φ(δ) =
·
= 2−N +1
1 − 1/2 δ + 1
δ+1
n=N
for infinitely many k so fk doesn’t converge to f in ρ, a contradiction. This finishes the
proof.
Folland 4.61
Theorem 4.43 remains valid for maps from a compact Hausdorff space X into a complete
metric space Y provided the hypothesis of pointwise boundedness is replaced by pointwise
total boundedness.
Proof. Let > 0. Since F is equicontinuous, for each x ∈ X there is open Ux of x such that
ρ(f (x), f (y)) < for all y ∈ Ux and f ∈ F. Since X is compact, we can choose x1 , ..., xn ∈ X
4
S
such that n1 Uxj = X. By pointwise total boundedness, {f (xj ) : f ∈ F, 1 ≤ j ≤ n} is a
totally bounded subset of Y being a finite union of totally bounded subsets of Y . Then we can
28
pick finite {z1 , ..., zm } that is
finite. For each φ ∈ B A let
-dense in it. Let A = {x1 , ..., xn } and B = {z1 , ..., zm }, then B A is
4
for 1 ≤ j ≤ n}
4
clearly ∪φ∈B A = F, and we claim that each Fφ has diameter ≤ , so we obtain a finite -dense
subset 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) < and ρ(g, φ) < on A and we have ρ(f, g) < on A. If x ∈ X,
4
4
2
we have x ∈ Uxj for some j, and then
Fφ = {f ∈ F : ρ(f (xj ), φ(xj )) <
ρ(f (x), g(x)) ≤ ρ(f (x), f (xj )) + ρ(f (xj ), g(xj )) + ρ(g(xj ), g(x)) < This shows that F is totally bounded. Then F is totally bounded. Since Y is complete, by exercise 4.22 C(X, Y ) is complete. Being a closed and totally bounded subset of complete metric
space C(X, Y ), F is compact.
Folland 4.63
Let K ∈ C([0, 1] × [0, 1]). For f ∈ C([0, 1]), let T f (x) =
C([0, 1]), and {T f : ||f ||u ≤ 1} is precompact in C([0, 1]).
R1
0 K(x, y)f (y)dy. Then T f
∈
Proof. We first show that T f ∈ C([0, 1]). Let > 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)| < . Then for x1 , x2 ∈ [0, 1] such that |x1 − x2 | < δ,
Z 1
Z 1
|T f (x1 ) − T f (x2 )| ≤
|K(x1 , y) − K(x2 , y)|f (y)dy < f (y)dy
0
0
Observe that f ∈ C([0, 1]), so by extreme value theorem |f | ≤ M for some M > 0. Then
|T f (x1 ) − T f (x2 )| < M . Since is arbitrary, T f is uniformly continuous on [0, 1] and thus
T f ∈ C([0, 1]).
We then show that {T f : ||f ||u ≤ 1} is precompact in C([0, 1]). For convenience, we use F to
denote the family {T f : ||f ||u ≤ 1}. Let > 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 > 0
such that |K(x) − K(y)| < 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]. When
x 2 ∈ U x1 ,
Z 1
Z 1
|T f (x2 ) − T f (x1 )| ≤
|K(x2 , y) − K(x1 , y)|f (y)dy < f (y)dy ≤ · 1 = 0
0
The last inequality is validated since |f | ≤ 1. Then F is equicontinuous at x1 and thus equicontinuous. Since K is continuous on a compact set, |K| ≤ M for some M > 0 by extreme value
R1
theorem. Then for x ∈ [0, 1], |T f (x)| ≤ 0 |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 precompact
as desired.
29
Folland 4.64
Let (X, ρ) be a metric space. A function f ∈ C(X) is called Holder continuous of exponent
α if the quantity
|f (x) − f (y)|
Nα (f ) = sup
ρ(x, y)α
x6=y
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. For
our convenience, we define F := {f ∈ C(X) : ||f ||u ≤ 1 and Nα ≤ 1}. Let > 0, x ∈ X consider
Ux = B(1/α , x).3 Thus for any f ∈ F, y ∈ Ux ,
|f (y) − f (x)| ≤ Nα (f )ρ(x, y)α ≤ ρ(x, y)α < Thus F is equicontinuous at x and is thus equicontinuous. Furthermore F is clearly pointwise
bounded since it is uniformly bounded by 1. By Arzela-Ascoli, F is precompact, so it suffices to
show that F is closed. Suppose f ∈ F, then for every > 0, there is f 0 ∈ F such that ||f 0 −f ||u <
and Nα (f 0 ) ≤ 1. Then ||f ||u < ||f 0 ||u + ≤ 1 + . Since is arbitrary, ||f ||u ≤ 1. Also we have
|f 0 (x) − f 0 (y)|
Nα (f 0 ) = supx6=y
≤ 1. For x 6= y ∈ X, |f (x) − f (y)| ≤ |f (x) − f 0 (x)| + |f 0 (x) −
ρ(x, y)α
f 0 (y)| + |f 0 (y) − f (y)| < 2 + ρ(x, y)α . But is arbitrary, so |f (x) − f (y)| ≤ ρ(x, y)α and thus
|f (x) − f (y)|
Nα (f ) = supx6=y
≤ 1. Thus f ∈ F and thus F is closed. This means that F = F is
ρ(x, y)α
compact and this finishes the proof.
Folland 4.68
Let X and Y be compact Hausdorff spaces. The algebra generated by functions of the
form 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 ). Then
f = 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 Hausdorff
and therefore normal, and {x1 } and {x2 } are disjoint closed sets in X since singletons are closed
in 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 . By
Stone-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 coordinate
maps 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 = hxα iα∈A and y = hyα iα∈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 separates points and closed sets.
Proof. Since X is second countable, let B be a countable base of X. For each pair (U, V ) ∈ B × B
such that U ⊂ V , by Urysohn’s lemma, there is some function f : X → I such that f (U ) = 0
and 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 the
collection 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 ⊂ X
closed and x ∈ E c , since E c is open, there is some V ∈ B such that x ∈ V ⊂ E c . Then E =
(E c )c ⊂ V c . We claim that there is some U 0 open such that x ∈ U 0 ⊂ U 0 ⊂ V . To show the
claim, we notice that {x} and V c are disjoint closed sets, so by normality there are disjoint open
V 0 ⊃ V c and U 0 3 x. Given U 0 ∩ V 0 = ∅, U 0 ⊂ (V 0 )c and thus U 0 ⊂ (V 0 )c since (V 0 )c is closed.
Then x ∈ U 0 ⊂ U 0 ⊂ (V 0 )c ⊂ (V c )c = V , as desired. Since U 0 is open, we have U ∈ B such that
x ∈ U ⊂ U 0 . Then x ∈ U ⊂ U 0 ⊂ V for U, V ∈ B. By definition there is some f ∈ F such that
f (U ) = 0 and f (V c ) = 1. Remember that x ∈ U and E ⊂ V c , so f (x) = 0 6∈ {1} = f (E), as
desired.
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 {Tn x}n is Cauchy
for each x since kTn x − Tm xk ≤ kTn − Tm kkxk → 0 as n, m → ∞ since kTn − Tm k → 0 as
n, m → ∞. Thus {Tn x} converges. Define T x := limn→∞ Tn x. We first show that T ∈ L(X , Y).
Let > 0. Since {Tn } is Cauchy, there is some N > 0 such that when m, n ≥ N , ||Tn − Tm || < .
Then ||Tn x − Tm x||/||x|| < for all x 6= 0. Sending m to infinity, we have ||Tn (x) − T (x)||/||x|| < for all x 6= 0. In particular, ||TN (x) − T (x)|| < ||x|| for all x. Since TN is bounded, suppose
||TN x|| ≤ CN ||x|| for all x. Then
||T (x)|| ≤ ||TN (x)|| + ||T (x) − TN (x)|| ≤ (CN + )||x||
for all x and thus T ∈ L(X , Y). Furthermore, for the and N given above, when n > N , ||Tn (x)−
T (x)||/||x|| < for all x 6= 0 and thus ||Tn − T || = sup{||Tn (x) − T (x)||/||x|| : x 6= 0} < , showing
that ||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 measures
on (X, M). Then ||µ|| = |µ|(X) is a norm on M (X) that makes M (X) into a Banach
space.
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 positiveR measure ν such
R that dµ = f dν.
Then d(λµ) = λdµ = λf dν 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 absolutely convergent
series in M (X) converges
P∞
P∞under this norm. Suppose we have
P∞µ1 , µ2 , ... ∈
M (X)
such
that
||µ
||
<
∞.
Then
|µ
|(X)
<
∞.
Define
ν
:=
µn . Notice
n
n
n=1
n=1
Pn=1
Pm
Pm
∞
|µ|(X)
for
all
m
and
thus
sending
m
→
∞
gives
µ
(X)
≤
that
n (X) ≤
n=1
n=1
Pm n=1 n
Pµm
|µ|(X) < ∞, showing that ν P
is a finite signed measure. Thus limm→∞ kν − n=1 µn k =
n=1P
∞
kv − ∞
ν
k
=
0,
showing
that
n
n=1
n=1 converges to ν ∈ M (X), as desired. Thus M (X) is a
Banach space.
Folland 5.9
Let C k ([0, 1]) be the space of functions on [0, 1] possessing continuous derivatives up to
order k on [0, 1], including one-sided derivatives at endpoints.
(a) If f ∈ C([0, 1]), then f ∈ C k ([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.
P
(b) kf k = k0 kf (j) ku is a norm on C k ([0, 1]) that makes C k ([0, 1]) into a Banach space.
(a) If f ∈ C k ([0, 1]), then clearly f ∈ C k (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 ∈ C k [0, 1]. Since f ∈ C k (0, 1), it suffices to show that f is continuously
differentiable at 0 and 1. So we proceed by induction. The base case where j = 0 is
true since f ∈ C[0, 1]. Suppose f is j times continuously differentiable at 0. Notice that
f (j) (x) − f (j) (0)
= limx↓0 f (j) (c) for some c ∈ (0, x] by mean value theorem, and
limx↓0
x
f (j) (x) − f (j) (0)
thus c ↓ 0 as x ↓ 0. Thus limx↓0
= limc↓0 f (j) (c) which is assumed to
x
exist. Thus f (j+1) (0) = limc↓0 f (j) (c), showing that f is j + 1 times continuously differentiable at 0. Then it follows by induction that f is C k at 0. Similarly we can show f is C k
at 1. Then f ∈ C k ([0, 1]), as desired.
P
P
P
(b) We first show that kf k = k0 kf (j) ku is a norm. kf + gk = k0 k(f + g)(j) ku = k0 kf (j) +
P
P
P
g (j) ku ≤ k0 (kf (j) ku + kg (j) ku ) = k0 kf (j) ku + k0 kg (j) ku = kf k + kgk. Also kλf k =
Pk
Pk
Pk (j)
(j) k = |λ|
k(λf )(j) ku =
ku = |λ|kf k. kf k = 0 implies kf ku ≤
u
0 kλf
0 kf
Pk0 (j)
kf
k
=
0
and
thus
f
≡
0
since
f
is
continuous.
We then show that this norm makes
u
0
k
C ([0, 1]) into a Banach space. Pick a Cauchy sequence {fn } in C k ([0, 1]) and let > 0.
P
(j)
(j)
Then there is N > 0 such that m, n > N implies kfn − fm k = kj=0 kfn − fm ku < .
In particular, kfn − fm k < for n, m > N and thus {fn } is uniformly Cauchy. Since fn
is continuous and C([0, 1]) is complete, fn → f for some f ∈ C([0, 1]). We now claim
(j)
that f ∈ C k ([0, 1]) and fn → f (j) uniformly for all j ≤ k. We prove the claim by induction. For k = 0, f ∈ C([0, 1]) and fn → f uniformly. Suppose f ∈ C l ([0, 1]) and
Proof.
32
(j)
fn → f (i) uniformly for all j ≤ l, we try to show the result for l + 1. Fix the and
l+1 k < for all m, n > N . Thus {f l+1 } is uniformly Cauchy. Since
m, n above, kfnl+1 − fm
n
k
l+1
l+1 are continuous and thus f l+1 → g for some g ∈ C([0, 1]).
fn , fm ∈ C ([0, 1]), fn andRfm
n
x
Notice that fnl (x)−fnl (0) = 0 fnl+1 (t)dt. Since fnl+1 → g uniformly and fnl+1 is continuous,
Rx
by undergraduate analysis, sending n → ∞ we get f l (x) − f l (0) = 0 g(t)dt. By fundamental theorem of calculus, f l+1 (x) = g(x), showing the result, and the claim follows by
P
(j)
(j)
induction. By the claim, kfn − f k = kj=0 kfn − fm ku → 0 as n → ∞ and f ∈ C k ([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 :
T x = 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 ||.
(a) Since T ∈ L(X , Y), T is continuous. Since {0} is closed in Y, N (T ) = T −1 ({0}) is
closed 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 welldefined. If x + N (T ) = x0 + N (T ), x0 = x + y for some y ∈ N (T ). Thus
Proof.
S(x0 + N (T )) = S(x + y + N (T )) = T (x + y) = T x + T y = T x = 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 bounds
T also bounds S. To show this claim, we suppose ||T x||Y ≤ C||x||X for all x ∈ X . Then
for any y ∈ N (T ), ||S(x + N (T ))||Y = ||T x||Y = ||T x + T y||Y = ||T (x + y)||Y ≤ C||x + y||X
and thus ||S(x + N (T ))||Y ≤ C · inf{||x + y|| : y ∈ N (T )} = C||x + N (T )||, proving 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 ). Then
S1 (x + N (T )) = S1 ◦ π(x) = S2 ◦ π(x) = S2 (x + N (T )), showing that S1 ≡ S2 and that
S 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 : ||T x||Y ≤ C||x||X for all x} = ||T ||.
Conversely we have ||T || = ||S ◦ π|| ≤ ||S|| · ||π|| = ||S|| and thus ||T || = ||S||, finishing the
proof.
Folland 5.20
If M is a finite-dimensional subspace of a normed vector space X , there is a closed subspace 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 , where
K is C or R and x1 is a non-zero vector in M. We may assume kx1 k = 1, otherwise we just do
33
some scaling to make this happen. Then we define f : M → K by f (λx1 ) = λ. Since kf k =
sup{||f (x)|| : kxk = 1} = sup{kf (λx1 )k : kλx1 k = 1} = sup{|λ| : |λ|kx1 k = 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 is
bounded linear by assumption and is thus continuous. Since {0} is closed, N = F −1 ({0}) is also
closed. Moreover, for any λx1 in M, λx1 ∈ N iff F (λx1 ) = f (λx1 ) = λ = 0 iff λx1 = 0, so
M ∩ 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 + N
and thus M + N = X . Then our claim is true.
We now finish the proof by induction. Suppose the results holds for dimension ≤ n, we show
that it holds for dimension n + 1. Suppose we have M0 with dimension n + 1, we may choose
M ⊂ M0 an n-dimensional subspace of M0 . By induction hypothesis we can choose a closed
N ⊂ X such that M ∩ N = {0} and M + N = X . Let y ∈ M0 \ M, we have y = m + x
for m ∈ M and x ∈ N . Clearly x 6∈ M, and thus dim(M + Kx) = dim(M) + dim(Kx) =
n + 1 = dim(M0 ). Since M + Kx ⊂ M0 , M + Kx = M0 . Since Kx is a one-dimensional
subspace of N 0 , we use the same technique as above to choose some closed N 0 ⊂ N such that
Kx ∩ N 0 = {0} and Kx + N 0 = N . Since N 0 is closed in N and N is closed, N 0 is closed in
X . Now we claim that N 0 is a closed subspace such that M0 ∩ N 0 = {0} and M0 + N 0 = X .
M0 ∩ N 0 = (M + Kx) ∩ N 0 . If m + kx ∈ N 0 ⊂ N for m ∈ M, since kx ∈ N , m ∈ N and thus
m = 0. Then kx ∈ N 0 and kx = 0. Thus m + kx = 0 and M0 ∩ N 0 = (M + Kx) ∩ N 0 = {0}.
Moreover, M0 + N 0 = M + Kx + N 0 = M + (Kx + N 0 ) = M + N = X . The result follows by
induction.
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 of
α. 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) where
f and g are defined at the very beginning. Then α ◦ β(h) = h. Thus β = α−1 is the two-sided
inverse. The inverse of a linear map is linear, so it remains to verify that β is bounded. Observe
that 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,
kα(f, g)k = sup{kα(f, g)(x, y)k : k(x, y)k = 1} = sup{kα(f, g)(x, y)k : k(x, y)k = 1}
= sup{kf (x) + g(y)k : k(x, y)k = 1} ≥ sup{kf (sgnf · x) + g(sgng · y)k : k(x, y)k = 1}
= sup{k(sgnf )f (x) + (sgng)g(y)k : max(kxk, kyk) = 1}
= sup{k(sgnf )f (x) + (sgng)g(y)k : max(kx||, kyk) = 1}
= sup{(sgnf )f (x) + (sgng)g(y) : max(kxk, kyk) = 1}
= sup{kf (x)k + kg(y)k : max(kxk, kyk) = 1} ≥ sup{kf (x)k + kg(y)k : kxk = 1, kyk = 1}
= sup{kf (x)k : kxk = 1} + sup{kg(y)k : kyk = 1} = kf k + kgk = k(f, g)k
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 corresponding x0 such that ||x0 || = 1 and ||f (x)|| ≤ ||f (x0 )||. 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)|| and
thus α is an isometry.
A Better Proof Idea. We first show that α is isometric and then show that it is bijective. If α is
isometric, α−1 is also isometric and automatically bounded and thus α is an isomorphism. This
avoids 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 kT + k = kT k. 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
ˆ and Ŷ in X ∗∗ and Y ∗∗ , then T ++ |X =
are identified with their natural images cX
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) kT + f k ≤ kT kkf k and thus kT + k = sup{kT + f k : kf k = 1} ≤ kT k < ∞, meaning
that T + is bounded. It remains to show that kT + k ≥ kT k. Let > 0, choose x ∈ X
such that kxk = 1 and kT xk > kT k − . If kT xk = 0, then automatically kT + k ≥ kT k.
Otherwise we can choose f ∈ Y ∗ such that kf k = 1 and f (T x) = kT xk. Then
kT + k ≥ kT + f k ≥ kT + f xk = kf (T x)k = kT xk ≥ kT k − and kT + k ≥ kT k since is arbitrary.
(b) T ++ f = f ◦ T + . Given x̂ ∈ X̂ , (T ++ x̂)(f ) = x̂ ◦ T + (f ) = x̂(T + f ) = T + f (x) = f (T x).
Then T ++ x̂ = (Tˆx) and thus T ++ x = T x if we identify T x with its canonical image (Tˆx).
(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 there
is some f ∈ Y ∗ such that f (y) := inf z∈T (X ) ky − zk. 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 injective. Conversely, suppose T is injective, since X̂ = X ∗∗ , T ++ = T , and thus T ++ is
injective. 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 result
is 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
1 and |fn (xn )| ≥ 2 ||fn ||. We claim that the linear combinations of {xn }∞
1 are dense in X . To
prove the claim, we first define M to be the closure of linear combinations of {xn }∞
1 . Then M
is a closed subsapce of X . Suppose there is x ∈ X \ M, then there is some f ∈ X ∗ such that
f (x) := inf y∈M ||x − y||. Let > 0. By denseness of {fn }∞
1 we have some fn such that ||f − fn || <
. In particular |fn (xn ) − f (xn )| < for the corresponding xn and thus |fn (xn )| < since xn ∈ M
and thus f (xn ) = 0. Since |fn (xn )| ≥ 21 ||fn ||, ||fn || < 2 and ||f || < ||fn || + < 3. Since is
arbitrary, ||f || = 0 and thus f ≡ 0. But this means that 0 = f (x) = inf y∈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 }∞
with
coefficients
whose
real
and
imaginary
parts are
1
∞
both 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 of
coefficients). 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 combination endows a space structure, so I should consider the whole space spanned by {xn }∞
1 . The solution 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 there
is a meager subset of Im for all m whose complement in Im has Lebesgue measure zero. To show
this, we first claim that for every n > 1, there is a generalized Cantor set Kn on Im such that
m(Im \ Kn ) = 1/n. To show the claim we define Kn this way: Kn,0 = Im , and suppose we have
defined Kn,j , define Kn,j+1 by removing the middle αj th content from every interval that makes
36
up Kn,j , where αj = 1/(n + j − 1)2 for j ≥ 1. Thus
∞
Y
1
1
1−
...
m(Kn ) =
(1 − αj ) = 1 − 2
n
(n + 1)2
i=1
1 − 1/n
1 − 1/(n + 1)
·
...
1 − 1/(n + 1) 1 − 1/(n + 2)
1 − 1/n
1
= lim
=1−
k→∞ 1 − 1/(n + k)
n
=
and thus m(Im \ Kn ) = m(Im ) − m(Kn ) = 1S− (1 − 1/n) = 1/n, showing that the claim is true.
We know that Kn is nowhere dense, so K := ∞
n=1 Kn is meager. Meanwhile 1 ≥ m(K) ≥ 1 − 1/n
for all n and thus m(K) = 1. Thus m(Im \ K)S= 0. Therefore, on every Im we have Mm ⊂ Im
meager such that m(Im \ Mm ) = 0. Let M = m∈Z Mm . Then M is meager and
[
[
m(R \ M ) = m
(Im \ Mm ) ≤
m(Im \ Mm ) = 0
m∈Z
m∈Z
and thus m(R \ M ) = 0, as desired.
Another Proof. Equivalently we prove that there is a residual set of measure 0.SLet {qn }n be an
enumeration of Q. Let > 0, define Bn, := B(2−n , qn ). It is clear that B := n Bn,Tis an open
dense subset of R. Then (B )c is nowhere dense and thus B is residual. Define B := n B1/n ,
then B is residual as a countable intersection of residual sets. Moreover m(B) ≤ m(B1/n ) = 2/n
for every n and thus m(B) = 0, as desired.
Folland 5.29
P
Let Y ∈ L1 (µ), where µ is counting measure on N, and let X = {f ∈ Y : ∞
1 n|f (n)| <
1
∞}, equipped with the L norm.
(a) X is proper dense subspace of Y, hence X is not complete.
(b) Define T : X → Y by T f (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.
(a) Let > 0, g ∈ Y. We want to find some f ∈ X such that ||g − f ||1 < .
Recall thatP
simple functions on N are dense in L1 (µ), so we can pick some simple funcn
tion f :=
i=1 ck χEk , where Ek is a measurable subset of N and ck < ∞, such that
||f − g||1 < . We claim that
each Ek is finite. Suppose not, there is some Ek that has
R
infinite cardinality. Then |f |dµ ≥ |ck |µ(Ek ) = ∞, contradicting f ∈PL1 (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 kg − f k1 < , as desired. This shows that RX is a dense
of Y.
Psubspace
∞
2 , we know that
2 < ∞
Now consider f on N such that
f
(n)
=
1/n
|f
|dµ
=
1/n
1
P
P∞
and thus f ∈ Y. However, ∞
n|f
(n)|
=
1/n
=
∞,
so
f
∈
6
X
.
Thus
X is a proper
1
1
6 X , so X is not closed. Since a complete subspace of
dense subspace of Y. Then X = Y =
a 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) be
a limit point in Γ(T ), we have {(fn , gn )} ∈ γ(T ) such that (fn , gn ) → (f, g) in the product
norm. We want to show g = T f . Let > 0, by convergence there is some large N such
that when n > N , k(fn , gn ) − (f, g)k < . This means that max(kfn − f k1 , kgn − gk1 ) < Proof.
37
for large enough n. Then
kg − T f k1 ≤ kg − gn k1 + kgn − T fn k1 + kT fn − T f k1
Z
≤ + 0 + m|fn (m) − f (m)|dµ
R
R
R
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,
R
sending nP→ ∞ we get kg − T f k < . Since is arbitrary, kg − T f k = 0. Then |g −
T f |dµ = ∞
1 |g(n) − T f (n)| = 0 and thus g(n) = T f (n) for all n, from which we conclude
g = T f , 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 kfn k1 = |f (n)| =
1 for all n. Then sup{kT f k : kf k1 = 1} ≥ supn kT fn k = supn nf (n) → ∞ and thus T is
unbounded.
(c) To make S well-defined, we need to show that T P
is bijective. For
R f (n) :=
Pg∞ ∈ Y, define
|g(n)|
=
|g|dµ < ∞
n|f
(n)|
=
g(n)/n for all n. (Here we assume 0 6∈ N) Then ∞
1
1
and thus f ∈ X . Also the most importantly T f (n) = nf (n) = g(n) and thus g = T f ,
showing that T is surjective. Suppose f1 6= f2 , f1 (n) 6= f2 (n) for some n. Then T f1 (n) =
nf1 (n) 6= nf2 (n) = T f2 (n) and thus T f1 6= T f2 , showing that T is injective. Then T
is bijective as desired and S is well-defined. We now claim that S is defined such that
Sg(n) = g(n)/n. To show the claim, observe that T Sg(n) = ng(n)/n = g(n) and thus
T S is the identity. Similarly we can show that ST is also the identity. Then the claim is
true. We need to show that S is bounded. This is true since
Z
Z
g(n)
dµ : |g|dµ = 1
sup{kSgk : kgk = 1} = sup
n
Z
Z
≤ sup
|g|dµ : |g|dµ = 1 = 1
Also S is surjective since T is bijective. Eventually, if S = T −1 is open, T is continuous
and 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 ∗ for
every 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−1
f (Γ(f ◦
T
−1
T )) = {(x, y) : (x, f (y)) = (x, f ◦ T (x))} is closed. We claim that f ∈Y ∗ hf (Γ(f ◦ T )) = Γ(T ).
To show this claim, we first observe that Γ(T ) ⊂ h−1
f (Γ(f ◦ T )) for every f and thus Γ(T ) ⊂
T
T
−1
−1
f ∈Y ∗ hf (Γ(f ◦ T )). Conversely, suppose (x, y) ∈
f ∈Y ∗ hf (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
T is some g such
that g(y) 6= g ◦T (x), a contradiction. Then (x, y) ⊂ Γ(T ), showing our claim. f ∈Y ∗ h−1
f (Γ(f ◦T ))
−1
is closed since each hf (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. Therefore 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 might
not 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 and
apply 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 lim Tn x
exists for every x ∈ X. Let T x = lim Tn x; then T ∈ L(X , Y).
Proof. Since addition and multiplication respect limits, T is linear. {Tn x} converges and in particular bounded for each x. Since X is a Banach space, by uniform boundedness supm kTm k < M
for some M > 0. Then
kT xk = lim kTm xk ≤ sup kTm kkxk ≤ M kxk
n→∞
m
and thus T is bounded, as desired.
Remark. This is a direct application of uniform boundedness principle. Uniform boundedness
principle 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 piecewise 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 therefore uniformly continuous on [0, 1]. Let > 0, we know that there is some δ > 0 such
, δ). Then conthat when |x − y| < δ, |f (x) − f (y)| < 2 . We define δ 0 := min( 4n
sider a partition 0 = x0 ≤ x1 ≤ · · · ≤ xn = 1 such that xj − xj−1 < δ 0 for all
j. We construct a piecewise linear function g by connecting f (xj−1 ) and f (xj ) for every 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 it
with a bell-shaped graph, i.e. a wedge such that the ascending piece has slope 2n and descending piece has slope −2n. The refined partition still satisfies all the assumptions mentioned above. Now notice that for the g we just constructed, in every [xj−1 , xj ], for any
x, y ∈ [xj−1 , xj ], |g(x) − g(y)| ≤ min( 2 , 2n|x − y|) ≤ min( 2 , 2nδ 0 ) = 2 . It remains to show
that supx∈[0,1] |f (x) − g(x)| < . 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 )| < + = 2 2
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 our
claim, for any > 0 and f ∈ En , we can find g 6∈ En such that supx∈[0,1] |f (x) − g(x)| < 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 C c is
meager. Suppose f ∈ C c , then f is differentiable at some x0 ∈ [0, 1]. We define φ(x) :=
f (x) − f (x0 )
where φ(x0 ) := f 0 (x0 ). and claim that it is continuous on [0, 1]. It is clear
x − x0
that φ is continuous at x ∈ [0, 1] \ {x0 }, so it suffices to show continuity at x0 . This is
true since
f (x) − f (x0 )
= f 0 (x0 ) = φ(x0 )
lim φ(x) = lim
x→x0
x→x0
x − x0
Since φ is continuous, it is bounded on [0, 1] by extreme value theorem. Then f ∈ Em for
S
S
◦
some m. Then C c ⊂ n En and C c = n (En ∩C c ). Since each En is nowhere dense, E n = ∅
◦
◦
and thus En ∩ C c ⊂ E n = ∅. Then En ∩ C c is nowhere dense and C c 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(k)
ogy with respect to which fn → f iff fn → 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 because 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 −
(l)
(l)
f ) = sup|x|≤k |fn (x) − f (l) (x)| → 0 iff fn → f (l) uniformly on all compact subsets since any compact 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 Hausdorff
topological vector space. Let f 6= 0, there is some x0 ∈ R such that f (x0 ) 6= 0. Then consider
a 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
(l)
sequence hfn in in C ∞ (R), then pk,l (fn − fm ) = sup|x|≤k |fnl (x) − fm (x)| → 0 as m, n → ∞. Then
(l)
(l)
hfn |[−k,k] i is uniformly Cauchy for all l and k. Since each fn |[−k,k] is continuous and C[k, k] is
(l)
complete, fn |[−k,k] uniformly converges to some gk,l ∈ C[−k, k]. Now we define gl on R such that
(l)
gl (x) = gk,l (x) where x ∈ [−k, k]. This function is well-defined since gk,l (x) = limn→∞ fn (x)
for all k which means that the definition of gk,l (x) is independent of k. Eventually we observe
(l)
(l)
that fn → gl locally uniformly by our definition of gl , and we claim that gl = g0 . To show
this claim, we shall do an induction on l as in problem 5.9. The base case is trivial since g0 = g0 .
(l)
Suppose we have gl = g0 ,
Z x
gl (x) = lim fn(l) (x) = lim
fn(l+1) (t)dt
n→∞
n→∞ 0
(l+1)
Notice that fn
theorem,
is continuous and therefore bounded on [0, x], so by dominated convergence
Z x
Z x
Z x
(l+1)
(l+1)
gl (x) = lim
fn
(t)dt =
lim fn
(t)dt =
gl+1 (t)dt
n→∞ 0
0 n→∞
40
0
and by fundamental theorem of calculus we get gl0 (x) = g0l+1 (x) = gl+1 , and the claim follows by
(k)
(k)
induction. Thus fn → g0 locally uniformly for every k and thus fn → g0 by the previous part
(l)
of the problem. g0 = 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 : kxk ≤ 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. Suppose
y ∈ B c . Since B is norm closed, by Hahn-Banach there is a linear functional f such that
f (y) = δ where δ := inf x∈B kx − yk. Then f −1 [B(y, δ)] is a weakly open ball excluding B.
Hence B c is weakly open and B is weakly closed.
(b) Since E is bounded with respect to norm, E ⊂ B = {x ∈ X : kxk ≤ M } for some
M ≥ 0. By appropriate scaling if necessary, we may assume M = 1. Then by a) B is
weakly closed and hence the weak closure of E is contained in B. It follows that the weak
closure 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 may
assume F ⊂ B, where B is the norm closed unit ball as defined in a). By Alaoglu, B is
compact in the weak* topology on X and hence closed. Then the weak* closure of F is
also contained in B and it follows that it is weak* bounded.
(d) Let hfn in 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 kfn (x) − f (x)k → 0 as n, m → ∞. Thus
hfn (x)in is Cauchy for each x and thus converges since K = {R, C} is complete. We set
f (x) := limn fn (x), then f ∈ X ∗ and fn (x) → f (x) for all x ∈ X , showing that fn → f in
weak* 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 the
complement of V is open. Suppose x ∈ V c , then since V is norm closed, it follows by HahnBanach that there is a linear functional f such that f (x) = δ where δ = inf y∈V kx − yk. Then
f −1 [B(δ, f (x))] is a weakly open neighborhood of x excluding V [weakly open because f as a linear functional is assumed to be continuous in weak topology and B(δ, f (x)) is open]. Thus V c is
weakly open and V is weakly closed. Conversely, suppose V ⊂ X is weak closed. Let x ∈ V in
the norm topology, then there is hxn in such that xn → x in the norm topology, i.e. kxn − xk → 0
as n → ∞. For any f ∈ X ∗ , |f (xn − x)| ≤ kf k · kxn − xk → 0 as n → ∞, so xn → x in weak
topology 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 that
Tn → T strongly and Sn → S strongly.
(a) If {xn } ⊂ X and kxn − xk → 0, then kTn xn − T xk → 0.
(b) Tn Sn → T S strongly.
Proof.
(a) Notice that
kTn xn − T xk ≤ kTn xn − Tn xk + kTn x − T xk
= kTn (xn − x)k + kTn x − T xk
≤ kTn kkxn − xk + kTn x − T xk (∗)
For any x ∈ X , since {Tn } converges strongly, {Tn x} converges and in particular bounded
in the norm metric, i.e. supn kT xk < ∞ for every x. Since X is a Banach space, by uniform boundedness we have supn kTn k < M for some large M > 0. Then (∗) ≤ M kxn −
xk + kTn x − T xk. Sending n to infinity, kxn − xk → 0 and kTn x − T xk → 0 (strong
convergence). Thus kTn xn − T xk → 0, as desired.
(b) For every x ∈ X , Tn x → T x and Sn → Sx since Tn → T and Sn → S strongly. Then
Tn xSn x → T xSx or equivalently Tn Sn x → T Sx. Since x is arbitrary, Tn Sn → T S
strongly. 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 positive
measure, and Lq 6⊂ Lp iff X contains sets of arbitrarily large finite measure. What about
the 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
Z
∞ > kf kpp ≥ kf χEn kpp = |f χEn |p dµ > np µ(En )
and thus µ(En ) < kf kpp /np . Therefore µ(En ) → 0 as n → ∞ since f ∈ Lp and thus kf kpp <
∞. Now it suffices to show that µ(En ) > 0 for each n. If in the contrary µ(En ) = 0, let
Fn := Enc , we have
Z
Z
kf kpp = |f |p dµ = |f |p χFn dµ < ∞
notice that also we have
Z
Z
Z
q
p
q−p
p
q−p
q−p
kf kq = |f | |f | dµ = |f | |f | χFn dµ ≤ n
|f |p χFn dµ = nq−p kf kpp < ∞
contradicting f 6∈ Lq . Conversely, suppose X contains sets of arbitrarily small positive
measure, then we can pick a disjoint family of sets {En }n such that 0 < µ(En ) < 2−n .
−n
This is because
S∞we can take a family {Fn }n of sets such that µ(Fn ) < 2 , and
P defining
En := Fn \ n+1 Ei creates the desired subsets {En }n . Now we define f := ∞
1 an χEn ,
42
where an = µ(En )−1/p . Notice that
p
Z X
Z
Z X
∞
∞
p
−1/p
−1/p
p
p
µ(En )
χEn dµ
µ(En )
χEn dµ =
kf kp = |f | dµ =
n=1
n=1
= lim
m→∞
Z X
m
−1/p
µ(En )
p
χEn
dµ = lim
Z X
m
m→∞
n=1
χEn dµ = lim
m→∞
n=1
m
X
µ(En )
(1)
n=1
P
−n < ∞,
since En s are disjoint. Then by our assumption 0 < (1) < limm→∞ m
n=1 2
p
showing that f ∈ L . The same thing won’t happen on kf kq , since
Z X
Z
Z X
∞
∞
q
−1/p
q
q
µ(En )−q/p χEn dµ
µ(En )
χEn dµ =
kf kq = |f | dµ =
n=1
n=1
≥ lim
m→∞
m
X
(1/µ(En ))q/p−1 ≥ lim
m→∞
n=1
m
X
(2q/p−1 )n ≥ lim m = ∞
n=1
m→∞
which fails to converge. Thus kf kq = ∞, showing that f 6∈ Lq .
q
L 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 ∞P
> µ(F1 ) ≥ 1, and
then suppose we have chosen Fn , choose Fn+1 such that µ(Fn+1 ) ≥ 2 n1 µ(Fn ). Thus by
P
S
Ei we get the desired subsets. Consider f := ∞
considering En := Fn \ n−1
1 an χEn ,
1
−1/q
where an = µ(En )
. 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
first part of
Pthe
∞
the statement, we use the same proof as above, except setting f =
1 an χEn , where
an = µ(En )−1/(p+1) . For the second part of the statement, if X has infinite measure, we
can 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 the
following silly example: let X = {0, 1} such that 0 has infinite measure and 1Rhas zero
measure. For every f ∈ L∞ , f must be actually bounded, and thus kf kp = |f |p =
|f (1)|p < ∞, showing that f ∈ Lp .
Folland 6.9
Suppose 1 ≤ p < ∞. If kfn −f kp → 0, then fn → f in measure, and hence some subsquence
converges to f a.e. On the other hand, if fn → f in measure and |fn | ≤ g ∈ Lp for all n,
then kfn − f kp → 0.
Proof. First of all suppose kfn → f kp → 0 for some 1 ≤ p < ∞. Let > 0. We define En, := {x :
|fn (x) − f (x)| ≥ }. Then
1/p
Z
1/p Z
p
p
kfn − f kp =
|fn − f |
≥
|fn − f |
≥ (p µ(En, ))1/p = µ(En, )1/p
En,
and thus µ(En, ) ≤ −p (kfn − f kp )p → 0 as n → ∞ since kfn → f kp → 0 as n → ∞. This
means that fn → f in measure, as desired. In particular, since fn → f in measure, there is a subsequence of hfn in that converges to f pointwise a.e. On the other hand, suppose fn →
meaR f in
p
p
p
1/p
sure and |fn | ≤ g ∈ L for all n. We first make the observation that g ∈ L implies ( |g| )
<
43
R
∞ and thus |g|p < ∞, meaning that g p ∈ L1 . Now since fn → f in measure, there is some
subsequence hfnk ik that converges to f a.e. Since hfnk ik is a subsequence, we also have |fnk | < g
for all k. Sending k → ∞ we also obtain |f | ≤ g a.e. Therefore, |fnk − f |p ≤ (|fnk | + |f |)p ≤ (2g)p .
Remember our observation that g p ∈ L1 , we have |fnk − f |p ≤ (2g)p = 2p g p ∈ L1 . By dominated
convergence theorem, we have
Z
Z
p
lim |fnk − f |p = 0
lim
|fnk − f | =
n→∞
n→∞
and hence taking 1/pth power on each side we get kfnk − f kp → 0. We now claim that actually
kfn − f kp → 0. Suppose not, there is some δ > 0 such that kfn − f kp ≥ δ for infinitely many
fn . We arrange them to a new sequence, which for our convenience we call hgn in . Since hgn in is
essentially a subsequence of hfn in , 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 hgnk in
such that kgnk − f kp → 0, meaning that for large enough k we have kgnk − f kp < δ, contradicting
our assumption. Thus our claim is true, meaning that kfn − f kp → 0, as desired.
Folland 6.13
Lp (Rn , m) is separable for 1 ≤ p < ∞. However, L∞ (Rn , m) is not separable.
p
Proof. We first show
Pnthat L (R, m) is separable for 1 ≤ p < ∞. Let F be the family of simple
functions of form j=1 aj χFj 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 all
we need to show that F is countable. Notice that there are countably many such sets Fj since
there are countably many such measurable rectangles, and countably many such aP
j , so there are
countably many characteristic functions of the form aj χRj . Identifying each sum nj=1 aj χFj
as (a1 χF1 , ...) we know that there are countably many of them. Thus F is countable. We proceed to show that F is dense. Let f ∈ Lp (Rn , m) and > 0. Since simple
functions are dense
P
in Lp (Rn , m), we may assume that f is simple and can be written as n1 bj χEj . Fix the . By
denseness of rationals, we have some aj such that |aj − bj | < [/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(Ej 4Fj ) < /|aj |p . Then
Z X
1/p X
1/p
n
n Z
p
(bj χEj − aj χFj ) dm
=
|bj χEj − aj χFj |p dm
(1)
j=1
j=1
For each j,
Z
|bj χEj − aj χFj |p dm ≤
Z
|(bj − aj )χEj |p dm +
Z
|aj (χEj − χFj )|p dm
≤|bj − aj |p m(Ej ) + |aj |p m(Ej 4Fj ) < 2
and thus (1) < (2n)1/p . Since 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 kf − gk∞ ≥ 1 for all f, g ∈ F with f 6= g. This is because if we take an open ball of
radius 1 around each f ∈ F, we obtain an uncountable disjoint collection of open balls. Then for
any countable subset of L∞ (R, m), we must have some open ball in this collection not containing
any 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 = χBr0 ,
we must have r 6= r0 and we may assume r < r0 . Then for x ∈ Br0 \ Br , a positive measure
44
set, we have |f (x) − g(x)| = 1, and thus kf − gk∞ ≥ 1. The family is clearly uncountable, as
desired.
Folland 6.19
P
Define φn ∈ (l∞ )∗ by φn (f ) = n−1 n1 f (j). Then the sequence {φn } has a weak* cluster
point 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 of
counting measure this means that f is actually bounded by M . φn (f ) takes the arithmetic mean
of the first n terms, and is thus bounded by M for all n. Therefore
kφn k = sup{|φn (f )| : kf k∞ = 1} ≤ 1
by above reasoning. This means that φn ∈ B ∗ , the unit ball of (l∞ )∗ . By Alaoglu, B ∗ is compact
in the weak* topology. We define Km := {φn : n ≥ m}, and notice that hKm im is a family
of closed sets (here T
it means weak* closure) T
with finite intersection property. Since each Km ⊂
∗
∗
B = B , we have m Km 6= ∅. Pick φ ∈ m Km , we claim that φ is a weak* cluster point of
hφn in . 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 hφin is
frequently 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 ) =
R
ϕg (f ) :=R f gdµ 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 hφn in , φ(fk ) is a cluster point of hφn (fk )in . Then a subsequence of
hφn (fk )in 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 and
therefore φ(f ) = 1 ≥ 0, a contrasiction. Then φ doesn’t arise from l1 .
Folland 6.20
Suppose supn kfn kp < ∞ 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 is
σ-finite and weak convergence is replaced by weak* convergence.
Proof.
(a) Let 1 < p < ∞ and q the conjugated of p. By Riesz representation theorem, to
prove fn R→ f weakly in Lp , it suffices to prove that ϕg (fn ) → ϕg (f ) for all g ∈ Lq , where
ϕg (f ) = f g. Given g ∈ Lq and let > 0, we have the following observations, which we
shall prove at the end of this question:
R
1. There is some δ > 0 such that E |g|q < whenever
µ(E) < δ.
R
2. There is an A ⊂ X such that µ(A) < ∞ and X \ A |g|q < .
3. There is B ⊂ A such that µ(A \ B) < and fn → f uniformly on B.
45
Based on the observations, we have
Z
Z
Z
Z
|ϕfn (g) − ϕf (g)| =
(fn − f )g ≤
(fn − f )g +
(fn − f )g +
(fn − f )g
B
A \ B
X \ A
Z
Z
Z
|(fn − f )g|
|(fn − f )g| +
|(fn − f )g| +
≤
X \ A
A \ B
B
Z
1/p Z
1/q Z
p
q
≤
|(fn − f )|
|g|
+
B
B
q
|g|
A \ B
1/q
Z
q
kf − fn kp +
|g|
1/q
kfn − f kp
X \ A
Z
1/p
p
<
|fn − f |
kgn − gkq + 4M 1/q ≤ kfn χB − f χB ku µ(B) + 4M 1/q
(1)
B
since fn → f uniformly on B, kfn χB − f χB ku → 0. Also notice that µ(B) ≤ µ(A) < ∞.
Therefore, sending n → ∞, we get limn→∞ |ϕfn (g) − ϕf (g)| < 4M 1/q . Since is arbitrary,
this shows that limn→∞ |ϕfn (g) − ϕf (g)| = 0, as desired. Thus it remains to show the
observations to finish the proof.
R
1. Since |g|q is measurable, ν(E) := E |g|q dµ is a well-defined measure, and it is clear
that ν is absolutely continuous to µ since integrating on a measure-zero set we get 0.
Notice that g ∈ L1 , so ν(E) ≤ kgkq is a finite signed measure. Then the result follows
by the − δ definition
R of absolute continuity.
2. S
Since |g|q ∈ L+ and |g|q < ∞, K := {x : |g|q > 0} is σ-finite.
Sn Then we can write K =
∞
E
where
each
E
has
finite
measure,
and
define
K
:=
n
n
1 En for our convenience.
1
Rn
q
Notice that ν(K) = K |g| dµ = limn→∞ ν(Kn ) by continuity from below, where ν is
defined as above. Since ν(K) < ∞, this means that there is some N ∈ N such that
|ν(K) − ν(KN )| < . Take A = KN , we have
Z
|ν(X) − ν(A)| = |ν(K) − ν(A)| < =⇒ |ν(X \ A)| < =⇒
|g|p dµ < X \ A
as desired.
3. Since µ(A) < ∞ and fn χA → f χA 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 of
all, in L1 (R, m) consider fn = n1 χ(0,n) .
Z
Z
1
1
χ(0,n) = m(0, n) = 1
|fn |dm =
n
n
for all n and thus supn kfn k1 < ∞. Also it is clear
that fn → 0 a.e. However, for g ≡ 1 ∈
R
L∞ (since ∞ is the conjugate of 1), ϕg (fn ) = fn dm = 1 for all n and thus ϕg (fn ) 6→ 0 =
ϕg (0). Thus fn 6→ f weakly in L1 . Next we consider fn = n1 χ{1,...,n} ∈ l1 , and we use µ for
counting measure. Notice that
Z
Z
1
1
fn dµ =
χ
= ·n=1
n {1,...,n} n
∞
for all n and
R thus supn kf k1 < ∞. Also fn → 0 a.e. However, for g ≡ 1 ∈ l , we have
ϕg (fn ) = fn dµ = 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 show
that ϕfn → ϕf , and this yields to show that ϕfn (g) → ϕf (g) for all g ∈ L1 .
Z
Z
Z
|ϕfn (g) − ϕf (g)| =
fn g − f g ≤ |fn g − f g| (1)
46
Suppose supn kfn k∞ < M , we have |fn g − f g| ≤ 2M |g| ∈ L1 for all n. Also fn → f a.e.
and g essentially bounded implies fn g → f g 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 to
R
show that ϕg (fn ) := fn g converges to ϕg (0) = 0 for all g ∈ L2 . To this end, let g ∈
2
L
Pnand > 0. By denseness of simple functions, we can choose a simple function φ :=
1 ai χEi such that kg − φk2 < . 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 call
Fi , such that m(F
/(n|ai |) for each i. For convenience, let’s call the redefined
P i 4Ei ) < P
n
b
χ
(=
function φ0 := m
1 ai χFi ), where Ui are intervals with end points ai and bi .
1 i Ui
Now
Z
Z
Z bi
m
m
X
X
0
0
|ϕφ (fn )| =
φ cos 2πnx =
bi
cos 2πnx ≤
bi
cos 2πnx
Ui
i=1
≤
m
X
i=1
bi
bi
sin 2πnx
≤
2πn
ai
m
X
i=1
|bi |
→0
πn
i=1
ai
as n → ∞
By assumption we have kg − φk2 ≤ , and
Z
n
X
kφ − φ0 k2 = |φ − φ0 | ≤
|ai |m(Fi 4Ei ) < i=1
so by Minkowski inequality we have kg − φ0 kR< 2. Since the assignment g 7→ ϕg is isometric, we have kϕφ0 − ϕg k < 2. Since kfn k2 = [0,1] cos 2πnx ≤ 1, it follows that
|ϕφ0 (fn ) − ϕg (fn )| ≤ kϕφ0 − ϕg kkfn k2 ≤ kϕφ0 − ϕg k < for all n and sending n → ∞ we have limn→∞ |ϕg (fn )| < . Since is arbitrary, we have
limn→∞ |ϕg (fn )| = 0. Since g is arbitrary, this shows that fn → 0 weakly. Moreover,
we show that no subsequence of hfn in converges to 0 a.e. Let hfnk ik be a subsequence of
hfn in , suppose fnk → 0 a.e., we have |fn2k | = | cos2 (2πnk x)| ≤ χ[0,1] ∈ L1 ([0, 1], m).
R
Applying dominated convergence we should get cos2 2πnk x → 0 as k → ∞. However,
Z
Z
1
1
cos 4πnx + 1
1
1
2
cos (2πnk x) =
=
sin 4πnx + = 6= 0
2
8πn
2
2
0
a contradiction. It immediately follows that fn 6→ 0 a.e. If fn → 0 in measure, there is a
subsequence 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 n1 < x and hence
fn (x) = 0. fn → 0 in measure since for every ≥ 0, µ{|fn | ≥ } ≤ n1 → 0 as n → ∞. We
then show that fn 6→ 0 weakly in Lp for any p. Let p ∈ (1, ∞) consider
itsRconjugate q.
R
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 a
well-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,
Z Z
ZZ
|K(x, y)f (y)|dν(y) dµ(x) ≤
|K(x, y)f (y)|dµ(x)dν(y)
Z Z
Z
=
|K(x, y)|dµ(x) |f (y)|dν(y) ≤ C |f (y)|dν(y) (1)
R
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 T f (x) converges absoluetely for a.e.-x and also
Z
Z Z
Z
|T f (x)|dµ(x) ≤
|K(x, y)f (y)|dν(y) dµ(x) ≤ C |f (y)|dν(y) < ∞(2)
by (1), showing that T f is defined in L1 (µ). Eventually, we can read from (2) that kT f k1 ≤
Ckf k1 . 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, if
f ∈ (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 αp λf (α) < ∞. We
may assume αp λf (α) ≤ M for some M > 0. Given that α > 0, we have λf (α) ≤ M/αp . Since
0 < p < ∞, by proposition 6.24,
Z 1
Z ∞
Z ∞
q−1
q−1
q
kf kq =
α λf (α)dα = q
α λf (α)dα + q
αq−1 λf (α)dα
0
0
1
Z 1
Z ∞
= qC
αq−1 dα + qM
αq−p−1 dα
0
1
= Cαq
1
+ qM
0
∞
1
αq−p
q−p
1
(1)
∞
1
qM
1
Since q > 0, Cαq 0 = C, and for q < p, q − p < 0 and thus qM ( q−p
αq−p 1 ) = − q−p
. It follows
q
q
that kf kq = (1) < ∞ for all q < p, showing that f ∈ L 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
Z ∞
Z β
β
1
1
q
q−1
kf kq = q
α λf (α)dα = qM
αq−p−1 dα = qM
αq−p = qM
β q−p < ∞
q
−
p
q
−
p
0
0
0
for all q ∈ (p, ∞). Since we already assume f ∈ L∞ , f ∈ Lq for all q > p. This finishes the
proof.
48
Remark. The key technique in this problem is splitting up an integral into different scales, which
is a standard technique when integrals behave differently at different scales. In part 1, the integral is nice in large scale, so we bound it in small scale; in part 2 the integral is nice in small
scale, so we bound it in large scale.
Folland 6.45
If 0 < α < n, define an operator Tα on functions on Rn by
Z
Tα f (x) = |x − y|−α f (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 the
same thing as in the problem. Observe that
β q λK(x,·) (β) = β q m{y : |x − y|−α > β} = β q m{y : |x − y| < β −1/α }
≤ β q mB(x, β −1/α ) ≤ β q C1 + β −n/α = β q C1 β −q = C1
1/q
for some constant C1 > 0. Therefore, [β q λK(x,·) (β)]1/q ≤ C1
for all β and thus [K(x, ·)]q ≤
1/q
C1 for some C1
> 0. By symmetry of K(x, y), we can similarly prove that [K(·, y)]q ≤ C2
1/q
for some C2 > 0 for all y. By setting C = max(C1 , 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 we
conclude the proof.
Folland 6.37
If f is a measurable function and A > 0, let E(A) = {x : |f (x)| > A}, and set
hA = f χX \ 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 < A
0
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{|f χX \ E(A) + A(sgnf )χE(A) | > α} (3)
We now claim that {|f χX \ E(A) + A(sgnf )χE(A) | > α} = {|f χX \ E(A) | > α} ∪ {|A(sgnf )χE(A) | >
α}. To prove the claim, we first prove the forward inclusion. x ∈ {|f χX \ E(A) + A(sgnf )χE(A) | >
α} means that |f χX \ E(A) (x) + A(sgnf )χE(A) (x)| > α. Since X \ E(A) and E(A) are disjoint,
we have either
|f χX \ E(A) (x) + A(sgnf )χE(A) (x)| = |f χX \ E(A) (x) + 0| = |f χX \ E(A) (x)| > α
49
or
|f χX \ E(A) (x) + A(sgnf )χE(A) (x)| = |0 + A(sgnf )χE(A) (x)| = |A(sgnf )χE(A) (x)| > α
meaning that x ∈ {|f χX \ E(A) | > α} ∪ {|A(sgnf )χE(A) | > α}. Conversely, suppose x ∈
{|f χX \ E(A) | > α} ∪ {|A(sgnf )χE(A) | > α}. If x ∈ {|f χX \ E(A) | > α}, x ∈ X \ E(A) and
hence A(sgnf )χE(A) (x) = 0. It follows that
|f χX \ E(A) (x) + A(sgnf )χE(A) (x)| = |f χX \ E(A) (x) + 0| = |f χX \ E(A) (x)| > α
and x ∈ {|f χX \ E(A) + A(sgnf )χE(A) | > α}. Similarly we can show the result for the case where
x ∈ {|A(sgnf )χE(A) | > α}. Therefore we have shown the claim. By the claim and the fact that
{|f χX \ E(A) | > α} and {|A(sgnf )χE(A) | > α} are disjoint by the disjointness of X \ E(A) and
E(A), we obtain
(3) = µ{|f χX \ E(A) | > α} + µ{|A(sgnf )χE(A) | > α}
If α < A, x ∈ {|f χX \ E(A) | > α} iff x ∈ X \ E(A) and |f (x)| > α iff α < |f (x)| ≤ A, and
(3) = µ{|f χX \ 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 {|f χX \ E(A) | > α} = ∅. Moreover
{|A(sgnf )χE(A) | > α} = {χE(A) > α/A ≥ 1} = ∅
so (3) = µ{|f χX \ E(A) | > α} + µ{|A(sgnf )χE(A) | > α} = 0. This finishes the proof.
Folland 6.38
P
kp
k
f ∈ Lp iff ∞
−∞ 2 λf (2 ) < ∞.
Proof. Suppose f ∈ Lp . Then
∞
∞
∞
X
X
X
2k−1 2(k−1)(p−1) λf (2k )
2kp λf (2k ) =
2k 2k(p−1) λf (2k ) = 2p
−∞
(1)
−∞
−∞
Notice that on [2k−1 , 2k ), α ≥ 2k−1 and λf (α) ≥ λf (2k ) (since λf is a decreasing function), so
αp−1 λf (α) ≥ (2k−1 )p−1 λf (2k ). Hence
Z
Z
p−1
α λf (α)dα ≥
(2k−1 )p−1 λf (2k ) = 2k−1 2(k−1)(p−1) λf (2k ) (2)
[2k−1 ,2k )
[2k−1 ,2k )
and it follows by additivity that
(1) ≤ 2
p
∞ Z
X
−∞
αp−1 λf (α)dα
(3)
[2k−1 ,2k )
Notice that the collection of intervals {[2k−1 , 2k )}k are pairwise disjoint, and that
(0, ∞). We have
∞ Z
X
p
(3) ≤ 2
αp−1 λf (α)χ[2k−1 ,2k ) dα (4)
−∞
50
S∞
k−1 , 2k ) =
−∞ [2
Since |αp−1 λf (α)χ[2k−1 ,2k ) | ≤ αp−1 λf (α) ∈ L1 by proposition 6.24, by dominated convergence
theorem,
Z
Z X
∞
p−1
p
p
αp−1 λf (α)χS∞
α λf (α)χ[2k−1 ,2k ) dα = 2
(4) = 2
k−1 ,2k ) dα
−∞ [2
−∞
2p
kf kpp < ∞
p
P∞ kp
P
k
kp
k
and the result follows. Conversely, suppose ∞
−∞ 2 µ{|f | > 2 } < ∞. For
−∞ 2 λf (2 ) < ∞,
our convenience, define Kn := {2n < |f | ≤ 2n+1 },
Z
∞
∞
∞
X
X
X
|f |p ≤
2(n+1)p µ(Kn ) ≤
2(n+1)p λf (2n ) ≤ 2p
2np λf (2n ) < ∞
= 2p
Z
αp−1 λf (α) ≤
−∞
−∞
−∞
showing that f ∈ Lp , as desired. This finishes the proof.
Folland 6.41
−1
−1
p
RSuppose 1 R< p ≤ ∞ and p + qp =q 1. If T is a bounded operator on L such that
(T f )g = f (T g) for all f, g ∈ L ∩ L , 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 functions
that vanish outside a set of finite measure. For f ∈ Lp ∩ Lq , T f ∈ Lp and is thus measurable.
Moreover, for any g ∈ Σ, clearly g ∈ Lp ∩ Lq , and thus kg(T f )k1 ≤ kgkq kT f kp < ∞ by Holder’s
inequality, meaning that g(T f ) ∈ L1 . Also, by assumption
Z
Mq (T f ) := sup
g(T f ) : g ∈ Σ, kgkp = 1
Z
≤ sup
f (T g) : g ∈ Σ, kgkp = 1
≤ sup{kf (T g)k1 : g ∈ Σ, kgkp = 1}
≤ sup{kf kq kT gkp : g ∈ Σ, kgkp = 1}
(Holder)
≤ sup{kf kq kT kop kgkp : g ∈ Σ, kgkp = 1}
≤ sup{kf kq kT kop : g ∈ Σ, kgkp = 1}
= kT kop kf kq < ∞
We
notice that since f ∈ Lp and T is a bounded operator on Lp , T f ∈ Lp . If p < ∞,
R also
p
|T f | < ∞. Hence {|T f |p 6= 0} = {T f 6= 0} is σ-finite. If p = ∞, µ the background measure is assumed to be semifinite. Now applying theorem 6.14 on Folland we obtain that T f ∈ Lq .
That is, we prove that T f ∈ Lq for any f ∈ Lp ∩ Lq .
Observe that Lp ∩ Lq is dense in Lq since simple functions, which are dense in Lp , are contained
in Lp ∩ Lq . Therefore for g ∈ Lq we can choose {fn }n ⊂ Lp ∩ Lq such that fn → g in Lq . Now
with a little abuse of notation we extend T to Lq by defining
T g = lim T fn
n→∞
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 of
the sequence {fn }.
51
Proof of Claim. Linearity of T easily follows from linearity of T on Lp ∩ Lq and linearity of limit.
Notice that kT fm − T fn kq ≤ Cq kfm − fn kq → 0 as m, n → ∞, so {T fn }n is Cauchy and converges
since Lq is complete. Since the limit is unique, it must be T g. Hence T g ∈ Lq and kT g−T fn kq →
0. This shows that for any fn → g in Lq , T fn → T g in Lq . Then this definition indeed defines a
linear operator on Lq and the definition is independent of the choice of the sequence. (Since for
any such sequence {fn }n the limit in Lq will be T g.
Hence for f + g ∈ Lp + Lq where f ∈ Lp and g ∈ Lq , T (f + g) = T f + T g ∈ Lp + Lq and thus
T is a linear operator on Lp + Lq . Moreover, we showed above that T is of strong type (p, p) (by
assumption) and strong type (q, q) (by claim). By Riesz-Thorin applied with p and q as given in
the 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 0 , h ∈ Lr .
Without loss of generality suppose r ∈ [p, q], h = f + g for some f ∈ Lp and g ∈ Lq . Then
T 0 h = T 0 (f + g) = T 0 f + T 0 g = T f + T 0 g
Choose gn ∈ Lp ∩ Lq such that gn → g ∈ Lq . Then T 0 g = limn→∞ T gn = T g and hence
T 0 h = T f + T g = T h. 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 comRplex numbers. If for each finite set B ⊂ A there is µB ∈ M (X) such that RkµB k ≤ 1 and
fα dµB = cα for α ∈ B, then there is µ ∈ M (X) such that kµk ≤ 1 and fα dµ = 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) a
normed vector space. Then by Alaoglu’s theorem B ∗ := {Iµ ∈ C(X)∗ : kIµ k ≤ 1} is compact
in the
R weak* topology of C(X). We define Mα to be the set of measures µ such that kIµ k ≤ 1
and fα dµ = 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 T
is some
R
∗
µB ∈ B such that kIµBTk ≤ 1 and fα dµB = cα for each α ∈ B, meaning that µB ∈ α∈B Mα
and thus showing that α∈B Mα 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 ) =
−1
cα } = B ∗ ∩ fˆα ({cα )} where fˆα : µ 7→ µ(fα ) is the canonical embedding. Since fˆα is bounded
−1
and thus continuous, fˆα ({cα }) is closed and therefore each Mα is closed as a closed subset of a
compact set. ThenT
{Mα }α∈A is a family of T
closed subsets of B ∗ compact
with finite intersection
R
property and thus α∈A Mα 6= ∅. Let µ ∈ α∈A Mα , kµk ≤ 1 and fα dµ = cα for all α ∈ A, as
desired.
8. Chapter 8-Elements of Fourier Analysis
Folland 8.4
If f ∈ L∞ and kTy f − f k∞ → 0 as y → 0, then f agrees a.e. with a uniformly continuous
function.
4In this proof I may use µ and I
µ interchangably, which is common in this situation.
52
Proof. The main goal of this proof is to show that h(x) := limn→∞ A1/n f (x) (where Ar f (x) =
R
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. Ar f is uniformly continuous for any r > 0.
Proof of Claim. Let r > 0. For our convenience we use gr to denote Ar f , then
Z
Z
1
f (z)dz
f (z)dz −
kτy gr − gr ku = sup |gr (x) − gr (x − y)| = sup
x
x m(B(r, x))
B(r,x−y)
B(r,x)
Z
1
= sup
f (z − y) − f (z)dz
x m(B(r, x)
B(r,x−y)
Z
1
≤ sup
|τy f (z) − f (z)|dz
x m(B(r, x)) B(r,x−y)
Z
1
≤ sup
kτy f − f k∞ dz
x m(B(r, x)) B(r,x−y)
= sup kτy f − f k∞ = kτy f − f k∞
which tends to 0 as y → 0 and thus gr = Ar f is uniformly continuous.
Claim 2. Ar f is uniformly Cauchy as r → 0.
Proof of Claim. Let > 0. Since kτy f − f k∞ → 0 as y → 0, there is some δ > 0 such that when
|y| < δ, kτy f − f k∞ < . Therefore, if r1 , r2 < δ,
Z
Z
1
1
f (y)dy − f (x) ≤
|f (y) − f (x)|dy
m(B(r1 , x)) B(r1 ,x)
m(B(r1 , x)) B(r1 ,x)
Z
1
=
|f (x − (x − y)) − f (x)|dy < m(B(r1 , x)) B(r1 ,x)
Since |x − y| ≤ r1 < δ. Analogously we can prove that
Z
1
f (y)dy − f (x) < m(B(r2 , x)) B(r2 ,x)
and therefore
Z
1
f (y)dy −
f (y)dy
m(B(r2 , x)) B(r2 ,x)
B(r1 ,x)
Z
Z
1
1
≤ sup
f (y)dy − f (x) +
f (y)dy − f (x)
m(B(r1 , x)) B(r1 ,x)
m(B(r2 , x)) B(r2 ,x)
x
< sup 2 = 2
1
kAr1 f − Ar2 f ku = sup
m(B(r1 , x))
x
Z
x
and thus Ar f is uniformly Cauchy as r → 0.
We now prove that h is uniformly continuous. Since {A1/n f }n is uniformly Cauchy by claim 2,
{A1/n f (x)}n is a Cauchy sequence for each x and therefore converges since R is complete. Let
> 0, there is some N ∈ N such that when n, m > N , |A1/n f (x) − A1/m f (x)| < for all x.
Sending m → ∞ gives us |A1/n f (x) − h(x)| < , and since x is arbitrary {A1/n }n must converges
uniformly to h.
Claim 3. h is uniformly continuous.
Proof of Claim. Let > 0. Since {A1/n f }n → h uniformly, there is some N large enough such
that kA1/N f − hku < /3. By claim 1 A1/N f is uniformly continuous, so there is some δ > 0 such
53
that when |y| < δ, |A1/N f (x − y) − A1/N f (x)| < /3 for all x. Now for all x and |y| < ,
|h(x − y) − h(x)| ≤ |h(x − y) − A1/n (x − y)| + |A1/n (x − y) − A1/n (x)| + |A1/n (x) − h(x)|
< + + =
3 3 3
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 is
locally integrable. By theorem 3.18 on Folland, h agrees with f a.e. Since h is uniformly continuous, this finishes the proof.
Folland 8.8
Suppose that f ∈ Lp (R). If there exists h ∈ Lp (R) such that
lim ky −1 (τ−y f − f ) − hkp = 0
y→0
we call h the strong Lp derivative of f . If f ∈ Lp (Rn ), Lp partial derivatives of f are defined similarly. Suppose that p and q are conjugate exponents, f ∈ Lp , g ∈ Lq , and the Lp
derivative ∂j f exists. Then ∂j (f ∗ g) exists (in the ordinary sense) and equals (∂j f ) ∗ g.
Proof. First of all we show that the definition (∂j f ) ∗ g makes sense. Since ∂j f is the Lp derivative, it lies in Lp , and g ∈ Lq by assumtion. It follows from proposition 8.8 that k∂j f ∗ gku ≤
k∂j f kp kgkq < ∞, showing that (∂j f ) ∗ g is well defined. Behold that
Z
f (x − y + hej ) − f (x − y)
−1
− ∂j f (x − y) g(y) dy (∗)
|(h (τ−h (f ∗ g) − f ∗ g) − ∂j f ∗ g)(x)| =
h
f (x+he )−f (x)
j
where ej is the unit vector on the jth component. For our convenience, we denote
h
as Fh (x) and ∂j f (x) as F (x). Note that Fh ∈ Lp for each h since f ∈ Lp , and F = ∂j f ∈ Lp , so
F − Fh ∈ Lp . Now
(∗) = |(Fh − F ) ∗ g(x)| ≤ k(Fh − F ) ∗ gku ≤ kFh − F kp kgkq
and sending h → 0, our assumption limy→0 ky −1 (τ−y f − f ) − hkp = 0 tells us exactly that
kFh − F kp tends to 0 and hence
f ∗ g(x + h) − f ∗ g(x)
− ∂j f ∗ g
h→0
h
0 = lim |(h−1 (τ−h (f ∗ g) − f ∗ g) − ∂j f ∗ g)(x)| = lim
h→0
showing that ∂j (f ∗ g) exists in an ordinary sense and that it equals ∂j f ∗ g, as desired.
Folland 8.9
If f ∈ Lp (R), the Lp derivative of f (which we call h) exists iff f is absolutely continuous on every bounded interval (perhaps after modification on a null set) and its pointwise
derivatice f 0 is in Lp , in which case h = f 0 a.e.
Proof. Only If. Let [a, b] be a bounded interval since addition or deletion of one single point
doesn’t affect the result Rof integration. It is well known that we can construct a g ∈ Cc∞ such
that 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
1 + max(|a|, |b|) 2
−2
C(1 + |x|) = M
≥ M ≥ |g(x)|
1 + |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 )0 = 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 theorem
8.14 a), h ∗ gt → h in Lp [a, b], and in particular {kn } is bounded in Lp [a, b]. Since [a, b] has finite
measure b − a, by proposition 6.12 we have
kkn χ[a,b] k1 ≤ kkn χ[a,b] kp (b − a)1−1/p
and hence {kn } is bounded in L1 [a, b]. Notice that
Z b
Z x
Z x
0
kn χ[a,x]
h ∗ g1/n
=
(f ∗ g1/n )0 =
f ∗ g1/n (x) − f ∗ g1/n (a) =
a
a
a
and sending n to infinity and apply dominated convergence theorem (since |kn χ[a,x] | ≤ |kn | is
bounded in L1 [a, b]) we obtain
Z
x
f (x) − f (b) =
h(t)dt
b
showing that f is abosolutely continuous on [a, b] with a possible modification on a null set and
that f 0 = h a.e.
If. For y > 0, notice that
Z
1 y 0
f (x + y) − f (x)
[f (x + t) − f 0 (x)]dt
− f 0 (x) =
y
y 0
Therefore
Z y
p 1
p
f (x + y) − f (x)
1 0
0
0
− f (x) =
[f (x + t) − f (x)]dt dx
y
0 y
p
1
Z y Z
p
f 0 (x + t) − f 0 (x) p
dx dt
≤
y
0
1
Z y Z
p
1
0
0
p
≤
|f (x + t) − f (x)| dx dt
y 0
Z
1 y
=
kτ−t f 0 − f 0 kp dt
y 0
where the first inequality is obtained by Minkowski’s inequality for integrals. Since kτ−t f 0 kp =
kf 0 kp , by triangular inequality we have kτ−t f 0 − f 0 kp χ[0,y] ≤ 2kf 0 kp χ[0,y] ∈ L1 for all y > 0 since
f 0 ∈ Lp . Moreover, by proposition 8.5,
lim kτ−t f 0 − f 0 kp = 0
t→0
55
(1)
so applying dominated convergence theorem, we get
Z
f (x + y) − f (x)
1 y
0
kτ−t f 0 − f 0 kp dt
lim
− f (x) = lim
+ y
y
y→0+
y→0
0
p
Z
1 y
=
lim kτ−t f 0 − f 0 kp dt
y 0 y→0+
(2)
Let > 0. By (1) we know that there is some δ > 0 such that when |t| < δ, kτ−t f 0 − f 0 kp < .
Thus when y < δ,
Z
y
dt = (2) < y
0
showing that
lim
y→0+
f (x + y) − f (x)
− f 0 (x) = 0
y
p
The case where y < 0 is the same is we replace every [0, y] with [y, 0]. It follows that
lim
y→0
f (x + y) − f (x)
− f 0 (x) = 0
y
p
meaning that the Lp derivative of f exists and equals to its usual derivative.
Folland 8.14
(Wirtinger’s Inequality) If f ∈ C 1 ([a, b]) and f (a) = f (b) = 0, then
Z
Z b
b−a 2 b 0
2
|f (x)|2 dx
|f (x)| dx ≤
π
a
a
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 = 21 , i.e. for f ∈ C 1 ([0, 21 ]) and f (0) =
f ( 12 ) = 0,
2 Z 1/2
Z 1/2
1
2
|f (x)| dx ≤
|f 0 (x)|2 dx (1)
2π
0
0
Now, given an f ∈ C 1 ([a, b]) with f (a) = f (b) = 0, we consider g(x) := f (2(b − a)x + a). Since
2(b − a)x + a ∈ [a, b], x ∈ [0, 21 ], meaning that g is defined on [0, 12 ]. Also since 2(b − a)x + a is
clearly a C 1 function g is still C 1 as f . Thus g ∈ C 1 ([0, 21 ]) and g(0) = g( 21 ) = 0. Applying our
assumption (1) we get
2 Z 1/2
Z 1/2
1
2
|g 0 (x)|2 dx
|g(x)| dx ≤
2π
0
0
Notice that by change of variable we have
Z 1/2
Z b
1
LHS =
|f (2(b − a)x + a)|2 dx =
|f (x)|2 dx
2(b
−
a)
0
a
and since g 0 (x) = [f (2(b − a)x + a)]0 = 2(b − a)f 0 (2(b − a)x + a), we also have
2 Z 1/2
1
RHS =
[2(b − a)]2 f 0 (2(b − a)x + a)dx
2π
0
2
Z b
1
1
2
=
[2(b − a)]
|f 0 (x)|2 dx
2π
2(b − a) a
56
and putting them together we get
Z
Z b
b−a 2 b 0
|f (x)|2 dx
|f (x)|2 dx ≤
π
a
a
as desired.
Step 2. Now that we have assume without loss of generality that f is defined on [0, 12 ], we extend f to [− 21 , 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 ∈ C 1 (T).
We use f 0 ( 21 ) to denote the left one-sided derivative of f at 21 , and use f 0 (− 21 ) to denote the right
one-sided derivatice of f at − 12 . For h > 0, we have
−f (− 12 ) + f (− 12 + h)
f ( 12 ) − f ( 12 − h)
0 1
0 1
−f
=
−f
h
2
h
2
and thus
f ( 21 ) − f ( 12 − h)
f (− 12 + h) − f (− 12 )
0 1
0 1
0 = lim
−f
= lim
−f
h→0
h→0
h
2
h
2
showing that f 0 (− 21 ) = f 0 ( 12 ), and we denote this common quantity by L. Since f is C 1 ,
lim f 0 (x) = L =
x→1/2−
lim
x→−1/2+
f 0 (x)
2n+1
and the way we extend f makes sure that this result holds for [ 2n−1
2 , 2 ] for every n ∈ Z. It follows that f is C 1 at the endpoint of T. f is C 1 in the interior of T by assumption, so f ∈ C 1 (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 0 ∈ L2 (T).
By Parseval’s identity, kf k2 = kfˆk2 and kf 0 k2 = kfˆ0 k2 . Thus
Z 1/2
Z
X
2
|f (x)| dx =
|f (x)|2 dx = kfˆk22 =
|fˆ(k)|2
0
T
k∈Z
and using integration by parts we have
Z
Z 1/2
−2πikx
ˆ
f (k) =
f (x)e
dx =
f (x)e−2πikx
0
T
1/2
Z 1/2
1
1 0
−2πikx
f (x)e
+
f (x)e−2πik dx
=
2πik
2πik
0
0
Z 1/2
1
1 ˆ0
=0+
f 0 (x)e−2πikx =
f (k)
2πik 0
2πik
1
implying |fˆ(k)| = | 2πk
||fˆ0 (k)| and thus
Z 1/2
X
X 1 2
X 1 2
2
2
2
2
0
ˆ
ˆ
ˆ
|f (x)| dx = kf k2 = kf k2 =
|f (k)| ≤
|f (k)| ≤
|fˆ0 (k)|2
2πk
2π
0
k∈Z
k∈Z
k∈Z
2
2 Z 1/2
2
1
1
1
=
kfˆ0 k22 =
kf 0 k22 =
|f 0 (x)|2 dx
2π
2π
2π
0
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
Γ( 1 (n + 1))
φ(x) = (n+1)/2 2
π
(1 + |x|2 )−(n+1)/2
R
∞
(a) If β > 0, e−β = π −1 −∞ (1 + t2 )−1 e−iβt dt.
R
2
∞
(b) If β ≥ 0, e−β = 0 (πs)−1/2 e−s e−β /4s ds.
(c) Let β = 2π|ξ| where ξ ∈ Rn ; the the formula in (b) expresses e−2π|ξ| as a superposition of dilated Gauss kernels. Use proposition 8.24 again to derive the asserted
formula for φ.
Proof.
1
(a) By (8.37), we have φ(x) = π(1+x
2 ) and therefore
Z
e−2πiξt
−2π|ξ|
dt (1)
e
= Φ(ξ) = φ̂(ξ) =
2
R π(1 + t )
If β > 0, β = 2πξ for some ξ > 0, and then plugging in (1) we get
Z
Z ∞
e−iβt
−1
e−β =
dt
=
π
(1 + t2 )−1 e−iβt dt
2)
π(1
+
t
R
−∞
as desired.
(b) Notice that
Z ∞
Z
Z
1
1 ∞ −iβt
1
−(1+t2 )s
−iβt
e
e
ds
dt
e
=
e−β =
π
1 + t2
π 0
0
Z Z
1 ∞ ∞ −(1+t2 )s −iβt
=
e
e
dtds (1)
π 0
−∞
2
where the last step holds since e−(1+t )s e−iβt ∈ L+ and Tonelli’s theorem justifies the
interchange of order of integrals. Now substituting z = βt/2π, we obtain
Z
Z
Z
Z
1 ∞ −s ∞ 2π −s 4πβ22z2 −2πiz
2 ∞ −s ∞ 4πβ22 s z 2 −2πiz
e
dzds
(1) =
e
e
e
dtds =
e
e
π 0
β 0
−∞ β
−∞
−1/2
Z
Z ∞
2
β2
2 ∞ −s
4πs
2
− 4π2 s z 2
−s
=
e F(e β
)(1)ds =
e
e−π 4πs ds (Prop 8.24)
2
β 0
β 0
β
2 1/2
Z ∞
Z
∞
β2
2
β
2
=
e−s
e− 4s =
(πs)−1/2 e−s e−β /4s ds
β 0
4πs
0
showing the result as desired.
(c) Since β = 2π|ξ|, now
Z ∞
Z ∞
4π 2 |ξ|2
π 2 |ξ|2
−β
−2π|ξ|
−1/2 −s − 4s
e =e
=
(πs)
e e
ds =
(πs)−1/2 e−s e− s ds
0
0
And computing the inverse Fourier transform we have
Z
Z Z ∞
π 2 |ξ|2
−2π|ξ| ∨
−2π|ξ| 2πiξ·x
(e
) (x) =
e
e
dξ =
(πs)−1/2 e−s e− s e2πiξ·x dsdξ
Rn
Rn 0
58
(2)
Note that |(πs)−1/2 e−s e−
Z
π 2 |ξ|2
s
e2πiξ·x | ∈ L+ , so by Tonelli’s theorem we have
|(πs)−1/2 e−s e−
Rn ×[0,∞)
Z ∞Z
|(πs)−1/2 e−s e−
=
π 2 |ξ|2
s
π 2 |ξ|2
s
e2πiξ·x |ds ⊗ ξ
e2πiξ·x |dξds
Rn
0
Z ∞Z
π 2 |ξ|2
|(πs)−1/2 e−s e− s |dξds
n
Z
Z0 ∞ R
π 2 |ξ|2
−1/2 −s
− s
=
|(πs)
e |
|e
|dξ ds
Rn
Z0 ∞
s
=
|(πs)−1/2 e−s |(π · 2 )2/n ds
π
0
n/2
Z ∞
s
ds (since s > 0)
|πs|−1/2 |e−s |
=
π
0
Z ∞
n+1
1+n
n+1
− 1+n
2
=π
s 2 −1 e−s ds = π − 2 Γ(
)<∞
2
0
=
π 2 |ξ|2
where the last step is by proposition 2.55. Then (πs)−1/2 e−s e− s e2πiξ·x ∈ L1 , and we
can apply Fubini-Tonelli to interchange the order of integral in (2). Thus we get
Z
Z ∞
π 2 |ξ|2
(2) =
(πs)−1/2 e−s
e− s e2πiξẋ dξ ds
Rn
Z0 ∞
π 2 |ξ|2
=
(πs)−1/2 e−s (e− s )∨ (x)ds
0
Z ∞
=
0
=
1
e
n/2
s
2
e−s|x| ds
π
n−1
2
−1/2 −s
(πs)
Z ∞
(Prop 8.24)
s 2 e−s(1+|x| ) ds
(3)
0
π
Substituting z = −s(1 + |x|2 ), we get
n+1 Z ∞
2
Γ( 21 (n + 1))
n+1
1
1
−1 −s
2
(3) = n+1
z
e
dz
=
1 + |x|2
π (n+1)/2 (1 + |x|2 )−(n+1)/2
0
π 2
as desired. This finishes the proof.
n+1
2
Folland 8.30
If f ∈ L1 (Rn ), f is continuous at 0, and fˆ ≥ 0, then fˆ ∈ L1 .
59
Proof. Observe that
kfˆk1 =
Z
|fˆ(ξ)|dξ =
Z
Z
fˆ(ξ)dξ
2
lim fˆ(ξ)e−π|tξ| e2πiξ·0 dξ
Z
2
≤ lim fˆ(ξ)e−π|tξ| e2πiξ·0 dξ
=
t→0
t→0
(Fatou0 s Lemma)
= f (0) < ∞
The last equality holds by Theorem 8.35 and the fact that Gauss kernel fits the theorem; the last
inequality 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 distribution
derivative f 0 ∈ D0 (U ) coincides with the pointwise (a.e.-defined) derivative of f .
Proof. Let f 0 be the distribution derivative and g pointwise a.e.-defined derivative. For any φ ∈
Cc∞ (U ), Suppφ = K for some K compact and thus f is absolutely continuous on K. It follows
that
Z
Z
Z
Z
Z
φf 0 = −
f φ0 = −
f dφ = φdf =
φg (1)
K
K
K
K
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 + n1 , b − n1 ]. Taking
φ = χKn in (1) we obtain f 0 = g a.e. on Kn . We may suppose En ⊂
Kn is the set on which f 0
S∞
0
and g don’t
S∞ agree, and hence En is a null set for0 each n. Since U = 1 Kn , f and g disagree on
at most 1 En which is still a null set. Thus f and g agree a.e. on U , as desired.
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