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Math 426: Homework 5 Solutions
Mary Radcliffe
due 14 May 2014
In Bartle:
6H. Let X = Z+, and let λ be the measure on X which has measure 1n2 at the
point n. Show that λ(X) < ∞. Let f be defined on X by f(n) = √n.
Show that f ∈ Lp if and only if 1 ≤ p < 2.
Solution. We have λ(X) =
∑∞
n=1 λ(n) =
pi2
6 , which is finite. Moreover,∫
|f |p dλ =
∞∑
n=1
|f(n)|pλ(n)
=
∞∑
n=1
np/2n−2
=
∞∑
n=1
np/2−2,
which converges if and only if p/2− 2 < −1, if and only if p < 2. But as
p is assumed to be at least 1, since otherwise we do not have a norm, we
have that the integral is finite if and only if 1 ≤ p < 2.
6I. Modify the previous exercise to obtain a function on a finite measure space
which belongs to Lp if and only if 1 ≤ p < p0.
Solution. Let X = Z+, and let λ be the measure on X with λ(n) = n−p0 .
Note that as p0 > 1, we have that λ(X) =
∑∞
n=1 n
−p0 <∞.
Take f(n) = n1−1/p0 . Then we have∫
|f |p dλ =
∞∑
n=1
|f(n)|pλ(n)
=
∞∑
n=1
np−p/p0n−p0
=
∞∑
n=1
np((p0−1)/p0)−p0 ,
which converges if and only if p((p0 − 1)/p0) − p0 < −1, if and only if
p < (p0 − 1)(p0/(p0 − 1)) = p0, as desired.
6N. Let (X,F , µ) be a measure space, and let f belong to both Lp1 and Lp2 ,
with 1 ≤ p1 < p2 < ∞. Prove that f ∈ Lp for any value of p such that
p1 ≤ p ≤ p2.
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Solution. Define E1 = {x ∈ X | |f(x)| < 1} and E2 = {{x ∈ X | |f(x)| ≥
1}. Note that if x ∈ E1, then |f(x)|p ≤ |f(x)|p1 , and if x ∈ E2, then
|f(x)|p ≤ |f(x)|p2 . Therefore we have∫
|f |p dµ =
∫
E1
|f |p dµ+
∫
E2
|f |p dµ
≤
∫
E1
|f |p1 dµ+
∫
E2
|f |p2 dµ
≤
∫
|f |p1 dµ+
∫
|f |p2 dµ <∞
by hypothesis. Therefore, f ∈ Lp.
6P. Let f ∈ Lp(X,F , µ), 1 ≤ p < ∞, and let � > 0. Show that there exists
a set E� ∈ F with µ(E�) < ∞ such that if F ∈ F and F ∩ E� = ∅, then
‖fχF ‖p < �.
Solution. For n ∈ Z+, define Gn = {x ∈ X | |f(x)|p > 1n}. Then |f(x)|p ≥
1
nχGn , and thus ∞ >
∫ |f |p dµ ≥ ∫ 1nχGn dµ = 1nµ(Gn), so µ(Gn) < ∞
for all n.
Now, |f |pχGn is an increasing sequence of functions with limit |f |p, and
thus by the MCT, we have that
∫ |f |pχGn dµ → ∫ |f |p dµ. Let N ∈ Z+
such that
∣∣∫ |f |pχGN dµ− ∫ |f |p dµ∣∣ < �p. Then as ∫ |f |pdµ = ∫ |f |pχGN dµ+∫ |f |pχX\GN dµ, we have that ∫ |f |pχX\GN dµ < �p. Let GN = E�.
Now, let F ⊂ F with F ∩ E� = ∅, so F ⊂ X\GN . Then
‖fχF ‖pp =
∫
|f |pχF dµ ≤
∫
|f |pχX\GN dµ < �p,
and the desired result follows.
6Q. Let fn ∈ Lp(X,F , µ), 1 ≤ p < ∞, and let βn be defined for E ∈ F by
βn(E) =
(∫
E
|fn|p dµ
)1/p
. Show that |βn(E) − βm(E)| ≤ ‖fn − fm‖p.
Hence, if {fn} is a Cauchy sequence in Lp, then limβn(E) exists for each
E ∈ F .
Solution. Note that βn(E) = ‖fnχE‖p. Moreover, by the triangle inequal-
ity, we have that for any two functions f and g, ‖f‖p ≤ ‖f−g‖p+‖g‖p, and
thus ‖f‖p − ‖g‖p ≤ ‖f − g‖p. Wolog, assume that ‖fnχE‖p > ‖fmχE‖p.
Then
|βn(E)− βm(E)| = |‖fnχE‖p − ‖fmχE‖p|
= ‖fnχE‖p − ‖fmχE‖p
≤ ‖(fn − fm)χE‖p
≤ ‖fn − fm‖p,
as desired.
6R. Let fn, βn be as in Exercise 6Q. If {fn} is a Cauchy sequence and � > 0,
then there exists a set E� ∈ F with µ(E�) < ∞ such that if F ∈ F and
F ∩ E� = ∅, then βn(F ) < � for all n ∈ Z+.
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Solution. By problem 6P, such a set exists for each fn; for all � > 0, let
E
(n)
� be such that if F ∩ E(n)� = ∅, then ‖fnχE(n)� ‖p < �.
Now, fix � > 0. Let N be sufficiently large that if n,m ≥ N , then ‖fn −
fm‖p < �2 . Put E� =
N⋃
k=1
E
(k)
�/2.
Let F ∈ F with F ∩ E� = ∅. If n ≤ N , then F ⊂ Ec�/2, so ‖fnχF ‖p =
βn(F ) < �/2. If n > N , then |βn(F ) − βN (F )| ≤ ‖fn − fN‖p < �2 , and
thus βn(F ) <
�
2 + βN (F ) < �.
6T. If f ∈ L∞(X,F , µ), then |f(x)| ≤ ‖f‖∞ for almost all x. Moreover,
if A < ‖f‖∞, then there exists a set E ∈ F with µ(E) > 0 such that
|f(x)| > A for all x ∈ E.
Solution. For all � > 0, there exists a set N� such that µ(N�) = 0 and
S(N�) < ‖f‖∞ + �. Take N = ∪∞k=1N1/k. Then µ(N) = 0 by countable
additivity, and if x /∈ N , then x /∈ N1/k for all k. Thus, |f(x)| ≤ S(N1/k)
for all k, and thus |f(x)| < ‖f‖∞ + 1/k for all k, so |f(x)| ≤ ‖f‖∞ for all
x /∈ N . Therefore, |f(x)| ≤ ‖f‖∞ for almost all x.
Suppose there exists some A < ‖f‖∞ such that there is no E ∈ F with
µ(E) > 0 and |f(x)| > A for all x ∈ E. Let N = {x ∈ X | |f(x)| >
A}. Then by hypothesis, µ(N) = 0, and thus ‖f‖∞ ≤ S(N) ≤ A, a
contradiction. Thus, the result holds.
Additional Exercises:
1. Let L(Rn,Rm) denote the space of linear transformations from Rn to Rm.
Recall that in 425, we defined the norm of a transformation as follows:
‖A‖ = sup
‖x‖2=1
‖Ax‖2 (see Baby Rudin, page 208). We proved that this
satisfies the properties of a norm in Theorem 9.7 (again, page 208.) This
is usually called the Euclidean norm, the `2 norm, or the Frobenius norm
for matrices, and is more properly denoted by ‖A‖2.
(a) Let V be a vector space with norm ‖ · ‖, and let L(V ) be the space
of linear transformations from V to itself. Define N : L(V ) → R by
N(A) = sup
‖v‖=1
‖Av‖. Show that N defines a norm on V . (Usually
we denote this as N(A) = |||A|||). Thus we can define a p-norm for
operators as well.
(b) Show that if |||A||| is defined from a vector norm on V as in part (a),
then we have the inequality |||AB||| ≤ |||A||||||B||| for all A,B ∈ L(V ).
This property is usually called submultiplicativity.
(c) Let A be an n × n real matrix. Define |||A|||p (sometimes we write
just ‖A‖p) to be the matrix norm derived from the Lp norm on Rn
as in part (a). This is usually called the p-norm on linear operators.
Determine an expression for |||A|||1 and |||A|||∞ in terms of only A.
Solution. (a) Clearly, N(A) ≥ 0 for all A ∈ L(V ), and N(0) = 0. Now,
suppose that A 6= 0. Then there exists some x ∈ V such that Ax 6= 0,
so sup
‖v‖=1
‖Av‖ ≥ ‖Ax‖/‖x‖ > 0. Thus, N(A) = 0 if and only if A = 0.
Next, if α ∈ R, we have N(αA) = sup
‖v‖=1
‖αAv‖ = sup
‖v‖=1
|α|‖Av‖ =
|α|N(A), since ‖ · ‖ is a norm on V .
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Finally, let A,B ∈ L(V ). Then we have
N(A+B) = sup
‖v‖=1
‖(A+B)v‖
≤ sup
‖v‖=1
(‖Av‖+ ‖Bv‖)
≤ sup
‖v‖=1
‖Av‖+ sup
‖v‖=1
‖Bv‖ = N(A) +N(B).
Therefore, N is a norm on L(V ).
(b) Consider |||AB||| = sup
‖v‖=1
‖(AB)v‖. Let y = Bv. Then we have
|||AB||| = sup
‖v‖=1
‖(AB)v‖
= sup
‖v‖=1
‖Ay‖‖y‖‖y‖
= sup
‖v‖=1
‖Ay‖
‖y‖ ‖Bv‖
≤ sup
y∈V
‖Ay‖
‖y‖ sup‖v‖=1
‖Bv‖
= sup
‖y‖=1
‖Ay‖ sup
‖v‖=1
‖Bv‖ = |||A||||||B|||.
(c) We first consider |||A|||1. Note that if v ∈ Rn has ‖v‖1 = 1, then we
have
∑ |vi| = 1. Thus, if A1, A2, . . . , An are the columns of A, we
have ‖Av‖ = ‖v1A1 + · · ·+ vnAn‖ ≤
∑ |vi|‖Ai‖1 ≤ max1≤i≤n ‖Ai‖1.
Moreover, if i attains the maximum, taking v = ei yields that ‖Aei‖ =
max1≤i≤n ‖Ai‖1. Therefore, the 1-norm of A is the maximal 1-norm
of the columns of A.
On the other hand, if ‖v‖∞ = 1, then max |vi| = 1. Moreover, if
a1, a2, . . . , an are the rows of A, then Av = [a1 · v, a2 · v, . . . , an · v]T ,
so ‖Av‖∞ = max1≤i≤n |ai · v|. If S =
∑n
j=1 |aij | is the maximal row
sum, then, it is clear that ‖Av‖∞ ≤ S. Moreover, taking v to be a
vector of all ±1 such that aijvj > 0 for all j, we have ‖Av‖∞ = S.
Therefore, |||A|||∞ is the maximal 1-norm of the rows of A.
2. Let (X,F , µ) be a measure space.
(a) Prove that the simple functions with finite support are dense in
Lp(X) if 1 ≤ p < ∞. That is, for all f ∈ Lp and �> 0, there
exists a simple function φ with finite support and ‖f − φ‖p < �.
(b) Prove that the simple functions with finite support are dense in L∞
if X is of finite measure. Give an example of an L∞ function that
cannot be approximated by simple functions with finite support if
µ(X) =∞.
Solution. (a) Let f ∈ Lp, � > 0. Let En = {x ∈ X | |f(x)| ≥ 1/n}. Note
that as |f |pχEn increases to |f |p, by the MCT there exists n such
that
∫ |f |p dµ− ∫ |f |pχEn dµ < ( �2)p. Let g = |f |χEn , so g has finite
support and g ∈ Lp.
By Theorem 2.11, there exists a simple function φ such that |g−φ| <
�
2(µ(En))1/p
for all x, and φ = 0 whenever g = 0, so φ has finite
support. Then
∫ |g − φ|p dµ ≤ ∫
En
(
�
2(µ(En))1/p
)p
=
(
�
2
)p
.
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Therefore, we have
‖f − φ‖p =
(∫
|f − φ|p dµ
)1/p
=
(∫
En
|f − φ|p dµ+
∫
Ecn
∫
|f − φ|p dµ
)1/p
≤
(∫
En
|g − φ|p dµ+
( �
2
)p)1/p
≤
(
2
( �
2
)p)1/p
= 21/p
�
2
< �,
as desired.
(b) The first part is immediate by applying Theorem 2.11 to f . For the
second part, if f(x) = 1, where f ∈ L∞(R), it is clear that if g is
any finitely supported function, then ‖f − g‖∞ ≥ 1, so f cannot be
approximated by finitely supported simple functions.
3. Consider the real space under Lebesgue measure, (R,B, λ). Recall that
C(R) is the family of continuous functions on R. Define Cc(R) to be the
family of continuous, compactly supported functions on R.
(a) Show that for 1 ≤ p < ∞, Cc(R) is dense in Lp(R). (Hint: Since
we know the simple functions are dense by problem 2, it suffices
to approximate simple functions by compactly supported continuous
functions.)
(b) Does this result hold in L∞? Explain.
Solution. (a) First, we claim that if λ(E) < ∞, then for all � > 0 there
exists a continuous function f� such that ‖f − χE‖p < �.
Let En = E ∩ [−n, n]. As En ⊂ En+1 and ∪En = E, we have that
λ(En) → λ(E). Choose N sufficiently large that λ(En) ≥ λ(E) −(
�
4
)p
.
Let U be an open set such that En ⊂ U and λ(U\En) ≤
(
�
4
)p
. By
possibly intersecting with the open set (−n− δ, n+ δ), we can ensure
that U is contained in a compact set. Note, moreover, that as U
is open U can be written as an (at most) countable union of open
intervals. Write U = ∪Ik, where Ik = (ak, bk).
Define fk as follows. Fix some δk <
�p(p+1)
2kp+2
. Then put
fk(x) =

1 ak ≤ x ≤ bk
1− 1δk (x− bk) bk < x ≤ bk + δk
1 + 1δk (x− ak) ak − δk ≤ x < ak
0 else
.
Then we have fk is supported on (ak − δk, bk + δk), fk is continuous,
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and
‖χIk − fk‖pp =
∫
|χIk−fk |p dλ
=
∫ ak
ak−δk
∣∣∣∣1 + 1δk (x− ak)
∣∣∣∣p dx+ ∫ bk
ak
0pdx+
∫ bk+δk
bk
∣∣∣∣1− 1δk (x− bk)
∣∣∣∣p dx
=
∫ 1
0
δku
pdu+
∫ 0
1
−δkupdu
= 2
δk
p+ 1
< 2
�p(p+1)
2kp+2
p+ 1
=
( �
2k+1
)p
Let f =
∑
k fk. Then f is continuous, and we have
‖f − χU‖p = ‖
∑
k
(f − χIk)‖p
≤
∑
k
‖f − χIk‖p
<
∑
k
�
2k+1
= �/2
Therefore, we have
‖χE − f‖p ≤ ‖χE − χEn‖p + ‖χEn − χU‖p + ‖χU − f‖p
≤ µ(E\En)1/p + µ(U\En)1/p + �
2
≤ �
4
+
�
4
+
�
2
< �.
Now, let φ be an arbitrary finitely supported simple function. Then
we may write φ =
∑n
k=1 akχEk , where λ(Ek) <∞ and ak 6= 0 for all
k. Then put fk to be a continuous function such that ‖χEk − fk‖p ≤
�
|ak| . Taking f =
∑
akfk yields a continuous function satisfying the
desired property by the triangle inequality.
(b) No. We have the same problem with compact (aka finite) support
here as in 2b.
Extras:
6C. Let N be a norm on a linear space V and let d be defined for u, v ∈ V by
d(u, v) = N(u− v). Show that d is a metric on V .
Solution. Clearly d(u, v) ≥ 0 for all u, v, and d(u, v) = 0 iff u = v. Also,
d(v, u) = N(v−u) = |−1|N(u−v) = N(u−v) for all u, v, and for all u, v, w,
we have d(u,w) = N(u−w) = N(u− v + v −w) ≤ N(u− v) +N(v −w)
by the triangle inequality. Thus, d is a metric.
6D. Whoops, this is a part of additional exercise 2.
6E. If f ∈ Lp, 1 ≤ p <∞, and if E = {x ∈ X | |f(x)| 6= 0}, then E is σ-finite.
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Solution. Define En = {x ∈ X | |f(x)|p ≥ 1n}. Notice then that ∪En = E,
and as f ∈ Lp, we have ∞ > ∫ |f |p dµ ≥ ∫
En
|f |p dµ ≥ 1nµ(En), and thus
µ(En) <∞ for all n. Therefore, E is σ-finite.
6F. If f ∈ Lp and if En = {x ∈ X | |f(x)| ≥ n}, then µ(En)→ 0 as n→∞.
Solution. Note that En ⊂ En+1, so µ(En) ≤ µ(En+1). By the same logic
as problem 6E, it is clear that µ(En) is finite, and thus {µ(En)} is a
bounded monotonic sequence in R, and it thus has a limit. Suppose the
limit is not 0, so that µ(En)→ C > 0. Then for all n, we have
∫ |f |p dµ ≥∫
En
|f |p dµ ≥ Cnp, and thus the integral is infinite, a contradiction.
6K. If (X,F , µ) is a finite measure space and f ∈ Lp, then f ∈ Lr for 1 ≤ r ≤ p.
Apply Ho¨lder’s Inequality to |f |r in Lp/r and g = 1 to obtain the inequality
‖f‖r ≤ ‖f‖pµ(X)s, where s = 1/r − 1/p.
Solution. Let E1 = {x ∈ X | |f(x)| < 1}. Then if x ∈ Ec1, |f(x)|r ≤
|f(x)|p. Therefore,∫
|f(x)|r dµ ≤
∫
E1
|f(x)|r dµ+
∫
Ec1
|f(x)|r
≤
∫
E1
1 dµ+
∫
Ec1
|f(x)|p dµ
< µ(E1) + ‖f‖pp <∞.
Thus, f ∈ Lr.
Now, since |f |r ∈ Lp/r and 1 ∈ Lp/(p−r), which are conjugate indices,
Ho¨lder’s Inequality yields
‖|f |r‖1 ≤ ‖|f |r‖p/r‖1‖p/(p−r).
Note that for any x, y, we have ‖|f |x‖y = ‖f‖xxy, and thus we may rewrite
the above inequality as
‖f‖rr ≤ ‖f‖rpµ(X)(p−r)/p.
Taking 1/r power on both sides yields the given inequality.
6L. Suppose that X = Z+ and µ is the counting measure on Z+. If f ∈ Lp,
then f ∈ Ls for 1 ≤ p ≤ s <∞, and ‖f‖s ≤ ‖f‖p.
Solution. If f ∈ Lp, then we have
∑ |f(n)|p < ∞. Then by comparison
test, for n sufficiently large |f(n)| < 1, and we have |f(n)|s < |f(n)|p, and
thus f ∈ Ls and ‖f‖s =
∑ |f(n)|s ≤∑ |f(n)|p = ‖f‖p.
6U. If f ∈ Lp, 1 ≤ p ≤ ∞, and g ∈ L∞, then the product fg ∈ Lp and
‖fg‖p ≤ ‖f‖p‖g‖∞.
Solution. Let M = ‖g‖∞. If there exists N ∈ F with µ(N) = 0 and
g(x) > M on N , replace g with a function h that takes value 0 on N
and agrees with g elsewhere, so g = h a.e. and h is bounded. Then∫ |fg|p dµ = ∫ |fh|p dµ ≤ ∫ Mp|f |p dµ. The result follows.
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