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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. 1 https://www.coursehero.com/file/9770580/Homework-5-Spring-2014-Solutions/ Th is s tud y r eso urc e w as sha red vi a C ou rse He ro. com 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+. 2 https://www.coursehero.com/file/9770580/Homework-5-Spring-2014-Solutions/ Th is s tud y r eso urc e w as sha red vi a C ou rse He ro. com 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 . 3 https://www.coursehero.com/file/9770580/Homework-5-Spring-2014-Solutions/ Th is s tud y r eso urc e w as sha red vi a C ou rse He ro. com 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 . 4 https://www.coursehero.com/file/9770580/Homework-5-Spring-2014-Solutions/ Th is s tud y r eso urc e w as sha red vi a C ou rse He ro. com 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, 5 https://www.coursehero.com/file/9770580/Homework-5-Spring-2014-Solutions/ Th is s tud y r eso urc e w as sha red vi a C ou rse He ro. com 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. 6 https://www.coursehero.com/file/9770580/Homework-5-Spring-2014-Solutions/ Th is s tud y r eso urc e w as sha red vi a C ou rse He ro. com 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. 7 https://www.coursehero.com/file/9770580/Homework-5-Spring-2014-Solutions/ Th is s tud y r eso urc e w as sha red vi a C ou rse He ro. com Powered by TCPDF (www.tcpdf.org)
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