Functional Analysis, Stein-Shakarchi Chapter 1

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1 Functional Analysis, Stein-Shakarchi Chapter 1 L p spaces and Banach Spaces Yung-Hsiang Huang Abstract Many problems are cited to my solution files for Folland [4] and Rudin [6] post here. 1 Exercises 1. The details and generalizations are discussed in my solution files for Folland [4, Exercise 6.6] and Rudin [6, Exercise ].. Proof. (a) follows from the hint by noting the concavity of x p on (0, 1), 0 < p < 1. (b) Assume l 0, then there exists f L p such that l(f) 1. Let g 1 = fχ ( inf,x) such that g 1 p p = 1 f p p, g = f g 1, then l(g i ) 1 for some i. Let f 1 = g i, then f 1 p = 1 1/p f p and l(f 1 ) 1. Repeat the previous process inductively, we have f n with f n p = 1 1/p f n 1 p = = n(1 1/p) f p 0 and 1 l(f n ) 0 by the continuity of l, which is a contradiction. 3. Track back the condition for the equality in (), page 4. See also Folland [4, Exercise 6.1]. 4. Proof. (a) If g = 0 a.e., then we are done. If not, then apply Hölder s inequality to f p = (fg) p g p (b) Try to mimic the proof for p 1 and use the inequality in Hint of Exercise (a). (c) The triangle inequality is proved by using the same inequality we used in (b). 5. Proof. For each k N, there exists n k such that f nk f p k. We may assume n k increasing and note that for each k, f nk f nk+1 p f nk f p + f nk+1 f p 1 k. Define h(x) = k=1 f n k f nk 1, then by the proof in completeness of L p, it s finite a.e. and the partial sum f nk converges to h a.e. and in L p. So h = f and f nk converges to f a.e. Department of Math., National Taiwan University. d041001@ntu.edu.tw 1

2 6. As Hints, (b) is proved by apply LDCT. 7. Proof. (Sketch) (a) Cut-off the domain, approximation by simple function, using outer and inner regularity of Lebesgue measure, and then use Urysohn s lemma. functions you construct in (a). See Folland [4, Proposition 7.9 and 8.17] (b) Mollify the C c 8. As Hints. For p = consider χ [0,1] d. 9. Suppose is a measure space and 1 p 0 < p 1. (a) Consider L p 0 L p 1 equipped with f L p 0 L p 1 = f L p 0 + f L p 1. Show that (L p 0 L p 1, L p 0 L p 1 ) is a Banach space. (b) Show that (L p 0 + L p 1, L p 0 +L p 1 ) is a Banach space, where f L p 0 +L p 1 := inf{ f 0 L p 0 + f 1 L p 1 : f = f 0 + f 1 L p 0 + L p 1 }. (c) Show that L p L p 0 + L p 1 if p 0 p p 1. Remark 1. Also see Exercise 4 (as an example of Orlicz space) and Folland [4, Exercise and papers cited there]. Proof. (c) f = fχ { x 1} + fχ { x <1}. (a) it s easy to see this is a norm on the vector space. you can check the completeness through the one of L p. (b) Scaling property is trivial. If f = 0, then for every n N, there are f n 0 L p 0 and f n 1 L p 1 with the L p norm less than 1 n f n k 0 0 a.e. and a further subsequence f n k j 1 0 a.e. So does f. respectively. Then there is a subsequence such that Given f, g L p 0 + L p 1, for each n N, there exist decompositions f = f n 0 + f n 1, g = g n 0 + g n 1 such that f n 0 + f n 1 < f + 1 n, gn 0 + g n 1 < g + 1 n. Therefore, f + g f n 0 + g n 0 + f n 1 + g n 1 f n 0 + g n 0 + f n 1 + g n 1 < f + g + n. Letting n, we see f + g f + g. Given {f k } be a Cauchy sequence in L p 0 +L p 1, and given ɛ > 0, then there exists N = N(ɛ) such that f k f j < ɛ for all k, j > N and hence there exist fk 0, f j 0, fk 1, f j 1 such that fk 0 f j 0 p0 + fk 1 f j 1 p1 < ɛ. By completeness of L p, there exist f 0 L p 0, f 1 L p 1 such that fk 0 f 0 in L p 0 and fk 1 f 1 in L p 1. Then 0 f k (f 0 + f 1 ) fk 0 f 0 p0 + fk 1 f 1 p1 0.

3 10. See my solution files for Folland [4, Exercise 6.13, Additional Exercise 6.1 and Exercise 5.5]. 11. The same as Exercise Proof. (a)(b) are standard. (c) Since L p is separable, there exists a dense subset {x n } in L p. For x 1, since (f k, x 1 ) = f k x 1 M x 1, where M = sup f n p, by Bolzano-Weierstrauss, there exists a subsequence f k,1 such that (f k,1, x 1 ) a 1, some constant. Inductively in x n, we conclude that for each k, there exists {f k,n } {f k 1,n } such that (f k,n, x n ) a n. Choose {f k,k } as the desired subsequence and we are going to prove that it converges weakly. Given l (L p ), by Riesz s theorem, it is represented by some x L p. {x n } x j x in L p. Note that Then there are l(f k,k ) l(f m,m ) (I) + (II) + (III) := x(t)f k,k (t) x j (t)f k,k (t) + x j (t)f k,k (t) x j (t)f m,m (t) + x(t)f m,m (t) x j (t)f m,m (t). Given ɛ > 0, there exists J = J(ɛ) such that x x J < ɛ. By Hölder, (I) and (III) are less M than ɛ. For this J, we can choose a N = N(ɛ) such that (II) < ɛ for all k, m > N. This implies that, for each l (L p ) = L p, l(f k,k ) c(l) as k. Note that the map l c(l) is linear and c(l) = lim k l(f k,k ) M l p. By Riesz representation theorem, there is some f L p such that lim k l(f k,k ) = c(l) = l(f) for all l L p. Remark. The weak compactness theorem is true for any reflexive Banach space. converse also hold, known as Eberlein- Smulian theorem, whose proof is put in my additional exercise 6..5 for Folland [4]. Also see Brezis [, Theorem 3.19 and its remarks] 13. Proof. (a) This is Riemann-Lebesgue lemma, I think it s weak-* convergence in L, not weak convergence; I put my comment on various proofs below: (1) I think the easy way to understand is through Bessel s inequality. But this method does not work on L p, except L. () The second way is through integration by parts, and one needs the density theorem (simple functions or C c The functions). This method is adapted to our claim, except for weak convergence in L 1. But for L 1 case, it s a simple consequence of squeeze theorem in freshman s Calculus. (b) Apply Hölder s inequality and use the absolute continuity of the indefinite integral of L 1 function. (c) is easy. 14. Suppose is a measure space, 1 < p <, and suppose {f n } is a sequence of functions with f n L p M <. 3

4 (a) Prove that if f n f a.e. then f n f weakly. (b) Show that the above result may fail if p = 1 (c) Show that if f n f 1 a.e. and f n f weakly, then f 1 = f a.e. Proof. (a) s hint is given in Folland [4, Exercise 0(a)] by using Egoroff s theorem. (b) f n = χ (n,n+1) in L 1 (R, m). (c) We don t need to bother whether f n p M is a necessary assumption, since by Uniform Boundedness Principle, a weakly convergent sequence is bounded (cf: Exercise 4.13). So by (a), f n f 1, that is, f n f 1 0. Next, we show the weak limit is unique as follows: Note that f n f 1 f f 1, too. Now let E j = { f f 1 > 1 } which has finite measure by j Chebyshev s inequality and hence sgn(f f 1 )χ Ej 0 implies E j has zero measure for each j and hence f 1 = f a.e. L p. The weak convergence to f f 1 and 15. Proof. 16. Multiple Hölder inequality is proved by mathematical induction. 17. (a) Prove Young s convolution inequality, including the measurability, integrability of f(x )g( ) for each x R d (b) A version of (a) applies when g is replaced by a finite Borel measure µ: and show that f µ L p f L p µ (R d ) (c) Prove that if f L p (f µ)(x) = f(x y) dµ(y), R d and g L q, where p, q are conjugated exponents, then f g L f L p g L q. Moreover, f g is uniformly continuous on R d, and if 1 < p <, then (f g)(x) 0 as x. Remark 3. The last assertion is not true for endpoint exponents, e.g. g(x) = f(x) f(x), then (f g)(x) f 1. However, it s true for the case f L 1 and g L 1 L. See Exercise.4 in Book III. Proof. (a) For p = 1, this is the Fubini-Tonelli theorem. We consider 1 < p now. If f, g 0, then Tonelli s theorem implies f g(x) exists a.e. and measurable. For 1 < p <, by Hölder s inequality, we have (f g)(x) g 1/p 1 (f p g)(x) 1/p. Then we integrate both sides and use Minkowski s integral inequality to conclude the desired inequality; for p =, 4

5 (f g)(x) f g 1. For general f, g, then by Minkowski s inequality f g L p (R d ). And then f g < a.e., that is, for a.e. x, f(x )g( ) L 1 and hence f g exists and finite a.e. To see its measurability, Consider f N = fχ { x <N} L 1, then f N g is measurable by Tonelli s theorem. Fixed x, f N (x t)g(t) f(x t)g(t) a.e. t and f N (x )g( ) f(x )g( ) L 1. By LDCT, f N g(x) f g(x) for a.e. x, and therefore f g is measurable. (b) (c) The inequality is just Hölder s inequality. WLOG, we assume f L p with p <. Given ɛ > 0, by translation is continuous in L p norm (Exercise 8), there is δ > 0 such that, f( + h f( ) L p < ɛ/ g L q whenever h < δ. Then for all x R d, (f g)(x + h) (f g)(x) f( + h f( ) L p g L q < ɛ whenever h < δ. The vanishing properties is proved as follows: One consider the case of functions with compact support and then use the density argument. Remark 4. On , I saw this problem. One of the answer states a less explicit fact: Salem and Zygmund proved that convolution map L 1 (T) L 1 (T) L 1 (T) is onto. This was shown to hold for all locally compact groups by Paul Cohen in This result was the starting point of an entire industry establishing factorization theorems. A nice survey on this topic is Jan Kisynski, On Cohen s proof of the Factorization Theorem, Annales Polonici Mathematici 75, (000), Proof. 19. Proof. 0. Proof. 1. Proof.. The proof for Young s inequality for products is immediate if you draw a correct picture. 3. Let (, µ) be a measure space and suppose 0 Φ(t) is a continuous convex and increasing function on [0, ), with Φ(0) = 0. Define the Orlicz space L Φ := {f measurable : Φ( f(x) /M) dµ < for some M > 0}, 5

6 and f Φ = inf Φ( f(x) /M) dµ 1. M>0 Prove that : (a) L Φ is a vector space. (b) L Φ this norm. Note that in the special case Φ(t) = t p, 1 p <, then L Φ = L p. is a norm. (c) L Φ is complete in Remark 5. I think we should assume Φ 0. one also note that the property that there is A > 0 so that Φ(t) At for all t 0 may not be true for small t, e.g. Φ(t) = t. Proof. (a)(b) Given f, g L Φ and s K, where K = R or C. Then it s obvious that sf L Φ and sf L Φ = s f L Φ. Moreover, for each ɛ > 0, there exists 0 < M f < f L Φ + ɛ and 0 < M g < g L Φ + ɛ such that Φ( f(x) /M f) dµ 1, and Φ( g(x) /M g) dµ 1. Then by the convexity and monotone increasing property, f(x) + g(x) f(x) + g(x) Φ( ) dµ Φ( ) dµ = Φ( f(x) M f + g(x) M g ) dµ M f + M g M f + M g M f M f + M g M g M f + M g M f Φ( f(x) M g ) dµ + Φ( g(x) ) dµ 1. M f + M g M f M f + M g M g That is, f + g L Φ and f + g L Φ f L Φ + g L Φ + ɛ. Since ɛ > 0 is arbitrary, f + g L Φ f L Φ + g L Φ. Finally, if f L Φ = 0, then for each k N, there exists 0 < ɛ k < 1 k such that f(x) Φ( ɛ k ) dµ 1. Since Φ(0) = 0, for each λ > 0 and m N, pick k large such that λɛ k < 1 m Φ(λ f(x) ) dµ = Φ(λɛ k f(x) ɛ k + (1 λɛ k )0) dµ λɛ k So Φ(λ f(x) ) dµ = 0, then Φ(λ f(x) ) = 0 for µ-a.e. x. Φ( f(x) ɛ k ) dµ < 1 m. Since Φ 0, A = sup{x : Φ(x) = 0} [0, ). Therefore λ f(x) A for µ-a.e. x for all λ > 0. Hence f(x) = 0 for µ-a.e. x. (c)(simplfied the proof on Rao-Ren [5, Theorem ].) Given {f n } be a Cauchy sequence in L Φ, then there exists 0 < f n f m L Φ n, m (we omit the trivial case that f n f m L Φ = 0 for all large n, m) such that Φ( f n f m ) dµ 1. 0 as For each ɛ > 0, we note that, by monotonicity of Φ, { f m f n ɛ} {Φ( fm fn ) Φ( ɛ )}. Hence µ({ f m f n ɛ}) µ({φ( f m f n ) Φ( ɛ )}) 6 1 Φ( ɛ ) Φ( f m f n ) dµ 1 Φ( ɛ )

7 Let A be the same as the above. Note that 0 < Φ(A) = Φ(t A t + 0 (1 A t )) A t Φ(t) for t A, so Φ(t) t Φ(A) A as t. So µ({ f m f n ɛ}) 0 as n, m. Therefore, {f n } is Cauchy in measure and hence there exists a measurable function f and a subsequence {f ni } such that f ni f a.e. on. Since f n L Φ f m L Φ f n f m L Φ, { f n L Φ} forms a Cauchy sequence in R, and then f n L Φ ρ for some ρ 0. If ρ > 0, then f L Φ since by Fatou s lemma, the definition of L Φ and the continuity and convexity of Φ, Φ( f ) dµ lim inf ρ i f ni Φ( ) dµ 1. f ni L Φ If ρ = 0, then there exists 0 < ɛ ni 0 as i such that Φ( fn i (x) ɛ ni ) dµ 1. Using Fatou s lemma, we have f L Φ since 0 Φ( f(x) ) dµ lim inf i lim inf ɛ n i Φ( i f n i (x) ) dµ 0. ɛ ni Φ( f ni (x) ) = lim inf i Φ(ɛ ni f ni (x) ɛ ni + (1 ɛ ni )0) dµ Finally, given {f nk } k {f n }, there is a further subsequence {f nkm } m that converges to f a.e. (the limit function must be f). For each k, j N, Φ(k f nkm f nkj ) Φ(k f f nkj ) as m. Moreover, since f nm f nj L Φ 0 as m, j, there exists N(k) N such that if m, j > N(k), there exists ɛ m,j < 1 k such that Φ( fn km fn k j ɛ m,j ) 1. So f f nkj L Φ < 1 k if j > N(k) since Φ( f f n kj 1/k ɛ m,j 1/k lim inf m ) dµ lim inf m Therefore f f nkj L Φ Φ( f n km f nkj ɛ m,j ) 1. Φ( f n km f nkj ) dµ = lim inf 1/k m Φ( f n km f nkj ɛ m,j ɛ m,j 1/k ) dµ 0 as j. Since every subsequence of {f n } contains a further convergent subsubsequence, f n converges to f in L Φ. 4. Proof. 5. A Banach space is a Hilbert space if and only if it satisfies the parallelogram law. As a consequence, prove that if L p (R d ) with Lebesgue measure is a Hilbert space, then p =. Generalizations of parallelogram law are stated in Problem 6. 7

8 Proof. (Sketch) Use standard polar identity. For L p part, see my discussion in Folland [4, Exercise 6.1] which works not only for Lebesgue measure but also for general measure space. 6. Proof. 7. Proof. 8. Proof. 9. Proof. 30. Proof. 31. Proof. 3. If the dual space B of a Banach space B is separable, then B is separable. (The converse is not true, see Exercise 11.) Proof. Let {y n} be a countable dense subset of B and x n B such that x n = 1 and y n(x n ) 1 y n. Let us denote by L the vector space over Q generated by {x n }, which is easy to see it s countable. Suppose L is not dense in B, then by Hahn-Banach Theorem, there is y B such that y (L) = {0} and y = 1. Then for each n N, 1 y n y n(x n ) = y n(x n ) y (x n ) y n y. Hence 1 = y y y n + y n 3 y n y 0 by picking a suitable subsequence, which is a contradiction. Therefore L is dense in B, that is, B is separable. 33. Proof. 34. Proof. 35. Proof. 36. Proof. 8

9 Problems 1. Proof.. Proof. 3. Proof. 4. Proof. 5. Proof. 6. There are generalizations of the parallelogram law for L (see Exercise 5) that hold for L p. These are the Clarkson inequalities: (a) For p the statement is that f + g p L + f g p p L 1 p (b) For 1 < p the statement is that f + g q L + f g q p L 1 p ( f pl p + g pl p ). ( f pl p + g pl p ) q p, where p 1 + q 1 = 1. This is trickier to prove than (a). (c) As a result, L p is uniformly convex when 1 < p <. This means that there is a delta function δ = δ p (ɛ), with 0 < δ < 1 and δ p (ɛ) 0 as ɛ 0, so that whenever f L p = g L p = 1, then f g L p ɛ implies that f+g 1 δ. This is stronger than the conclusion of strict convexity in Exercise 7. (d) Using (c) to prove the Radon-Riesz Theorem: suppose 1 < p <, and the sequence {f n } L p or arbitrary uniformly convex Banach space, converges weakly to f. If f n f, then f n f 0 as n. Remark 6. Milman-Pettis Theorem states every uniformly convex space is reflexive. See [, Section 3.7]. Remark 7. A related result for (d) is the Brezis-Lieb theorem (refined Fatou lemma), see [3]. Remark 8. More discussions of Clarkson inequalities can be found in [1]. Proof. (c) is easy. (d) We may assume f 0. Let F n = f n 1 f n and F = f 1 f. So F n F weakly. It follows that 1 = F lim inf F n + F 1. Then the uniform convexity implies F n F 0 as n. So f n f 0 as n. 9

10 7. Proof. 8. Proof. 9. Proof. References [1] Keith Ball, Eric A Carlen, and Elliott H Lieb. Sharp uniform convexity and smoothness inequalities for trace norms. Inventiones mathematicae, 115(1):463 48, [] Haim Brezis. Functional Analysis, Sobolev Spaces and Partial Differential Equations. Springer Science & Business Media, 010. [3] Haïm Brézis and Elliott Lieb. A relation between pointwise convergence of functions and convergence of functionals. Proceedings of the American Mathematical Society, 88(3): , [4] Gerald B Folland. Real analysis: Modern Techniques and Their Applications. John Wiley & Sons, nd edition, [5] Malempati Madhusudana Rao and Zhong Dao Ren. Theory of Orlicz spaces, volume 146. CRC Press, [6] Walter Rudin. Real and Complex Analysis. Tata McGraw-Hill Education, third edition,