REAL ANALYSIS II TAKE HOME EXAM. T. Tao s Lecture Notes Set 5
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1 REAL ANALYSIS II TAKE HOME EXAM CİHAN BAHRAN T. Tao s Lecture Notes Set 5 1. Suppose that te 1, e 2, e 3,... u is a countable orthonormal system in a complex Hilbert space H, and c 1, c 2,... is a sequence of complex numbers. a) Show that the series ř 8 n 1 c ne n is conditionally convergent in H if and only if c n is square-summable. b) Also if c n is square summable, show that ř 8 n 1 c ne n is unconditionally convergent in H, i.e. every permutation of the c n e n sums to the same value. c) ř Finally: what condition on the sequence tc n u is equivalent to the series 8 n 1 c ne n being absolutely convergent (i.e. the sum of the norms of the terms converges)? a) Let S N ř N n 1 c ne n. Then the series ř 8 n 1 c ne n is conditionally convergent in H if and only if the sequence ts N u is convergent in H if and only if ts N u is Cauchy in H since H is complete. For M ď N we have S N S M 2 Nÿ 2 Nÿ c n e n c n 2 n M n M thus ts N u is Cauchy in H if and only if c n is square summable. b) Suppose c n is square-summable. Since ř 8 n 1 c n 2 is a series with non-negative real terms, it is convergent if and only if every rearrangement of it converges, that is, every permutation of c n is convergent. And since any permutation of e n s is still an orthornormal system, by (a) every permutation of c n e n sums to the same value. c) The series ř 8 ř n 1 c ne n being absolutely convergent means by definition that the series 8 n 1 c ne n of real numbers converge. Since c n e n c n, this is equivalent to saying that c n are absolutely summable, that is, ř 8 n 1 c n ă 8. Folland p.165, 38. Let X and Y be Banach spaces, and let tt n u be a sequence in LpX, Y q such that lim T n x exists for every x P X. Let T x lim T n x; then T P LpX, Y q. For every x P X, by assumption tt n xu n is convergent, hence bounded. Because X is Banach, by the uniform boundedness principle C : sup n T n ă 8. So for every x P X we have T x lim T n x nñ8 lim T n x ď C x nñ8 thus T is bounded. 1
2 REAL ANALYSIS II TAKE HOME EXAM 2 p.168, 41. Let X be a vector space of countably infinite dimension (that is, every element is a finite linear combination of members of a countably infinite linearly independent set). There is no norm on X with respect to which X is complete. (Given a norm on X, apply Exercise 18b and the Baire category theorem.) X has a countable algebraic basis, say te 1, e 2,... u. For every n P N, let X n be the subspace of X spanned by e 1,..., e n. Then we have 8ď X n 1 since every element of X can be written as a finite linear combination of e j s, so it lies in one of the X n s. By Exercise 18b each X n is closed in X. Suppose is a norm on X. We claim that each X n has empty interior. Fix n. Now for every ε ą 0, the vector y n ε 2 e n`1 is outside X n : otherwise we would get X n 2 ε y n e n`1 P X n spante 1,..., e n u, contradicting the fact that te 1,..., e n, e n`1 u is linearly independent. So for every x P X n, the vector x y n lies outside X n but x px y n q y n ε 2 ă ε. In other words, for every ε ą 0 and x P X n the ε-ball around x contains an element outside X n. Thus each X n is nowhere dense, hence X Ť X n is meager. So X cannot be complete with respect to. As was arbitrary, we are done. p.170, 48. Suppose that X is a Banach space. a. The norm-closed unit ball B tx P X : x ď 1u is also weakly closed. (Use Theorem 5.8d.) b. If E Ď X is bounded (with respect to the norm), so is its weak closure. c. If F Ď X is bounded (with respect to the norm), so is its weak* closure. d. Every weak*-cauchy sequence in X converges. (Use Exercise 38.) a. By Theorem 5.8d, there is a linear isometry ι embedding X inside X. Since X is Banach, it embeds in X as a closed subset. Thus B embeds in B tα P X : α ď 1u, in fact once we identify X with ιpxq we have B X X B. By Alaoglu s theorem, B Ď X is compact in the weak* topology of X. Since weak* topology on a dual space is always Hausdorff, B is weak*-closed in X. So it suffices to show that the subspace topology on X induced by the weak* topology on X is exactly the weak topology on X. So suppose that tx α u is a net in X such that ιpx α q Ñ ιpxq in X with the weak* topology. This means that for every f P X we have fpx α q ιpx α qpfq Ñ ιpxqpfq fpxq. By definition, this means that x α Ñ x in X weakly.
3 REAL ANALYSIS II TAKE HOME EXAM 3 b. By assumption there exists a real number R ą 0 such that E Ď B R p0q tx P X : x ď Ru. let γ : X Ñ X be the map x ÞÑ Rx. Clearly γ is a bounded linear map with the bounded inverse x ÞÑ R 1 x. So in particular γ is a homeomorphism with respect to the norm topology. Moreover we claim that γ is continuous with respect to the weak topology. Indeed, if x α Ñ x weakly, then fpx α q Ñ fpxq for every f P X. Thus fpγpx α qq fprx α q R fpx α q ÝÑ R fpxq fprxq fpγxq for every f P X, so γpx α q Ñ γpx α q weakly. Similarly γ 1 is continuous, hence γ is a homeomorphism with respect to the weak topology on X as well. Because we have γ 1 peq Ď B, denoting the weak closure of E by r E, we get γ 1 p r Eq Č γ 1 peq Ď r B B by part (a). Thus r E Ď γpbq B R p0q, so r E is bounded. c. The proof is similar to (b). We have F Ď B Rp0q for some R ą 0. Consider the map δ : X Ñ X given by f ÞÑ Rf. As in (a), δ is a homeomorphism with respect to the norm topology of X. Let s show that δ is continuous with respect to the weak* topology. Suppose f α Ñ f in X in the weak* topology. So for every x P X, we have f α pxq Ñ fpxq, hence pδf α qpxq prf α qpxq f α prxq ÝÑ fprxq prfqpxq pδpfqqpxq so δpf α q Ñ δpfq in the weak* topology. Denoting the weak*-closure by r, we get δ 1 p r F q Č δ 1 pf q Ď Ă B B where B is the norm unit ball in X. The last equality above follows by Alaoglu s theorem (which says B is weakly-compact in X, so weakly-closed as X is Hausdorff by Hahn-Banach). Hence r F Ď δpb q B Rp0q, so r F is bounded. d. Let tf n u be a weak*-cauchy sequence in X. Fix x P X. Then tf n pxqu is a Cauchy sequence of scalars by the definition of weak* topology. Thus tf n xu converges. So by Exercise 38, defining fpxq lim f n pxq for every x P X, we get f P X. And again by the definition of weak* convergence, we get that f n Ñ f weakly in X. p.178, 66. Cpr0, 1sq. Let M be a closed subspace of L 2 pr0, 1s, mq that is contained in a. There exists C ą 0 such that f u ď C f L 2 for all f P M. (Use the closed graph theorem.) b. For each x P r0, 1s there exists g x P M such that fpxq xf, g x y for all f P M, and g x L 2 ď C. c. The dimension of M is at most C 2. (Hint: If tf j u is an orthonormal sequence in M, ř f j pxq 2 ă C 2 for all x P r0, 1s.) a. We know that L 2 pr0, 1s, mq with the L 2 -norm L 2 is a Hilbert space and Cpr0, 1sq with the uniform norm u is a Banach space. Also Cpr0, 1sq as a set can be embedded
4 REAL ANALYSIS II TAKE HOME EXAM 4 inside L 2 pr0, 1s, mq because if two continuous functions are equal almost everywhere then they are in fact equal. So we have the inclusion i : Cpr0, 1sq Ñ L 2 pr0, 1s, mq of sets. We claim that i is bounded (equivalently continuous) when Cpr0, 1sq and L 2 pr0, 1s, mq are endowed with u and L 2, respectively. So suppose tf n u is a sequence in Cpr0, 1sq that converges to f P Cpr0, 1sq uniformly. We want to show that f n Ñ f in the L 2 -norm. Note that both the target and the domain of i are Banach spaces, so by the closed graph theorem we may assume that there exists g P L 2 pr0, 1s, mq such that f n Ñ g in the L 2 -norm. Hence tf n u has a subsequence tf nk u such that f nk Ñ g almost everywhere. But f nk Ñ f almost everywhere too, so f g almost everywhere. This means f g in L 2 pr0, 1s, mq. As i is bounded, there exists a C ą 0 such that for every f P Cpr0, 1sq we have f u ď C f L 2. b. The closed subspace pm, L 2q of L 2 pr0, 1s, mq is a Hilbert space. Now fix x P r0, 1s. The map ev x : M Ñ C f ÞÑ fpxq is evidently linear. Moreover for every f P M we have ev x pfq fpxq ď f u ď C f L 2 hence ev x P M. Thus by Theorem 5.25, there is a unique g x P M such that fpxq ev x pfq xf, g x y for all f P M. And by the proof of Theorem 5.25, for some h P M with h L 2 1. Therefore g x ev x phqh hpxqh g x L 2 hpxq ď h u ď C. c. Let tf j u jpj be an orthonormal basis of pm, L 2q. Then for every x P X, by Parseval s identity we have ÿ f j pxq 2 ÿ xf j, g x y 2 g x 2 L ď C 2. 2 jpj jpj Since C does not depend on x, we conclude that ÿ f j 2 u ď C2. jpj Suppose that J ą C 2. Then by the above inequality, there exists j P J such that f j 2 u ă 1. So there exists ε ą 0 such that f j 2 u ď 1 ε. But then 1 f j 2 L 2 ż 1 0 f j pxq 2 dx ď a contradiction. Hence dim M ď J ď C 2. ż 1 p.255, 15. Let sinc x psin πxq{πx (sinc 0 1). a. If a ą 0, pχ r a,as pxq qχ r a,as pxq 2a sinc 2ax. 0 f j 2 u dx f j 2 u ď 1 ε,
5 REAL ANALYSIS II TAKE HOME EXAM 5 b. Let H a tf P L 2 : fpξq p 0 (a.e.) for ξ ą au. Then H a is a Hilbert space and t? 2a sincp2ax kq : k P Zu is an orthonormal basis for H a. c. (The Sampling Theorem) If f P H a, then f P C 0 (after modification on a bill set), and fpxq ř 8 8 fpk{2aq sincp2ax kq, where the series converge both uniformly and in L 2. (In the terminology of signal analysis, a signal of bandwidth 2a is completely determined by sampling its values at a sequence of points tk{2au whose spacing is the reciprocal of the bandwith.) a. If x 0, we have Since sincp2axq sin 2πax 2πax ż a pχ r a,as pxq e 2πixt dt a 1 e 2πixt t a t a 2πix 1 pe 2πiax e 2πiax q 2πix 1 2πix peip2πaxq e ip2πaxq q 1 2i sinp2πaxq 2πix 1 πx sinp2πaxq., we see that pχ r a,aspxq 2a sincp2axq. And pχ r a,as p0q ż a a dt 2a 2a sinc 0 so in fact for every x we have pχ r a,as pxq 2a sincp2axq. For the second part, observe that qχ r a,as pxq pχ r a,as p xq 2a sincp 2axq 2a sinp 2πaxq 2πax 2a sinp2πaxq 2πax 2a sincp2axq. b. As L 2 is a Hilbert space, it suffices to show that H a is closed in L 2. So assume tf n u is a sequence in L 2 which converges to f in the L 2 -norm. Then there is a subsequence tf nk u such that f nk Ñ f almost everywhere. p.255, 18. Suppose f P L 2 prq. a. The L 2 derivative f 1 (in the sense of Exercises 8 and 9) exists iff ξf p P L 2, in which case f p1 pξq 2πiξfpξq. p b. If the L 2 derivative f 1 exists, then ż j ż ż fpxq 2 dx ď 4 xfpxq 2 dx f 1 pxq 2 dx.
6 REAL ANALYSIS II TAKE HOME EXAM 6 (If ş the integrals on the right are finite, one can integrate by parts to obtain f 2 2 Re ş xff 1.) c. (Heisenberg s Inequality) For any b, β P R, ż ż px bq 2 fpxq 2 dx pξ βq 2 fpξq p 2 dξ ě f π. 2
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