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Humboldt-Universit¨at zu Berlin Institut f¨ur Mathematik

C. Wendl, S. Dwivedi, L. Upmeier zu Belzen

Funktionalanalysis

WiSe 2020–21

Problem Set 4

Due: Thursday, 3.12.2020 (22pts total)

Problems marked with p˚q will be graded. Solutions may be written up in German or English and should be submitted electronically via the moodle before the ¨Ubung on the due date. For problems without p˚q, you do not need to write up your solutions, but it is highly recommended that you think through them before the next Tuesday lecture. You may also use the results of those problems in your written solutions to the graded problems.

Convention:Unless otherwise stated, you can assume in every problem thatpX, µqis an arbitrary measure space and functions inLppXq:“LppX, µqtake values in a fixed finite- dimensional inner product spacepV,x, yqover a fieldKwhich is eitherRorC. Whenever X is a subset of Rn, you can also assume by default that µis the Lebesgue measurem.

Problem 1 p˚q

Assume 1†p, q †8with 1p` 1q “1. Prove:

(a) For any closed subspace E Ä LppXq with E ‰ LppXq, there exists a function g P LqpXqzt0u that satisfies≥

Xxg, fydµ“0 for everyf PE.

Hint: SinceLppXqis uniformly convex, there exists a closest point inE to any given point in LppXqzE. [6pts]

(b) A linear subspace E ÄLppXqis dense if and only if every bounded linear functional

⇤:LppXq ÑK that vanishes on E is trivial. [3pts]

Comment: The result of this problem is often used in applications and cited as a con- sequence of the Hahn-Banach theorem, which implies a similar result for subspaces of arbitrary Banach spaces. However, the uniform convexity of LppXqfor 1† p†8 makes the use of the Hahn-Banach theorem (which relies on the axiom of choice) unnecessary in this setting. You should not use it in your solution either, since we have not proved it yet.

Problem 2

Show that iff PL8pXqsatisfies |f|†}f}L8 almost everywhere, then ˇˇ

ˇˇ ª

Xxg, fydµ ˇˇ

ˇˇ†}g}L1¨ }f}L8 for every gPL1pXqzt0u,

i.e. there is strict inequality. Give an example f PL8pr0,1sq satisfying this condition.

Hint: What can you say about ≥

Xpc´ |f|q|g|dµ if|f|†c almost everywhere?

Comment: The Hahn-Banach theorem implies that for every nontrivial element x in a Banach space E, there exists a bounded linear functional ⇤ P E˚ with }⇤} “ 1 and

⇤pxq “ }x}. For E “ L8pXq, it follows that this ⇤ P E˚ cannot be represented as

g “≥

Xxg,¨ydµfor anyg PL1pXq. This is one way of seeing that the Riesz representation theorem is false for p“ 8.

1

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Problem Set 4

Problem 3

(a) Show that ifpM, dqis a metric space containing an uncountable subsetS ÄM such that every pair of distinct points x, y PS satisfies dpx, yq• ✏ for some fixed ✏° 0, thenM is not separable.

(b) SupposepX, µqcontains infinitely many disjoint subsets with positive measure. Show thatL8pXqcontains an uncountable subsetS ÄL8pXq, consisting of functions that take only the values 0 and 1, such that }f ´g}L8 “1 for any two distinct f, gPS.

Conclude thatL8pXqis not separable.

Hint: If you’ve forgotten or never seen the proof via Cantor’s diagonal argument that R is uncountable, looking it up may help.

(c) Let LpHq denote the Banach space of bounded linear operators H Ñ H on a separable Hilbert spaceHoverKP tR,Cu. Show that any orthonormal basistenu8n1

ofH gives rise to a continuous linear inclusion :`8 ãÑLpHq,

where `8 denotes the Banach space of bounded sequences t n PKu8n1 with norm }t nu}`8 :“supnPN| n|, and pt nuq PLpHqis uniquely determined by the condi- tion pt nuqej :“ jej for all jPN.

Comment: It is not hard to show that every subset of a separable metric space is also separable. Since `8 “L8pN,⌫q for the counting measure⌫, parts (b) and (c) thus imply thatLpHqis not separable.

Problem 4 p˚q

This problem deals with weak convergence xn á x. Assume H is a separable Hilbert space over K P tR,Cu with orthonormal basis tenu8n“1, and consider a sequence of the form xn :“ nenPH for some nPK. Prove:

(a) xná0 whenever the sequence n is bounded. [3pts]

(b) If the sequence n is unbounded, then xn is not weakly convergent. [5pts]

Hint: Show that limnÑ8xej, xny “ 0 for every j P N and conclude that if xn á x then x“0. Then associate to any subsequence with | nj|•j for j “ 1,2,3, . . . an element of the form v “∞8

j1ajenj PH such thatxv, xnjy Û0 as jÑ 8.

Remark: We will later use a general result called the “uniform boundedness principle”

to show that weakly convergent sequences must always have bounded norms. But you should not use that result here, since we have not proved it.

(c) If | n|§?

n for all nPN, then every weakly open neighborhood of 0 PH contains infinitely many elements of the sequence xn. [5pts]

Comment: If the weak topology onHwere metrizable, then one could deduce from part (c) that a subsequence of ?

nen converges weakly to 0, contradicting part (b). It follows the- refore that the weak topology on an infinite-dimensional Hilbert space is not metrizable.

Problem 5

Find a sequence fn P LppRq for 1 † p † 8 that converges weakly to 0 but satisfies }fn}Lp “1 for alln, and deduce thatfn has noLp-convergent subsequence.

2

n

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Problem gelsslion 5

a Go through the proof of Riesz rep them for

x

E C LPC x closed Choose he LPCx IE LMD is uniformly convex F fo EE

minimizing the distance to h

Suppose f EE arbitrary

11h Ho tf HIt o o

O t Ih folk 24 h fo f du

b

2

D J th fol th fo f du O f f e E

C 14 31 of

f o because he LPCx IE and fo EE

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Iq L g L f I

11h folk 2 h fo 1h folk

h fo c LPCx th fol da ca

to th folk 4h fo c LNG l o

Da

Suppose Hahn Banach

X E E X no C XIE

span E Xo y e 12

0

b Suppose F is dense

A LPCx k s't ACE O

het f c L'Kx fn

n

n p f

E

Attn Act but Nfn 0

1 0

on

LMD

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conversely suppose E is not dense

E f Mx

contradicts problem

a

as we can find non zero

gelkx s't g E 0

E must be dense

D

G E L x I of 3 some A CX

w

MCA o

s t g fado

1 49 f du EJKg.FI dae Jl9llfIdae

A A A

lying da z Jiglifldal

A

Adding 11g Hull the

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11811L Http 1µg day's 1181L Http

fight dal

0 A

o

e.g fix X on Oil

HfHµ ess sup f L

Ifl HfHp on 94 measure zero

i Ifl f Il ftp are

a Mid metric space

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Suppose M is separable F countable dense set g Mn new

F SES Bezels n mm new 10

so choose ncs C IN

s

t Mncs C Beg as

Si ncs is injective

if

n

na

Mn Mnet C Bez Cns 7 BE Xt

Ns Ht

S N contradiction as 5 is uncountable

M is not separable

b S set of sequences containing only O's and l's

is uncountable

Cantor's Diagonal Argument

Suppose S is countable

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F bijection b

w

S and N

S O 1 I O l

S2 O O O L L t 1100

S3 I 1 I

X

0

S4 L L L L L

5 O O I 0

look at Sii th entry in Si

t S because j t Sj for j

then the j th entry of t Sjj twenty

of Sj

t S is uncountable

f c Lol x

Xilie disjoint subsets of X

w

positive

Xx

measure

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c en n o n basis of Tl

Define Yo la LCH by dela

11

in

X d N N by

en An en clearly

an

inclusion

Check linear

1145112 sup 1145,43112

11h11 L

msn.IE HII AnanenH where h Einen

h I

yup Hnl little E Knies

is an inclusion of P BCN

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a un then An elk

het see Jl be an antibitrary Then

n.LI Lx.x n7l2EMllxlI2dnen

where M is the bound

on 5km9

Ken 12 0 Len x O

0 0

Un 0 If tnf is bounded

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b Hnj I Z j Fj L 2

a

a 2 j enj c H

j I

is absolutely convergent

1 v xnj I

j Hnj Z I 2

O

fo N n Y

Suppose U E th is

a

weakly open nbd

Of OEM

finitely many hi Az An C te't

s t

o e see H It G 121 tried n9

Kui xj 121 EU

If a topology is generated by a given sets

then open set in that topology

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arbitrary union of finite

intersections of elements S

By Riesz rep H nm

Ano n Lv x for some vie

Hi L in

their

II with EIJI It vi g IT

II Kri g it La

D there must exist infinitely many jew

set I.zkv.i.es TE Tn

Otherwise the above series will diverge

Then I Ijf Ej

IN xD I vi 1 g I Hjlkvi.es X

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a F

g zel

infinitely many j tr i L in

Nj

e

for infinitely many j's

In e int t E C it it

O Otherwise

clearly Http L

Use the density of polynomials mi Lt to

note that for any f ELP F a polynomial p s't

yf theft

Also note that as n a

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Hn Hy o

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