Compactness in Banach spaces – a crash course

DefinitionsAssume that A is a subset at a Banach space XA set FEA is called E dense in A where Eso

if for every a cA tree exists some xeF with ka ell E

A is called totally bounded if for every Eso

there exists a finite s dense subset of A

Proposition A subset A of a Banach space istotally bounded of and only if every sequencein A has a convergent subsequenceNote We do not claim that the limit ofthe subsequence belongs to A Sona Xis a Bored space it is equivalent tosaythat each sequence in A has a Cauchysubsequence

Proaf First assume that every sequence in Ahes a Cauchy subsequence To show that A is totallybounded let Eso e d pick x GA Assumingthat we have picked x an c A pick Xue c itso that hint xjK E fer j 42 n n i't possibleIf this can be done for all n 42,3 n then Cxncannot have a Cauchy subsequence since any

two members of the sequence are at least adistance e apart So the process must stopThen X x is E dense in A

Conversely assume that A is totally boundedLet Cxn be a sequence in A If E o endFEA is Edewee then few each n pick ane Fwith Mxn anti Etz Sina F is finite at leastone member of F must occur infinitelyoftenamong the en I e there a fee with an _finfinitely often This the subsequence of Euincluding only the n fer which an _f satisfiesME Enke E fer ell j kNow start oneri Pick subsequencesetch with Hxf NIK I fa ell j hetches with Ux D x K E for dlj.h

nDJefCxEDuithHxf3l xlEksffadlj.hand so on e d look at the diagonal sequeaeyn innThen if jin n then Hyj gullet so Cynis a Cauchy sequence This sequence is a

subsequence et CenThis completes the proof

Compactness A subset k of a Banach epee Xis called compact of it is closed and totallybounded Or equivalently if every sequence in Khas a subsequence converging to some point in K

It is easy to see that the closure of anytotally wounded set is totally boundedand hence coupe et For this reason totallyfounded sets are also caked precompeet

Proposition compact sets are almostfinitedimensional A subset A of e Benach speakis totally bounded if and only if A is boundedand for each Eso there exists a finitedimensional subspace4 ex so that distca 4 E

fr ell a e AProof First if A is totally bounded e d Eso

pick a e dense finite set FEA Let 4 bethe linear span of F It satisfies the statedconditionConversely if A is bounded sayHallEM fer eGA

e d e o e a 4 is a finite dimensional subspacewith distca 4 a e f ell a c A let G bea finite e dense subset of yet 1 KylieMt ePossible because 4 is finite dimensional

For any ac A there is some yG4 with kagheeThen HyHE Hall1kg all EMte so there is some GGGwith Kygu E hence ka ga 2eCreate a subset F of A es follows for each GGGif there exists some a c A with Ila glidepick on such a e d include it in FNow if a c A there is some gag with ka glideThen there is f GF with Hf ghc 2e So

Ha f HE ka gilt g f ke 4 e i e F is GE decree

This shows that A is totally bounded

Definition Let X and Y be Benach spacesand L X 4 a linear op Then L iscalled compact if 2 14 114 is totally bounded

Bellick's the one states thatthe inclusion map H Tn L2 Tn is

compect provedby Fourier series

Another variant says the same aboutHfcr es Leer

if r is a bounded regionproved by reducing it to the first case