Lemma II.1 (Baire) 1nnX

nIntX

Let X be a complete metric space and a seq. of closed sets. Assume that for each n . Then

1)(

nnXInt

Remark 1

1nnX

1nnXX

0nIntX

Baire’s Lemma is usually used inthe following form. Let X be a nonempty complete metric space and a seq. of closed setssuch that . Then there is such that 0n

Baire’s Category Theorem

First Category

M

nn

n XXM ,

XM

X: metric space

, M is nonwhere dence in X i.e. has no ball in X.

is nonwhere dense in X.M is called of first category.

By Baire’s Category TheoremNo complement metric space is of first Category.

is nonwhere dence in X.nn

n XXX ,1

Theorem II.1(Banach Steinhaus)Let E and F be two Banach spaces and a family of linear continuous operators from E to FSuppose (1)

then (2)

IiiT

ExxTiIi

)(sup

iIiTsup

IiExxcxTi ,)(

In other words, there is c such that

Application of Banach Steinhaus

)()(,,ˆ ffCfCE

)(max

,xff

x

kxikxexeZk ikxk sincos)(,

,Cf

dxexfkf ikx)(

21)(ˆ

Fourier Series ik

n

nkn ekffS )(ˆ),(

k

ikekff )(ˆ~)(

is called Fourier series of f

is called Fourier nth partial sum of f),( fSn

If f is real valued, then

where

10 sincos

21)(ˆ

kkk

k

ik kbkaaekf

,2,1,0cos)(1 kdxkxxfak

,2,1sin)(1 kdxkxxfbk

n

kkkn kbkaafS

10 sincos

21),(

proved in next page

11

11

11

0

)(21)(

21

)(21

)(21)(

21

)(21

)(21)(

21

)(21

)(21

)(ˆ

k

ikikx

k

ikikx

k

ikikx

k

ikikx

k

ikikx

k

ikikx

xi

k

ikikx

k

ik

edxexfedxexf

dxxf

edxexfedxexf

dxxf

edxexfedxexf

dxexf

edxexf

ekf

1

)()(

11

))((21

)(21

)(21)(

21

)(21

k

xikxik

k

ikikx

k

ikikx

dxeexf

dxxf

edxexfedxexf

dxxf

1

1

1

sin)sin)((cos)cos)((21

)(21

)sinsincos)(cos(1

)(21

)(cos2)(21

)(21

k

k

k

kxkxdxxfkxkxdxxf

dxxf

dxkxkkxkxf

dxxf

dxxkxf

dxxf

Lebesque Theorem],[ˆ Cf

such that

)0,(suplim fSnn

dxxkxf

dxexf

dxexf

edxexf

ekffS

n

k

n

nk

xik

n

nk

xik

n

nk

ikikx

n

nk

ikn

1

)(

)(

)(cos21)(21

)(21

)(21

)(21

)(ˆ),(

t

ttnth

ttn

tktk

ktttht

thenktthLet

n

k

n

k

n

k

21sin2

21sin)

21sin(

)(

21sin)

21sin(

21

)21sin()

21sin(

21

cos21sin)(

21sin

,cos)(

1

1

1

2sin

)21sin(

21)(

2sin

)21sin(

2sin

2sin)

21sin(

1

)(21cos211

t

tntDLet

t

tn

t

ttn

thkt

n

n

k

Dirichlet kernel

dxxD

dxxDf

dxxDxfT

EfxDxffSfT

NneachforwhereETsequenceaconsiderCEOn

dxxDxffS

n

n

fEf

n

fEf

n

nnn

n

nn

)(

)(sup

)()(sup

)()()0,()(

,],,[ˆ

)()(),(

1

1

)0,(sup

)(sup

sup

ln~)(:

fS

EfsomeforfTThmSteinhausBanachBy

T

ndxxDTClaim

nn

nn

nn

nn

II.4 Topological Complementoperators invertible on right(resp. on left)

Theorem II.8Let E be a Banach space and let G andL be two closed vector subspacessuch that G+L is closed . Then there exists constant

such that0c

(13) any element z of G+L admits a decomposition of the form z=x+y with

zcyandzcxLyGx ,,,

GL x

yz

TheoremmappingOpensurjectiveandlinearcontinuousisTHence

yxyxyxyxTyxyxTLGLGT

LG

yxyxLG

E

,,,,

),(:

,

zCyandzCx

thenc

CLet

zc

yxyxLyGxwhere

yxzasressedbecanzallumentogeneityBy

yxyxLyGxwhereyxzasressedbecanzthen

czwithLGziftsoc

,1

1,,,

exparghom

1,,,exp

..

Corollary II.9Let E be a Banach space and let G andL be two closed vector subspacessuch that G+L is closed . Then there exists constant

such that0c

(14)

ExLxdisGxdiscLGxdis ,),(),(),(

GL

x

LGbbaabacbandbacababatscandLbGaexiststhere

baztoApplyLxdisbxGxdisax

tsLbandGathenandExLet

,..0,,

;)13(),(,),(

..,0

bxaxcLGxdis

axcbxcLGxdishaveweSimilarly

bxcaxcLGxdis

bxcaxc

bxaxcax

bacax

aax

aaxLGxdis

221),(

)1(),(,

)1(),(

)1(

)(),(

),(),(),(

0,221

2),(),(221

),(),(221),(

LxdisGxdisCLGxdis

havewelettingbycCLet

LxdisGxdisc

LxdisGxdiscLGxdis

Remark

0csomefor

Let E be a Banach space and let G andL be two closed vector subspaces with

Then G+L is closed.

ExLxdisGxdiscLGxdis ,),(),(),(

Exercise

Topological ComplementLet G be a closed vector subspace ofa Banach space E. A vector subspace L of E is calleda topological complement of G if

(i)L is closed.(ii)G∩L={0} and G+L=E

see next page

In this case, all

can be expressed uniquely as z=x+ywith

It follows from Thm II.8 that theprojections z→x and z→y are linearcontinuous and surjective.

Ez

LyGx ,

Example forTopological Complement

E: Banach spaceG:finite dimensional subspace of E; hence is closed.Find a topological complement of G

see next page

ExxxtsREextensionanhas

ThmBanachHahnby

sublinearisxxpSince

GxxxcontinuouslinearisRG

niForexexxGx

GofbasisabeeeeLet

Gii

ii

Gi

Gii

i

nn

n

)(ˆ..:ˆ

,)(

)(:

,,1)()(

.,,,

11

21

0

ˆkersin,00)(ˆ

sin,)(

,0)2(

.ˆker

,ˆker)1(:

.log:

.ˆker

111

1

1

1

LGHence

Lxceeex

Gxceexx

thenLGxIfLGthatshowTo

closedisL

closedisSincePf

GofcomplementicaltopoaisLClaim

LLet

n

iii

n

ii

n

ii

i

n

ii

n

ii

i

n

ii

LGEHence

LGyxyzLyzthen

zz

ezz

ezzyz

niFor

ezzyz

thenGezylet

EzanyForLGEthatshowTo

ii

n

iiiii

n

iiiii

n

iii

n

iii

)(

0)(ˆ)(ˆ

)(ˆ)(ˆ)(ˆ

)(ˆˆ)(ˆ

,,1

)(ˆ

,)(ˆ

.)3(

00

00

00

1

1

0

1

1

Remark

On finite dimensional vector space, linear functional is continuous.

Prove in next page

continuousisboundedis

exexx

exexx

EexxFor

EonfunctionallinearabeandEforbasisabeeeeLet

nEwithspacevectorabeELet

n

ii

n

ii

n

iii

n

iiii

n

ii

i

n

ii

n

.

)(

)(

.,,,

.dim

21

1

221

1

2

1

11

1

21

Remark

Let E be a Banach space. Let G be a closed v.s.s of E with codimG < ∞, thenany algebraic complement is topological complement of G

Typial example in next page

Let

then

be a closed vector subspace of E and

codimG=p

pNEN dim,

NfxfExG ,0,

Prove in next page證明很重要

)()(\,

,,,,)(

::

,1,

..,,,:

.,,,

0

1

21

21

FormGeometricSecondThmBanachHahnbyandERxnotSuppose

surjectiveisthatshowTo

Exxfxfx

bydefinedREmaptheConsiderpf

pjief

tsEeeearethereClaim

NforbasisabefffLet

p

p

p

ijji

p

p

dependentlinearareff

f

Exxfxf

Exx

Exxx

tsfindcanwe

p

p

iii

p

iiii

p

ii

p

,,

0

,,0

0)(

)(

..0,,,

1

1

11

0

21

.log

,,

,,

,1,

1,,,0)(

0,,1,0)(

0,,0,1)(

..,,

1

1

2

1

1

Gofcomplementicaltopotheis

eebygeneratedspacethe

tindependenlinearareee

pjief

e

e

e

tsEeeThen

p

p

ijji

p

p

Question

FET :

Does there exist linear continuousmap from F to E such that FidST

Let E and F be two Banach spacesis linear continuous surjective

S is called an inverse on right of T

Theorem II.10

FET :

The folloowing properties are equivalent :

Let E and F be two Banach spacesis linear continuous surjective

)0()( 1TTN

(i) T admits an inverse on right

(ii)

admits a topological complement

Prove in next page

0)()(0)0()(

)())(()(0)(

)()(0)()()1(:

)(log)()(:

.)()(

TNSRHenceSfSx

TNxffSTxTFfsomeforfSx

TNSRxTNSRthatshowTopf

TNofcomplementicaltopoaisFSSRClaimTofrightoninverseanbeSLet

iii

.)()())((

))(()()(

)())((,)(,)()(.)()2(

closedisSRHenceSRxTSx

xTSfSxTf

xTfSTthenxfSSRfSIf

closedisSRthatshowTo

n

n

n

nn

)(log)()3(~)1(

)()()()())(())((

)())((0)))(((

)()))(((

)()()3(

TNofcomplementicaltopoaisSRby

TNSREHenceSRTNxTSxTSxx

TNxxTSxxTSTxTxTST

ExanyForTNSREthatshowTo

FidSTandcontinuouslinearisSthatverifytoeasyisitandcontinuousisS

fPcxPxPfSxofchoicetheoftindependenisSthatNote

xPfSLetfcxwithfxTtsEx

FfFortsc

TheoremmappingopenByoperatorsurjectivecontinuouslinearaisP

thenLontoprojectionthebePLetTNofcomplementicaltopoabeLLet

iii

)()(.

)()()(..

..0

.,

).(log)()(

inverse on left

FET :

If S is a linear continuousoperator from F onto E such that

EidTS

Let E and F be two Banach spacesis linear continuous injective

S is called an inverse on left of T

Theorem II.11

FET :

The following properties are equivalent :

Let E and F be two Banach spacesis linear continuous injective

)()( ETTR

(i) T admits an inverse on left

(ii)

is closed and admits a topological complement.

Prove in next page

continuousisSHence

fPTfPTxfS

continuousisT

IIcorollaryByidTSthenxfSlet

fPxTtsExFfPFfForTRontoFfrom

projectivecontinuousthebePLetiii

vertifytoeasyisItiii

TRTR

TR

E

)(1

)(1

)(1

))(()(

,6.,)(

)()(..!)(,

).(

)()(.)()(