lemma ii.1 (baire)
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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.,)(
)()(..!)(,
).(
)()(.)()(
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