lemma ii.1 (baire) let x be a complete metric space and a seq. of closed sets. assume that for each...
TRANSCRIPT
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
)(n
nXInt
Remark 1
1nnX
1nnXX
0n
IntX
Baire’s Lemma is usually used in
the following form. Let X be a
nonempty complete metric space
and a seq. of closed sets
such 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 Theorem
No complement metric space is
of first Category.
is nonwhere dence in X.n
nn 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)(
2
1)(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
2
1)(ˆ
kkk
k
ik kbkaaekf
,2,1,0cos)(1
kdxkxxfak
,2,1sin)(1
kdxkxxfbk
n
kkkn kbkaafS
10 sincos
2
1),(
proved in next page
11
11
11
0
)(2
1)(
2
1
)(2
1
)(2
1)(
2
1
)(2
1
)(2
1)(
2
1
)(2
1
)(2
1
)(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
))((2
1
)(2
1
)(2
1)(
2
1
)(2
1
k
xikxik
k
ikikx
k
ikikx
dxeexf
dxxf
edxexfedxexf
dxxf
1
1
1
sin)sin)((cos)cos)((2
1
)(2
1
)sinsincos)(cos(1
)(2
1
)(cos2)(2
1
)(2
1
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)(2
1
)(2
1
)(2
1
)(2
1
)(ˆ),(
t
ttnth
ttn
tktk
ktttht
thenktthLet
n
k
n
k
n
k
21
sin2
21
sin)21
sin()(
2
1sin)
2
1sin(
2
1
)2
1sin()
2
1sin(
2
1
cos2
1sin)(
2
1sin
,cos)(
1
1
1
2sin
)21
sin(
2
1)(
2sin
)21
sin(
2sin
2sin)
21
sin(1
)(21cos211
t
tntDLet
t
tn
t
ttn
thkt
n
n
k
Dirichlet kernel
dxxD
dxxDf
dxxDxfT
EfxDxffSfT
NneachforwhereET
sequenceaconsiderCEOn
dxxDxffS
n
n
fEf
n
fEf
n
nnn
n
nn
)(
)(sup
)()(sup
)()()0,()(
,
],,[ˆ
)()(),(
1
1
)0,(sup
)(sup
sup
ln~)(:
fS
EfsomeforfT
ThmSteinhausBanachBy
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
TheoremmappingOpen
surjectiveandlinearcontinuousisTHence
yxyxyxyxT
yxyxT
LGLGT
LG
yxyxLG
E
,,
,,
),(
:
,
zCyandzCx
thenc
CLet
zc
yxyxLyGxwhere
yxzasressedbecanzall
umentogeneityBy
yxyxLyGxwhere
yxzasressedbecanzthen
czwithLGziftsoc
,1
1,,,
exp
arghom
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
LGbbaa
bacbandbacababa
tscandLbGaexiststhere
baztoApply
LxdisbxGxdisax
tsLbandGa
thenandExLet
,
..0,,
;)13(
),(,),(
..
,0
bxaxc
LGxdis
axcbxcLGxdis
haveweSimilarly
bxcaxcLGxdis
bxcaxc
bxaxcax
bacax
aax
aaxLGxdis
2
21),(
)1(),(
,
)1(),(
)1(
)(),(
),(),(),(
0,2
21
2),(),(2
21
),(),(2
21),(
LxdisGxdisCLGxdis
havewelettingbyc
CLet
LxdisGxdisc
LxdisGxdisc
LGxdis
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=Esee next page
In this case, all
can be expressed uniquely as z=x+y
with
It follows from Thm II.8 that theprojections z→x and z→y are linearcontinuous and surjective.
Ez
LyGx ,
Example forTopological Complement
E: Banach space
G:finite dimensional subspace of E;
hence is closed.
Find a topological complement of G
see next page
Exxx
tsREextensionanhas
ThmBanachHahnby
sublinearisxxpSince
Gxxx
continuouslinearisRG
niFor
exexxGx
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
thenLGxIf
LGthatshowTo
closedisL
closedisSince
Pf
GofcomplementicaltopoaisLClaim
LLet
n
iii
n
ii
n
ii
i
n
ii
n
ii
i
n
ii
LGEHence
LGyxyz
Lyzthen
zz
ezz
ezzyz
niFor
ezzyz
thenGezylet
EzanyFor
LGEthatshowTo
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
continuousis
boundedis
exexx
exexx
EexxFor
Eonfunctionallinearabe
andEforbasisabeeeeLet
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
FormGeometricSecondThmBanachHahnby
andERxnotSuppose
surjectiveisthatshowTo
Exxfxfx
bydefinedREmaptheConsiderpf
pjief
tsEeeearethereClaim
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p
p
p
ijji
p
p
dependentlinearareff
f
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Exx
Exxx
tsfindcanwe
p
p
iii
p
iiii
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p
,,
0
,,0
0)(
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..0,,,
1
1
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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 continuous
map from F to E such that FidST
Let E and F be two Banach spaces
is 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 spaces
is 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)()(:
.
)()(
TNSRHence
SfSx
TNxffSTxT
FfsomeforfSx
TNSRx
TNSRthatshowTopf
TNofcomplement
icaltopoaisFSSRClaim
TofrightoninverseanbeSLet
iii
.)(
)())((
))(()(
)(
)())((
,)(,)()(
.)()2(
closedisSRHence
SRxTSx
xTSfS
xTf
xTfST
thenxfSSRfSIf
closedisSRthatshowTo
n
n
n
nn
)(
log)()3(~)1(
)()(
)()())(())((
)())((
0)))(((
)()))(((
)()()3(
TNofcomplement
icaltopoaisSRby
TNSREHence
SRTNxTSxTSxx
TNxxTS
xxTST
xTxTST
ExanyFor
TNSREthatshowTo
FidSTandcontinuouslinearisSthat
verifytoeasyisitandcontinuousisS
fPcxPxPfS
xofchoicetheoftindependenisSthatNote
xPfSLet
fcxwithfxTtsEx
FfFor
tsc
TheoremmappingopenBy
operatorsurjectivecontinuouslinearaisP
thenLontoprojectionthebePLet
TNofcomplementicaltopoabeLLet
iii
)()(
.
)()(
)(..
..0
.
,
).(log
)()(
inverse on left
FET :
If S is a linear continuous
operator from F onto E such that
EidTS
Let E and F be two Banach spaces
is 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 spaces
is 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
IIcorollaryBy
idTSthenxfSlet
fPxTtsEx
FfPFfFor
TRontoFfrom
projectivecontinuousthebePLet
iii
vertifytoeasyisItiii
TRTR
TR
E
)(1
)(1
)(1
))(()(
,6.
,)(
)()(..!
)(,
).(
)()(
.)()(