chapter 4 hilbert space. 4.1 inner product space
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Chapter 4 Hilbert Space
4.1 Inner product space
Inner product
CEE :),(
0,0),()( xholdxxi
Exforlinearisxii ),()(
E : complex vector space
is called an inner product on E if
Eyxxyyxiii ,),(),()(
Inner product space
),(
xxx ,
E : complex vector space
is an inner product on E
With such inner product E is called
inner product space. If we write
,then is a norm on E and hence
E is a normed vector space.
Show in next some pages
Schwarz Inequality
Eyxyxyx ,,
E is an inner product space
yxyxTherefore
yxyx
havewebyThenyxyxayax
andathenyx
yxaTaking
CaEyxyaxyax
EyxyxyxHence
xy
yx
y
yxx
y
yx
y
yxx
havewey
yxtTaking
ytyxtxtyxtyx
Rtanyfor
andyxthenyxIfCase
yxyxthenyxIfCase
),(
),(
(*).),(),(),(
1,),(
),(
(*),,),(Re
,),(Re
),(Re
),(Re),(Re),(Re20
,),Re(
),Re(2),(0
0,0,0),(:2
),(,0),(:1
222
222
222
2
2
2
2
22
2
2
2
22
2
222
Triangular Inequality for ∥ ．∥
Eyxyxyx ,
E is an inner product space
yxyx
yx
InequalitySchwarzbyyyxx
yyxx
yyxx
yyxx
yxyxyx
EyxanyFor
2
22
22
22
22
2
2
,2
,Re2
,Re2
,
,
Example 1 for Inner product space
nCE
nnn Cinzzzandzzz ,,,, 11
n
iiizzzz
1,
Let
For
Example 2 for Inner product space
221
2 ;,,)(Ni
izzzNE
)(,, 22121 Nzzzandzzz
Ni
iizzzz,
Let
For
Example 3 for Inner product space
),,(2 LE
),,(2 Lgandf
gdfgf ,
Let
For
Exercise 1.1 (i)
)(, 2 Nzz
summableiszzNjjj
zz ,
For Show that
and hence
is absolutely convergent
Njj
Njj
Njjj
jNj
j
Njjj
zz
zz
zz
zz
22
22
2
1
2
1
Exercise 1.1 (ii)
)(2 NShow that is complete
Nixz
sayCinsequenceconvergentaiszthen
CinsequenceCauchyaisz
Niforthen
Ninkmforzz
nkmforzz
nkmforzz
tsNnanyFor
zzz
zzz
withNinsequenceCauchyabezLet
inin
ni
ni
kimi
Nikimi
km
n
lim
,
,,
,
,
..,0
,,
,,
)(
22
22212
12111
2
.)(
)(lim
2
,,
max
2..
0
2
2
2
0
21
10
completeisNTherefore
NinxzHence
xzxz
nnforthen
xxxLet
nnTake
nnxztsNn
andNiFor
nn
Nii
Niinin
ii
iiinii
Hilbert space
),( E
.)(2 casespecialareCandN n
),,(2 L
An inner product space E is called
is complete Hilbert space if
is a Hilbert space of which
Exercise 1.2
Define real inner product space and
real Hilbert space.
4.2 Geometry for Hilbert space
Theorem 2.1 p.1
Ex
E: inner product space
M: complete convex subset of E
Let
then the following are equivalent
Theorem 2.1 p.2
zxyxsatisfiesMyMz
min
My
MzzyxysatisfiesMy 0),Re(
(1)
satisfing (1) and (2).
(2)
Furthermore there is a unique
0),Re(
,0
),Re(2
)0()(0
)0(),Re(2
)()(
)1()(
10
)2()1(
2
222
2
2
zyxy
havewelettingBy
zyzyyx
ff
fzyzyyxyx
zyyx
zyxflet
andMzanyFor
zxyxHence
zxyx
zxyxzxyxyx
xzyxyx
yxxzyx
yzyx
MzFor
Mz
min
),Re(
),Re(
),Re(
),Re(0
,
)1()2(
2
2
21
122211
122211
21212
21
21
0
,Re,Re
,,
,0
),2()1(
:
yythen
yyxyyyxy
yyxyyyxy
yyyyyy
thenandsatisfyyandyIf
yofUniqueness
nn
n
nm
nmnmnm
nmnmnm
nmnm
nmnm
n
nn
Mz
zxyxthen
nasyztsMycompleteisMSince
nmasnmnm
zz
xzz
zxzxzz
xzxzzxzxxzz
xzxzzxzx
zxxzzz
sequenceCauchyaiszClaimn
zxtsMzConsider
zxLet
yofExistence
lim
0..,
],011
241
21
2
2422
,Re22
4
,Re2
[
:
1..
.inf
:
2222
2222
222
22
22
222
Projection from E onto M
MEt :
Mt
The map
of Thm 1 is called the projection from E onto M.
y is the unique element in M which satisfies (1)
defined by tx=y, where
and is denoted by
Corollary 2.1
)()( 2 idempotentisttti
Mtt
)()( econtractivistyxtytxii
Let M be a closed convex subset of a Hilbert
has the following
properties:
space E, then
)(0),Re()( monotoneistyxtytxiii
0),Re((*),)(
),Re(
(*)),Re(
)),(Re(0
0),Re(
,0),Re()(
.)(
2
2
2
tytxtytxyxByiii
yxtytx
tytxyx
tytxyxtytx
tytxtytxyx
tytxtytxyx
txtyyty
tytxxtxii
obviousisi
Convex Cone
0, MxMx
A convex set M in a vector space is called
a convex cone if
Exercise 2.2 (i)
tIs
MxxyEyN 0),Re(;
NM tsandtt
Let M be a closed convex cone in a Hilbert
Put
Show that
space E and let
I being the identity map of E.
tIs
Xxtxxsx
txytxxtx
txyxtx
ytxxtx
ytxxxtxx
Mtxcetxy
Mcetxxtxtxxtx
NyandExanyFor
0),Re(),Re(
),Re(
),Re(
),Re(
sin,0),Re(
0sin,0)0,Re(),Re(
,
Exercise 2.2 (ii)
0)( iftxxt
( t is positive homogeneous)
txxtthen
Mzce
ztxxtx
ztxxtxztxxtx
thenIfCase
Mcexttxt
thenIfCase
MzandXxanyFor
)(
1sin,0
)1
,Re(
),Re(),Re(
,0:2
0sin),(0)0()(
,0:1
,
Exercise 2.2 (iii)
Exsxtxx ,222
222
22
22
22
),Re(2
,
]0),Re(
0)0,Re(
sin,),Re(
sin,),Re(0[
0),Re(:
,
txsxx
txsx
txsxtxsx
txsxxtxsxx
tsItIsSince
sxtxthen
txxtx
tIscetxxtx
NsxandMtxcesxtx
sxtxClaim
ExanyFor
Exercise 2.2 (iv)
0; txExN
0; sxExM
0;
0
)2(
0;
0
)1(
sxExMHence
sx
txsxtx
xtxMx
txExNHence
tx
txsxsx
xsxNx
Exercise 2.2 (v)
sxztxy ,
;0),Re( sxtxxandsxtx
,0),Re(,, zyandNzMyzyx
then
conversely if
sxzSimilarly
txythen
zw
zwywyzwyxy
Mwanyforthen
zyandNzMyzyxifConversly
sxtx
sxtxsxtxsxtxsxtx
sxtxxandsxtxxSince
sxtxxtxxsx
ExanyfortIsSince
,
0),Re(
),Re(),Re(),Re(
,
0),Re(,,,,
0),Re(
),Re(2
,
22222
222
Exercise 2.2 (vi)
MxxyEyMN 0),(::
M is a closed vector subspace of E. Show that
In the remaining exercise, suppose that
Mz
Myyzhence
Myyz
yz
My
yz
yz
yz
yz
subspacevectorclosedaisMce
My
iyz
iyz
yz
yz
MyyzNz
obviousisIt
0),(
0),Im(
0),Re(
0),Im(
0),Im(
0),Re(
0),Re(
sin
,
0),Re(
0),Re(
0),Re(
0),Re(
0),Re(""
.""
Exercise 2.2 (vii)
both t and s are continuous and linear
.
,
.
)(
)(
)()(
)()(
,,
222
22112211
22112211
22112211
221122112211
222111
2121
continuousaresandtthen
xsxandxtxsxtxxSince
lineararesandt
sxxsxxs
andtxxtxxt
NsxxsandMtxxtwhere
sxxstxxtxx
sxtxxandsxtxx
CandExxanyFor
Exercise 2.2 (viii)
tsENstEM ker;ker
tsENhaveweSimilarly
KersysyMy
sMthatshowTo
MytEy
tEyytyMy
tEMthatshowTo
ker,
0
ker)2(
""
"''
)1(
Exercise 2.2 (ix)
Eyxtyxytx ,),(),(
),(),(
),(),(),(
),(),(),(
tyxytx
tytxtysxtxtyx
tytxsytytxytx
Exercise 2.2 (x)
MzandMy
such that x=y+z
tx and sx are the unique elements
zsxSimilarly
ytx
tNMzcetytx
linearistcetztytx
MzandMywherezyx
,
kersin,
sin,
,
4.3 Linear transformation
We consider a linear transformation from
vector space Y over the same field R or C.
a normed vector space X into a normed
Exercise 1.1
T is continuous on X if and only if
T is continuous at one point.
.
)(
,
)(
..0,0
.""
.""
0
00
0
XoncontinuosisTHence
TsTx
sxTsx
Xsanyforhence
xxT
TxTxxx
tsanyforthen
xatcontinuousisTthatAssume
obviousisIt
Theorem 3.1
XxxcTx
0c
T is continuous if and only if there is a
such that
XxxcTx
thencchoosewe
xTx
xx
Txx
XxanyFor
Txxts
xatcontinuousisT
continuousisT
XoncontinuousisTExerciseBy
xatcontinuousisTthatobviousisIt
,1
1
1
0
1..0
0
""
.,1.3
.0""
Theorem 3.3Riesz Representation Theorem
Xy 0
X
Xxxyx ,)( 0
Let X be a Hilbert space and
Furthermore the map
such that then there is
0y
is conjugate linear and 0y
)(),(
,)(
)()(),)(
(
),(),(
)()()(
.
,
,
.dim,ker
0
0
020
00
0020
0
20000
00
0
0
xxy
havewethenxx
xyletweifHence
xxxxx
x
xxvxxx
xxvx
scalaraand
MinelementnonzerofixedaisxMvwhere
xvxwritecanweXxanyFor
ensionaloneisMthenMLet
thatassumemayWe
4.4 Lebesgue Nikondym Theorem
Indefinite integral of f
fd
,,
AfdAA
)(
Let
Suppose that
and f a Σ –measurable function on Ω
be a measurable space
has a meaning;
then the set function defined by
is called
the indefinite integral of f.
Property of Indefinite integral of f
nA0)(
0)(0)( AwheneverA
ν is σ- additive i.e. if
nn
nn AA )(
is a disjoint sequence, then
Absolute Continuous
,,,,, spacesmeasurebeandLet
0)(0)( AwheneverA
ν is said to be absolute continuous
w.r.t μ if
Theorem 4.1Lebesgue Nikodym Theorem
and
spacesmeasurebeandLet ,,,,
AhdAA
,)(
with
),,(1 Lh
Suppose that νis absolute continuous w.r.t.
μ, then there is a unque
such that
Furthermore ..0 eah
)1.4()1(
)(
..
Re,
)(
1)(
,)(
),,(
,
)(2
1
212
12
2
2
Xffgddgf
fgdfgdfgdfdf
tsXg
ThmonpresentatiRieszbyX
f
ddfdffSince
Xffdfbydefined
Xonfunctionallineartheconsiderand
LXspaceHilbertrealtheConsider
Let
L
0)(
0)(
0)(
0)()1(0
),1.4(
0)(0)(0)(
)1()(0
),1.4(
1)(;0)(;[
..1)(0:1
2
2
2
2
111
1
21
22
2
11
1
A
A
A
Adgdg
haveweinfTake
AAA
dgdgA
haveweinfTake
xgxAandxgxALet
onxeaxgClaim
AA
A
AA
A
]lim)1(lim)1(
)1.4(
0)1()1(0
.,.001
,2,1
[
..
)1.4(:2
fgdgdfdgfdgf
andThmeConvergencMonotonefromthen
fggfandgfgf
eagandgSince
nnfflet
andfunctionasuchbefLet
functionsenonnegativea
andfunctionmeasurableallforholdClaim
nn
nn
nn
n
.
),,(
,)(
)(
),1.4(,
..0,1
1
),1.4(1
,..
1
obviousisuniqueishsuchThat
Lhhence
hdknowweSince
hdhddA
theninztakeweAanyFor
hdzdz
andeahtheng
ghLet
dg
gzdz
thening
zfChoose
zfunctionenonnegativeaandmeasurableaFor
AAA
A
4.5 Lax-Milgram Theorem
Sesquilinear p.1
),(),(),( 22112211 xxBxxBxxxB
CXxxxfor 2121 ,,,,
CXXB :),(
Let X be a complex Hilbert space.
),(),(),( 22112211 xxBxxBxxxB
is called sesquilinear if
Sesquilinear p.2
EyxyxryxB ,),(
B is called bounded if there is r>0 such that
XxxxxB 2),(
B is called positive definite if there is ρ>0 s.t.
Theorem 5.1The Lax-Milgram Theorem p.1
1S
XyxySxByx ,),(),(
Let X be a complex Hilbert space and B a
a bounded, positive definite sesquilinear
functional on X x X , then there is a unique
bounded linear operator S:X →X such that
and
Theorem 5.1 The Lax-Milgram Theorem p.2
1S
rS 1
Furthermore
exists and is bounded with
1
1
2
*
*2
*1
*2
*1
2*2
*1
*2
*1
*2
*1
*2
*1
*2
*1
*
**
,,
,
,
0
),(0
0),(
),(),(
mindet)0(
),(),(..;
S
xSx
SxxSxxSxSxBSx
DonlinearisS
andsubspacelinearisDarsesquilineisBSince
xSxletDxFor
xxxx
xxxxxxB
XyyxxB
XyyxByxB
atederuniquelyisxandDDthen
XyyxByxtsXxXxDLet
*
*
*
**
*
),(),(lim),(lim,
0
),(),(),(
)(
.[
:
xSxandDxthen
yxBySxByxyxHence
nasyxSxr
yxSxByxBySxB
xSxthen
XinCauchyisSxboundedisS
xxSxxSSxSx
XxxandDxLet
closedisDthatshowtoFirst
XDClaim
nn
nn
n
nn
n
n
mnmnmn
nn
ioncontradictay
yyyByx
Dx
XxxyBxxxtsXx
ThmonpresentatiRieszBy
X
XxxyBx
bydefinedbeletandDyLet
DthenXDthatSuppose
,0
),(,0
),()(,..
,Re
),()(
0,
0
200000
0
000
0
0
rSxrxS
xSxr
xSxBxSxSSBxSxSxS
andexistSHence
XxxyBxx
tsXxbeforeasXy
ontoisSthatshowTo
xxBxx
thenSxIf
isSthatshowTo
11
1
1111121
1
00
00
,,,
),(),(
..
00),0(),(
,0
11
4.7 Bessel Inequality and parseval Relation
Propositions p.1
kEk tt
ne
neLet
Ut
be an orthogonal system in a
Hilbert space X, and let U be the closed vector
subspace generated by
Let be the orthogonal projection onto U
and where kk eeE ,,1
Proposition (1)
Xxexextk
jjjk
1),(
k
jjjk
ikikkii
i
k
jjjiki
k
jjjk
exextHence
xtextxtexeBut
eexte
kiFor
thenextLet
1
1
1
,
),(1,,
),(),(
,,1
,
Proposition (2)
xtxtXxFor Ukk
lim,
kUkUk
k
UUkk
U
kk
kk
kkk
k
N
iii
tttcextxt
xtxtt
UxtExanyFor
Uyifyytthen
yzzy
yzztyt
yzztytyyt
zztNk
zytsezncombinatiolinearaisthere
andUyanyFor
sin,lim
lim
,
lim
2
..
,0
1
Proposition (3)
).,(),(),(1
yexeytxt j
k
jjkk
For each k and x,y in X
),(),(
),(,),(
),(,),(),(
1
11
11
xexe
exeexe
exeexeytxt
j
k
jj
k
iiij
k
jj
k
iii
k
jjjkk
Proposition (4)
).,(),(),(1
yexeytxt jj
jUU
For any x,y in X
),(),(
),(),(lim
),(lim),(
0
),(),(
),(),(
),(),(),(
),(),(
),(),(
1
1
yexe
yexe
ytxtytxtthen
kasytytxyxtxt
ytytxtytxtxt
ytytxtytxtxt
ytxtytxtytxtxt
ytxtytxtxtxt
ytxtytxt
ii
i
i
k
ii
k
kkk
UU
kUkU
kUkUkU
kUkUkU
kkUkUkU
kkUkkU
kkUU
Proposition (5)
Xxxxei
i
,),(
2
1
2
Bessel inequality
22
1
2),( xxtxe U
jj
Proposition (6)
XUXxxxei
i
,),(
2
1
2
An orthonormal system ne
is called complete if U=X
( Parseval relation)
XUHence
oncontraditiaxext
xtxtx
xxttsXx
thenXUthatSuppose
xxtxe
thenXUIf
iiU
UU
U
Ui
i
,),(
)1(
..
,""
),(
,""
1
2
222
22
1
2
Separable
A Hilbert space is called separable
if it contains a countable dense subset
Theorem 7.1
2
A saparable Hilbert space is isometrically
nCisomorphic either to for some n
or to
.inf.inf
.,
,
.
.
1
initeisxthatassumeweHenceiniteis
xwhencasetheofthatofimitationeasyanis
xofycardinalittheisnwhereCtoisometric
llyisometricaisXthatproofthefiniteisxIf
xzthatsuchxesubsequenc
tindependenanzfromextractcanOne
Xindenseiswhich
elementsofsequenceabezLet
spaceHilbertseparableabeXLet
k
k
kn
k
kkk
k
kk
..
,),6(),(
),(
:
,)1.7(),(),(),(
)4(,
;,,
,
2
1
22
12
1
1
obviousislinearisThatisometryisso
xxbyxexSince
xewhere
xlettingbyXmaptheDefine
tIbecauseyexeyx
fromhaveweXinyxforthus
XUtheneUletbeforeAs
exthatsuche
systemorthonomalanxfromconstructcanwe
procedurelizationorthonormaSchmidtGramFrom
jj
kk
kk
Uj
jj
k
kkkk
k
)1.7(
.
,lim,
,lim.
0
.
int,
.
2
11
1
22
1
21
2
fromfollowsisomophismanisThat
ontoisTherefore
xhenceandeexethen
xxLetsequenceCauchyaisxHence
mastotendswhich
xxhavewemnFor
sequenceCauchyaisxthatclaimwe
exlet
negerpositiveeachforLet
ontoisthatnowshowWe
kki
n
jjji
ni
nn
n
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n
n
jjjn
kk
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