Transcript Chapter 4 Hilbert Space

Chapter 4
Hilbert Space
4.1 Inner product space
Inner product
E : complex vector space
(,) : E  E  C
is called an inner product on E if
(i) ( x, x)  0,  hold  x  0
(ii) ( x,) is linear for x  E
(iii) ( x, y )  ( y , x )  x, y  E
Inner product space
E : complex vector space
(,)
Show in next
some pages
is an inner product on E
With such inner product E is called
inner product space. If we write
,then

x 
 x, x 
is a norm on E and hence
E is a normed vector space.
Schwarz Inequality
E is an inner product space
 x, y  
x  y  x, y  E
Case 1 :
If
( x , y )  0,
then
Case 2 :
for any
If
( x , y )  0,
tR
then
2
0  ( x  ty, x  ty )  x
t  
Tak ing
0 x
2

y

Re 2 ( x , y )
y
Hence
2
2
 x
2
 x
2
( x, y )
,
( x, y )
Therefore
2
y
2
2
 x
2

Re 2 ( x , y )
y
2
2
2
y
y
2
then
( ax, y )  a ( x , y )  ( x , y )
( x, y )
and
have
Re 2 ( x , y )

Re 2 ( x , y )  x
a 
we
y
 Re 2 ( ax, y )  ax
Tak ing
,
2
2 Re 2 ( x , y )
x  0, y  0
 2t Re( x , y )  t 2 y
Re( x , y )
y
( x, y )  x  y
2
( x, y )  x
y
2
 x, y  E
 x, y  E ,
a 1
.Then
a C
(*)
and
by
(*)
we
have
Triangular Inequality for ∥ ．∥
E is an inner product space
x  y  x  y  x, y  E
For any x, y  E
x  y  x  y, x  y 
2
 x  2 Re  x, y   y
2
 x  2 Re  x, y   y
2
2
2
 x  2  x, y   y
2
2
 x 2 x  y  y
x  y
 x y  x  y

2
2
2
by Schwarz Inequality
Example 1
for Inner product space
Let
For
E C
z  z1,, zn 
n




n
and z  z1 ,, zn in C
n
z, z  
i 1

zz
i i
Example 2
for Inner product space
Let
For


2
E   ( N )  z1 , z2 ,;  zi   


iN
2




2
z  z1, z2  and z  z1 , z2    ( N )
z, z  
iN

zz
i i
Example 3
for Inner product space
Let
For
E  L (, ,  )
2
f and g  L (, ,  )
2
 f , g    f gd
Exercise 1.1 (i)
For
z, z   ( N )
2
z z 
j j jN
is summable
z, z
Show that
and hence
is absolutely convergent
 z j z j
jN
  z j  z j
jN
1


zj
 
2 jN 
1
   zj
2  jN

2
2
 z j
2



  z j
jN
2


Exercise 1.1 (ii)
Show that
2
 (N )
is complete
zn  be a
z1   z11, z12 ,
z2   z21, z22 ,
Let
Cauchy sequence in  2 ( N ) with

For any   0,  n  N s.t.
zm  zk  
  zmi  zki
for m, k  n
2
2
for m, k  n
iN
 zmi  zki  
for m, k  n, i  N
then for i  N
zni  is a Cauchy sequence in C
then
zni 
lim zni  xi
n 
is a convergent sequence in C , say
i  N
For i  N and   0
 ni  N s.t. zni  xi 

2
n  ni
i



Tak e n0  max ni i 1
Let x   x1 , x2 ,
then for n  n0
zn  x 
 zni  xi  
2

iN 2
iN
i

Hence lim zn  x in  2 ( N )
n 
2
Therefore  ( N ) is complete.
Hilbert space
An inner product space E is called
(E,  )
Hilbert space if
L ( ,  ,  )
2
2
is complete
is a Hilbert space of which
 ( N ) and C
n
are special case.
Exercise 1.2
Define real inner product space and
real Hilbert space.
4.2 Geometry
for Hilbert space
Theorem 2.1
p.1
E: inner product space
M: complete convex subset of E
Let
xE
then the following are equivalent
Theorem 2.1
(1)
p.2
y  M satisfies x  y  min x  z
zM
(2)y  M satisfies Re( y  x, y  z)  0  z  M
Furthermore there is a unique
satisfing (1) and (2).
yM
(1)  (2)
For any z  M and 0    1
let f ( )  x  (1   ) y  z
2
 ( x  y)   ( y  z)
2
 x  y  2 Re( x  y , y  z )   y  z  f (0)
2
2
f ( )  f (0)
0

 2 Re( x  y , y  z )   y  z
By letting   0, we have
Re( y  x, y  z )  0
2
2
( 2)  (1)
For z  M ,
0  Re( x  y , z  y )
 Re( x  y , z  x  x  y )
 x y
2
 Re( x  y , z  x )
 x y
2
 Re( x  y , x  z )  x  y  x  z
 x y  xz
Hence
x  y  min x  z
zM
Uniqueness of y :
If y1 and y2 satisfy (1) and ( 2), then
0  y1  y2
2
  y1  y2 , y1  y2 
  y1  x, y1  y2    y2  x, y2  y1 
 Re  y1  x, y1  y2   Re  y2  x, y2  y1 
0
then y1  y2
Existence of
y:
Let   inf x  z .
zM
Consider zn   M s.t.   x  zn
Claim :
[ zm  zn
2
zn 
2
 x  zm
 zm  zn
2
1
 
n
2
is a Cauchy sequence
 zm  x  x  zn
zm  zn
4
x
2
2
2
2
 x  zn
2
 2 Re zm  x, zn  x 
2
 x  zm
 2 x  zm
2
2
 x  zn
 2 x  zn
2
2
 2 Re zm  x, zn  x 
zm  zn
4
x
2
2
 2 1
 2 1
 1 1
2
 zm  zn  2     2     4  2    0 as m, n   ]
m
n


m n
Since M is complete,  y  M s.t. zn  y  0 as n  
2
then x  y  lim x  zn  
n 
Projection from E onto M
The map t : E  M defined by tx=y, where
y is the unique element in M which satisfies (1)
of Thm 1 is called the projection from E onto M.
and is denoted by
tM
Corollary 2.1
Let M be a closed convex subset of a Hilbert
space E, then
t  tM
has the following
properties:
(i ) t  t (t is idempotent)
2
(ii) tx  ty  x  y (t is contractive)
(iii) Re(tx  ty, x  y)  0 (t is monotone)
(i ) is obvious.
(ii) Re( tx  x, tx  ty )  0,
Re( ty  y , ty  tx)  0
 0  Re( x  y  (tx  ty ), tx  ty )
 Re( x  y , tx  ty )  tx  ty
2
(*)
2
 tx  ty  Re( x  y , tx  ty )
 x  y  tx  ty
 tx  ty  x  y
2
(iii) By (*), Re( x  y , tx  ty )  tx  ty  0
Convex Cone
A convex set M in a vector space is called
a convex cone if
x  M
 x  M ,  0
Exercise 2.2 (i)
Let M be a closed convex cone in a Hilbert
space E and let
N  y  E; Re( y, x)  0 x  M 
Put
t  tM and s  tN
Show that
s  I t
I being the identity map of E.
For any x  E and y  N ,
Re( tx  x, tx)  Re( tx  x, tx  0)  0 , sin ce 0  M
Re( y , tx)  0, sin ce tx  M
 Re( x  tx  x, x  tx  y )
 Re( tx, x  tx  y )
 Re( tx  x  y , tx)
 Re( tx  x, tx)  Re( y , tx)  0
 sx  x  tx x  X
 s  I t
Exercise 2.2 (ii)
t(x)  tx if   0
( t is positive homogeneous)
For any x  X and z  M ,
Case 1 : If   0, then
t (x )  t (0)  0  t ( x ), sin ce 0  M
Case 2 : If   0, then
Re( tx  x, tx  z )   Re( tx  x, tx  z )
1
 Re( tx  x, tx  z )

1
 0 , sin ce z  M

then t (x )  tx
Exercise 2.2 (iii)
2
2
2
x  tx  sx , x  E
For any x  E ,
Claim : Re( tx, sx )  0
[0  Re( tx, sx ) , sin ce tx  M and sx  N
 Re( tx, x  tx) , sin ce s  I  t
  Re( tx  x, tx  0)  0
then Re( tx, sx )  0 ]
Since s  I  t , I  s  t
 x  sx  tx  x
 x
2
 sx
2
 tx
2
2
 sx  tx
 sx
2
 sx
2
2
 tx
2
 tx
2
 2 Re( sx , tx)
Exercise 2.2 (iv)
N  x  E; tx  0
M  x  E; sx  0
(1) x  N  sx  x
 sx  sx  tx
 tx  0
Hence N  x  E ; tx  0
( 2) x  M  tx  x
 tx  sx  tx
 sx  0
Hence M  x  E ; sx  0
Exercise 2.2 (v)
Re(tx, sx)  0 and x  tx  sx;
conversely if
x  y  z, y  M , z  N and Re( y, z)  0,
then
y  tx, z  sx
Since s  I  t , for any x  E
sx  x  tx  x  tx  sx
2
2
Since x  tx  sx
2
2
2
2
and x  tx  sx
2
2
tx  sx  tx  sx  tx  sx  2 Re( tx, sx )
 Re( tx, sx )  0
Conversly, if x  y  z, y  M , z  N , and Re( y , z )  0
then for any w  M ,
Re( y  x, y  w)  Re(  z, y  w)   Re( y  w, z )
 Re( w, z )  0
then y  tx
Similarly, z  sx
Exercise 2.2 (vi)
In the remaining exercise, suppose that
M is a closed vector subspace of E. Show that
N  M : y  E : ( y, x )  0 x  M 

" " It is obvious.
" " z  N  Re( z, y )  0 y  M
 Re( z, y )  0
 Re( z, y )  0


y  M ,
 Re( z, iy )  0
 Re( z,iy )  0
sin ce M is a closed vector subspace
 Re( z, y )  0
 Re( z, y )  0


y  M
 Im( z, y )  0
 Im( z, y )  0
 Re( z, y )  0

y  M
 Im( z, y )  0
hence ( z, y )  0 y  M
 zM 
Exercise 2.2 (vii)
both t and s are continuous and linear
For any x1, x2  E and 1, 2  C
x1  tx1  sx1 and x2  tx2  sx2
 1 x1   2 x2  (1t x1   2 tx2 )  (1s x1   2 sx2 )
where (1t x1   2 tx2 )  M and (1s x1   2 sx2 )  N
 t (1 x1   2 x2 )  1t x1   2 tx2 and
s(1 x1   2 x2 )  1s x1   2 sx2
 t and s are linear.
2
2
2
Since x  tx  sx , tx  x and sx  x
then t and s are continuous.
Exercise 2.2 (viii)
M  tE  ker s; N  sE  ker t
(1) To show that M  tE
' ' " y  M  ty  y  y  tE
" " y  tE  y  M
( 2) To show that M  ker s
y  M  sy  0  y  Kers
Similarly, we have N  sE  ker t
Exercise 2.2 (ix)
(tx, y)  ( x, ty)  x, y  E
(tx, y )  (tx, ty  sy )  (tx, ty )
( x, ty )  (tx  sx , ty )  (tx, ty )
 (tx, y )  ( x, ty )
Exercise 2.2 (x)
tx and sx are the unique elements
y  M and z  M
such that x=y+z

x  y  z , where y  M and z  M
 tx  ty  tz , sin ce t is linear

 tx  ty , sin ce z  M  N  ker t
 tx  y
Similarly, sx  z

4.3 Linear transformation
We consider a linear transformation from
a normed vector space X into a normed
vector space Y over the same field R or C.
Exercise 1.1
T is continuous on X if and only if
T is continuous at one point.
" " It is obvious.
" " Assume that T is continuous at x0 .
then for any   0 ,   0 s.t.
x  x0    Tx  Tx0  
 T ( x  x0 )  
hence for any s  X ,
x  s    T ( x  s)  
 Tx  Ts  
Hence T is continuos on X .
Theorem 3.1
T is continuous if and only if there is a
c0
such that
Tx  c x
 x X
" " It is obvious that T is continuous at x  0.
By Exercise 3.1, T is continuous on X .
" " T is continuous
 T is continuous at x  0
   0 s.t.
x    Tx  1
For any 0  x  X
 
x    T 
x   1
x
 x 
1
 Tx 
x


we choose c 
Tx  c x
1

, then
 x X
Theorem 3.3
Riesz Representation Theorem
Let X be a Hilbert space and
then there is
y0  X
 X 
such that
( x)   y0 , x  x  X
Furthermore the map
  y0
is conjugate linear and
  y0
We may assume that   0
Let M  ker , then M  is one dim ensional.
For any x  X , we can write x  v  x0
where v  M , x0 is a fixed nonzero element in M 
and  a scalar.
( x )  ( v  x0 )  ( x0 )
( x0 , x )  ( x0 , v  x0 )   x0
(
( x0 )
x0
2
2
x0 , x )  ( x0 )  ( x )
Hence if we let y0 
( y 0 , x )  ( x )
( x0 )
x0
2
x0 , then we have
4.4 Lebesgue Nikondym
Theorem
Indefinite integral of f
Let
,,  
be a measurable space
and f a Σ –measurable function on Ω
Suppose that
 fd 
has a meaning;
then the set function defined by
 ( A)  A fd   A  
is called
the indefinite integral of f.
Property of
Indefinite integral of f
 ( )  0
ν is σ- additive i.e. if
An  
is a disjoint sequence, then


   An    ( An )
n  n
 ( A)  0 whenever ( A)  0
Absolute Continuous
Let , ,   and, ,  be measure spaces,
ν is said to be absolute continuous
w.r.t μ if
 ( A)  0 whenever ( A)  0
Theorem 4.1
Lebesgue Nikodym Theorem
Let , ,   and , ,  be measure spaces
   and    
with
Suppose that νis absolute continuous w.r.t.
μ, then there is a unque

(
A
)

hd

,
A


such
that

A
h  L (, ,  )
1
Furthermore
h  0   a.e.
Let      ,
Consider the real Hilbert space X  L (, ,  )
and consider the linear functional  on X
2
defined by ( f )   fd  , f  X

Since ( f )   f d   f d
  ()
1
2
f
2
  1d 
1
1
2

L2 (  )
  X , by Riesz Re presentation Thm
 g  X s.t.
( f )   fd    fgd   fgd   fgd
  f (1  g )d   fgd
 f X
( 4.1)
2
  a.e. x on 
[ Let A1  x  ; g ( x )  0 and A2  x  ; g ( x )  1
Take f   A in ( 4.1), we have
Claim1 : 0  g ( x )  1
1
0   ( A1 )  A (1  g ) d  A g d
1
1
  ( A1 )  0   ( A1 )  0   ( A1 )  0
Take f   A in ( 4.1), we have
2
0  A (1  g ) d  A g d   ( A2 )  0
2
  ( A2 )  0
  ( A2 )  0
  ( A2 )  0
2
Claim2 : ( 4.1) hold for all   measurable function and
  a.e. nonnegative functions
[ Let f be such a function and
let f n  f  n  n  1,2,
Since 1  g  0 and g  0   a.e.,
0  f n (1  g )  f (1  g ) and 0  f n g  fg
then from Monotone Convergence Thm and ( 4.1)
f n (1  g )d  lim  f n gd   fgd ]
 f (1  g )d  nlim



n 
For a   measurable and   a.e. nonnegative function z,
z
Choose f 
in ( 4.1), then
1 g
g
 zd   z 1  g d
g
Let h 
, then h  0   a.e. and
1 g
 zd   zhd
For any A  , we take z   A in ( 4.1), then
 ( A)    Ad    Ahd  A hd
Since  ()  , we know  hd  
hence h  L1 (, ,  )
That such h is unique is obvious.
4.5 Lax-Milgram Theorem
Sesquilinear
p.1
Let X be a complex Hilbert space.
B(,) : X  X  C
is called sesquilinear if
for x, x1, x2  X , 1, 2 C
B( x, 1x1  2 x2 )  1B( x, x1)  2 B( x, x2 )
B(1 x1  2 x2 , x )  1B( x1 , x )  2 B( x2 , x )
Sesquilinear
p.2
B is called bounded if there is r>0 such that
B( x, y )  r x  y x, y  E
B is called positive definite if there is ρ>0 s.t.
B ( x, x )   x
2
x  X
Theorem 5.1
The Lax-Milgram Theorem
p.1
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
( x, y)  B(Sx, y)
S 
x, y  X
1
and
Theorem 5.1
The Lax-Milgram Theorem
Furthermore
S
1
exists and is bounded with
S
1
r
p.2

Let D  x  X ; x *  X s.t. ( x, y )  B ( x * , y ) y  X

then D   ( 0  D ) and x * is uniquely det er min ated
 B ( x1* , y )  B ( x2* , y )  y  X

  B ( x1*  x2* , y )  0  y  X

  0  B ( x1*  x2* , x1*  x2* )   x1*  x2*

  x*  x*  0  x*  x*
1
2
1
2





2




For x  D, let Sx  x *
Since B is sesquilinear, D is linear subspace and
S is linear on D
 Sx  B Sx, Sx    x, Sx   x Sx
2
 Sx   1 x
 S   1
Claim : D  X
[ First to show that D is closed.
Let xn   D and xn  x  X
Sxn  Sxm  S ( xn  xm )  S xn  xm
S is bounded  Sxn  is Cauchy in X
then Sxn  x*
B( Sxn , y )  B( x* , y )  B( Sxn  x* , y )
 r Sxn  x*  y  0
Hence
as n  
 x, y   lim ( xn , y )  lim B( Sxn , y )  B( x
n 
then x  D and Sx  x
n 
*
*
, y)
Suppose that D  X , then D  0

Let y0  D

( x )  B ( y0 , x )
  X 
and let  be defined by
x X
By Riesz Re presentation Thm,
 x0  X s.t.  x0 , x   ( x )  B( y0 , x )
 x0  D
 0   x0 , y0   B( y0 , y0 )   y0
 y0  0, a contradiction
2
x X
To show that S is 1  1
Sx  0, then
If
( x, x )  B (0, x )  0  x  0
To show that S is onto
y0  X as before x0  X s.t.
( x0 , x )  B ( y0 , x )  x  X
Hence S
1
S x
2
1

exist and
1
1


1


1
 S x, S x  B SS x, S x  B x, S x
 r x S 1 x
1
1
 S x r x
 S
1
r

4.7 Bessel Inequality and
parseval Relation
Propositions p.1
Let
en 
be an orthogonal system in a
Hilbert space X, and let U be the closed vector
subspace generated by
Let
and
tU
en 
be the orthogonal projection onto U
tk  t E
where
k
Ek  e1 ,, ek
Proposition (1)
k
tk x   ( e j , x ) e j
j 1
 x X
k
Let tk x    j e j , then
j 1
For i  1,, k
k
(ei , tk x )  (ei ,   j e j )  i
j 1
But ei , x   ei , tk x  1  tk x   ( ei , tk x )  i
Hence tk x   e j , x e j
k
j 1
Proposition (2)
For x  X , lim tk x  tU x
k 
For any y U and   0,
N
there is a linear combination z   i ei s.t. y  z  
i 1
k  N  tk z  z
 tk y  y  tk y  tk z  z  y
 tk y  tk z  z  y
 y  z  z  y  2
then lim tk y  y
k 
if y U
For any x  E , tU x U
 lim tk tU x  tU x
k 
 lim tk x  tU x , sin ce tk  tU  tk
k 
Proposition (3)
For each k and x,y in X
( t k x , tk y ) 
k
 (e j , x )(e j , y ).
j 1
k
 k

(tk x, tk y )    ( e j , x )e j ,  ( ei , x )ei 
i 1
 j 1



  ( e j , x ) e j ,  ( ei , x )ei 
j 1
 i 1

k
k
k
  (e j , x ) (e j , x )
j 1
Proposition (4)
For any x,y in X

(tU x, tU y )   (e j , x )(e j , y ).
j 1
(tU x, tU y )  (tk x, tk y )
 (tU x  tk x  tk x, tU y )  (tk x, tk y )
 (tU x  tk x, tU y )  (tk x, tU y )  (tk x, tk y )
 (tU x  tk x, tU y )  (tk x, tU y  tk y )
 (tU x  tk x, tU y )  (tk x, tU y  tk y )
 tU x  tk x  y  x  tU y  tk y  0
then (tU x, tU y )  lim (tk x, tk y )
k 
k
 lim  ( ei , x ) ( ei , y )
k   i 1

  ( ei , x ) ( ei , y )
i 1
as k  
Proposition (5)

 (ei , x )  x , x  X
2
i 1
Bessel inequality
2

2
 (e j , x )  tU x  x
j 1
2
2
Proposition (6)

 (ei , x )  x , x  X  U  X
2
2
i 1
( Parseval relation)
An orthonormal system
is called complete if U=X
en 
" " If U  X , then

 (ei , x )  tU x  x
2
2
2
i 1
" " Suppose that U  X , then
 x  X s.t. tU x  x
2
2
 x  tU x  (1  tU ) x

2
 tU x   ( ei , x ) , a contradition
2
i 1
Hence U  X
Separable
A Hilbert space is called separable
if it contains a countable dense subset
Theorem 7.1
A saparable Hilbert space is isometrically
isomorphic either to
or to

2
C
n
for some n
Let X be a separable Hilbert space.
Let
zk k 1
be a sequence of elements
which is dense in X .
One can extract from
subsequence
If
xk 
xk 
zk 
such that
an independent
zk 
 xk 
is finite, the proof that X is isometrically
xk .
when xk 
isometric to C n , where n is the cardinality of
is an easy imitation of that of the case
is inf inite. Hence we assume that xk  is inf inite.
From Gram  Schmidt orthonormalization procedure,
xk  an orthonomal
ek k 1 such that xk   ek 
As before, let U  ek  , then U  X ;
we can construct from
system
thus for x, y in X we have from ( 4)
( x, y ) 

 ( e j , x )( e j , y )
(7.1)
j 1
Define the map  : X  
2
, because I  tU
by letting



x   k k 1
where  k  ( ek , x )
Since x
2


 (e j , x )
j 1
2
by (6),
x  x
2
,
so  is isometry. That  is linear is obvious.
We show now that  is onto  2 .
Let
 k k 1  2 ,
for each positive int eger n
n
let xn    j e j
j 1
we claim that
xn 
For n  m we have
is a Cauchy sequence.
xn  xm
2
n
  j
2
j  m 1
which tends to 0 as m  
Hence xn  is a Cauchy sequence. Let x  lim xn ,
n
n


then ei , x   lim  ei ,   j e j    i and hence x   k 
k 1
n 
 j 1

Therefore  is onto  2 .
That  is an isomophism follows from (7.1)