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Chapter 6
Compact operators
Spectral Decomposition of
self-adjoint compact operators
VI.1 Definition. Elementary
Properties Adjoint
Compact Operator
Let E and F be Banach spaces
An operator
T  LE, F 
is called compact if
T BE 
every seq.
has conv.
subseq.
is precompact
K(E,F) is the family of compact operators
in L(E,F); K(E,E)=K(E)
Theorem VI . 1
K(E,F) is a closed vector subspace of L(E,F).
for the norm

L( E , F )
K E , F  is obviously vector subspace of L( E , F ).
Suppose Tn   K ( E , F ), T  L( E , F ) with Tn  T
L( E ,F )
0
Since F is complete, it is sufficient to show that
for any   0, T BE  is cov ered by a finite number
of ball B  f i ,  . Fix n such that Tn  T
 
Tn BE    B f i ,  where I is finite
2
iI 
 T BE    B  f i ,  
iI
 T is compact .
L( E ,F )


2
and
Finite Rank
An operator
finite rank if
T  L( E , F )
is called
dim R(T )  
A continuous operator T of finite rank is
compact.
Corollary VI. 2
Let
Tn 
be a sequence of continuous
operators of finite rank from E to F and
T  L( E , F )
Tn  T

0
L( E , F )
such that
then
T  K (E, F )
Remark 1
p.1
The family << problem of approximation>>
(Banach, Grothendieck) concerns the
converse of Corollary VI.2 . Given a compat
operator T, does there exist a sequence
Tn 
Tn  T
of operators of finite rank such that

0
L( E , F )
?
Remark 1
p.2
In general, the answer is negative
(Enflo, 1972) – even for certain closed
vector subspace of
 (1  p  , p  2)
p
see for example Lindenstrauss- Tzafriri [2]
However the answer is affirmative in many
cases; for example of F is Hilbert
Let K  T BE , being compact , can be cov ered by
 B fi ,   , I is finite. Let G be the vector subspace
iI
generated by f i ' s and
let T  PG  T (T is of
finite rank )
T is compact . To show that T  T  2
If x  BE , then i0  I s.t. Tx  f i   (1)
0
then PG  Tx  PG f i  
0
 T x  f i  
0
( 2)
(1)  ( 2)  T x  Tx  2
 T  T  2
 x  BE
Remark 1
p.3
From this proof, it follows that if F is
Schauder basis; then the answer is
affirmative.
Remark 2
p.1
Let us indicate also a technique quite useful
in nonlinear analysis which permits to
approximate a continuous map (linear or
nonlinear) by non-linear map of finite rank.
Remark 2 p.2
Let X be a Topology space, F is a Banach
space and
T:X F
is continuous
map such that T(X) is precompact in F.
Then for any
map
  0,
there is a continuous
T : X  F
of finite rank s.t.
T ( x )  T ( x )    x  X
In
fact, K  T ( X ), being compact , can be cov ered
by a
finite number of

K   B ( f i , ) , I is
2
iI
Definite T by
T ( x ) 
 qi ( x ) f i
iI
 qi ( x )
balls.
finite
 x X
iI
where qi ( x )  max   Tx  f i ,0
 T ( x )  T ( x ) 
 qi ( x ) f i
iI
 qi ( x )
iI

 qi ( x ) f i  T ( x )
iI
 qi ( x )
iI

 T ( x) 
 qi ( x )( f i  T ( x ))
iI
 qi ( x )
iI
Proposition VI.3
Let E, F and G be Banach spaces.
If
T  L( E , F )
[ resp. T  K ( E , F )
then
and
and
S  K (F ,G)
S  L( F , G )
S  T  K ( E,G)
]
Equicontinuous
M  C (S )
M is called equicontinuous if for any ε>0,
there is a δ>0 s.t.
s, t  S
with d(s,t)<δ
 f ( s )  f (t )    f  M
Theorem (Arzelá-Ascoli)
A closed set
M  C (S )
if and only if
(i) M is bounded on C(S)
(ii) M is equicontinuous
is compact
Theorem VI.4 (Schauder)
If
T  K (E, F )
And conversely .
then
T  K ( F , E)
*
To show that T * ( BF  ) is precompact. Let vn   BF  ,
we show that one can extract a subsequence s.t
T *v n
k
converges in E . Let K  T ( BE ), compact in F ,
Let H  C ( K ) defined by
H   n : x  K  vn , x ; n  1,2,
H satisfies the conditions of Artzela ' Ascoli Theorem
(i ) sup  n ( x )  sup vn , x  sup vn
xK
xK
xK
F
x  vn
F
T  T
(ii) Equicontinuous :
 n ( x )   n ( y )  vn , x  y  vn x  y  x  y


There is a subsequence  n ( x ) of H s.t.
k
 n converges in C ( K ) to a function  ( x ).
k
In particular,
sup vn , Tu   (Tu )  0
k
uBE
as k  
 sup vn , Tu  vn , Tu  0

k
uBE
 sup T vn  T vn , u  0
*
*

k
uBE
 T vn  T vn
*
*
k

E
0
k
as k ,   
as k ,   
 T vn converges in E .
*
as k ,   
Conversely, sup pose that T  K ( F , E )
*
Pr evious proof  T **  K ( E , F )
 T ( BE  ) is precompact in F .
**
In particular, T ** ( BE ) is precompact in F .
Since
v, Tu  T *v, u  v, T **u
 u  BE , v  F  ,
T ( BE )  T ** ( BE )
 T ( BE )  T ( BE )
**
is compact in F 
Since F is closed , T ( BE )  F
Hence T ( BE ) is precompact in F
Therefore T is compact .
VI.2 The Riesz-Fredholm
Theory
Lemma VI.1 (Riesz-Lemma)
Let E be a normed vector space and
be a closed vector subspace with
Then
u 1
  0 , u E
and
M E
M E
such that
dist(u, M )  1  
Let v  E , v  M and d  dist( v, M )  0
d
Given   0, choose m0  M s.t.d  v  m0 
(*)
1 
v  m0
Take u 
, then u  1 and for m  M
v  m0
v  m0  m v  m0
v  m0
um 
m 
v  m0
v  m0
d

v  m0
, sin ce m0  m v  m0  M
 1   , by (*)
Hence dist(u, M )  1  
Remark
If dimM<∞ (or more generally if M is reflexive)
we can choose
 0
in Lemma VI.1;
But not in general case (see [BT])
Theorem VI.5 (Riesz)
Let E be a normed vector space such that
BE
is precompact. Then E is of finite
dimension.
Suppose the contrary. dim E  
There is a sequence En  of subspaces of
finite
dim ension such that En 1  En ,

Riesz Lemma  un   E s.t. un  En \ En 1
1
un  1 and dist(un , En 1 ) 
2
1
Then un  um 
if m  n
2
Hence un  contains no convergent subsequence.
Let
Theorem II.18
A : D( A)  F
D ( A)  E
closed linear operator with
then the following properties are equivalent
(i) R(A) is closed
(ii) R(A*) is closed
* 
(iii)
R( A)  N ( A )
(iv)
R( A )  N ( A)
*

Topological Complement
Let G be a closed vector subspace of
a Banach space E.
A vector subspace L of E is called
a topological complement of G if
(i)L is closed.
(ii)G∩L={0} and G+L=E
see
next
page
In this case, all
zE
can be expressed uniquely as z=x+y
with
x  G, y  L
It follows from Thm II.8 that the
projections z→x and z→y are linear
continuous and surjective.
Example for
Topological Complement
E: Banach space
G:finite dimensional subspace of E;
hence is closed.
Find a topological complement of G
see
next
page
Let e1 , e2 ,, en  be a basis of G.
x  G  x  1 ( x )e1     n ( x )en
For i  1,, n
 i : G  R is linear continuous
i ( x )  i
G
 x G
x
Since p( x )   i
G
x is sublinear,
by Hahn  Banach Thm
 i has an extension ˆi : E  R s.t.
ˆi ( x )   i
G
x
xE
n
Let L   ker ˆi .
i 1
Claim : L is a topo log ical complement of G.
Pf :
(1) Since ker ˆi is closed ,
n
L   ker ˆi is closed .
i 1
( 2)To show that G  L  0
If x G  L , then
n
x   i ( x )ei
, sin ce x  G
i 1
n
n
i 1
i 1
 ˆi ( x )ei   0 ei  0
Hence G  L  0
n
, sin ce x  L   ker ˆi
i 1
(3) To show that E  G  L.
For any z  E
n
let y  ˆi ( z )ei  G , then
i 1
n
z  y  z  ˆi ( z )ei
i 1
For i0  1,, n
n


ˆi ( z  y )  ˆi  z  ˆi ( z )ei 
 i 1

0
0
n
 ˆi ( z )  ˆi ( z )ˆi ( ei )
0
i 1
 ˆi ( z )  ˆi ( z )  0
0
0
then z  y  L
 z  y  ( x  y)  G  L
Hence E  G  L
0
Remark
On finite dimensional vector space,
linear functional is continuous.
Prove in
next page
Let E be a vector space with dim E  n.
Let e1 , e2 ,, en  be a basis for E and
 be a linear functional on E.
n
For x   xi ei  E
i 1
n
 n
 ( x )     xi ei    xi ei 
 i 1
 i 1
  ( x) 
n
 xi ei 
i 1
  is bounded .
  is continuous
12

2
   xi 
 i 1 
n
12


2
  ei 
 i 1

n
Remark
Let E be a Banach space.
Let G be a closed v.s.s of E with
codimG < ∞, then
any algebraic complement is
topological complement of G
Typial example
in next page
Let
N  E , dim N  p
then
G  x  E f , x  0, f  N 
be a closed vector subspace of E and
codimG=p
Prove in
next page
證明很重要
Let
f1, f 2 ,, f p  be
a basis for N .
Claim : there are e1 , e2 ,, e p  E s.t.
f i , e j   ij
 1  i, j  p
pf : Consider the map  : E  R p defined by

 ( x )  f1 , x ,, f p , x

xE
To show that  is surjective
Suppose not ,  x0  R p \  ( E ) and
by Hahn  Banach Thm ( Second Geometric Form)
we can find   1 , 2 ,, p   0 s.t.
   ( x )    x0
    ( x)  0
xE
xE
p
p
i 1
i 1
 0  i fi , x  i fi , x
xE
p
 i fi  0
i 1
 f1 ,, f p are linear dependent 
Then  e1 ,, e p  E s.t.
 ( e1 )  1,0,,0
 ( e2 )  0,1,,0

 ( e p )  0, ,,1
 f i , e j   ij
 1  i, j  p
 e1 ,, e p are linear independent
 the space generated by e1 ,, e p
is the topo log ical complement of G.
Lemma VI.6
(Fredholm Alternative)
Let E be a Banach space and T  K ( E )  K ( E , E )
then (a) N(I-T) is of finite dimension
(b) R(I-T) is closed and more precisely
* 
R( I  T )  N ( I  T )
(c)
N ( I  T )  0  R( I  T )  E
(d)
dim N ( I  T )  dim N ( I  T )
*
(a ) Let E1  N ( I  T ), then BE  T ( BE )
1
Since T  K ( E ), T ( BE ) is precompact
then BE is precompact
1
Thm VI .5  dim E1  
(b) Suppose that f n  un  Tun  I  T un  f as n  
To show that f  R( I  T )
Since N ( I  T ) is of finite dim ension,
there exist vn  N ( I  T ) such that
d n  un  vn  dist(un , N ( I  T ))
we have f n  (un  vn )  T (un  vn )
Claim :
 un  vn  is
( 4)
bounded .
[ Suppose the contrary. There is a subsequence
such that un  vn
k
k
un  v n
 . Let wn 
un  v n
( 4)  wn  Twn  0
k
k
Extracting a subsequence ( still denoted by wn )
k
We may assume that Twn  z
k
Then wn  z and z  N ( I  T )
k
On the other hand ,
dist(un , N ( I  T ) un  vn
dist( wn , N ( I  T ) 

1
un  vn
un  v n
In the lim it dist( z, N ( I  T ))  1, which is absurb.]
un  vn  is bounded sequence.


Then there is a subsequence s.t. T un  vn  
( 4)  un  vn  f  . Put g  f  
k
k
then g  Tg  f i.e. f  R ( I  T )
Hence R ( I  T ) is closed .
k
k
We have shown I  T is of closed image.
We can apply Theorem II .18
* 
R( I  T )  N ( I  T ) ;
R( I  T )  N ( I  T )
*

(c ) " "
Suppose the contrary. E1  R( I  T )  E
Then E1 is a Banach space and T ( E1 )  E1
Since T is compact , T
E1
is compact
(b)  ( I  T ) E1 is a closed subspace of E1
Furthermore, ( I  T ) E1  E1 because I  T is injective.
Let E2  ( I  T ) E1 , E3  ( I  T ) E2 ,, En  ( I  T ) En 1 ,
Then E  E1  E2  E3 , because I  T is injective.




By Riesz Lemma , un  En such that un  1 and
1
dist(un , En 1 ) 
2
For n  m, En 1  En  Em 1  Em
Tun  Tum  un  Tun   um  Tum   un  um
Since  un  Tun   um  Tum   un  Em 1 ,
1
Tun  Tum  dist(um , Em 1 ) 
2
Then Tun  contains no converging subsequence,
a contradiction.
 R( I  T )  E.
( c ) " "
Assume that R ( I  T )  E
Then N ( I  T * )  R ( I  T )   0, by Corollary II .17
Since T  K ( E ), by Schauder Theorem T  K ( E )
*
and then by previous proof we have R ( I  T )  E 
*
Again by Corollary II .17
N ( I  T )  R( I  T * )   0.
( d ) Let d  dim N ( I  T ) and d *  dim N ( I  T * )
First to show that d *  d
Suppose d  d .
*
N ( I  T ) admits a topo log y complement
and  continuous projection P
from E onto N ( I  T ).
* 
R( I  T )  N ( I  T )
 R ( I  T ) has co dim d
*
and then admits
a topo log y complement F of dim ension d * .
 a linear injection  : N ( I  T )  F .
Let S  T    P. Since T  K ( E ) and
P is finite rank , S  k ( E )
Claim : N ( I  S )  0
[0  u  Su  u  Tu    P u
 u  Tu  0 and   P u  0
 u  N ( I  T ) and then u  0
 u  0]
( c )  R( I  S )  E , which is impossible
because d  d and so f  F , f  R(  )
then the equation u  Su  f has no solution
Otherwise, u  Su  u  Tu    P u  f  F
then   P u  f 
*
Apply this to T
*
dim N ( I  T )  dim N ( I  T )  dim N ( I  T )
*
**
But N ( I  T )  N ( I  T )
**
T : E  E ,
T : E   E ) and E  E 
**
Hence dim N ( I  T )  dim N ( I  T )
*

Lemma VI.1 (Riesz-Lemma)
Let
For any
 
fixed , apply Green’s second
v

1( x,  ) in the domain
u  C ,  : C  domain, bounded
 \ B   and then let   0 we have
identity to u2and

u
( x,  ) 
u    ( x,  )udx   ( x,  )
u
ds

n x
n x 
