Compact operators

1
Chapter 6

• Compact operators
• Spectral Decomposition of
• self-adjoint compact operators

2
VI.1 Definition. Elementary Properties Adjoint
3
Compact Operator
every seq. has conv. subseq.
Let E and F be Banach spaces
An operator
is precompact
is called compact if
K(E,F) is the family of compact operators
in L(E,F) K(E,E)K(E)
4
Theorem VI . 1
K(E,F) is a closed vector subspace of L(E,F).
for the norm
5
(No Transcript)
6
Finite Rank
is called
An operator
finite rank if
A continuous operator T of finite rank is
compact.
7
Corollary VI. 2
be a sequence of continuous
Let
operators of finite rank from E to F and
such that
then
8
Remark 1 p.1
The family ltlt problem of approximationgtgt
(Banach, Grothendieck) concerns the
converse of Corollary VI.2 . Given a compat
operator T, does there exist a sequence
of operators of finite rank such that
9
Remark 1 p.2
In general, the answer is negative
(Enflo, 1972) even for certain closed
vector subspace of
see for example Lindenstrauss- Tzafriri 2
However the answer is affirmative in many
cases for example of F is Hilbert
10
(No Transcript)
11
Remark 1 p.3
From this proof, it follows that if F is
Schauder basis then the answer is
affirmative.
12
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.
13
Remark 2 p.2
Let X be a Topology space, F is a Banach
space and
is continuous
map such that T(X) is precompact in F.
Then for any
there is a continuous
map
of finite rank s.t.
14
(No Transcript)
15
Proposition VI.3
Let E, F and G be Banach spaces.
If
and
resp.

and
then
16
Equicontinuous
M is called equicontinuous if for any egt0,
with d(s,t)ltd
there is a dgt0 s.t.
17
Theorem (Arzelá-Ascoli)
is compact
A closed set
if and only if
(i) M is bounded on C(S)
(ii) M is equicontinuous
18
Theorem VI.4 (Schauder)
If
then
And conversely .
19
(No Transcript)
20
(No Transcript)
21
(No Transcript)
22
VI.2 The Riesz-Fredholm Theory
23
Lemma VI.1 (Riesz-Lemma)
Let E be a normed vector space and
be a closed vector subspace with
Then
such that
and
24
(No Transcript)
25
Remark
If dimMlt8 (or more generally if M is reflexive)
we can choose
in Lemma VI.1
But not in general case (see BT)
26
Theorem VI.5 (Riesz)
Let E be a normed vector space such that
is precompact. Then E is of finite
dimension.
27
(No Transcript)
28
Theorem II.18
be an unbounded
Let
closed linear operator with
then the following properties are equivalent
(i) R(A) is closed
(ii) R(A) is closed
(iii)
(iv)
29
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

1. L is closed.
2. GnL0 and GLE

see next page
30
In this case, all
can be expressed uniquely as zxy
with
It follows from Thm II.8 that the projections z?x
and z?y are linear continuous and surjective.
31
Example forTopological Complement
E Banach space Gfinite dimensional subspace of
E hence is closed.
Find a topological complement of G
see next page
32
(No Transcript)
33
(No Transcript)
34
(No Transcript)
35
Remark
On finite dimensional vector space, linear
functional is continuous.
Prove in next page
36
(No Transcript)
37
Remark
Let E be a Banach space. Let G be a closed
v.s.s of E with codimG lt 8, then any algebraic
complement is topological complement of G
Typial example in next page
38
Let
then
be a closed vector subspace of E and
codimGp
Prove in next page ?????
39
(No Transcript)
40
(No Transcript)
41
(No Transcript)
42
Lemma VI.6 (Fredholm Alternative)
Let E be a Banach space and
then
(a) N(I-T) is of finite dimension
(b) R(I-T) is closed and more precisely
(c)
(d)
43
(No Transcript)
44
(No Transcript)
45
(No Transcript)
46
(No Transcript)
47
(No Transcript)
48
(No Transcript)
49
(No Transcript)
50
(No Transcript)
51
(No Transcript)
52
(No Transcript)
53
Lemma VI.1 (Riesz-Lemma)
Let
For any
fixed , apply Greens second
identity to u and
in the domain
we have
and then let
54
(No Transcript)