Page 347 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes 31.1.8 Compact Operators

Definition: A subset A in a normed linear space N is said to be relatively compact if its closure A
is compact.
A linear transformation T of a normed linear space N into a normed linear space N is said to be
a compact operator if it maps a bounded set of N into a relatively compact set in N , i.e.

T : N N is compact of every bounded set B N, T(B) is compact in N .

31.1.9 Properties of Compact Operators

1. Let T : N N be a compact operator. Then T is bounded (continuous) linear operator. For,
let B be a bounded set in N. Since T is compact, T(B) is compact in N . So T(B) is complete
and totally bounded in N . Since a totally bounded set is always bounded, T(B) is bounded
and consequently T(B) is bounded, since a subset of a bounded set is bounded.
T is a bounded linear transformation and it is continuous.

2. Let T be a linear transformation on a finite dimensional space N. Then T is compact
operator. For, N is finite dimensional and T is linear T(N) is finite dimensional. Since any
linear transformation on a finite dimensional space is bounded. T(B) is bounded subset of
T(N) for every bounded set B N. Now if T(B) is bounded so is T(B) and is closed. T(N) is
finite dimensional, any closed and bounded subset of T(N) is compact, so that T(B) is
compact, being closed and bounded subset of T(N).
3. The operator O on any normed linear space N is compact.
4. If the dimension of N is infinite, then identity operator I : N N is not compact operator.

For consider a closed unit sphere.
S = {x N : x 1} then S is bounded.
Since N is a infinite dimensional.

I (S) = S = S is not necessarily compact.
Hence I : N N is not compact operator. But I is a bounded (continuous) operator.
Theorem: A set A in a normed linear space N is relatively compact every sequence of points
in A contains a convergent sub sequence.
Proof: Let A is relatively compact.

Since A A , every sequence in A is also sequence in A . Since A is compact, such a sequence
in A contains a convergent subsequence. Hence every sequence in A has a convergent
subsequence.

Conversely, let every sequence in A has a convergent subsequence.
Let (y ) be a sequence of points in A . Since A is dense in A , a sequence (x ) of points of A s.t.
n n
1
x y … (1)
n n
n

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