Compact Action Spaces

Definition¹

Let $X$ and $Y$ be normed spaces, and let $T : X \to Y$ be an operator between these spaces. If for every bounded subset $M \subset X$ , the image $T(M)$ of the operator $T$ is precompact, then $T$ is called a compact operator.

Explanation

That $T(M)$ is precompact means its closure $\overline{T(M)}$ is compact. In other words, a compact operator is an operator that maps bounded sets to precompact sets.

The necessity to study compact operators stems from solving integral equations such as: $(T - \lambda I)x(s) = y(s) \qquad \text{where} \qquad Tx(s) = \int_{a}^{b} k(s, t)x(t) dt.$ where $\lambda \in \mathbb{C}$ is a constant, $y$ and the kernel $k$ are given functions. The unknown, i.e., the function to be found, is $x$ . David Hilbert discovered that the solvability of the above integral equation depends not on the integral form of $T$ but solely on the compactness of $T$ .

A compact operator is also called a completely continuous operator. This nomenclature comes from the following theorem. Generally, (a) does not imply (b), with (b) being a counterexample.

Theorem

Continuity Theorem: Let $X$ and $Y$ be normed spaces.

(a) Every compact linear operator $T : X \to Y$ is bounded. That is, it is continuous.

(b) If $X$ is infinite-dimensional, then the identity operator $I : X \to X$ is not compact (though continuous).

Proof

(a)

The unit ball $U = \left\{ x \in X : \left\| x \right\| = 1 \right\}$ is bounded. Assuming $T$ is compact, by the definition of a compact operator, $\overline{T(U)}$ is compact. Being compact implies boundedness, and according to the lemma below, the boundedness of $\overline{T(U)}$ means there exists a $c$ such that for all $Tx \in \overline{T(U)}$ , $\left\| Tx \right\| \le c$ holds.

Lemma
The following two propositions are equivalent.
The subset $M \subset X$ of the normed space $X$ is bounded.
There exists a positive $c \gt 0$ such that for all $x \in M$ , $\left\| x \right\| \le c$ is satisfied.

Therefore,

$\sup\limits_{x \in U} \left\| Tx \right\| \lt \infty$

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(b)

Assume $\dim X = \infty$ . The closed ball $B = \left\{ x \in X : \left\| x \right\| \le 1 \right\}$ is bounded. $\overline{I(B)} = \overline{B}$ cannot be compact according to Riesz’s lemma. Therefore, $I : X \to X$ is not a compact operator.

Riesz’s Lemma
For a normed space $X$ ,
$X$ is finite-dimensional. $\iff$ $\overline{ B ( 0 ; 1 ) }$ is compact.

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Properties

Compact Condition

Learn More

Let $X$ and $Y$ be normed spaces, and let $T : X \to Y$ be a linear operator. Then the following two propositions are equivalent.

$T$ is a compact operator.
$T$ maps “every bounded sequence in $X$ ” to “a sequence in $Y$ that has a converging subsequence”.

Vector Spaces