# Compact operators on normed spaces

1. Aug 3, 2011

### AxiomOfChoice

Is it easy to show that $T: X \to Y$ is a compact linear operator -- i.e., that the closure of the image under $T$ of every bounded set in $X$ is compact in $Y$ -- if and only if the image of the closed unit ball $\overline B = \{x\in X: \|x\|\leq 1\}$ has compact closure in $Y$? One direction is (of course) easy, since $\overline B$ is bounded, but I'm having trouble with the other one. Can someone help please? Thanks!

2. Aug 3, 2011

### micromass

Hi AxiomOfChoice!

Let A be a bounded set, then there exists an $\alpha>0$ such that

$$\alpha A\subseteq \overline{B}$$

(with of course $\alpha A:=\{\alpha a~\vert~a\in A\}$).

Then

$$\alpha T(A)=T(\alpha A)\subseteq T(\overline{B})$$

which means that the closure of $\alpha T(A)$ is compact, which implies that T(A) is compact.

I'll leave it to you to work out the details.

3. Aug 3, 2011

### zhentil

The definition way might be even easier. Take a bounded sequence x_n. We have to show that T(x_n) has a convergent subsequence. Take a subsequence x_j with x_j non-zero. Look at T(x_j/||x_j||). This has a convergent subsequence. Then T(x_j) = ||x_j|| T(x_j/||x_j||). The ||x_j|| have a convergent subsequence in R by boundedness, hence T(x_n) has a convergent subsequence.