Compact and bounded if and only if $X$ is finite dimensional

I tried to prove the theorem

Let $X$ be a Banach space. Then $K(X) = B(X)$ if and only if $X$ is finite-dimensional.

Please could someone check my proof?

(The implication where $X$ is finite dimensional is clear.)

For the other direction: I will show that if $X$ is an infinite dimensional Banach space then $K(X) \subsetneq B(X)$ by giving an example of $T$ such that $T: X \to X$ is bounded but not compact. Take $T$ to be the identity operator. Then $T$ is bounded but it does not map the unit ball to a relatively compact set because the closed unit ball is compact if and only if $X$ is finite dimensional.

• You need to show that for every infinite-dimensional $X$ you have $K(X)\neq B(X)$. Of course the identity is not compact for any infinite-dimensional $X$, but showing it for the example $\ell^\infty(\mathbb{N})$ does not suffice. Apr 19, 2014 at 14:01
• Have you learned Riesz's lemma yet? Apr 19, 2014 at 14:09
• @DanielFischer But then couldn't I argue like this: If $X$ is infinite dimensional then the identity is an example of a bounded operator that is not compact. Therefore $B (X) \subsetneq K(X)$? Apr 20, 2014 at 9:06
• Is the other part of the proof correct? Apr 20, 2014 at 9:07
• Yes, the first part is correct. As a side remark (not sure whether you know), for a finite-dimensional (and Hausdorff) $X$, all linear operators are continuous (bounded). Apr 20, 2014 at 22:48

1. $X$ is finite dimensional.
2. Closed unit ball of $X$ is compact in the norm topology.
3. The identity operator on $X$ is compact.
Thus, $B(X)=K(X)$ implies $X$ is finite dimensional.