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

Question

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.

Answer 1

Yes, the proof is correct. The following are equivalent:

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.

The equivalence of 1 and 2 is classical; the equivalence of 2 and 3 is a tautology.

Thus, $B(X)=K(X)$ implies $X$ is finite dimensional.

Conversely; in a finite dimensional case 2 holds, which implies that every bounded operator is compact.