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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.

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  • $\begingroup$ 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. $\endgroup$
    – Daniel Fischer
    Apr 19, 2014 at 14:01
  • $\begingroup$ Have you learned Riesz's lemma yet? $\endgroup$
    – user137301
    Apr 19, 2014 at 14:09
  • $\begingroup$ @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 $B (X) \subsetneq K(X)$? $\endgroup$
    – student
    Apr 20, 2014 at 9:06
  • $\begingroup$ Is the other part of the proof correct? $\endgroup$
    – student
    Apr 20, 2014 at 9:07
  • $\begingroup$ 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). $\endgroup$
    – Daniel Fischer
    Apr 20, 2014 at 22:48

1 Answer 1

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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.

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