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Suppose $X$ is Banach and $T\in B(X)$ (i.e. $T$ is a linear and continuous map and $T:X \to X$). Also, suppose $\exists c > 0$, s.t. $\|Tx\| \ge c\|x\|, \forall x\in X$. Prove $T$ is a compact operator if and only if $X$ is finite dimensional.

"$X$ is finite dimensional $\implies$ $T$ is compact" is easy to show. To prove the other side, at first, I made a mistake, thinking $X$ is reflexive. Then this work can be easily done by the fact that any sequence of a reflexive linear space has a weakly convergent subsequence and $T$ is completely continuous (since $T$ is compact). But this is not the situation.

So how to prove "$T$ is compact $\implies X$ is finite dimensional"?

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Here is an idea: let $T(X) \ni y_n=T(x_n)\to y\in \overline{T(X)}$. Then, since $\|Tx_n\| \ge c\|x_n\|,\ x_n\to x\in X$ and continuity of $T$ implies that $y_n=T(x_n)\to T(x)$ and so $y=T(x)$. Therefore, $T$ has closed range.

$T(B_X)$ contains a ball in the Banach space $T(X)$, by the open mapping theorem, and since $\overline {T(B_X)}$ is compact, $T(X)$ is locally compact, hence finite dimensional. But $T$ is injective, so $X$ must be finite dimensional ,too.

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Hint: if not, the image of the unit ball of $X$ contains a ball in an infinite-dimensional space.

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Suppose that $X$ is infinite-dimensional. Then its unit ball is not compact, so there exists a sequence $\{x_n\}$ in the unit ball that admits no convergent subsequence; by replacing with a subsequence if necessary, we may assume that there exists $\delta>0$ with $\|x_n-x_m\|\geq\delta$ for all $n\neq m$. Now $$ \|Tx_n-Tx_m\|=\|T(x_n-x_m)\|\geq c\|x_n-x_m\|\geq c\delta>0, $$ so $\{Tx_n\}$ does not admit a convergent subsequence.

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  • $\begingroup$ That's not true. For example, the sequence $1,1,2,2,3,3,\ldots$ admits no convergent subsequence, but there is no such $\delta$. $\endgroup$ – Robert Israel May 19 at 3:52
  • $\begingroup$ The key word is "unit ball" here. I'll clarify, since it was not obvious. $\endgroup$ – Martin Argerami May 19 at 3:54
  • $\begingroup$ That doesn't help. In the unit ball of a Hilbert space with orthonormal basis $u_n$, the sequence $u_1, u_1, u_2, u_2, u_3, u_3, \ldots$ admits no convergent subsequence, but there is no such $\delta$. $\endgroup$ – Robert Israel May 19 at 16:05
  • $\begingroup$ What is true is that in the unit ball of an infinite-dimensional normed space there is a sequence $x_n$ such that $\|x_n - x_m\| \ge \delta$ for all $n \ne m$ (and you can take any $\delta$ with $0 < \delta < 1$). But it's not just any sequence that admits no convergent subsequence. $\endgroup$ – Robert Israel May 19 at 16:15
  • $\begingroup$ Yes, good point. I have edited the answer. $\endgroup$ – Martin Argerami May 19 at 17:23

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