A sufficient and necessary condition for a special linear operator to be compact

Question

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"?

Answer 1

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.

Answer 2

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

Answer 3

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.