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

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"?

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.

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

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.

• That's not true. For example, the sequence $1,1,2,2,3,3,\ldots$ admits no convergent subsequence, but there is no such $\delta$. – Robert Israel May 19 at 3:52
• The key word is "unit ball" here. I'll clarify, since it was not obvious. – Martin Argerami May 19 at 3:54
• 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 $\delta$. – Robert Israel May 19 at 16:05
• 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. – Robert Israel May 19 at 16:15
• Yes, good point. I have edited the answer. – Martin Argerami May 19 at 17:23