Reflexive Banach spaces, compactness

Question

Let $X$ be a reflexive Banach space. Then, consider a linear and compact operator $T \colon X \to X$.

Prove that if:

$\inf \left\{ \|Tx\| : x \in X\quad \text{s.t.}\quad \|x\| = 1 \right\} > 0$,

then $S = \{x \in X \colon \|x\| = 1 \}$ is compact.

My idea is to prove that $S$ is sequentially compact; since $X$ is reflexive, every sequence in $S$ admits a weakly convergent subsequence...

But, how can I switch to strong convergence??

Thanks, any hint is appreciated...

Answer 1

I don't see the point of reflexivity here, $inf\{\|T(x)\|, \|x\|=1\}>0$ implies that the spectrum of $T$ does not contain zero. Since $T$ is compact, $X$ must be finite dimensional so $S$ is compact.

Answer 2

Reflexivity is superfluous. You can use the simple fact that if $T:X\rightarrow X$ is a compact map on a Banach space $X$, with $\dim(X)=\infty$, then $0\in \sigma(T)$.