# Reflexive Banach spaces, compactness

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...

• I don't understand why it matters that there exists $T$ with non-zero norm (which is what the statement is saying I believe) – Tom83B Jan 30 '16 at 22:12
• also sphere should always be compact I think (I checked here): math.stackexchange.com/questions/382211/compactness-of-a-sphere (don't take my comments as something, that is necessarily true. I'm studying for my functional analysis exam right now so I don't understand it very well) – Tom83B Jan 30 '16 at 22:15

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.
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 $0\in \sigma(T)$.
The assumption translates in the following way: there exists a positive constant $c$ such that for all $x\in X$, $$\tag{*} \left\lVert Tx\right\Vert\geqslant c\left\lVert x\right\Vert.$$ Let $\left(x_n\right)_{n\geqslant 1}$ be a sequence of elements of $S$. Since $S$ is bounded and $T$ is compact, we can extract a subsequence $\left(x_{n_k}\right)_{k\geqslant 1}=\left(x'_k\right)_{k\geqslant 1}$ such that the sequence $\left(Tx'_k\right)_{k\geqslant 1}$ converges in norm to some $y$. Apply (*) to $x=x'_m-x'_n$ to show that the sequence $\left(x'_n\right)_{n\geqslant 1}$ is Cauchy.