Let $f$ be a bounded linear operator $X \to X$ s.t that $\|f(x)\|\geq m\|x\|$ for some $m$, $\forall x \in V$. Prove that $f$ cannot be compact.

Question

Let $V$ is an infinite dimensional subpace of a Banach space and let $f$ be a bounded linear operator $X \to X$ s.t that $\|f(x)\|\geq m\|x\|$ for some $m$, $\forall x \in V$. Prove that $f$ cannot be compact. My solution relies ont he fact that we can find a sequence in $V$ s.t $\|e_n-e_m\|>\frac{1}{2}$. Then if $f$ is compact we know $f(e_{n_k})$ has a convergent subsequence. But that is a contradiction as that would imply $e_{n_k}$ converges.

I was wondering whether there is a different approach to the problem, one that relies on the fact that $f(B_1)$ would be precompact perhaps. I do not like using the existence of the sequence $e_n$.

Answer 1

By considering the restriction $f: V \mapsto f(V)$ we may assume that we are in the situation where $f:X \to Y$ is a surjective linear operator between infinite dimensional spaces and $\|f(x)\| \geq m\|x\|$ for all $x$.

The given assumption implies that $f$ is injective and that its algebraic inverse is continuous so that $f$ is invertible.

If $f$ is compact then $\operatorname{Id}: X \to X$ is also compact since $\operatorname{Id} = f^{-1} \circ f$ and the composition of a compact operator with a bounded operator is again compact. However, on an infinite dimensional space, the identity operator is not compact. Hence $f$ cannot be compact.