Let $K(H,H)$ be a linear bounded operator. Is it true that given an orthonormal basis $\{e_n\}_n$ if $Ke_n\to 0$ then $K$ is compact?
I know that in an Hilbert space $K$ is compact iff it is weak-strong convergent, so the question can be also expressed as:
If an operator is strong convergent on an orthonormal basis is it always weak-strong convergent?
Best Answer
The answer is no. For a counterexample you can take this one: on $L^2(0,1)$, define $$Kf: x \mapsto \frac{1}{x}\int_0^x f(y)dy.$$ Consider the following orthonormal basis $e_n :=\sqrt{2} \sin(n \pi x)$, $n\in \mathbb{N}$. After some calculations you get $$K e_n \to 0.$$ However $K$ is not compact as proved in the answers of the above link.