Characterization of compact operators on Hilbert spaces

Question

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?

Is there an $l^2$ sequence $(c_1,c_2,\dots)$ with $c_1+\dots+c_n = \Omega(\sqrt{n})$? — mathworker21, Feb 11 '20 at 14:43

score 5 · Accepted Answer · answered Feb 11 '20 at 15:19

5

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.

answered Feb 11 '20 at 15:19

S. Maths

2,238

+1. would you mind letting me know the answer to the question I asked in the comment above? I'm curious... – mathworker21 Feb 11 '20 at 15:34
+1. If $Ke_n\to 0$ for $every$ orthonomal basis ${e_n}_n,$ is $K$ compact? – DanielWainfleet Feb 12 '20 at 03:23
3

@DanielWainfleet Yes. See e.g. https://math.stackexchange.com/questions/66329/a-question-on-compact-operators – Sten Feb 12 '20 at 07:16

Characterization of compact operators on Hilbert spaces

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