Let $H$ be an infinite-dimensional separable Hilbert space and $(e_n)$ an orthonormal basis of $H$.
$T\in B(H)$ is called a Hilbert-Schmidt operator if $\sum_{n=1}^{\infty} ||Te_n||^2 < \infty$.
I want to show that every Hilbert-Schmidt operator is compact.
I am not getting the idea how to use the separability and if it doesn't matter whether the space is infinite or finite-dimensional.
But I know that if there is a series of compact finite rank operators that converges to T, then T is compact.
*I saw many versions of similar questions with different contexts.
Best Answer
Finite or infinite dimensional is not a big deal (in finite dimensions all linear operators are compact, so that is for free).
We use the separability of the space to ensure that the orthonormal basis is countable. To get compactness, we define the following sequence of finite rank operators:
$$ T_n(x)=\sum_{j=1}^n \langle x, e_j\rangle T(e_j).$$
Can you show that $T_n\rightarrow T$? Cauchy-Schwarz inequality and Parseval's identity are quite useful :) also the fact that $B(H)$ is a Banach space with respect to the operator norm is pretty handy.