Prove that $T$ is compact if and only if $\mu_n \to \infty$ when $n \to \infty$

Question

Let $\{e_n\}$ be an orthonormal basis for a Hilbert space $H$, $T: H \rightarrow H$ be a bounded linear mapping and $$\mu_n = \sum_{x \perp \{e_1, \cdots, e_n\}, x \neq 0}\frac{\|T(x)\|}{\|x\|^2}$$ Prove that $T$ is compact if and only if $\mu_n \to \infty$ when $n \to \infty$.

I've done this exercise reasoning by contradiction, but the proof is very cumbersome and doesn't look very good. For this reason I would like to know if there is any direct proof for these two implications without having to reason by absurdity. I appreciate any help.

Answer 1

Let $P_n$ denote the projection on the orthogonal complement of $e_1,e_2,\ldots,e_n.$ Define $$\mu_n=\sup_{x\perp e_1,\ldots, e_n, \ \|x\|=1}\|Tx\|$$Then $\mu_n=\|TP_n\|.$ The sequence $P_n$ tends to $0$ pointwise. Hence, if $T$ is compact, $TP_n$ tends to $0$ with respect to the operator norm, i.e. $\mu_n$ tends to $0.$

On the other hand $$\|TP_n\|=\|T-T(I-P_n)\|=\mu_n.$$ If $\mu_n$ tends to $0,$ the operator $T$ is a limit of finite dimensional operators $T(I-P_n).$ Hence it is compact.