The vanishing property of a compact operator acting on an orthonormal basis of a Hilbert space

Question

The question is:

Suppose $H$ is a Hilbert space, $A \in K(H)$, and $\{e_n\}$ is an orthonormal basis of $H$. Prove $\lim\limits_{n \to \infty} \langle Ae_n, e_n \rangle = 0$.

$\langle \cdot, \cdot \rangle $ is the inner product defined on $H$.

$A \in K(H)$ means $A$ is a compact operator (thus linear and continuous) mapping $H$ to itself.

There are numerous ways to define a compact operator. Suppose $X, Y$ are Banach spaces and $T:X\to Y$ is a linear operator. The followings are equivalent:

1. $T$ is compact
2. the closure of the image of any bounded set (or just consider the unit ball in $X$) is compact (i.e. let $E$ be bounded in $X$, then $\overline{TE}$ is compact in $Y$)
3. the image of any bounded set (or just consider the unit ball in $X$) is totally bounded (i.e. $TE$ is totally bounded in $Y$ if $E$ is bounded in $X$)
4. the image of any bounded sequence in $X$ has a convergent subsequence in $Y$ (i.e. if $\{x_n\}$ is bounded in $X$, then $\{Tx_n\}$ has a convergent subsequence

I totally have no idea how to prove this amazing effect.

Answer 1

Take $\varepsilon>0$. As $A$ is compact, there is a finite rank operator $T$ such that $\|A-T\|\leq\varepsilon$. Then, $$|\langle Ae_n,e_n\rangle|\leq|\langle Te_n,e_n\rangle|+\varepsilon$$

But as $T$ has finite rank, you know that $\mathrm{Im} T\subset \mathrm{Vect}(e_1,\cdots,e_N)$ for some $N\geq 0$. This will imply that $$\langle Te_n,e_n\rangle=0$$ as soon as $n> N$.

This shows what we wanted.