Let $T:H\to H$ be a continuous operator on a Hilbert space.
Assume there exists an orthonormal base $(e\_j)\_{j\in\mathbb N}$, such that the sequence $Te\_j$ tends to zero.


Must $T$ be compact?



Criterion for compactness
$T$ is not necessarily compact. Let me produce a counterexample.


Let $H$ be any infinite dimensional real or complex separable Hilbert space. Let $(f\_{j,k})\_{1\leq k\leq j},(e\_{j})\_{j=1}^{\infty}$ be orthonormal bases for $H$.
Then let $T:H\rightarrow H$ be the bounded linear operator defined by letting
$T(f\_{j,k})=\frac{1}{\sqrt{j}}\cdot e\_{j}$ whenever $1\leq k\leq j$. We observe that $T(f\_{j,k})\rightarrow 0$ as $j\rightarrow\infty$.


We observe that $$\biggl\|\sum\_{j=1}^{\infty}\sum\_{k=1}^{j}a\_{j,k}f\_{j,k}\biggr\|^{2}=\sum\_{j=1}^{\infty}\sum\_{k=1}^{j}a\_{j,k}^{2}
=\sum\_{j=1}^{\infty}\sum\_{k=1}^{j}a\_{j,k}^{2}
\geq\sum\_{j=1}^{\infty}\frac{1}{j}\biggl(\sum\_{k=1}^{j}a\_{j,k}\biggr)^{2}$$
$$=
\sum\_{j=1}^{\infty}\biggl(\frac{1}{\sqrt{j}}\sum\_{k=1}^{j}a\_{j,k}\biggr)^{2}
=\biggl\|\sum\_{j=1}^{\infty}\biggl(\frac{1}{\sqrt{j}}\sum\_{k=1}^{j}a\_{j,k}\biggr)e\_{j}\biggr\|^{2}=\biggl\|T\biggl(\sum\_{j=1}^{\infty}\sum\_{k=1}^{j}a\_{j,k}f\_{j,k}\biggr)\biggr\|^2.$$


Therefore, $T:H\rightarrow H$ is a bounded linear operator with norm $1$.


However, if $y\in H$, then $y=\sum\_{j}a\_{j}e\_{j}$, then let
$x=\sum\_{j=1}^{\infty}\sum\_{k=1}^{j}\frac{1}{\sqrt{j}}\cdot a\_{j}f\_{j,k}$.


Then $$\|x\|^{2}=\sum\_{j=1}^{\infty}\sum\_{k=1}^{j}(\frac{1}{\sqrt{j}}\cdot a\_{j})^{2}=\sum\_{j=1}^{\infty}\sum\_{k=1}^{j}\frac{1}{j}a\_{j}^{2}=\sum\_{j=1}^{\infty}a\_{j}^{2}=\|y\|^{2}.$$


However, $$T(x)=\sum\_{1\leq k\leq j}\frac{1}{j}a\_{j}e\_{j}=\sum\_{j=1}^{\infty}a\_{j}e\_{k}.$$


Therefore, $T$ cannot be compact since $T$ maps the unit ball of $H$ surjectively onto the unit ball of $H$.


In fact, $T$ maps the orthonormal set $(\frac{1}{\sqrt{j}}\sum\_{k=1}^{j}f\_{j,k})\_{j=1}^{\infty}$ to the orthonormal basis $(e\_{j})\_{j=1}^{\infty}$.


