TITLE: Non-compact operator $T\in\mathcal{B}(\ell_2)$ with $\|Te_n\|\rightarrow 0$
QUESTION [5 upvotes]: I'm looking for an example of a non-compact linear operator in $\mathcal{B}(\ell_2)$ such that, for an orthonormal basis $\{e_1,e_2,\ldots\}$ of $\ell_2$, we have $\lim_{n\rightarrow\infty}\|Te_n\|=0$. I've been trying shift operators but I'm hitting some snags. A forward unilateral shift is compact iff its weights converge to $0$ so I'm thinking such operators won't do the trick.

REPLY [3 votes]: If we think of $\{e_n\}$ as the canonical basis and represent $T$ in matricial form, we have that $Te_n$ is the $n^{\rm th}$ column of $T$. 
Consider $T=\bigoplus_k T_k$, where $T_k$ is the $k\times k$ matrix
$$
T_k=\frac1k\,\begin{bmatrix}1&\cdots&1\\ \vdots&\ddots&\vdots\\ 1&\cdots&1\end{bmatrix}.
$$
That is, these blocks appear in the diagonal of $T$. The norms of the colums are
$$
\left(\frac1{k^2}+\cdots+\frac1{k^2}\right)^{1/2}=\frac1{\sqrt k}.
$$
As $n$ grows, so does $k$ (too lazy to have an explicit formula, but $k$ goes to infinity as $n$ goes to infinity). 
And $T$ is not compact, because it is a projection but not finite-rank. It is self-adjoint and positive, though. 
It is worth noting that $T$ is an infinite-rank projection with infinite co-dimension. As any two such projections are unitarily equivalent, this shows that for any projection $P\in \mathcal B(\ell_2)$ with $\text{Tr}(P)=\text{Tr}(I-P)=\infty$, there exists a basis satisfying the conditions in the question.