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.


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.


Note that $T:l^2\rightarrow l^2$ is compact if and only if its adjoint $T^*$ is compact. Furthermore, since $$\|Tx\| = \sup\limits_{\|y\|\le1}{|\langle Tx,y\rangle|} = \sup\limits_{\|y\|\le1}{|\langle x,T^*y\rangle|}$$ it follows that $$\lim\limits_{n\rightarrow\infty}{\|Te_n\|} = 0\iff \lim\limits_{n\rightarrow\infty}{\sup\limits_{\|y\|\le1}{|\langle e_n,T^*y\rangle|}} = 0$$ i.e. $\|Te_n\|\rightarrow 0$ if and only if there exists a function $f:\mathbb{N}\rightarrow\mathbb{R}_{\ge 0}$ such that $f(n)\rightarrow 0$ as $n\rightarrow\infty$ and the modulus of the $n$th component of $T^*y$ is uniformly bounded by $f(n)$ over all $\|y\|\le 1$. We thus reduce our problem to the adjoint problem of finding a noncompact bounded linear operator satisfying the above property.

Let $S:l^2\rightarrow l^2$ be defined by $$ S(e_i) = \frac{1}{\sqrt{i}}\sum\limits_{j=\frac{i(i-1)}{2}+1}^{\frac{i(i+1)}{2}}{e_j}$$ i.e. \begin{align} S(e_1) &= e_1 \\ S(e_2) &= \frac{1}{\sqrt{2}}(e_2+e_3) \\ S(e_3) &= \frac{1}{\sqrt{3}}(e_4+e_5+e_6),\text{ etc.} \end{align} Then $\|S(e_i)\| = 1$ for all $i$, and since $\{S(e_i)\}_i$ is also an orthonormal sequence, it follows that if $x = \sum\limits_{i=1}^{\infty}{a_ie_i}$, then $$\|S(x)\|^2 = \left\|\sum\limits_{i=1}^{\infty}{a_iS(e_i)}\right\|^2 = \sum\limits_{i=1}^{\infty}{|a_i|^2\|S(e_i)\|^2} = \sum\limits_{i=1}^{\infty}{|a_i|^2} = \|x\|^2.$$ Hence, $S$ is bounded as well. Since $\{S(e_i)\}$ is an orthonormal sequence, it cannot have a convergent subsequence, so $S$ is not compact. Finally, note that for $x = \sum\limits_{i=1}^{\infty}{a_ie_i}$, the $n$th component of $S(x)$ is $\frac{a_i}{\sqrt{i}}$, where $i$ is the unique integer satisfying $\frac{i(i-1)}{2} < n < \frac{i(i+1)}{2}$. Since $i$ is a positive integer, for all $i$ we have $i^2\ge\frac{i(i+1)}{2}\ge n\implies i\ge \sqrt{n}$, and $|a_i|\le \|x\|$ for all $i$, so $$\sup\limits_{\|x\|=1}{|\langle e_n,S(x)\rangle|} \le \sup\limits_{\|x\|=1}{\frac{\|x\|}{n^{1/4}}} = \frac{1}{n^{1/4}} \rightarrow 0 $$ as $n\rightarrow\infty$. Hence, $S$ has all of the desired properties of the adjoint. It follows that $T = S^*$ is a noncompact bounded linear operator such that $\|Te_n\|\rightarrow 0$ as $n\rightarrow\infty$.

For completeness, the operator $T:l^2\rightarrow l^2$ is defined by $$ T(e_n) = \frac{1}{\sqrt{i}}e_i $$ where $i$ is the unique positive integer satisfying $\frac{i(i-1)}{2} < n \le \frac{i(i+1)}{2}$.


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