Let $T$ be a compact symmetric operator on $\ell^2$ and $T\vert\_{\ell^1}$ be bounded on $\ell^1$. Are there any non-trivial conditions that $T\vert\_{\ell^1}$ is compact as well (for example would $T$ belonging to some Schatten-class on $\ell^2$ be sufficient)?


The obvious proof estimating


$$\left\lVert \sum\_{i=0}^{\infty} \lambda\_i \langle \cdot ,\varphi\_n \rangle \varphi\_n - \sum\_{i=0}^{k} \lambda\_i \langle \cdot ,\varphi\_n \rangle \varphi\_n \right\rVert\_{L(\ell^1)} $$ does not work as the eigenvectors, we get from the $\ell^2$ representation, are not necessarily bounded in $\ell^1$ norm.


Recall also that a set $M \subset \ell^1$ is compact if it is bounded, closed and $\lim\_{n \rightarrow \infty} \operatorname{sup}\_{x \in M} \sum\_{k=n}^{\infty} \left\lvert x\_k \right\rvert=0$ 


EDIT: Since I received an answer that outlined to me that it does not work in the non-symmetric case, I thought it would be good to explain why I think that symmetry could help. In this case, I can show that $\sigma(T) \subset \sigma(T\_{\vert\_{\ell^1}})$ and the point spectra coincide. Moreover, also all finite-dimensional eigenspaces are the same for each element of the point spectrum. In other words, the spectrum of $T\vert\_{ \ell^1}$ also contains the spectrum that one would assume to have for a compact operator. However, I could so far not exclude any continuous spectrum for $T\vert\_{ \ell^1}$ which would be a necessary condition for compactness. 



Compact operators on $\ell^1$
Here's an example showing that $T$ can be trace-class but $T|\_{\ell^1}$ is not compact.


Let $(x\_n)$ be a sequence of vectors in $\ell^2$ with disjoint supports, $\sum\_n \|x\_n\|\_2 \leq 1$ and $\|x\_n\|\_1=1$ for all $n$. Define
$$ T(\xi) = \sum\_n \xi\_n x\_n \qquad (\xi\in\ell^2). $$
Then $\| T(\xi) \|\_1 \leq \sum\_n |\xi\_n| \|x\_n\|\_1 \leq \|\xi\|\_1$ so $T$ is bounded on $\ell^1$. However, $(T(e\_n)) = (x\_n)$ has no convergent subsequence so $T$ is not compact on $\ell^1$. As $\sum\_n \|x\_n\|\_2\leq 1$, $T$ is trace-class.


An example of such a sequence is as follows. Let $N(n)$ be a rapidly increasing sequence of integers, and choose $x\_n$ to be the sequence $(0,\cdots,0,N(n)^{-1},\cdots,N(n)^{-1},0,\cdots)$ where we repeat $N(n)^{-1}$ exactly $N(n)$ times, and we place the non-zero terms so that the $x\_n$ have disjoint support. Then $\|x\_n\|\_1=1$ but $\|x\_n\|\_2 = N(n)^{-1/2}$ so as long as $N(n)$ increases fast enough that $\sum\_n N(n)^{-1/2} \leq 1$ we're done.


This $T$ is not self-adjoint, but notice that $T^\*$ is trace-class, and $T^\*$ is still bounded on $\ell^1$. Furthermore, $S=T+T^\*$ is seen to still be such that $S(e\_n)$ has no convergent subsequence in $\ell^1$.


