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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.

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    $\begingroup$ To get an example it is enough to observe that you can find vectors $x$ in $\ell_1$ s.t. the ratio of $\|x\|_1 \cdot \|x\|_\infty $ to $\|x\|_2^2$ is arbitrarily large. Consider, for example, $x_n :=n^{1/2} e_1 + \sum_{k=2}^n e_k$; the desired ratio is about $n^{1/2}$. Take $\sum_{n\in S} n^{-3/2} x_n\otimes x_n$ where the $S$ is infinite, $\sum_{n\in S} n^{-1/2} < \infty$, and the sum is a direct sum. $\endgroup$ Jul 14, 2017 at 16:33
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    $\begingroup$ Regarding your edit, I think the problem is that the spectral containment you mention doesn't really say anything about the multiplicity of points in the spectrum. So the points which are non-zero eigenvalues for your compact operator on $\ell^2$ might still be eigenvalues for the operator on $\ell^1$, but with "infinite multiplicity" now. $\endgroup$
    – Yemon Choi
    Jul 14, 2017 at 21:07

1 Answer 1

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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 $\ell^1$.

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    $\begingroup$ Is this $T$ self-adjoint (as an operator on $\ell^2$)? $\endgroup$
    – Yemon Choi
    Jul 14, 2017 at 13:20
  • $\begingroup$ In this construction the columns look like disjoint, large blocks. Could you "inflate horizontally" and then rescale to get something block-diagonal which has the right properties? I guess this comes down to comparing various norms on the $k\times k$ matrix whose entries are all $1$. $\endgroup$
    – Yemon Choi
    Jul 14, 2017 at 13:37
  • $\begingroup$ I think I fixed this quite easily in the symmetric case, just by an obvious "symmetrisation" procedure. As the columns of T are long, flat, disjoint blocks, the rows of $T^*$ are the same, and so don't affect (much) the $\ell^1$ norm and don't interact badly with $T$. $\endgroup$ Jul 15, 2017 at 9:53

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