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.

  • 4
    $\begingroup$ 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
  • 1
    $\begingroup$ 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


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 $\ell^1$.

  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.