Let $\mathcal H$ be a separable Hilbert space and let $(e_i)$ be some orthonormal basis. Let $K$ be a compact operator on $\mathcal H$ with matrix elements $K_{ij}=\langle K e_i,e_j\rangle$.

My goal is to compare the $l^p$ norm of the matrix $(K_{ij})$ with the $p$-Schatten norm $\|K\|_{S_p}$ of $K$. More precisely, assume that $$\sum_{i,j=1}^\infty |K_{ij}|^p <\infty .$$ Can one conclude that $K$ is in the $p$-Schatten class and that $$\|K\|_{S_p}\leq C\left(\sum_{i,j=1}^\infty |K_{ij}|^p\right)^{\frac 1 p}\;?$$ If not, can one conclude that $\|K\|_{S_q}<\infty$ for some $q>p$?

The answer is clearly positive for $p=2$, but for general $p$ it seems less obvious to me.

  • $\begingroup$ What doesn't sit right with me regarding this inequality is that as $p\to\infty$, the l.h.s. goes to the usual operator norm of $K$ and the r.h.s. goes to the largest matrix entry of $K$ in the chosen basis so the inequality would read $\|K\|_\text{op}\leq C|\langle e_i,Ke_j\rangle|$. However, the largest matrix entry of an operator with $\|K\|_\text{op}=1$ can become arbitrarily small (choose, e.g., something like $K=\langle x,\cdot\rangle x$ with $x=(\frac1{\sqrt n},\ldots,\frac1{\sqrt n},0,0,\ldots)$) so no constant $C$ can save this for all $K$. $\endgroup$ – Frederik vom Ende Apr 20 at 19:03
  • $\begingroup$ (2/2) Of course this is no direct counterexample but rather my intuition saying that for $p$ large enough (prob. even $p>2$) things might go wrong. I hope this can be of use somehow. $\endgroup$ – Frederik vom Ende Apr 20 at 19:05

I agree with the comments and I'm sure now that the answer to both my questions is negative. The simplest thing you could think of, $A_{mn}=\frac{1}{\sqrt{mn}}$, actually provides a counterexample. For arbitrary $p\in\mathbb N$, one has

$$ \|A\|_{2p} = \text{Tr}((A^*A)^{p})$$

$$ =\sum_{i_1,i_2,\dots,i_{2p}}A_{i_1i_2}\,\overline{A_{i_3i_2}}\, A_{i_3i_4}\, \overline{A_{i_5i_4}}\cdots A_{i_{2p-1}i_{2p}} \, \overline{A_{i_1i_{2p}}} $$

$$ = \sum_{i_1,i_2,\dots,i_{2p}} \frac{1}{\sqrt{i_1}}\frac{1}{\sqrt{i_2}}\frac{1}{\sqrt{i_3}}\frac{1}{\sqrt{i_2}}\cdots $$

$$=\underbrace{\sum_{i_2} \frac{1}{i_2}}_{\text{divergent!}}\sum_{i_1,i_3,\dots,i_{2p}}\cdots$$ No matter how large one chooses $p$, the above sum will always diverge for this $A$.


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