0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $A$.

$\endgroup$

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.