# Relation between Schatten-$p$-norm and $l^p$ norm of operator matrix

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.

• 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 $K$. – Frederik vom Ende Apr 20 at 19:03
• (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. – 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 $$A$$.