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.
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 $A$.