Suppose $\mathcal H$ is a separable Hilbert space and $T$ is a compact self-adjoint operator on $\mathcal H$. Let $\{e_n\}$ be an orthonormal basis for $\mathcal H$. Fix $1<p<2$.

Does $T\in$Schatten p-class imply that $\displaystyle\sum_{m,n}|\langle Te_n,e_m\rangle|^p<+\infty$

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    $\begingroup$ $Te_n$ can be anything, and in particular it can have arbitrary $\ell^2$ expansion coefficients wrt $\{ e_m\}$, so there's no reason for the sum (over just $m$) to be finite for $p<2$. $\endgroup$ – Christian Remling May 20 '15 at 19:10
  • $\begingroup$ If I remember correctly, from a paper of Kehe Zhu, for $p>2$ this implication holds. But usually the case $p\in(1,2]$ is different. $\endgroup$ – BigM May 20 '15 at 19:16
  • $\begingroup$ Do you mean $\sum_n|\langle Te_n,f_n\rangle|<\infty$ for any two ONB $(e_n)$, $(f_n)$? $\endgroup$ – user1688 May 20 '15 at 19:44
  • $\begingroup$ No its a double-sum over an ONB namely $\displaystyle\sum_{m,n}|\langle Te_n,e_m\rangle|^p<+\infty$ $\endgroup$ – BigM May 20 '15 at 19:46
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    $\begingroup$ @YemonChoi: This is no problem at all: Take a rank one $A=\langle e_1, \cdot \rangle f$ as above, with $f$ chosen such that $\sum |\langle f, e_m\rangle |^p=\infty$, and let $T=A+A^*$ (this won't mess up the divergence because $R(A)=L(e_1)$). $\endgroup$ – Christian Remling May 22 '15 at 22:43

This is an elaboration of Christian Remling's comment above. The answer is no. Not even for a finite rank operator. Take $\cal H$ to be $l^2({\mathbb N})$ and choose any sequence $(a_n)$ in $l^2({\mathbb N})$ which is not in $l^p({\mathbb N})$. Let $(e_n)$ be the usual standard ONB. Let $T(e_1)=(a_n)$ and $T(e_j)=0$ for $j>1$. This operator has rank one. But $$ \sum_{n,m}|\langle Te_n,e_m\rangle|^p=\sum_{m}|a_m|^p=\infty. $$

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    $\begingroup$ Isn't this exactly my comment from above? $\endgroup$ – Christian Remling May 21 '15 at 17:33
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    $\begingroup$ @ChristianRemling: yes, of course, but as the proposer didn't react, even answered to my question, I figured he might not have understood, so I expanded the answer. Besides, maybe he didn't take it, since it was not formulated as an answer, but as a doubt. $\endgroup$ – user1688 May 22 '15 at 16:35
  • $\begingroup$ See my comment above: $T$ is assumed to be self-adjoint $\endgroup$ – Yemon Choi May 22 '15 at 20:01
  • $\begingroup$ It seem self-adjointness can be resolved by considering $\sqrt{T^*T}$, instead of $T$ in Corbennick's example. unless I'm missing something in here. $\endgroup$ – BigM May 22 '15 at 20:17
  • $\begingroup$ @BigM: yes, or $T+T^*$. $\endgroup$ – user1688 May 23 '15 at 5:02

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