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Let $(\lambda_{mn})_{(m,\:n)\in\mathbb N^2}\subseteq[0,\infty)$ with $$\lambda_{mn}^2\le\mu_m\mu_n\;\;\;\text{for all }(m,n)\in\mathbb N^2\tag1$$ for some $(\mu_n)_{n\in\mathbb N}\subseteq[0,\infty)$ such that $\sum_{n\in\mathbb N}\mu_n$ exists in $\mathbb R$. Let $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of a separable $\mathbb R$-Hilbert space $H$ and $$e_m\otimes e_n:=\langle\;\cdot\;,e_m\rangle_He_n\;\;\;\text{for }(m,n)\in\mathbb N^2\;.$$ Let $\mathfrak L_1(H)$ denote the space of nuclear operators on $H$.

Can we show that $$A:=\sum_{(m,\:n)\in\mathbb N^2}\lambda_{mn}e_m\otimes e_n$$ is a bounded linear self-adjoint operator on $H$?

I've tried the following: Let $\operatorname{HS}(H)$ denote the space of Hilbert-Schmidt operators on $H$. Note that $(e_m\otimes e_n)_{(m,\:n)\in\mathbb N^2}$ is an orthonormal basis of $\operatorname{HS}(H)$ and hence $A$ exists in $\operatorname{HS}(H)$ if and only if $\sum_{(m,\:n)\in\mathbb N^2}\left\|\lambda_{mn}e_m\otimes e_n\right\|_{\operatorname{HS}(H)}^2$ exists in $\mathbb R$. Now, \begin{equation}\begin{split}\sum_{(m,\:n)\in\mathbb N^2}\left\|\lambda_{mn}e_m\otimes e_n\right\|_{\operatorname{HS}(H)}^2&=\sum_{(m,\:n)\in\mathbb N^2}\lambda_{mn}^2\underbrace{\left\|e_m\otimes e_n\right\|_{\operatorname{HS}(H)}^2}_{=\:1}\\&\le\sum_{(m,\:n)\in\mathbb N^2}\mu_m\mu_n\\&=\left(\sum_{m\in\mathbb N}\mu_m\right)\sum_{n\in\mathbb N}\mu_n\end{split}\tag2\end{equation} and hence $A$ exists in $\operatorname{HS}(H)$. However, I've no idea how I can show the self-adjointness.

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  • $\begingroup$ Start with looking at $\langle A e_r, e_s\rangle$ and $\langle e_r, Ae_s\rangle$. $\endgroup$ – Daniel Fischer May 28 '17 at 19:37
  • $\begingroup$ @DanielFischer I've started with that. And, clearly, $e_m\otimes e_n$ itself is not self-adjoint. $\endgroup$ – 0xbadf00d May 28 '17 at 19:39
  • $\begingroup$ Can you find a necessary condition on the family $\{ \lambda_{mn}\}$ for self-adjointness? $\endgroup$ – Daniel Fischer May 28 '17 at 19:41
  • $\begingroup$ @DanielFischer I know you cannot see how this question is related, but do you know if the "square-root" (which is uniquely determined for any nonnegative and self-adjoint bounded linear operator) exists even when the operator in question isn't self-adjoint? (I didn't find a proof and I've noticed that some authors call an operator $A$ nonnegative if it is self-adjoint and $\langle Ax,x\rangle_H\ge0$ for all $x\in H$, while others (including me) call it nonnegative if only the second condition is satisfied) $\endgroup$ – 0xbadf00d May 28 '17 at 20:01
  • $\begingroup$ So you're asking whether for all non-negative $A$, there is a (non-negative, presumably) $B$ such that $A = B^2$? Off the top of my head, I don't know. (If we only were dealing with complex vector spaces.) $\endgroup$ – Daniel Fischer May 28 '17 at 20:07
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You already know that $A$ is bounded, since it is Hilbert-Schmidt.

Knowing that $A$ is bounded, the following statements are equivalent:

  • $A$ is selfadjoint

  • $\lambda_{mn}=\lambda_{nm}$ for all $n,m$.

So, in general, your $A$ will not be selfadjoint.

The relevant computation is $$ \langle Ae_n,e_m\rangle =\lambda_{mn}, \ \ \ \ \langle A^*e_n,e_m\rangle=\lambda_{nm}. $$ together with the fact that it is enough to test the selfadjoint condition on elements of a fixed orthonormal basis.

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  • $\begingroup$ In fact, "my" $\lambda_n$'s satisfy your condition. Could you provide a proof for the equivalence? $\endgroup$ – 0xbadf00d May 28 '17 at 19:42
  • $\begingroup$ Yes, I added a bit to the answer. $\endgroup$ – Martin Argerami May 28 '17 at 19:43
  • $\begingroup$ And let me note the following: If you're right and $A$ is self-adjoint, then your deleted answer to my other question was correct! The self-adjointness ensures the existence of a unique square-root and this allows us to show that a bounded, linear, nonnegative and self-adjoint operator is nuclear iff it has finite trace. $\endgroup$ – 0xbadf00d May 28 '17 at 19:43
  • $\begingroup$ I think there is a typo: It should be $\langle Ae_n,e_m\rangle =\lambda_{mn}$. $\endgroup$ – 0xbadf00d May 28 '17 at 19:51
  • $\begingroup$ I don't follow. In what sense would a self-adjoint operator have a "unique" root? $\endgroup$ – Martin Argerami May 29 '17 at 1:29

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