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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.

0xbadf00d
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  • Start with looking at $\langle A e_r, e_s\rangle$ and $\langle e_r, Ae_s\rangle$. – Daniel Fischer May 28 '17 at 19:37
  • @DanielFischer I've started with that. And, clearly, $e_m\otimes e_n$ itself is not self-adjoint. – 0xbadf00d May 28 '17 at 19:39
  • Can you find a necessary condition on the family ${ \lambda_{mn}}$ for self-adjointness? – Daniel Fischer May 28 '17 at 19:41
  • @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) – 0xbadf00d May 28 '17 at 20:01
  • 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.) – Daniel Fischer May 28 '17 at 20:07
  • @DanielFischer If $H$ is complex, then a nonnegative operator is already self-adjoint. – 0xbadf00d May 28 '17 at 20:09
  • Yes, that's the point of that remark ;) – Daniel Fischer May 28 '17 at 20:09
  • @DanielFischer Let me note why I'm asking: I have no clue how I can prove the desired statement of my other question, unless I can show that the operator in question is nuclear (trace class) iff $\sum_{n\in\mathbb N}\langle Ae_n,e_n\rangle_H<\infty$ for all orthonormal bases $(e_n)_{n\in\mathbb N}$ of $H$. – 0xbadf00d May 28 '17 at 20:11
  • Have you tried computing $A^{\ast}A$ and then $\sqrt{A^{\ast}A}$? I won't promise that that works, but I lean towards expecting it will work. – Daniel Fischer May 28 '17 at 20:22
  • @DanielFischer Computing $A^\ast A$ for my specific $A$? I'm not sure, if I understood why this could help to prove the claim. Maybe you can explain your idea in more detail below the other question (or provide an answer ;)). – 0xbadf00d May 28 '17 at 20:47
  • If $$\sum \langle \sqrt{A^{\ast}A},e_r, e_r\rangle < +\infty,$$ then $A$ is trace class. – Daniel Fischer May 29 '17 at 10:59
  • @DanielFischer That's the definition of being trace class (at least in Hilbert spaces). – 0xbadf00d May 30 '17 at 15:05
  • Yes. And isn't there a relation between nuclear operators and trace class operators? – Daniel Fischer May 30 '17 at 17:20

1 Answers1

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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.

Martin Argerami
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  • In fact, "my" $\lambda_n$'s satisfy your condition. Could you provide a proof for the equivalence? – 0xbadf00d May 28 '17 at 19:42
  • Yes, I added a bit to the answer. – Martin Argerami May 28 '17 at 19:43
  • 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. – 0xbadf00d May 28 '17 at 19:43
  • I think there is a typo: It should be $\langle Ae_n,e_m\rangle =\lambda_{mn}$. – 0xbadf00d May 28 '17 at 19:51
  • I don't follow. In what sense would a self-adjoint operator have a "unique" root? – Martin Argerami May 29 '17 at 01:29
  • What I've meant was that a nonnegative and self-adjoint bounded linear operator $A$ admits a unique nonnegative and self-adjoint bounded linear operator $B$ with $A=B^2$. If $A$ is not self-adjoint, I'm not sure if we still can show that there is a (unique) nonnegative $B$ with that property. That's what I've asked fore here. – 0xbadf00d May 29 '17 at 11:00