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Consider the Banach space $\mathcal K=S_2(H)$ of Hilbert Schmidt operators on a Hilbert space $H$. I am looking for an example of two pairs of sequences $\{T^{(i)}\},\{\tilde T^{(j)}\}$ and $\{S^{(i)}\},\{\tilde S^{(j)}\}$ in the unit ball of $\mathcal K$ and an anti-linear operator $\phi:\mathcal K\to \mathcal K$ such that the both iterated limits exists but $$\lim_i\lim_j\sum_{r,s}T^{(i)}_{rs}\tilde T^{(j)}_{rs}\overline{\phi(S^{(i)}\star \tilde S^{(j)})_{rs}}\neq \lim_j\lim_i\sum_{r,s}T^{(i)}_{rs}\tilde T^{(j)}_{rs}\overline{\phi(S^{(i)}\star \tilde S^{(j)})_{rs}}$$

Where $T_{rs}$ denote the $r\times s$ entry in the matrix of $T$ and "$\star$" denotes the Schur product of operators(entrywise product of matrices).

(Or otherwise, prove that these limits are always equal irrespective of the choice of sequences in unit ball and $\phi$).

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  • $\begingroup$ I guess $H$ comes with a fixed orthonormal basis, which allows us to think of elements of $\mathcal K$ as matrices. I think the "anti-linear operator" a bit unnatural and unmotivated. Given $\phi$ cannot I define $\psi:\mathcal K\rightarrow\mathcal K$ by $\psi(T)_{rs} = \overline{\phi(T)_{rs}}$. Then $\psi$ is bounded exactly when $\phi$ is bounded (I guess $\phi$ is bounded?) and is linear. Why couldn't we work with $\psi$ to start with? $\endgroup$ Feb 26, 2020 at 21:12
  • $\begingroup$ Actually, I am dealing with the bilinear forms of type $m:\mathcal K\times\mathcal K\to \mathbb C$ such that $m(S,T)=\left<T,\phi(S)\right>=\sum_{r,s}T_{rs}\overline{\phi(S)_{rs}}$, hence I need it to be anti-linear bounded map. $\endgroup$
    – NewB
    Feb 27, 2020 at 2:04
  • $\begingroup$ But I guess what you said is true. A linear and bounded operator would suffice for a counterexample. $\endgroup$
    – NewB
    Feb 27, 2020 at 2:17
  • $\begingroup$ Is this the paper you refer to when you say (in the bounty) 'Need this for my paper"? arxiv.org/abs/2001.00830 $\endgroup$
    – Yemon Choi
    Mar 4, 2020 at 5:08
  • $\begingroup$ Yes it is @yemon $\endgroup$
    – NewB
    Mar 16, 2020 at 7:57

1 Answer 1

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The limits are always the same.

As $\mathcal K = S_2(H)$ is a Hilbert space, and as only the Schur product is involved anywhere, we can infact identify $\mathcal K$ with $\ell^2 = \ell^2(\mathbb N)$ and consider the pointwise product of vectors in $\ell^2$.

Let $x=(x_r)\in\ell^2$ and let $(y^{(i)})$ be a bounded sequence in $\ell^2$. Thus, for each $r$ the scalar sequence $(y^{(i)}_r)$ is bounded, so by moving to subsequences and using a diagonal argument, we may suppose that $y^{(i)}_r\rightarrow y_r$, say, as $i\rightarrow\infty$. Set $y=(y_r)$ and observe that for any $R$, $$ \sum_{r=1}^R |y_r|^2 = \sum_{r=1}^R \lim_i |y^{(i)}_r|^2 = \lim_i \sum_{r=1}^R |y^{(i)}_r|^2 \leq \lim_i \|y^{(i)}\|_2^2 < \infty. $$ As $R$ was arbitrary, we conclude that $y\in\ell^2$.
Then, for $R>0$, \begin{align*} \lim_i \|xy^{(i)} - xy\|_2^2 &= \lim_i \sum_r |x_r y^{(i)}_r - x_r y_r|^2 \\\\ &= \sum_{r=1}^R \lim_i |x_r y^{(i)}_r - x_r y_r|^2 + \lim_i \sum_{r>R} |x_r y^{(i)}_r - x_r y_r|^2 \\\\ &= \lim_i \sum_{r>R} |x_r y^{(i)}_r - x_r y_r|^2 \\\\ &\leq \lim_i \Big(\sum_{r>R} |y^{(i)}_r - y_r|^2\Big) \Big( \sup_{r>R} |x_r| \Big). \end{align*} We can make this arbitrarily small by choosing $R$ large. We conclude that $xy^{(i)} \rightarrow xy$ in norm.

Given bounded sequences $(x^{(i)}), (y^{(i)})$ in $\ell^2$, let $x,y$ be the pointwise limits, as in the previous paragraph. Let $z\in\ell^2$, and consider \begin{align*} \lim_i \sum_r x^{(i)}_r z_r y^{(i)}_r. \end{align*} As $x^{(i)} y^{(i)}$ is in $\ell^1$ and $(z_r)$ is bounded, this sum make sense, and we can copy the argument to show that \begin{align*} \lim_i \sum_r x^{(i)}_r z_r y^{(i)}_r = \sum_r x_r z_r y_r . \end{align*}

Thus, given bounded sequences $(x^{(i)}), (y^{(i)}), (a^{(i)}), (b^{(i)})$ in $ell^2$, let $x,y,a,b$ be the pointwise limits, as in the previous paragraph. Given a bounded linear map $\phi$ on $\ell^2$, we see that as $a^{(i)} b^{(j)} \rightarrow a b^{(j)}$ in norm, as $i\rightarrow\infty$, also $\phi(a^{(i)} b^{(j)}) \rightarrow \phi(a b^{(j)})$ in norm. Hence, by the same argument, \begin{align*} \lim_j \lim_i \sum_r x^{(i)}_r y^{(j)}_r \phi(a^{(i)} b^{(j)})_r = \lim_j \sum_r x_r y^{(j)}_r \phi(a b^{(j)})_r = \sum_r x_r y_r \phi(a b)_r. \end{align*} By symmetry, we get the same answer with the limits taken in the other order.

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