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 $\phi$).



Sequence of Hilbert Schmidt operators
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.


