# Sequence of Hilbert Schmidt operators

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 $$\phi$$).

• 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? – Matthew Daws Feb 26 at 21:12
• 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. – NewB Feb 27 at 2:04
• But I guess what you said is true. A linear and bounded operator would suffice for a counterexample. – NewB Feb 27 at 2:17
• Is this the paper you refer to when you say (in the bounty) 'Need this for my paper"? arxiv.org/abs/2001.00830 – Yemon Choi Mar 4 at 5:08
• Yes it is @yemon – NewB Mar 16 at 7:57

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.