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 $\phi$).

  • $\begingroup$ 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? $\endgroup$ – Matthew Daws Feb 26 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 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 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 at 5:08
  • $\begingroup$ Yes it is @yemon $\endgroup$ – NewB Mar 16 at 7:57

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.