Iterated limits equal?

Question

Consider the Banach algebra $B_2(H)$ of Hilbert Schmidt operators on a Hilbert space $H$ with Hilbert Schmidt norm. We know that $B_2(H)$ is a Hilbert space as well with $\left<A,B\right>=tr(B^*A)$.I am looking for an example of pair of sequences $A_i,\tilde A_j$ and $T_i,\tilde T_j$ in the closed unit ball of $B_2(H)$ and $B(B_2(H))$ respectively and a $D\in B_2(H)$ such that both the iterated limits of $\left<T_i\tilde T_j(A_i\tilde A_j),D\right>$ exists but $$\lim_i\lim_j \left<T_i\tilde T_j(A_i\tilde A_j),D\right>\neq \lim_j\lim_i \left<T_i\tilde T_j(A_i\tilde A_j),D\right>$$ Otherwise prove that if iterated limits exists then they are equal.

Answer 1

For $\xi,\eta\in H$ let $\theta_{\xi,\eta}$ be the rank-one operator $\theta_{\xi,\eta}(\gamma) = (\gamma|\eta) \xi$ for $\gamma\in H$.

Let $(e_i)$ be an orthonormal sequence in $H$, set $S_i = \theta_{e_1, e_i} \in B(H)$ and let $R_j$ be the projection onto the span of $\{e_1,e_2,\cdots,e_j\}$. Then $$ \lim_i \lim_j (S_iR_j(e_i)|e_1) = \lim_i (S_i(e_i)|e_1) = \lim_i (e_1|e_1) = 1, $$ while $$ \lim_j \lim_i (S_iR_j(e_i)|e_1) = \lim_j \lim_i (R_j(e_i)|e_i) (e_1|e_1) = 0. $$

Now we "embed" this into your example. Let $\tilde A_j = \theta_{e_1,e_1}$ for all $j$, and set $A_i = \theta_{e_i,e_1}$ so $A_i \tilde A_j = A_i$. Define $\tilde T_j(a) = \theta_{R_j(a(e_1)), e_1}$ for $a\in B_2(H)$; a simple estimate shows that $\tilde T_j$ is bounded. Define $T_i(a) = \theta_{S_i(a(e_1)),e_1}$. Then $$ \tilde T_j (a) (e_1) = \theta_{R_j(a(e_1)), e_1} (e_1) = R_j(a(e_1)) \qquad (a\in B_2(H)). $$ Thus $$ T_i \tilde T_j(a) = \theta_{S_i R_j(a(e_1)), e_1} \implies T_i \tilde T_j (A_i \tilde A_j) = T_i \tilde T_j(\theta_{e_i,e_1}) = \theta_{S_iR_j(e_i),e_1}. $$ If we now let $D=\theta_{e_1,e_1}$ then $$ \Big\langle T_i \tilde T_j (A_i \tilde A_j), D \Big\rangle = \operatorname{Tr}\Big( \theta_{S_i R_j(e_i),e_1} \Big) = (S_i R_j(e_i) | e_1). $$ As above, this has different iterated limits.