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So I am stuck at this situation . Suppose $m:B_2(H_1)\times B_2(H_2)\to \mathbb C$ be bilinear form given by $m(S,T)=\left<T,\phi(S)\right>$, where $\phi: B_2(H_1)\to B_2(H_2)$ be a bounded linear map and $B_2(H)$ denotes the space of Hilbert Schmidt operators on $H$ with inner product $\left<S,T\right>=Tr(T^*S)$. Let $\{S_i\}$,$\{\tilde S_j\}$ and$\{T_i\}$,$\{\tilde T_j\}$ be two pair of sequence in closed unit ball of $B_2(H_1)$ and $B_2(H_2)$ respectively such that the limit $$\lim_jm(S_i\tilde S_j,\phi(T_i\tilde T_j))<\infty$$ for each $i$

Which means, for each $i$ $$\lim_j \sum_{n\in I}\left<T_i\tilde T_je_n,\phi(S_i\tilde S_j)e_n\right><\infty$$ $\{e_n\}$ is orthonormal basis for $H_2$. Under what conditions can I take limit inside the summation ? I feel like using some variant of dominated convergence theorem with respect to counting measure but cannot figure out how. Are the limit and summation even interchangeable here?

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No, of course the sum and limit aren't interchangeable. E.g. take $H_1 = H_2 = l^2$, $\phi = {\rm id}$, $T = S =$ orthogonal projection onto the first coordinate, and $\tilde{T}_j = \tilde{S}_j =$ the rank 1 operator taking $e_j$ to $e_1$. (Note that the index $i$ does not come into the problem, it is a separate question for each $i$.)

In this example $\langle TT_je_n, SS_je_n \rangle = \delta_{jn}$, so each sum is $1$ and each limit is $0$.

It's a standard sort of counterexample. BTW $m$ is sesquilinear, not bilinear.

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