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?


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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