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?



interchanging limits and summation
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.


