Let $T:\mathcal{H}\rightarrow \mathcal{H}$ be a linear continuous operator between Hilbert spaces and $\{b_i\: |\: i \in I\}$ an orthonormal basis. Prove that if:
$$\sum_{i\in I}\lVert T b_i\rVert^2<\infty$$ Then, $T$ is compact.
Hilbert spaces are reflexive, as is easily seen from applying Riesz Representation Theorem twice. This means that if $x_n$ is a sequence whose norm is bounded by $M$, there is a subsequence such that $x_{n_j}\rightharpoonup x$. By continuity of $T$, we obtain that $T(x_{n_j})\rightharpoonup T(x)$. I want to show this convergence is strong. Because we have a basis it is clear that from Parseval we obtain:
$$\lVert T(x_{n_j})-T(x)\rVert^2=\sum_{i\in I}|\langle T(x_{n_j}-x),b_i\rangle |^2=\sum_{i\in I}|\langle x_{n_j}-x,T^*(b_i) \rangle|^2 $$
By our hypothesis there is also $I_o$ with $|I_o|<\infty$ such that $\sum_{i\in I_o^C}\lVert Tb_i\rVert^2<\varepsilon$. Using weak convergence for every $i\in I_o$ we have $|\langle x_{n_j }-x_j, T^*(b_i) \rangle|\rightarrow 0$, and thus one can take $j$ large enough such that:
$$\lVert T(x_{n_j})-T(x)\rVert^2< \varepsilon+4M^2\sum_{i\in I_o^C}\lVert T^*(b_i)\rVert^2$$
If we had a self-adjoint operator, we would be done. But that is not the case… Any thoughts on how to proceed?
Best Answer
I think you can do exactly the same proof, except that you try to show that $T^*$ is compact.
Let $x_n$ be a bounded sequence in $H$. By reflexivity, you find a weakly convergent subsequence $x_{n_j}$. Then $$||T^*(x_{n_j}) - T^*(x)||^2 = \sum_{i\in I} |\langle x_{n_j} - x, T(b_i)\rangle|^2$$ and you can apply your hypothesis to get that $T^*$ is compact, which is equivalent to $T$ itself being compact.