# Operator continuity on Hilbert space

Let $A: H \to H$ be a linear operator on Hilbert space $H$, and let $\{\alpha_n\}_{n = 1}^{\infty} \subset \mathbb{R}$ converges to nonzero number. Prove that if the series $\sum_{n = 1}^{\infty} \alpha_n\langle Ax_n, y\rangle$ converges for every $y \in H$ and every sequence $\{x_n\}_{n = 1}^{\infty} \subset H$ such that $\|x_n\| \leq n^{-2}, n \geq 1$ then $A$ is continuous.

• Do you have any thoughts or tries yourself? – mickep May 1 '16 at 17:59
• There was an equivalence in the problem statement, but one implication is almost obvious. For the second one i don't have any valuable ideas at this moment. – Rrr Rrr May 1 '16 at 19:36
• I think I can do it. I assume $\alpha_n \equiv 1$ (In general we can use the same argument). It's not hard to prove that if functional $f : H \to \mathbb{R}$ or $\mathbb{C}$ satisfies $\forall (x_n) \subset H$ $\|x_n\| \leq n^{-2}$ $\sum_{n = 1}^{\infty} f(x_n) < \infty$ then $f$ must be continuous. By Riesz every functional has the form $f(x) = \langle x, y \rangle$ for some $y \in H$ By the given condition $f \circ A$ is continuous for every $f \in H^{*}$ and so is A. Is it OK? – Rrr Rrr May 2 '16 at 15:21