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$H$ is a Hilbert space, $T_n$ is a family of bounded linear operators, satisfies: $\lim_{n\to\infty}\langle T_nx,y\rangle$ exist for any $x,y\in H$. Proof that exist a bounded linear operator $T$ satisfies $\lim_{n\to\infty}\langle T_nx,y\rangle=\langle Tx,y \rangle$

I think the condition is a bit weak, generally speaking, we use the property of weak convergence by acting with functional to get strong convergence. But in this situation, even $ T_n x$ is weak convergence, how do we study $T$?

I tried to proof $\sup_{n}\Vert T_n\Vert$ exist by uniform bounded them, but later I gave a counter-example: $T_n(f)=f(x)\sqrt{n}sin(nx)$ on the space $L^2([0,1])$

Trinifold
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  • See https://math.stackexchange.com/questions/3628832/wot-convergence-implies-norm-boundedness – Evangelopoulos Foivos Oct 29 '23 at 18:40
  • Your sequence ${T_n}$ does not converge. What happens is that a subnet converges. But, not being a sequence, the argument with the Uniform Boundedness Principle does not work (because the key is that a convergent sequence of numbers is bounded). – Martin Argerami Oct 29 '23 at 19:22

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