Uniform boundedness principle on a Hilbert space

Question

I can't prove following proposition:

Let $H$ be a Hilbert space and $L_n:H\to H$ be bounded, linear operators $(n=1,2,\cdots)$.

If, for all $x,y\in H$, a sequence $\{(L_nx,y)\}_{n=1}^{\infty}$ is bounded, then a sequence $\{||L_n||\}_{n=1}^{\infty}$ is bounded.

This proposition is similar to uniform boundedness principle, so I tried to prove in a same way, but I couldn't because I can't prove \begin{equation} H=\bigcup_{m=1}^{\infty}X_m\hspace{10pt} \left(X_m=\{x\in H:\forall n\in\mathbb{N}, ||L_nx||\leq m\}\right) \end{equation} so Baire category theorem fails to be applied.

Is there any way to prove the above equation or completely different way to prove the main proposition?

Thank you in advance.

Answer 1

Apply uniform boundedness twice. One application shows that $||T_nx||$ is bounded for every $x$ (that is, for every $x$ there exists $c_x$ such that $||T_nx||\le c_x$ for all $n$), and then another application shows that $||T_n||$ is bounded.

Answer 2

Thanks to David, I could prove the proposition.

For every $x\in H$, let $T_n:H\to\mathbb{R}$ be an operator such that

\begin{equation} T_ny=(y,L_n x) \end{equation}

Using the assumption on $L_n$ and uniform boundedness, we can deduce that a sequence $||T_n||\subset\mathbb{R}$ is bounded.

Here, since $||T_n||=||L_nx||$, we have the conclusion, applying uniform boundedness again in an opposite way.