$H$ is Hilbert Space, select $x_n\in H$ such that $||x_n||=n$ then $\exists x \in H$ such that $\sup_n |(x_n,x)|=\infty$

Question

$H$ is Hilbert Space and choose $x_n\in H$ such that $\|x_n\|=n$ then $\exists x \in H$ such that $\sup_n |(x_n,x)|=\infty$

Edit: I did the following : To get a contradiction, Assume $\forall y, \in H \sup_n |(x_n,y)| < M \ne \infty$

And by Riesz Rep. $\forall y\in H, T_n(y)=<y,x_n>$ such that $T_n \in H'$ so $T_n$ is bounded linear functional. Indeed, $\forall y\in H, T_n(y)$ is bounded. Now, applying Uniform Bounded Theorem I can say $||T_n||$ is bounded however $||T_n||=\sup_n \frac {||T_n (y)||}{||y||}\le \sup_n \frac{||y||||x_n||}{||y||}=\sup_n ||x_n||=\sup_n n$ which is not finite. (that doesnt work) And also $||T_n||=\sup_{||y||=1}||T_n (y)||=<y,x_n> \le ||y||||x_n||=||x_n||$ and I could not continue from here.

Answer 1

You said you know the uniform boundedeness theorem. Try to use its contraposition with the family of continuous linear forms of $H$, $L_n : x \mapsto \left<x_n,x\right>$.

Answer 2

Consider linear functionals $T_{n}(x)=\langle x, x_{n}\rangle$ then $||T_{n}||=||x_{n}||$. If for all $x\in H$ , there exists $C_{x}$ such that $\displaystyle\sup_{n\in\Bbb{N}}||T_{n}(x)||\leq C_{x}$ . Then by Uniform Boundedness Principle , there exists $C>0$ such that $\sup_{n}||T_{n}||\leq C$ . But this is not possible as $||T_{n}||=||x_{n}||=n$ which is unbounded.

Thus there must exist some $x\in H$ such that $\displaystyle\,\sup_{n}\,\langle x,x_{n}\rangle$ is unbounded.