Let $T_{n}:X \rightarrow Y$ bounded s.t $\left\|T_{n}\right\| \rightarrow \infty$, Then $\exists x_{0}\in X$ s.t $|T_{n}(x_{0})|\rightarrow \infty$

Question

Let $X$ be Banach space an $Y$ normed vector space. Let $T_{n}:X \rightarrow Y$ lineal and bounded operators such that $\left\|T_{n}\right\| \rightarrow \infty$ as $n\rightarrow \infty$. Then there exists $x_{0}\in X$ such that $$\left|T_{n}(x_{0})\right|\rightarrow \infty$$ as $n\rightarrow \infty$.

Remark: I tried to proceed by contradiction, my idea was to contradict the Uniform boundedness principle or Banach-Steinhaus theorem but I have failed since if we suppose that for all $x\in X$ $$\left|T_{n}(x)\right|\nrightarrow \infty$$ Then the most that I can hope for is that for every $x\in X$ exists a subsequence $n^{(x)}_{k}$ such that $\left|T_{n^{(x)}_{k}}(x)\right|$ is convergent. But the ideal would be to have $$\left\{\left|T_{n}(x)\right|\:|\: n\in\mathbb{N}\right\}$$ bounded, which I have not been able to infer.

Answer 1

Counter example: Consider $X=\ell^2(\mathbb{N}),Y=\mathbb{C}$ and let $T_n:\ell^2\rightarrow \mathbb{C}$ be the bounded linear functional $T_n(x)=\sqrt{n}x_n$. Then $\lVert T_n\rVert = \sqrt{n}\rightarrow \infty$. If $\lvert T_n(x)\rvert \geqslant 1$ for all $n\geqslant k$, then $\lvert x_n \rvert \geqslant 1/\sqrt{n}$ for $n\geqslant k$, but this contradicts $\sum_{n=1}^\infty \lvert x_n \rvert ^2 < \infty$.

Answer 2

You have to read carefully the uniform boundedness principle.

Suppose that for every $x$, the sequence $\|T_n(x)\|$ is bounded, then $sup_n{\|T_n\|}$ is bounded. Contradiction.

https://en.wikipedia.org/wiki/Uniform_boundedness_principle#Theorem