sequence of bounded linear operators Tn: X to Y where X, Y are Banach spaces, ||Tn|| goes to infinity. Show there exists x ||Tnx|| goes to infinity Suppose that $X,Y$ are Banach spaces. Let $T_n: X\to Y$ be linear operators for all $n\in \mathbb{N}$ such that $\lim_{n\to\infty} \|T_n\|=\infty$. Show that $\exists x_0\in X$ such that $\lim_{n\to \infty}\|T_nx_0\|=\infty$.
I know that if $\sup_{n\in\mathbb{N}}\|T_n\|=\infty$, then by Uniform Boundedness Prinnciple, $\exists x_0\in X$ such that $\sup_{n\in\mathbb{N}}\|T_nx_0\|=\infty$. But the question I am asking is stronger.
I want to proceed by contradiction. Suppose there is a subsequence $\{T_{n_k}\}$ such that $\|T_{n_k}x_0\|$ is bounded. How to reach a contradiction?
 A: The claim is not true: Not even in $\mathbb R^2$. The idea is to define $T_n$ as suitably scaled unit vectors. Then for every vector $v$ there will be $T_n$ almost orthogonal to $v$, so there is a subsequence with $T_{n_k}v\to0$.
Let $(q_n)$ be  a sequence of rational numbers in $[0,1]$ of the form
$$
(q_n) = \left(0,1, \frac12,\frac14,\frac34,\frac18,\dots\right),
$$
so it contains numbers of the form $\frac{2j+1}{2^k}$ sorted with respect to increasing $l$ and $k$.
Define the vector
$$
t(\theta):=\pmatrix{ \cos\theta&-\sin\theta}.
$$
Set
$$
T_n = k\cdot t(2\pi q_n) \quad \text{ for }\quad q_n=\frac{2j+1}{2^k}.
$$
Then clearly $\|T_n\|\to\infty$. Now let $v$ be a unit  vector of the type
$$
v=\pmatrix{\sin(2\pi\theta)\\ \cos(2\pi\theta)}
$$
with $\theta\in[0,1]$,
which yields $t(2\pi\theta)v=0$. Then there is a subsequence of $(q_n)$ converging to $\theta$: for given $k$ choose $j$ such that $\frac{2j+1}{2^k}$ is as close as possible to $\theta$. Denote this subsequence by $(r_k)$. Then $|r_k-\theta|\le 2^{1-k}$.
Take the corresponding subsequence of the operators $(k \cdot t(2\pi r_k))$.
Then
$$
t(2\pi r_k) v = (t(2\pi r_k)-t(2\pi\theta)) v + t(2\pi\theta) v
$$
which yields
$$
k |t(2\pi r_k) v|\le k |t(2\pi r_k)-t(2\pi\theta)|_2\le k \cdot 2\pi \cdot 4 \cdot |r_k-\theta|\le 8\pi \cdot k \cdot 2^{1-k}\to0.
$$
This shows that for every non-zero vector, there is a subsequence of operators $(T_{n_k})$ such that $T_{n_k}\to0$ while $\|T_n\|\to\infty$.
