$E$ normed linear spce, $T_n$ converges strongly to a bounded linear operator $T$. Can $\{\|T_n\|\}$ be unbounded?

Question

Let $E$ is a normed linear space, $T_n: E\to E$ is a bounded linear operator, $n=1,2,\cdots$. Suppose $T_n$ converges strongly to a bounded linear operator $T$ (i.e., $\forall x\in E, \lim\|T_nx-Tx\|=0$). Can $\{\|T_n\|\}$ be unbounded?

If $E$ is a Banach space, we can prove $\{\|T_n\|\}$ is bounded by uniformly bounded theorem. I think the complete condition cannot be removed but I have trouble finding a counterexample.

Appreciate any help!

Answer 1

For another, somewhat simpler example, consider the space of finitely non-zero real sequences with the $\ell ^2$ norm, as in Kavi's answer, and let $T_n$ be the operator defined by $$ T_n(x_1,\ldots , x_{n-1},x_n,\ldots )= (0, \ldots , 0, nx_n,0,0,\ldots ), $$ where the term $nx_n$ occurs in the $n^{th}$ position. Then $T_n$ converges strongly to zero, but clearly $\|T_n\|=n$.

Answer 2

Let $E$ be the space of finitely non-zero real sequences with the $\ell^{2}$ norm. Let $(e_n)$ be the canonical sequence and $T_n(x)=(x_{n+1}+x_{n+2}+\cdots +x_{2n})e_1$. Then $T_nx\to 0$ for every $x$ and $\|T_n\|\geq\sqrt n$.

[To prove that $\|T_n\|\geq \sqrt n $ consider the unit vector $\frac {e_{n+1}+e_{n+2}+\cdots+e_{2n}} {\sqrt n}$].