Let $A$ and $B$ be normed spaces with norms $||\cdot||_A$ and $||\cdot||_B$ respectively, and let $\mathcal L(A;B)$ be the normed space of linear transformations from $A$ to $B$, with the norm
$$||T|| := \sup_{||x||_A=1} ||Tx||_B$$
I am trying to prove or disprove the following: if the sequence $(T_n)$ converges pointwise to $T$, i.e. $||T_nx-Tx||_B\to 0$ for each $x\in A$, then $(T_n)$ converges in the norm above, i.e. $||T_n - T|| \to 0$.
The converse is true since:
$$||T_n - T|| = \sup_{||x||_A=1} ||T_nx - Tx||_B \geq ||T_nx - Tx||_B, \;( ||x||_A=1)$$ and for $x$ any size we can just consider $\epsilon/||x||_A$ and pick $N$ such that $n>N$ implies $||T_n - T|| < \epsilon/||x||_A$, proving pointwise convergence. I don't know about the other direction.
Note: Ok half-way through typing this question StackExchange suggested Pointwise convergence vs norm convergence in $X^*$ where it states that the other direction is not true. It doesn't include an example though. Can someone provide one?
Let $A=B=\ell^2$ and $T_n(x_1,x_2,...)=(0,...,0,x_{n+1},x_{n+2},...)$ for each $n\in\mathbb{N}$. For each $x\in\ell^2$ we have $||T_n(x)-0||^2=\sum_{i=n+1}^\infty |x_i|^2$. Since the tail of a convergent series tends to $0$ we conclude that $||T_n(x)=0||^2\to 0$ when $n\to\infty$. So the sequence $T_n$ converges pointwise to the zero operator. However, for each $n\in\mathbb{N}$ we have $||T_n||=1$. Hence $||T_n-0||\not\to0$.