If a sequence of bounded operator converges pointwise, then it is bounded in norm $\newcommand{\vertiii}[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 
    \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}}$
Let $\left(V, \|\cdot\|_V \right)$ be a Banach space and $\left(W, \|\cdot\|_W \right)$ a normed vector space. Show that if a sequence $(T_n)_n\subset \mathcal{B}(V,W)$ converges pointwise then $\sup \vertiii{T_n}<\infty$.
So we know that $\sup\limits_{x\in V} \frac{\|T_n x\|}{\|x\|}=\vertiii{T_n}$, thus
$$\lim_{n\to\infty}\sup\limits_{x\in V} \frac{\|T_n x\|}{\|x\|}=\sup\limits_{x\in V} \frac{\|T x\|}{\|x\|}=\vertiii{T}<\infty$$
I'm wondering if I didn't miss something, since this seems a bit too simple. Please let me know.

Update (version 3):

Since $V$ is a complete metric space, it is Banach, and $T$ is defined on $V$. Since $\{(T_n(x))\}_n$ converges pointwise, it is bounded, so that there exists a function $C:V\to \mathbb{R}$ such that
$$\sup_{n\in \mathbb{N}} \|T_n(x)\|<C(x)$$
for each $x\in V$. By the Uniform Boundedness Principle, we deduce that there exists $M>0$ such that
$$\sup_{n\in\mathbb{N}} \vertiii{T_n} \le M <\infty.$$

The last update contains a correct solution to the exercise. 
