Let $V$ be a vector space, $B$ be a {1||Banach space}, and $\{T_n\}$ a sequence in $\mathcal{B}(B,V)$. Then if for all $b\in B$ we have
{2||$$\sup_{n}\lvert \lvert T_{n}b \rvert \rvert < \infty$$}
(ie the sequence is {2||pointwise bounded}) then we have
{3||$$\sup_{n} \lvert \lvert T_{n} \rvert \rvert < \infty$$}
(ie the {3||operator norms are bounded})

For all $k\in \mathbb{N}$, define

Then each set is closed because if $\{b_n\}\subset C_k$ and $b_n\to b$ then we have $||b||=\underset{n\to \mathrm{\infty }}{lim}||b_n||=1$. And for all $m\in \mathbb{N}$ we have

since each of the operators $T_m$ are continuous. And $||T_m{b}_n||\le k$ because $b_n\in C_k$, so $C_k$ must be closed.

Now, we also have

because for any $b\in B$, there is no $k$ such that $\underset{m}{sup}||T_{m}b||\le k$.

So LHS is complete because it is a closed subset of $M$, and so by Baire's Theorem, one of the $C_k$ must contain an open ball $B(b_0,{\delta }_0)$.

Thus for any $b\in B(0,{\delta }_0)$, we have that $b_0+b\in B(b_0,{\delta }_0)\subset C_k$ so

Then since both $b_0,b_0+b\in B(b_0,{\delta }_0)$, we have

Thus, rescaling, we have for any $n\in \mathbb{N}$ and all $b\in B$ with $||b||=1$, we have

ie the operator norm of $T_n$ is bounded for all $n$, and therefore its $sup$ is bounded as well.