= Solution
The <Banach-Steinhaus convergence theorem> can be stated as follows. Let $X$ be a <normed space>, $Y$ a <Banach space>, and $A_n\in\mathcal B(X,Y)$. If
$$
\sup_n\|A_n\|<\infty
$$
and $A_nd$ converges in $Y$ for every $d$ in a <dense subset> $D\subseteq X$, then $A_nx$ converges for every $x\in X$. The pointwise limit $A$ is a <bounded linear operator> and
$$
\boxed{A_nx\longrightarrow Ax\quad(x\in X),\qquad\|A\|\leq\liminf_n\|A_n\|.}
$$
If $X$ is also a <Banach space>, the uniform <operator norm> bound is necessary: pointwise convergence implies pointwise boundedness, so the <Banach-Steinhaus theorem>, namely the <Uniform boundedness principle>, supplies it. The limit is convergence in the <strong operator topology>; it need not be convergence in <operator norm>.
To see the sufficient direction, write $C=\sup_n\|A_n\|$. For $d\in D$ the <triangle inequality> gives
$$
\|A_nx-A_mx\|\leq2C\|x-d\|+\|A_nd-A_md\|.
$$
First approximate $x$ by $d$, then let $m,n\to\infty$. Completeness of $Y$ gives the limit. Linearity passes to limits, and $\|Ax\|=\lim_n\|A_nx\|\leq(\liminf_n\|A_n\|)\|x\|$ gives the displayed bound. If a candidate <bounded linear operator> $A:X\to Y$ is already given and $A_nd\to Ad$ on $D$, the same approximation proves convergence to $Ax$ even when $Y$ is merely a <normed space>.
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