I am wondering about the following version of the Banach-Steinhaus theorem.


Let $A$ be a closed convex subset contained in the unit ball of a Banach space $X$ and consider bounded operators $T\_n \in \mathcal L(X).$ 


Assume we know that for every $x \in A$ the sequence $\left\lVert T\_n x \right\rVert$ is bounded uniformly in $n.$


Does this imply that $$\sup\_{x \in A} \left\lVert T\_n x \right\rVert$$ is bounded uniformly in $n$?


If $A=X$ then the theorem is undoubtedly true by the folklore Banach-Steinhaus theorem but I was wondering whether this version holds as well?



Version of Banach-Steinhaus theorem
The answer is **yes**, as a close inspection of the standard proof of the uniform boundedness principle/Banach-Steinhaus theorem shows. The standard proof (or at least the proof which I would consider to be the standard one) can e.g. be found on [Wikipedia](https://en.wikipedia.org/wiki/Uniform_boundedness_principle).


The details are a bit different here, so let me give them below. Throughout, let us replace the sequence $(T\_n)$ with a general subset $\mathcal{T} \subseteq \mathcal{L}(X)$.


**Proof.** By Baire's Theorem we can find an integer $m$ such that the set
\begin{align\*}
 B := \{x \in A: \, \|Tx\| \le m \text{ for all } T \in \mathcal{T}\}
\end{align\*}
has non-empty interior within $A$. Thus, we can find a point $x\_0 \in B$ and a real number $\varepsilon \in (0,1]$ such that each $x \in A$ which has distance at most $\varepsilon$ to $x\_0$ is contained in $B$.


Now, let $y \in A$. The vector $z := x\_0 + \frac{\varepsilon}{2}(y-x\_0)$ is contained in $A$ due to the convexity of $A$, and its distance to $x\_0$ is at most $\varepsilon$ since both $y$ and $x\_0$ have norm at most $1$. Thus, $\|Tz\| \le m$ for all $T \in \mathcal{T}$. Since
\begin{align\*}
 y = \frac{2}{\varepsilon}(z - x\_0) + x\_0,
\end{align\*}
we conclude that
\begin{align\*}
 \|Ty\| \le \frac{4m}{\varepsilon} + m
\end{align\*}
for all $T \in \mathcal{T}$. This bound does not depend on $y$.


