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?

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    $\begingroup$ In the last paragraph, shouldn't it be "if A is the unit sphere then..." ? $\endgroup$ – user111 Jul 10 '18 at 7:45
  • $\begingroup$ It is worth noting that the implication does not hold without the closeness assumption. If e.g. $A$ is the convex hull of the standard basis of $X:=\ell_2$ and $\phi_n:=\sum_{k=1}^n ke_k\in X$, then $(\phi_n,x)$ is point-wise but not uniformly bounded for $x\in A$ $\endgroup$ – Pietro Majer Jul 10 '18 at 15:16

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.

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 $y$.

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Instead of inspecting the Banach-Steinhaus proof as in Jochen Glueck's answer one can apply Banach-Steinhaus to the Banach space $[A]$ (the linear hull of $A$) endowed with the Minkowski functional $\|x\|_A=\inf\{t>0: x\in tA\}$ (completeness of this norm follows from completeness of the Banach space $X$ and $A=\overline A$).

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    $\begingroup$ A small detail: we need a balanced convex set to get a norm, so we should replace $A$ with $A':= \text{co}(A\cup (-A))=[-1,1]\cdot A$, and check that the hypotheses are still satisfied by $A'$ $\endgroup$ – Pietro Majer Jul 10 '18 at 7:22
  • $\begingroup$ (the last equality is not true of course) $\endgroup$ – Pietro Majer Jul 10 '18 at 14:06

I like Jochen Wengenroth's approach, and I think there is a point that it is worth to clarify. If we want to make a norm out of $A$, we need it to be a balanced set, so we'd like to pass to the bounded absolutely convex set $\overline{\operatorname{co}}\left(A\cup(-A)\right)$ or to $A-A$. Any family of linear operators which is point-wise bounded on $A$ is clearly also point-wise bounded on $A-A$. However these sets are in general not closed, so some care is needed, because a bounded absolutely convex not closed set $B$ in general would not produce a Banach disk on its linear span, and in fact in general the statement itself does not hold on such $B$ (see the example in the initial comment).

A cheap solution to make the argument work smoothly is to use the notion of $\sigma$-convexity (see e.g. this MO thread) which also generalize slightly the statement); in particular, it covers both the case of a closed and an open bounded convex set $A$. Recall that for a subset $A$ of a Banach space $X$ the following easy facts hold:

  • If $A$ is $\sigma$-convex, it is bounded;
  • If $A$ is $\sigma$-convex, $A-A$ is $\sigma$-convex and symmetric (that is, $\sigma$-absolutely convex);
  • If $A$ is $\sigma$-absolutely convex, it is a Banach disk, that is, its Minkowski functional is a Banach norm on the linear span of $A$.

As a conclusion, we can follow Jochen Wengenroth's reduction to the standard Banach-Steinhaus theorem. We thus have: Any family of linear operators on a Banach space, which is point-wise bounded on a $\sigma$-convex set $A$, is also uniformly bounded on $A$.

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