Non-complete space verifying uniform boundedness

Question

Recently, I have seen the so-called uniform boundedness theorem, which says:

Let $(X,∥⋅∥)$ be a Banach space and $(Y,∥⋅∥)$ be a normed linear space. Let $A⊂B(X,Y)$ be a pointwise bounded family of bounded linear transformations from $X$ to $Y$ . Then the family $A$ is uniformly bounded.

I was wondering if there is any 'simple' non-Banach space that verifies this property: that is, that for all choices of $Y$, a pointwise bounded family of bounded linear transformations is uniformly bounded.

I have thought that maybe some kind of sequence space might work, but I have not quite found one that does.

EDIT: I found in a book that the space consisting on the real sequences that take on finitely many distinct values is an example, but it does not include a proof. Is this proven easily?

Answer 1

Locally convex spaces which satisfy the uniform bounded principle, i.e., every pointwise bounded family of continuous linear maps (with values in any normed space) is equicontinuous, are called barrelled spaces.

Of course, the most prominent examples are those for which the uniform boundeness (or Banach-Steinhaus) theorem was proved originally, namely, Fréchet spaces. But compared to the class of Fréchet spaces, the class of barrelled spaces has much better permanence properties, for instance, it is (rather trivially) stable with respect to inductive limits in the category of locally convex spaces (in category theory, one would rather say colimits).

You can find much information (like the relation to completeness properties of the dual space) in the book Barrelled Locally Convex Spaces of Bonet and Pérez Carreras. In particular, 4.3.2 contains examples of barrelled normed spaces which are not complete.

Answer 2

My favorite example is the space of scalar-valued simple functions on a $\sigma$-algebra $\mathcal{A}$ (or a measurable space $(X,\mathcal{A})$ if you want).

Of course, we can endow this space with the uniform norm, and the completion of this space is the Banach space of bounded measurable functions.

The normed space of simple functions, while incomplete in general, is still barrelled (thus it does have uniform boundedness principle). This fact is a direct consequence of the Nikodym boundedness theorem, which can be regarded as a measure theory analogue of the uniform bounded principle. The proof I know of this theorem is based on the Baire category theorem, and is of a very similar flavor with the usual uniform boundedness principle for Banach spaces (or more generally, topological vector spaces which are not meager sets in themselves).

I wrote below a proof of the barrelledness of the space $\operatorname{Sp}(\mathcal{A})$ of simple functions, assuming Nikodym boundedness theorem for finitely-additive measures. The linked page only mentions the theorem for countably-additive measures, but a little bit of massage can lift it to finitely-additive measures. I am not sure what is the best reference for this fact, but this paper is one reference that I could quickly find from Google, which also mentions its connection to the usual uniform boundedness principle.

Proof. Let $A$ be a barrell in $\operatorname{Sp}(\mathcal{A})$. This means that $A$ is the polar of a weak-$*$ bounded subset $M$ of the dual space of $\operatorname{Sp}(\mathcal{A})$. The dual space of $\operatorname{Sp}(\mathcal{A})$, which is the dual space of its completion, is the space of finitely-additive measures on $\mathcal{A}$ equipped with the total variation norm. Then $M$ being weak-$*$ bounded simply means that $\sup_{\mu\in M}|\mu(A)| < \infty$ holds for all $A\in\mathcal{A}$. Hence, the Nikodym boundedness theorem shows $M$ is norm-bounded. (Note that here the weak-$*$ topology is the one coming from elements in $\operatorname{Sp}(\mathcal{A})$, not its completion, which is why Nikodym's theorem is relevant. If it were from the completion, than norm-boundedness of $M$ follows trivially by the usual uniform boundedness principle.) Therefore, the polar $A$ of the norm-bounded set $M$ must contain a ball in $\operatorname{Sp}(\mathcal{A})$, so $A$ is a neighborhood of $0$, which means that $\operatorname{Sp}(\mathcal{A})$ is barrelled. $\blacksquare$

Finally, note that:

EDIT: I found in a book that the space consisting on the real sequences that take on finitely many distinct values is an example, but it does not include a proof. Is this proven easily?

This is a special case of what's shown, with $\mathcal{A}=\wp(\mathbb{N})$. I would imagine that a direct proof of this special case would not be much easier than the general case, in the sense that it probably would involve essentially the same construction one would do for proving Nikodym's theorem.