# Is the filtered colimit topology on the space of signed Radon measures linear and locally convex?

Let $$X$$ be a compact Hausdorff space. In chapter 3 of Peter Scholze's Lectures on Analytic Geometry he considers the space of signed Radon measures on $$X$$ equipped with the filtered colimit (aka inductive limit) topology of the (in the weak$$^*$$-topology) compact absolutely convex subsets $$\mathcal{M}(X)_{\leq c}$$. Here $$\mathcal{M}(X)_{\leq c}$$ denotes the subset of measures with total variation norm less or equal than $$c$$ with $$c>0$$.

Then he states that the resulting topology is a locally convex vector topology. I was wondering if the subsets $$\mathcal{M}(X)_{\leq c}$$ form a neighborhood basis of the origin. If the answer is yes, then I do not see why the resulting topology is not the same as the one induced by the total variation norm. If the answer is no, then I do not see how to show that this topology is a locally convex vector topology.

Any clarification on this would be really appreciated.

This is a general, well-known fact about the dual of a Banach space. The finest topology which agrees with the weak$$\ast$$ topology on the bounded sets is locally convex. It is often called the bounded weak$$\ast$$ topology. It is complete and has the same convergent sequences as the weak$$\ast$$ topology. In non trivial situations, it is weaker than the norm topology—its dual is the original space. Yours is the case of the dual of a $$C(K)$$-space. For a general reference, look up the Banach-Dieudonné theorem in any standard text on Banach spaces.
• Yes, it displays the space of measures with the above structure as a kind of linearisation of $K$—this can be given a precise formulation by using the language of category theory. – user131781 Dec 2 '20 at 21:14
• This requires the concept of Waelbroeck spaces (for which see the monograph “Functors and Categories of Banach spaces” by Cigler, Losert and Michor which is readily available online). There is a natural forgetful functor from the category of such spaces to that of compacta, and the functor which takes $K$ to $M(K)$ is its adjoint. – user131781 Dec 3 '20 at 5:01