Let $X$ be a compact Hausdorff space. In chapter 3 of Peter Scholze's [*Lectures on Analytic Geometry*](https://www.math.uni-bonn.de/people/scholze/Analytic.pdf) 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.



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


