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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.

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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.

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    $\begingroup$ You mean the Banach-Dieudonné theorem instead of Banach-Steinhaus, don't you? $\endgroup$ – Jochen Wengenroth Dec 1 '20 at 21:57
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    $\begingroup$ Yes, of course, thank you. I have corrected it. $\endgroup$ – user131781 Dec 2 '20 at 5:06
  • $\begingroup$ @user131781 Thank you. I found a reference for this: An introducion to Banach Space Theory by Robert E. Megginson. I have another question. Is the construction mentioned by janacek in this question mathoverflow.net/questions/131420/… the same? $\endgroup$ – benjaminroos Dec 2 '20 at 16:55
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    $\begingroup$ 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. $\endgroup$ – user131781 Dec 2 '20 at 21:14
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    $\begingroup$ 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. $\endgroup$ – user131781 Dec 3 '20 at 5:01

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