Rudin functional analysis theorem 3.28, application of Reisz representation theorem.

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Suppose

(a) $X$ is a topological vector space on which $X^*$ separates points,

(b) $Q$ is a compact subset of $X$, and

(c) the closed convex hull $\overline{H}$ of $Q$ is compact

then $y \in \overline{H}$ iff there's a regular Borel probability measure $\mu$ on $Q$ such that $$ y = \int_Q x d \mu (x) $$

First part of the proof

Regard $X$ again as a real vector space. Let $C(Q)$ be the Banach space of all real continuous functions on $Q$, with the supremum norm. The Reisz representation theorem identifies the dual space $C(Q)^*$ with the space of all real Borel measure on $Q$ that are differences of regular positive ones.

I assume the Reisz representation theorem used is the following (Theorem 6.19 from Rudin's Real and Complex analysis).

If $X$ is a locally compact Hausdorff space, then every bounded linear functional $\Phi$ on $C_0(X)$ is represented by a unique regular complex Borel measure $\mu$, in the sense that $$ \Phi f = \int_X f d \mu $$ for every $f \in C_0(X)$. Moreover, the norm of $\Phi$ is the total variation of $\mu$: $$ \lVert \Phi \rVert = | \mu |(X) $$

The bit I don't get in the proof is "Borel measures on $Q$ that are differences of regular positive ones".

Where does the "differences" come from?

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You are using the wrong version of Riesz Representation Theorem In the chapter on Complex Measures there is a section on Riesz Representation Theorem where Rudin proves a more general version. You have to use that version.