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

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?

## 1 Answer

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.

• Changed the theorem with the one in the chapter you mention, but I still miss the bit "differences". – user8469759 Feb 28 '19 at 10:18
• The first version of the theorem is for positive linear functionals. The second version is for continuous linear functions for which you need complex measures (real measures in the real case). Any real measure is a difference of two positive finite measures. – Kavi Rama Murthy Feb 28 '19 at 10:23
• Your last statement is treated somewhere in the "Abstract Integration" chapter, is that right? Although is easy to prove on its own... but it's just for reference. – user8469759 Feb 28 '19 at 10:26
• No, it is in the chapter on Complex Measures. The main point here is this: the dual of the space of continuous functions is the space of complex measures and positive measures are not enough for this. The chapter on Abstract Integration deals only with positive measures. – Kavi Rama Murthy Feb 28 '19 at 10:29
• Is it in section 6.14? (Real and complex analysis). – user8469759 Feb 28 '19 at 10:31