# How to show that there's a continuous function separating convex sets of Radon measures?

First, the setup: $X$ is a compact set. By Riesz's representation theorem $C(X)^*=${all Radon measures on $X$}. $K$ is a convex, closed set of probability measures. $m$ is a probability measure out of $K$. In a paper I read, the author uses the fact that there's a continuous function $g$ s.t. $\int gdm>sup_{\mu \in K}\{\int gd\mu\}$

So far I was able to show that $\exists \psi\in C(X)^{**}$ s.t. $\psi(m)>sup_{\mu \in K}\{\psi(\mu)\}$

And also that the unit ball of $C(X)$ is dense in the weak-* topology in the unit ball of $C(X)^{**}$, meaning that $\exists \{g_n\}\subset B(C(X))$ s.t. $g_n\overset{w-*}{\longrightarrow} \psi$, where the canonical embedding is $g_n(\mu)=\int gd\mu$.

Making the final step of showing some continuous function actually fulfills the desired inequality, has been unsuccessful for me so far. Neither could I find any book that show a similar claim.

Any help would be appreciated, especially a simple quick solution that I missed.

Since the weak topology is Hausdorff, there is an open set containing $m$ and disjoint from $K$. So we can find $f_1,...,f_n$ and $\varepsilon$ such that the open set $$\{\mu \mid \int f_i dm -\varepsilon < \int f_i d\mu < \int f_i dm +\varepsilon \quad \forall \ i \}$$ is disjoint from $K$. Now project everything on ${\bf R}^n$ with the map $\mu \mapsto (\int f_1 d\mu,..., \int f_n d\mu)$.

The projections of $K$ and the open set are two disjoint convex sets that can be separated by an hyperplane $\{H=\sum a_i x_i +c=0\}$ with $H<0$ on the projection of $K$ and $H>0$ on the projection of the open set (which is just a small cube). Take $g=\sum a_i f_i +c$.

• Thanks for the answer,Im working on figuring what you wrote now- did you mean by "does not contain K" disjoint from K? – BOS Apr 24 '15 at 8:47
• thank you very much for your solution. I find it helpful, and even enlighting. – BOS Apr 24 '15 at 18:14

This question has already been answered but it might be useful to point that no detailed computations are required since it is an immediate consequence of the Hahn Banach theorem applied to the space of measures with the weak star topology. An advantage of this approach is that the fact that $K$ consists of probability measures is seen to be irrelevant and the result has nothing to do with the explicit form of the Banach space but applies in a suitable form to any closed, convex subset of the dual of a Banach space. Note also that you do not specify for which topology (norm or weak star) $K$ is assumed to be closed. The above shows that it must be the weak star topology as is tacitly assumed in the above answer.

The answer by corserine is somehow invalid here. The Hahn-Banach theorem states that there exists a linear operator in the dual space of $C(X)^*$ that separates the closed convex set and the point outside it. However, the dual of $C(X)^*$ is not necessarily the space of space of continuous functions, as asked in the question. Hence, a specific derivation is required (, i.e., by specifying the exact topology).

• The exact topology IS specified---see the third line of the answer referred to. – dalry Feb 10 '16 at 17:25
• Agree. But my comment still holds without the last sentence "i.e., by specifying the exact topology". – Yun Feb 10 '16 at 18:39
• We seem to be talking at cross purposes. The dual of $M(K)$ with the weak star topology is $C(K)$, just as the dual of any dual Banach space with the weak star topology coincides with the original space. – dalry Feb 10 '16 at 19:48
• Thank you for pointing out this. I realize that my original thought was with strong topology. Do you know how general this conclusion (with weak topology) is? I mean, how about if $X$ is noncompact? Do you have any reference explaining it? Thanks! – Yun Feb 10 '16 at 20:41
• If $X$ is not compact but, say, completely regular then one can consider the space of bounded, continuous functions with the so-called strict topology, which was introduced by R.C. Buck in the locally compact space and extended to the general case by several authors. Its dual is the space of bounded tight (or Radon) measures. Many features of the compact case carry over to this situation, but at the cost of replacing the concept of a Banach space by that of a Saks space. For references, google the latter phrase. – dalry Feb 10 '16 at 21:15