I do not know any background of this topic. Feel free to edit my post.
Let X be a compact metric space. We call $\mu \in C(X,\mathbb{R})^\ast$ a probability radon measure if it is a positive linear functional and satisfies $\mu (\mathbb{1}) = 1$,where $\mathbb{1}$ denotes the constant function which assumes 1 everywhere. Suppose there exists at least one such measure. Let $\phi : X \to X$ be a homeomorphism. Show that there exists a probability radon measure $\mu$ s.t. $\phi^\ast\mu = \mu$ where $\phi^\ast$ is defined as $(\phi^\ast\mu)(f)=\mu(f\circ \phi)$
I tried to solve this exercise by using Schauder fixed point theorem which states that a continuous function from a compact convex subset of a Banach space to itself has a fixed point. Let $PR$ be the set of probability radon measure. $PR$ is obviously convex and $\phi^\ast$ is continuous (because it is a bounded linear function and it maps $PR$ to $PR$ ). What I tried to do next is of course to show that $PR$ is compact but in the strong topology I do not know if it is true. By using Ascoli theorem, I could show that $PR$ is compact in the compact convergence topology, but again there is no guarantee that the uniform topology and the compact convergence topology agree on $PR$ or this approach is effective.
By Ascoli theorem, I refer to the following one which is proved in Munkres. Let Y be a metric space and X be a topological space and endow C(X,Y) with the compact convergence topology. Then a subset of C(X,Y) is precompact if it is pointwise precompact and equicontinuous.
In class, this problem was solved as follows. First apply Banach Alaoglu theorem and we have a weak* compact unit ball $B$ which includes $PR$. Let $F_\delta(f_1,f_2,\cdots f_n)=\left\{\mu\in PR: |(\phi^\ast\mu - \mu)(f_i)|<\delta , i =1,2,\cdots n\right\}$ then this kind of set is non-empty and closed and running $\delta\in\mathbb{R}^{+}, f_i \in C(X,\mathbb{R}), n\in\mathbb{Z}$ and taking the intersection we can show the existence of a probability radon measure invariant under $\phi^\ast$
My question is whether my idea can be corrected or not. If impossible, I welcome other nice proofs.
Let $Y$ denote the space of (signed) Radon measures over X, that is $C(X,\mathbb{R})^\ast$ equipped with the weak topology. This is a locally convex topological vector space (see the wikipedia article on weak topologies). Now the map $\Phi^\ast\colon Y \rightarrow Y$ is a continuous map, mapping the set $S := \{ \mu \in C(X,\mathbb{R})^\ast \,\colon\; \mu \ge 0, \|\mu\| \leq 1\}$ onto itself. Since $S$ is compact with respect to the weak topology, $\Phi^\ast\vert_S$ is a self-mapping of a convex compact subset of a locally convex topological vector space. Hence Tykhonov's Fixed Point Theorem (see here) implies that $\phi^\ast$ has a fixed point $\mu_0$. Set $\mu = \mu_0/\|\mu\|$, which proves the claim.