Let $(X, d)$ be a compact metric space and let $T:X \to X$ be a continuous mapping. Now, does there exist a probability measure $\nu$ such that $T_* \nu = \nu$ (the first thing is the image measure)?
Now I want to do this using a fixed-point theorem. So I start like this:
Define the operator $\psi_T$ from $C(X)$ into itself by $\psi_T f = f \circ T$. Now we also know that $C(X)^* = P(X)$ where $P(X)$ are the Borel probability measures on $X$. So we have a mapping $\psi_T^* : P(X) \to P(X)$. Further we have that:
$\langle f, \underbrace{\psi_T^* \mu}_{= \nu} \rangle = \langle \psi_T f, \mu \rangle = \int f \circ T \, d\mu.$ And $\nu$ is completely determined by $\int f \, d\nu = \int f \circ T \, d\mu$
Further, since $X$ is compact I know that $P(X)$ is compact (hence complete) with respect to the Bounded Lipschitz metric. This one is given by $d_\text{BL}(\mu, \nu) := \sup \left \{ \left | \int f \, d\mu - \int f \, d\nu \right | : f \in \text{BL}(X,d), \|f\|_\text{BL} \leq 1 \right \}.$
So everything is fine for the Banach fixed point theorem except one thing: is $d_\text{BL}$ a contraction? I don't think so. Does someone have an idea if I'm on the right track or does someone have a suggestion how to continue?
By the way, this is homework.