# Fix point theorem for measures? metric on space of measures?

I have the following problem: I consider a probability space $(\Omega, \mathcal{F}, \mu)$ where $\Omega= C_0([0,1])$ (space of continuous functions on $[0,1]$ starting from 0), $\mathcal{F}$ is a $\sigma$ algebra on $\Omega$ (for instance the one generated by point-evaluations) and $\mu$ a probability measure on $(C_0([0,1]),\mathcal{F})$.

Denote by $\mathcal{M}(\Omega)$ the space of probability measures on $(\Omega, \mathcal{F})$. I define a mapping from $\mathcal{M}(\Omega)$ onto itself by $$\mu \mapsto P^\mu = \int f d\mu$$ where $f$ (btw $f$ is not trivial, so that the mapping is not trivial) is fixed and it depends also on $\mu$ in fact. I can give more details if necessary.

I'm very interested in finding the measure $\mu^\ast$ such that $\mu^\ast = P^{\mu^\ast}$ if it exists and the first naiv thought was a "fix point theorem" but then the question is...

1) What topology/metric should I use in $\mathcal{M}(\Omega)$ so that it makes sense to consider this operator?

2) Having found a metric $d$, and proving $d(P^\mu, P^{\mu'})\leq d(\mu,\mu')$, would this imply there is $\mu^\ast$ such that $\mu^\ast = P^{\mu^\ast}$? (Banach fix point theorem)

In conclusion, I'd like to find a fix point of this mapping :( Is there any idea or standard trick? Thank you very much for your kind help!

## 1 Answer

Take a look at Contraction maps on probabilistic metric spaces. The nice thing about this paper is that it doesn't modify the definition of a contraction map or require specific triangle functions, it only stipulates a growth condition on the distance distribution functions (ddfs) between different metrics. These ddfs are non-decreasing, left continuous, and take the form $F : \mathbb{R}^+ \mapsto [0,1]$. $\triangle^+$ is the set of all possible ddfs.

Interpreting $F_{pq}(x)=\Pr[d(p, q) < x]$ will make the rest of the definitions more clear.

Basically, we define a probabilistic metric PM space by $(S, F, \tau)$ where $S = \mathcal{M}(\Omega)$, $F : S \times S \mapsto \triangle^+$ (Note that $F(p,q)=F_{pq}$ and must satisfy $F_{pq}$ is the unit step iff $p=q$, $F_{pq}=F_{qp}$, and $F_{pq}(x) \geq \tau(F_{pr}, F_{rq})(x)$ for all $x \geq 0, r \in S$), and $\tau$ is a binary operation on $\triangle^+$. Also, the PM space is complete if every Cauchy sequence (defined as a sequence $\{p_i\}_{i=1}^{\infty}$ where for all $t > 0$ there exists $N$ with $n, m > N$ that $F_{p_n p_m}(t) > 1 - t$) converges.

Then, any contraction map $M : S \mapsto S$ (which must satisfy $F_{M(p)M(q)}(x) \geq F_{pq}(\frac{x}{\alpha})$ for all $x\geq 0$) on the complete PM space $(S, F, \tau)$ is guaranteed to have a unique fixed point if $\int_0^\infty \ln(u) dF_{pq}(u)^1$ converges.

$^1$This strange integral is necessary as it then implies that $\lim_{x \to \infty}F_{pq}(x) = 1$, which is the necessary growth condition.

TL;DR

Once you find an appropriate map $F_{pq} = \Pr[d(p,q)]$, then there is an implication that $u^*$ exists and is unique.