Let $X$ denote a metric space. One can assume that $X$ is Polish if that helps, but I was trying to avoid to assume that $X$ is compact. Let $P(X)$ denote the set of Borel probability measures on $X$. The weak*-topology on $P(X)$ is defined as usual: a net $(p_{\alpha})$ in $P(X)$ converges to $p \in P(X)$ iff $|\int fdp_{\alpha}-\int fdp|$ converges to $0$.

Question: Is there any metric that metrizes the weak*-topology on $P(X)$ and has convex open balls?

  • $\begingroup$ Yes, it is tough. I am actually after a metric on $P(X)$, for $X$ Polish, that has the nice properties of the Prohorov metric (namely, that $P(X)$ is also Polish) and has convex open balls. I am not quite able to show that the Prohorov metric has convex open balls... $\endgroup$ – user3117 Nov 4 '10 at 17:29
  • $\begingroup$ @user3117: I'm curious: why do you need this? $\endgroup$ – Jonas Teuwen Nov 4 '10 at 20:42
  • $\begingroup$ @Jonas T: I was trying to extract a continuous selection from a correspondence, and I thought the existence of such a metric would help. But now I am curious about the question by itself... $\endgroup$ – user3117 Nov 4 '10 at 22:52

On a separable metric space $(X,d)$, weak convergence of probability measures is equivalent to convergence with respect to the Lévy–Prokhorov metric defined by $$ \beta(\mu,\nu) = \sup \left( \int_X f\ d(\mu-\nu): \|f\|_{BL}\leq 1\right),$$ where $$\|f||_{BL}=\sup_x |f(x)|+\sup_{x\neq y}|f(x)-f(y)|/d(x,y).$$

See Theorem 11.3.3 of R. M. Dudley's Real Analysis and Probability.

Let's show that $B=\{\nu: \beta(\mu,\nu)\leq\varepsilon \}$ is a convex set. Let $\nu_1,\nu_2\in B$, $\alpha\in(0,1)$, and $f$ with $\|f\|_{BL}\leq 1$. Then \begin{eqnarray*} \int f d(\mu-[\alpha\nu_1+(1-\alpha)\nu_2])&=&\alpha\int f d(\mu-\nu_1)+(1-\alpha)\int f d(\mu-\nu_2)\\[5pt] &\leq&\alpha\beta(\mu,\nu_1)+(1-\alpha)\beta(\mu,\nu_2)\\[5pt] &\leq &\alpha \varepsilon +(1-\alpha)\varepsilon\\[5pt] &=&\varepsilon. \end{eqnarray*} Taking the supremum over such $f$ gives $\beta(\mu,\alpha\nu_1+(1-\alpha)\nu_2)\leq \varepsilon$, so the closed ball is convex.

The open ball $\{\nu: \beta(\mu,\nu)<\varepsilon\}$ is the increasing union of the convex sets $\{\nu: \beta(\mu,\nu)\leq\varepsilon-1/n\}$, for $n>1/\varepsilon$, and so is itself convex.

  • $\begingroup$ Yes, okay, but why does it have open convex sets? $\endgroup$ – Jonas Teuwen Nov 5 '10 at 15:00
  • $\begingroup$ Fair enough. Yes, I mean open balls. $\endgroup$ – Jonas Teuwen Nov 5 '10 at 15:30

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