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I am wondering about extension of the the answer given here.

Namely, suppose $U$, $V$ are Polish spaces and $F:Uā†’V$ is uniformly continuous. Does this mean that the push-forward operator $F_*: \mathcal{P}(U) \rightarrow \mathcal{P}(V)$ is also uniformly continuous when each $\mathcal{P}(.)$ is endowed with weak* topology?

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Kass
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If you use the Levy-Prohorov metric (https://en.wikipedia.org/wiki/L%C3%A9vy%E2%80%93Prokhorov_metric) on both $\mathcal P(U)$ and $\mathcal P(V)$, then $F_*$ is indeed uniformly continuous. The proof is a matter of chasing through the definitions, the key being to observe that $$ \{x\in U: d_U[x, F^{-1}(A)]<\delta\}\subset \{x\in U: d_V[F(x),A]<\epsilon\}. $$ provided $A$ is a Borel subset sof $V$. Here, $\epsilon>0$, and $\delta$ is such that $$ d_U[x,y]<\delta \Longrightarrow d_V[F(x),F(y)]<\epsilon, $$ per the uniform continuity of $F$.

John Dawkins
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  • Thank you! For the special case that $U$ is compact, I realized also that one can use the fact that when $\mathcal{P}(U)$ has the weak* topology it is a compact space if $U$ is compact, and so a continuous operator on a compact space is uniformly continuous. But your solution is obviously better. ā€“ Kass Apr 13 '16 at 19:31
  • What is "uniformly continuous" supposed to mean, given that $U$ and $V$ have no uniform structures? A Polish space is a topological space homeomorphic to a separable, complete metric space, but notice that homeomorphisms do not preserve uniform (in particular metric) structures. So how is the question supposed to be understood? ā€“ Alex M. Mar 11 '18 at 15:44