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$h : \Bbb R \times S \to \Bbb R$ continuous. $x_n \to x$. $S$ compact. Does $h(x_n,.)\to h(x,.)$ uniformly ?

I know that pointwise convergence and equicontinuity implies uniform convergence on compact. But, I don't see here equicontinuity.

Anonymous
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  • Hint (as in the other post): Use that $L := {x_n \mid n} \cup {x}$ is compact, so that $L\times S$ is compact and use the uniform continuity of $h$ on $L\times S$. This works at least as long as $S$ is a metric space. – PhoemueX Jan 19 '15 at 13:47

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In the following, I will assume for simplicity that $S$ is a metric space. In the question Convergence of probability measures problem where you seem to need this statement, this is satisfied.

It is easy to see that $L := \{x_n \mid n \in \Bbb{N}\} \cup \{x\} \subset \Bbb{R}$ is compact. Hence, $L \times S$ is also compact when equipped with (e.g.) the metric

$$ d_1((x,s),(y,t)) := |x-y| + d(s,t), $$ where $d$ is the metric on $S$.

It is well-known that a continuous function on a compact metric space is uniformly continuous (see Continuous mapping on a compact metric space is uniformly continuous).

Let $\varepsilon > 0$. By uniform continuity, there is some $\delta > 0$ such that $d_1((x,s),(y,t)) < \delta$ for $(x,s),(y,t) \in L\times S$ implies $|h(x,s) - h(y,t)| < \varepsilon$.

Using $x_n \to x$, we see that there is some $N \in \Bbb{N}$ with $|x_n - x| < \delta$ for $n \geq N$. For arbitrary $s \in S$, this yields $d_1 ((x_n,s),(x,s)) = |x_n - x| < \delta$ and hence $$ |h(x_n, s) - h(x,s)| < \varepsilon $$ as soon as $n \geq N$.

But this yields

$$ \sup_{s \in S} |h(x_n, s) - h(x,s)| < \varepsilon $$

for $n \geq N$, which is nothing but the desired uniform convergence.

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