Let $M$ be a compact metric space, $V, W$ be Banach spaces, and $\alpha, \alpha_n : M \rightarrow V$, $n\in\mathbb{N}$, be a sequence of continuous functions such that $\lim_{n\rightarrow\infty}\alpha_n=\alpha$ uniformly on $M$.
Let further $\Psi : V \rightarrow W$ be continuous.
Question: Can we conclude that $\Psi\circ\alpha_n \rightarrow \Psi\circ\alpha$ uniformly on $M$ (as $n\rightarrow\infty$) even if $V$ is not necessarily compact?
(This is not a homework question.)
Best Answer
More generally let $V,W$ be metric spaces.
Notation: For any function $f$ and for any $T\subseteq dom(f)$ we write $f[T]=\{f(t): t\in T\}.$
Let $N=\cup_{n\in\Bbb N}\alpha_n[M].$
Let $S=(\,\alpha_{j(n)}(x_n)\,)_n$ be any sequence in $N.$ Let $(x_{i(n)})_n$ be a convergent sub-sequence of $(x_n)_n$ with limit $x\in M.$
(1). If for some $k$ we have $j(i(n))=k$ for infinitely many $n,$ then $(\,\alpha_{j(i(n))}(x_n)\,)_{[j(i(n)])=k]},$ that is, $(\,\alpha_k(x_{i(n)}\,)_{[j(i(n))=k]}$ is a subsequence of $S$ converging to $\alpha_k(x)\in N.$
(2). If $\{n: j(i(n))=k\}$ is finite for every $k,$ then we can find a subsequence $(i'(n))_n$ of the sequence $(i(n))_n$ such that $(\,j(i'(n))\,)_n$ is strictly increasing.
$(\bullet)$ The uniform convergence of $(\alpha_{j(i'(n))})_n$ to $\alpha$ then implies that $(\,\alpha_{j(i'(n))}(x_{i'(n)})\,)_n$ is a subsequence of $S$ converging to $\alpha(x).$ [Exercise for the reader.]
(3). From (1) and (2), every sequence in $N$ has a convergent subsequence, so $\overline N$ is compact.
(4). In the main Q we can replace $\Psi$ with its restriction $\Psi|_{\overline N}=^{def}F$ to the compact domain $\overline N.$ Now $F:\overline N\to W$ is uniformly continuous as both its domain and its image $F[\overline N]$ are compact. I leave it to the reader to show this implies that $(F\circ \alpha_n)_n$ converges uniformly to $F\circ \alpha.$
Hint for $(\bullet):$ I suggest a proof by contradiction, using the fact that $\alpha$ is continuous because it is a uniform limit of a sequence of continuous functions