$X$ compact, $A \subset X$, $(f_n)_n$ uniformly continous with $f_n:X\rightarrow \mathbb{R}$ $f_n\rightrightarrows f$, $g_n\rightrightarrows g$ with $g_n:A \rightarrow X$, then $f_n \circ g_n \rightrightarrows f \circ g$
I've done the following thing:
Let $\epsilon \gt 0$. Then, since $(f_n)_n$ are continous, exists $\delta \gt 0$ such that $d(x,y) \lt \delta \Rightarrow d(f_n(x), f_n(y)) \lt \epsilon/2$.
Since $g_n \rightrightarrows g$, exists $n_1 \in \mathbb{N}$ such that $n \geq n_1 \Rightarrow d_{\infty}(g_n,g) \lt \delta$
Since $f_n \rightrightarrows f$, exists $n_2 \in \mathbb{N}$ such that $n \geq n_2 \Rightarrow d_{\infty}(f_n,f) \lt \epsilon / 2$
So, taking $n_0 = max\{n_1,n_2\}$, if $n \geq n_0$ we have that $d_\infty(g_n,g) \lt \delta$, so for any $x \in A$, $d(f_n(g(x)), f_n(g_n(x))) \lt \epsilon/2$
So, we have the following:
$$|f(g(x)) - f_n(g_n(x))| \leq |f(g(x)) - f_n(g(x))| + |f_n(g(x))- f_n(g_n(x))| \lt \epsilon/2 + \epsilon/2 = \epsilon$$
Where the first is true since $f_n \rightarrow f$ and the second one is true since $f_n$ is continous and we have that, since $n \geq n_0$, $d(g(x),g_n(x)) \lt \delta$.
The difference with another solution I've seen, is that the compactness is used to prove that $f$ is continous, but here I didn't needed that (or I'm wrong with something)
Thanks
The resolution is not correct since the $\delta$ depends on $n$. The compactness should be used to find a $\delta$ that doesn't depend on $n$, but is quite simpler to say that, since $X$ is compact, and the $f_n$ are uniformly continous, then $f$ is uniformly continous, so we would have the following:
$$|f(g(x))−f_n(g_n(x))| \leq |f(g(x))−f(g_n(x))|+|f(g_n(x))−f_n(g_n(x))| \lt \epsilon$$
Where the first summand is $\lt \epsilon/2$ since $f$ is uniformly continous and the second summand is$\lt \epsilon/2$ since $f_n$ uniformly converges to $f$