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I'm hoping to make a generalization of the answer to this question.
Let's say that instead that we're composing two uniformly continuous function sequences, does this composition converge uniformly to $f circ g$.

Here's why I think it does.

Consider $left( f _ { k } right) _ { k = 1 } ^ { infty }$ and $( g_k ) stackrel { infty } { k } = 1$ to be sequences of continuous functions $[ 0,1 ] rightarrow [ 0,1 ]$ converging uniformly to $f$ and $g : [ 0,1 ] rightarrow mathbb { R }$ respectively.

By applying the Weierstrass M-Test, $ exists quad M > 0$ so that $|f_k(x)| leq M quad forall x in [0,1]$ and $k in mathbb{N}$

Now restrict the domain of $(g_k)$ to $[ - M , M ]$. Since the codomain of $(f_k)$ is $[0,1]$, this $[ - M , M ]$ will be within that domain. Therefore, there exists $N in mathbb{N}$ such that $(g_k)$ converges uniformly within this restricted domain.

Therefore

begin{equation}
left| g _ { k } ( f_k(x )) - g(f ( x )) right| < varepsilon text { for all } x in S text { and } k geq N
end{equation}
so $f _ { k } circ g _ { k }$ converge uniformly to $f circ g$.

Is it that simple, or am I missing something?

Suppose $g_n: [0,1] longrightarrow [0,1]$, $,f_n:[0,1] longrightarrow mathbb R$ $(n=1,2,ldots)$ are two sequences of continuous functions that converge uniformly to $g$ and $f$, respectively.

Let $varepsilon>0$ be given.
By the uniform limit theorem and by the Heine-Cantor theorem, $f$ is uniformly continuous on $[0,1]$. Thus there is $delta>0$ so that $|,f(u)-f(v)|<frac12 varepsilon$ whenever $|u-v|<delta$ and $u,v in [0,1]$. Since $g_n to g$ uniformly, there is a positive integer $N_1$ so that for all $x in [0,1]$ we have
$$|g_n(x)-g(x)|<delta text{ whenever } n geq N_1.$$
Since $f_n to f$ uniformly, there is a positive integer $N_2$ so that for all $x in [0,1]$ we have
$$|,f_n(x)-f(x)|<frac{varepsilon}{2} text{ whenever} n geq N_2.$$
Set $N=max{N_1,N_2}$. So if $n geq N$ and $x in [0,1]$, then we have
begin{equation} begin{split} | ,f_n(g_n(x)) - f(g(x))| &= |,f_n(g_n(x)) - f(g_n(x))+f(g_n(x)) - f(g(x)) |\
& leq |,f_n(g_n(x)) - f(g_n(x))|+|,f(g_n(x)) - f(g(x))| \
& <frac{varepsilon}{2} +frac{varepsilon}{2}\
& = varepsilon.
end{split}end{equation}

Therefore, $f_n circ g_n to f circ g$ uniformly on $[0,1]$ (definition of uniform convergence).