If $f_n,g_n: [0,1] \to [0,1]$ continuous $\forall n \in \mathbb{N}$ and converge uniformly prove that $f_n \circ g_n \xrightarrow{u.c} f\circ g$

Question

Assume that $f_n,g_n$ converge uniformly to $f,g$ and $[0,1]\subseteq (\mathbb{R},|\cdot|)$. My work so far: From the initial assumption we get that $|f_n(x)|\leq 1$ and $|g_n(x)|\leq 1$ $\forall n \in \mathbb{N}$. Since $f_n\xrightarrow{u.c} f$ and $g_n\xrightarrow{u.c} g$ there exists $n_0 \in \mathbb{N}$ such that

if $n\geq n_0 \Rightarrow |f_n(x)-f(x)|<\epsilon/2$ and $|g_n(x)-g(x)|<\epsilon/2 \ \forall x\in [0,1]$.

Now if $n\geq n_0$ we notice that: \begin{align*} |f_n\circ g_n(x) -f\circ g(x)|&=|f_n\ z g_n(x)-f_n\circ g(x)+ f_n\circ g(x)-f\circ g(x)| \\ &\leq |f_n \circ g_n(x)-f_n\circ g(x)|+|f_n\circ g(x)-f\circ g(x)| \end{align*} Can someone help me proceed? Is it safe to say that its part of the last sum is $<\frac{\epsilon}{2}$ and why so?

Thanks in advance

Answer 1

Instead of inserting $\pm f_n\circ g(x)$ in the middle, insert $\pm f\circ g_n(x):$

Since $f_n\xrightarrow{u.c} f$ there exists $n_0 \in \mathbb{N}$ such that

if $n\geq n_0$ then $|f_n(x)-f(x)|<\epsilon/2\ \forall x\in [0,1]$

and therefore, $$\begin{align} |f_n\circ g_n(x) -f\circ g(x)|&=|f_n\circ g_n(x)-f\circ g_n(x)+ f\circ g_n(x)-f\circ g(x)| \\ &\leq |f_n\circ g_n(x)-f\circ g_n(x)|+|f\circ g_n(x)-f\circ g(x)| \\ &\le\epsilon/2+|f\circ g_n(x)-f\circ g(x)|. \end{align}$$

Now, $f$ is continuous (as a uniform limit of continuous functions), hence uniformly continuous (since $[0,1]$ is compact), so you may choose $\eta>0$ such that $$\forall u,v\in[0,1]\quad(|u-v|\le\eta\implies|f(u)-f(v)|\le\epsilon/2),$$ and then, using that $g_n\xrightarrow{u.c} g,$ choose $n_1\ge n_0$ such that

if $n\geq n_1$ then $|g_n(x)-g(x)|<\eta,\ \forall x\in [0,1]$.

For every $n\ge n_1$ and every $x\in[0,1],$ you thus get $$|f\circ g_n(x)-f\circ g(x)|<\epsilon/2$$ and you are done.

Note that the continuity of the $g_n$'s was not used.

Answer 2

Actually one needs to be a bit more careful. What we need is uniform continuity of the limit function $f$ (you know it is continuous as the uniform limit of continuous functions, but it is also uniformly continuous since it is defined on a closed interval). Now given $\varepsilon>0$ there exists some $\delta>0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon$.

Moreover, since $g_n\to g$ and $f_n\to f$ uniformly we can guarantee that given $\varepsilon>0$ there exists some $n_0$ such that for all $n>n_0$ we have $|g_n(x)-g(x)|<\delta$ and $|f_n(x)-f(x)|<\varepsilon$.

Combining the two: $$|f_n\circ g_n(x)-f\circ g(x)|\leq |f_n(g_n(x))-f(g_n(x))|+|f(g_n(x))-f(g(x))|<\varepsilon+\varepsilon=2\varepsilon$$ Where the first part is less then $\varepsilon$ by the assumption on uniform convergence and the second part is true since $f$ is uniformly continuous and $|g_n(x)-g(x)|<\delta$ for all $x$.