composition of uniformly convergence sequence with continuous function, is uniformly convergence? Let $(f_n)$ be a series of functions in $C[0,1]$ that uniformly converge to a continuous function $f\in C[0,1]$.
a. Let $g: [0,1]\to [0,1]$ be a continuous function. Is it true that $f_n\circ g$ uniformly converges to $f\circ g$?
b. If no, what about the special case $g(x)=x^2$? It would also be differentiable, have bounded derivative, etc.
About my thoughts:
Intuitively I find it logical that a. would be true, and intuitively I'd guess that in order to prove such claim I'd have to go directly to the $\epsilon-\delta$ definition.
 A: This is actually very simple: 
Let $\epsilon>0$, then there is $N\in \mathbb N$ so that 
$$|f_n(y) - f(y) | < \epsilon$$
for all $y\in [0,1]$ and $n\geq N$. In particular, 
$$|f_n(g(x)) - f(g(x))| < \epsilon$$
for all $x\in [0,1]$ and $n\geq N$. Thus $f_n\circ g$ converges to $f\circ g$ uniformly. Note that continuity of $g$ is not needed. 
The harder one is the following:
Claim: If $g: \mathbb R \to \mathbb R$ is continuous and $f_n: [0,1] \to \mathbb R$ converges uniformly to $f$, then $g\circ f_n$ converges uniformly to $g\circ f$. 
Proof: First of all, as $f_n$ converges to $f$ uniformly, there is $M>0$ so that $|f_n(x)|\leq M$ for all $x\in [0,1]$ and $n\in \mathbb N$ (Why?). Then we consider $g:[-M, M]\to \mathbb R$. Let $\epsilon>0$. Then by uniform continuity of $g$ (as the domain is closed and bounded), there is $\delta>0$ so that $$ |g(y) - g(z) | < \epsilon$$
whenever $|y-z|<\delta$. Using this $\delta$ there is $N$ so that 
$$|f_n(x) - f(x)| < \delta$$
for all $x\in [0,1]$ and $n\geq N$. Thus 
$$|g(f_n(x)) - g(f(x))| <\epsilon$$
for all $x\in [0,1]$ and $n\geq N$. 
