Suppose $g:\mathbb{R}\to\mathbb{R}$ is continuous. Define $F:C[0,1] \to C[0,1]$ by $F(f) =g\circ f$ Suppose $g:\mathbb{R}\to\mathbb{R}$ is continuous. Define $F:C[0,1] \to C[0,1]$ by $F(f) =g\circ f$
a. Prove that $F$ is continuous
b. Prove that if $g$ is uniformly continuous, then $F$ is uniformly continuous
I was trying to use the definition of continuity to prove, but it didn't work out
 A: Point a)
Fix $f\in C[0,1]$.
Because $f$ is continuous and defined on a compact set, by Weierstrass theorem it attains its maximum, say $b\in\mathbb{R}$, and its minimum, say $a\in\mathbb{R}$.
Because $g:\mathbb{R}\rightarrow\mathbb{R}$ is continuous and $[a-1,b+1]$ is compact, by Heine-Cantor theorem, $g$ is uniformly continuous on $[a-1,b+1]$. 
Fix $\varepsilon>0$.
Let $\delta\in(0,1)$ such that $$\forall x,y\in[a-1,b+1], (|x-y|\le\delta)\implies(|g(x)-g(y)|\le\varepsilon).$$
Fix $\varphi\in C[0,1]$ such that $\|\varphi-f\|_\infty\le\delta$.
Fix $t\in[0,1]$.
Then $|\varphi(t)-f(t)|\le\delta$, so not only $f(t)\in[a,b]\subset[a-1,b+1]$ but also $\varphi(t)\in[a-1,b+1]$, and then: 
$$|F(\varphi)(t)-F(f)(t)|=|(g\circ \varphi)(t)-(g\circ f)(t)|=|g(\varphi(t))-g(f(t))|\le\varepsilon.$$
By the arbitrariness of $t$: $$\|F(\varphi)-F(f)\|_\infty\le\varepsilon.$$
By the arbitrariness of $\varphi$ and $\varepsilon$, $F$ is continuous in $f$.
By the arbitrariness of $f$, $F$ is continuous.
Point b)
Fix $\varepsilon>0$.
Let $\delta>0$ such that $$\forall x,y\in\mathbb{R}, (|x-y|\le\delta)\implies(|g(x)-g(y)|\le\varepsilon).$$
Fix $f_1,f_2\in C[0,1]$ such that $\|f_1-f_2\|_\infty\le\delta$.
Then $\forall t\in[0,1], |f_1(t)-f_2(t)|\le\delta$, so $$\forall t\in[0,1], |F(f_1)(t)-F(f_2)(t)|=|g(f_1(t))-g(f_2(t))|\le\varepsilon.$$
So: $$\|F(f_1)-F(f_2)\|_\infty=\sup_{t\in[0,1]}|F(f_1)(t)-F(f_2)(t)|\le\varepsilon.$$
By the arbitrariness of $f_1,f_2$ and $\varepsilon$, $F$ is uniformly continuous.
