Show $F:C[0,1] \rightarrow C[0,1]$ defined by $F(f) = g \circ f$ is continuous.

Question

Define $F:C[0,1] \rightarrow C[0,1]$ by $F(f) = g \circ f$, where $g:\Bbb R \rightarrow \Bbb R$ is continuous.

I am struggling to show that F is continuous. We use the supremum norm: $$ ||f|| = sup_{x\in[0,1]}|f(x)| $$

Let U be an open set in $C[0,1]$ (in the range of F). If $h \in U$, then $h$ is of the form $h = g \circ f$. The composition of continuous functions is continuous, so $h = g \circ f$ is continuous. Since $U$ open, $\exists \epsilon \gt 0$ such that $B_{\epsilon}(h) \subseteq U$.

Now, $F^{-1}(U) = \{f:g \circ f \in U\}$ Thus since $U$ is open, $F^{-1}(U)$ is open (how can I assert this)?

Any hints are greatly appreciated!

Answer 1

Let $\epsilon>0$ and fix $f_2\in C[0,1]$. Since $g$ is continuous in $\{|z|\leq \|f_2\|+1\}$, then it is also uniformly continuous. Therefore, there is $1>\delta>0$, such that if $|z_1-z_2|<\delta$ then $|g(z_1)-g(z_2)|<\epsilon$.

If we take $\|f_1-f_2\|<\delta$, then for all $x\in[0,1]$ we have $|f_1(x)-f_2(x)|<\delta$. Therefore, $|g(f_1(x))-g(f_2(x))|<\epsilon$ for all $x\in[0,1]$.

Hence $\|g\circ f_1-g\circ f_2\|<\epsilon$.

Answer 2

Let's see that $F$ is continuous at some $f_0$. Let $\varepsilon > 0$. By uniform continuity of $g$ in $B_{1+\|f_0\|}(0)$ we have $\delta > 0$ such that:

$$ |x - y| < \delta \Rightarrow |g(x) - g(y)| < \varepsilon $$

Now if $\|f - f_0\| < \min\{1,\delta\}$, $|f(x) - f_0(x)| < \delta$ for all $x$, so

$$ |F(f_0)(x) - F(f)(x)| = |g(f_0(x))-g(f(x))| < \varepsilon \quad (\forall x \in [0,1]) $$

and therefore $\|F(f) - F(f_0)\| \leq \varepsilon$. Here I use that $f(x),f_0(x) \in B_{1 + \|f_0\|}(0)$ for any $x$.