# show that a functional operator $F : X \to \mathbb{R}$ is continous

i have the following homework:

Let $g : \mathbb{R} \to \mathbb{R}$ be continous and $X = C([a,b])$ with the metric $$d(x,y) := \sup_{t \in [a,b]} |x(t) - y(t)|$$ show that $F: X \to \mathbb{R}$ with $$F(x) = \int_a^b g(x(t)) \mathrm{d}t$$ is a continous function.

I was able to show a much weaker proposition, namely that the functional $$F'(f) = \int_a^b f(t) \mathrm{d}t$$ is uniformly continous.

My proof: For $\varepsilon > 0$, set $\delta := \frac{\varepsilon}{b-a}$, if $d(f,g) < \delta$ then \begin{align*} |F'(f) - F'(g)| & = | \int_a^b f(t) \mathrm{d}t - \int_a^b g(t) \mathrm{d}t| \\ & = | \int_a^b f(t) - g(t) \mathrm{d}t | \\ & \le \int_a^b |f(t) - g(t)| \mathrm{d}t \\ & \le \int_a^b \delta \mathrm{d}t \\ & = \delta \cdot (b-a) = \varepsilon \end{align*} But i am not able to generalize my results, do you have any hints for me?

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Let $(x_i)_{i\in\mathbb{N}}$ be a convergent series in $X$. Since convergence in $X$ is pointwise convergence (due to the use of the supremum norm), and since $g$ is continous, you have $$g\left(\left(\lim_{i\to\infty}x_i\right)(t)\right) = \lim_{i \to \infty} g(x_i(t)) \text{ for every } t \in [a,b]$$
Now, all $x \in X$ are uniformly continuous on $[a,b]$ because they are continous and $[a,b]$ is compact. Also, every convergent series in $X$ must be bounded, i.e. if $\lim_{i\to\infty}x_i = x$, there is a compact set $K$ with $\text{ran }x_i \subset K$ for all $i \in \mathbb{N}$. Thus, you may replace $g$ with $g|_K$ ($g$ restricted to $K$), which is then uniformly continous, again because it's continous on a compact set. Together, you get that the limit converges uniformly (wrt $t$), which should be sufficient to swap it with the integral.
Hint: $g$ is uniformly continuous on the interval $[a,b]$.
This could be misinterpreted, because there are a fixed $a$ and $b$ in the problem, and $g$'s behavior on that particular interval isn't most relevant. – Jonas Meyer Oct 12 '12 at 19:37