# Example compact operator on space of continuous functions

compact-operatorscontinuityfunctional-analysisnormed-spaces

Suppose $$F:[a,b]^2\times\mathbb{R}\to\mathbb{R}$$ is a continuous
function. Define $$A:C([a,b])\to C([a,b])$$ by
$$(Ax)(t):=\int_a^bF(s,t,x(s))ds$$. Suppose $$-M\leq x(s)\leq M$$ for all
$$s\in[a,b]$$. Show $$A$$ is well defined and $$A$$ is a compact
operator.

I am quite new to functional analysis and I am not sure how to prove this exercise. The only thing that I notice is that since $$x$$ is bounded, we can consider $$F:[a,b]^2\times [-M,M]$$ and then $$F$$ is uniformly continuous. Also, I believe that I can use the Arzela-Ascoli to prove that $$A$$ is compact, but for that I need to show $$A$$ is equicontinuous. Any help will be appreciated.

By definition $$A$$ is compact if the image $$A(B)$$ is relatively compact for every bounded subset $$B \subseteq C([a,b])$$.
As you noticed correctly, you can use Arzelà-Ascoli to show this. Let $$B \subseteq C([a,b])$$ be bounded with $$B \subseteq B_M(0)$$. For pointwise boundedness observe that for every $$x \in B, t\in [a,b]$$ it is $$|(Ax)(t)| = \left| \int_a^b F(s,t, x(s)) ds \right| \leq \int_a^b |F(s,t,x(s)| ds \leq (b-a) C$$ with $$C = \max_{(s,t,x) \in [a,b]^2 \times [-M,M]} |F(s,t,x)|$$. This maximum exists since $$F$$ is continuous.
Equicontinuity can be obtained as follows. Let $$\epsilon > 0$$. Then, as $$F$$ is uniformly continous on $$[a,b]^2 \times [-M,M]$$, there is some $$\delta$$ such that for all $$t_1, t_2 \in [a,b]$$ it is $$|t_1 - t_2| < \delta \quad \Rightarrow \quad | F(s, t_1, x) - F(s, t_2, x)| < {\epsilon \over b-a}$$ for all $$s \in [a,b], x \in [-M,M]$$.
Hence, for $$t_1, t_2$$ as above one obtains \begin{align*} |(Ax)(t_1) - (Ax)(t_2)| &= \left| \int_a^b F(s,t_1, x(s)) - F(s, t_2, x(s)) ds \right| \\ &\leq \int_a^b | F(s,t_1, x(s)) - F(s, t_2, x(s))| ds \\ &\leq \int_a^b {\epsilon \over (b-a)} ds \leq \epsilon. \end{align*}