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*}