I have been stuck on this problem for a little while. I think the proof might be similar to that of proving the Fredhom integral of the second type, however I am not sure. I am a little confused on the idea behind this proof.
$[a,b]$ is a closed interval in $R$ and let $A$ and $K$ be continuous real-valued functions on $[a,b]$ and
{$(x,y) \in E^2: x,y \in [a,b]$}.
$W \in C([a,b])$, and $F(W) \in C([a,b])$ by $(F(W))(x$) = A(x) + $\int_{a}^{b} K(x,y)W(y)dy$
I have already shown the continuity of $F(W)$. However, I am now trying to show that if $|(b-a)K(x,y)| < 1$ for all $x,y \in [a,b]$ then $F$ is a contraction map and thus there exists a unique $p \in C([a,b])$ such that
p(x) = A(x) + $\int_{a}^{b} K(x,y)p(y)dy$
Proof for understanding would be much appreciated. I understand the concept but the proof is throwing me off.
Assume $(b - a)|K(x,y)| < 1$ for all $x,y\in [a,b]$. Let $C := (b - a)\sup_{x,y\in [a,b]} |K(x,y)|$ Since $K$ is continuous on the rectangle $[a,b]\times [a,b]$, there exist $x_0, y_0\in [a,b]$ such that $C = (b - a)|K(x_0,y_0)|$ Thus $0 \le C < 1$. For all $U, W\in C([a,b])$,
\begin{align}\|F(U) - F(W)\| &= \sup_{x\in [a,b]}\left|\int_a^b K(x,y)[U(y) - W(y)]\, dy\right|\\ &\le \sup_{x\in [a,b]}\int_a^b |K(x,y)| |U(y) - W(y)|\, dy \\ &\le \sup_{x\in [a,b]}\int_a^b \frac{C}{b-a} \|U - W\|\, dy \\ &= C\|U - W\|. \end{align}
Therefore, $F$ is a contraction mapping of $C[a,b]$.