I was reviewing another question on here (linked above), and I wanted to expand on it with another question (I hope this is the correct way to do such a thing).
Here is my question in full:
Let $[a,b]$ be a closed interval in $\Bbb R$ and let $A$ and $K$ be continuous real-valued functions on $[a,b]$ and $\{(x,y) \in E^2: a \le y \le x \le b \}$ respectively. How do you prove that there is a unique $\phi \in C([a,b])$ such that
$$\phi (x) = A(x) + \int_a^xK(x,y)\phi(y)dy$$.
I have been trying to imitate the proof in the answer to question linked above to prove this, but I am having trouble.
I think you must require some condition on how big $K$ can be to obtain this result with the Contraction Mapping Principle. Although I have not come up with a counterexample that shows we require some bound, the standard application of the Contraction Mapping Principle would not work. The weakest condition that we can apply on $K$ to make the standard argument work would be that
$$ \sup_{x \in [a,b]} \left|\int_{a}^{x}K(x,y) \;dy \right| < 1. $$
Now, define the function $F: C([a,b]) \rightarrow C([a,b])$ by $F(\phi)(x) = A(x) + \int_{a}^{x} K(x,y) \phi(y) \;dy.$ $F(\phi)$ will be a continuous function on $[a,b]$, because $A$, $K$, and $\phi$ are continuous and their differences on points are bounded above by their supremum norms. We see that $$ \left|F(\phi)(x) - F(\phi)(x')\right| \leq \left|A(x) - A(x')\right| + \left|\int_{x'}^{x} (K(x,y)-K(x',y))\phi(y) \; dy \right| $$ and $$ \left|\int_{x'}^{x} (K(x,y)-K(x',y))\phi(y) \; dy \right| \leq \left|x-x'\right| \left(\sup_{y \in [a,b]}|K(x,y)-K(x',y)|\right) \left(\sup_{y \in [a,b]} |\phi(y)|\right). $$ Since $K$ is uniformly continuous, we can make the first supremum as small as we like by taking a small enough $\delta$.
We can also verify that $F$ is a contraction with respect to the supremum norm on $C([a,b])$.
$$ \left|F(\phi)(x) - F(\psi)(x)\right| = \left| \int_{a}^x K(x,y) (\phi(y) - \psi(y))\; dy \right| \leq \left(\sup_{t \in [a,b]}\left|\int_{a}^{t}K(t,y) dy \right| \right) \left \|\phi - \psi \right\|_{\infty}. $$ But this will only be a contraction if this supremum is less than $1$, which we had to add as an additional hypothesis.
Since $F$ is a contraction, and $C([a,b])$ is complete, it must have a unique fixed point $\phi$ so that $\phi(x) = F(\phi)(x) = A(x) + \int_{a}^{x}K(x,y) \phi(y) \; dy$