# Prove that the Fredholm Integral Equation is a contraction

As a part of an exercise I have to prove that the Fredholm Integral Equation is a contraction, I have the following definitions and theorem:

Definition. Let $$(X,d)$$ be a metric space and $$G:X \rightarrow X$$. The mapping G is a contraction if  $$\exists$$ $$0 \le\theta < 1$$ s.t: $$d(G(x),G(y))\le \theta d(x,y), \forall x,y \in X.$$ Definition. Let  $$C([a,b])$$ be the space of bounded continuos funcions on [a,b].

Theorem (Banch Contraction-Mapping).  Let $$(X,d)$$ be a complete metric space and $$G$$ a contraction map of $$X$$. Then $$\exists!$$ fixed point of $$G$$ in $$X$$.

Note. We use the supremun norm.

Exercise Show that the Fredholm Integral Equation $$f(x)= \psi(x)+ \lambda \int_{a}^{b}K(x,y)f(y)dy$$ has a unique solution $$F \in C([a,b])$$, with $$\lambda$$ small enough, $$\psi \in C([a,b])$$ and $$K \in (C([a,b]) \times C([a,b]))$$.

My attempt: I want to use the Theorem, I know that I have a complete metric space, I still have to prove that it is a contraction. So i define the map $$Tf:=f$$, then I have to prove that  $$\exists$$ $$0 \le \theta < 1$$ $$s.t$$  $$d(Tf,T \tilde{f})\le \theta$$ $$d(f, \tilde{f}), \forall f, \tilde{f} \in C([a,b])$$.

$$d(Tf,T \tilde{f})= \underset{a \le x \le b}{\sup} | \psi(x)+ \lambda \int_{a}^{b}K(x,y)f(y)dy - \psi(x)- \lambda \int_{a}^{b}K(x,y) \tilde{f}(y)dy | \\ = \lambda \underset{a \le x \le b}{\sup} | \int_{a}^{b}K(x,y)f(y)dy - \int_{a}^{b}K(x,y) \tilde{f}(y)dy |= \lambda \underset{a \le x \le b}{\sup} | \int_{a}^{b}K(x,y)(f(y)-\tilde{f}(y))dy | \\ \le \lambda \underset{a \le x \le b}{\sup} \int_{a}^{b} | K(x,y)| |f(y)-\tilde{f}(y)|dy= \lambda \int_{a}^{b} \underset{a \le x \le b}{\sup} (| K(x,y)|) \underset{a \le x \le b}{\sup}(|f(y)-\tilde{f}(y)|)dy \\ = \lambda \int_{a}^{b} \underset{a \le x \le b}{\sup} | K(x,y)| \ d(f(y),\tilde{f}(y))dy.$$ There I'm stuck, I know that $$K, f, \tilde{f}$$ are bounded but I don't know how to extract $$f, \tilde{f}$$ from the intregal, since we integrating in function of $$y$$.

Here a similar question Understanding Fredholm integral equation and to proof it is a contraction on C[a,b] , but the answer does not go through the details

Any help or hint will we really appreciated, thank you in advance.

$$$$\begin{split} \lambda \underset{a \le x \le b}{\sup} \int_{a}^{b} | K(x,y)| |f(y)-\tilde{f}(y)|dy & \le \lambda \underset{a \le x \le b}{\sup} \int_{a}^{b} | K(x,y)| dy \underset{a \le y \le b}{\sup} |f(y)-\tilde{f}(y)| \\ & = \left (\lambda \underset{a \le x \le b}{\sup} \int_{a}^{b} | K(x,y)| dy \right )d(f, \tilde{f}). \end{split}$$$$
Then, for small $$\lambda$$, you have a contraction.
Define $$T: (C([a,b]),d) \rightarrow (C([a,b]),d)$$ by
$$(Tf)(x)= \psi(x)+ \lambda \int_{a}^{b}K(x,y)f(y)dy$$ where $$d$$ is the sup metric. We want to find $$f \in C([a,b])$$, such that $$Tf = f$$. But \begin{align} d(Tf,T \tilde{f}) &= \underset{a \le x \le b}{\sup} | \psi(x)+ \lambda \int_{a}^{b}K(x,y)f(y)dy - \psi(x)- \lambda \int_{a}^{b}K(x,y) \tilde{f}(y)dy | \\ &= |\lambda| \underset{a \le x \le b}{\sup} | \int_{a}^{b}K(x,y)f(y)dy - \int_{a}^{b}K(x,y) \tilde{f}(y)dy |\\ &= |\lambda| \underset{a \le x \le b}{\sup} | \int_{a}^{b}K(x,y)(f(y)-\tilde{f}(y))dy | \\ &\leq |\lambda| \underset{a \le x \le b}{\sup} \underset{a \le y \le b}{\sup} \left( | K(x,y)| |f(y)-\tilde{f}(y)| \right)(b-a)\\ &\leq |\lambda|(b-a) \underset{a \le x \le b}{\sup} \underset{a \le y \le b}{\sup} (| K(x,y)|) \underset{a \le y \le b}{\sup}(|f(y)-\tilde{f}(y)|) \\ &= |\lambda|(b-a) \underset{a \le x \le b}{\sup} \underset{a \le y \le b}{\sup} (| K(x,y)|) d(f,\tilde{f}). \end{align} Now I believe you should choose small $$\lambda$$, for example $$|\lambda|< \dfrac{1}{(b-a) \underset{a \le x \le b}{\sup} \underset{a \le y \le b}{\sup} (| K(x,y)|)+\varepsilon}$$ for small $$\varepsilon.$$