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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.

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2 Answers 2

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You almost got it. Note that

$$\begin{equation} \begin{split} \lambda \underset{a \le x \le b}{\sup} \int_{a}^{b} | K(x,y)| |f(y)-\tilde{f}(y)|dy & \le \lambda \underset{a \le x \le b}{\sup} \int_{a}^{b} | K(x,y)| dy \underset{a \le y \le b}{\sup} |f(y)-\tilde{f}(y)| \\ & = \left (\lambda \underset{a \le x \le b}{\sup} \int_{a}^{b} | K(x,y)| dy \right )d(f, \tilde{f}). \end{split} \end{equation}$$

Then, for small $\lambda$, you have a contraction.

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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 $\varepsilon.$

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