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.
 A: 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.
A: 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 $\varepsilon.$
