Prove that there exists a unique solution for all $|\lambda|$ $\lt$ $\frac{1} {M(b-a)}$
Where $f(x)=\phi(x)+ \lambda \int_{a}^{b}K(x,y;f(y)) dy$
And K satisfies a Lipschitz condition of the form $|K(x,y;r)-K(x,y;s)|\le M|r-s|$
And moreover what would the successive approximations to this solution be given by?
I have spent all day researching and attempting to do this however I am stumped.
Any help greatly appreciated.
Consider the Banach space $(C[a,b], ||*||)$, where $||*||$ is the $\sup$ -norm and let $T:C[a,b] \to C[a,b]$ be defined by
$(Tf)(x)=\phi(x)+ \lambda \int_{a}^{b}K(x,y;f(y)) dy$.
Its your turm to show, that
$||Tf-Tg|| \le \lambda M (b-a) ||f-g||$ for all $f,g \in C[a,b]$.
Hence, if $\frac{1} {M(b-a)}$, Banach's fix point theorem applies !