I am trying to solve the following problem in Kolmogorov's real analysis textbook:
Consider the nonlinear integral equation \begin{align*} f(x) = \lambda \int_a^b K(x, y; f(y)) \ dy + \varphi(x) \; \; \; (1) \end{align*} with continuous $K$ and $\varphi$, where $K$ satisfies a Lipschitz condition of the form \begin{align*} |K(x,y; z_1) - K(x,y; z_2)| \leq M|z_1 - z_2|. \end{align*} in its ``functional argument. Prove that (1) has a unique solution for all \begin{align*} |\lambda| < \frac{1}{M(b-a)}. \end{align*} Write the successive approximations to this solution.
I am struggling with where to even get started with this proof. Plugging in an appropriately small $\lambda$ didn't make things easier and there doesn't seem to be a particularly clear way to simplify equation $(1)$.
Any help would be greatly appreciated.
Define $T: C[a,b] \to C[a,b]$ by $Tf(x)=\lambda \int_a^{b} K(x,y,f(y))dy+\phi(x)$. Then $|Tf(x)-Tg(x)|=|\lambda \int_a^{b} [K(x,y,f(y))-K(x,y,g(y))]dy|\leq \lambda M \|f-g\| (b-a)$ (where $\|.\|$ is the supremum norm of $C[a,b]$). Hence $\|Tf-Tg\| \leq \lambda M \|f-g\| (b-a)$ . If $\lambda M(b-a) <1$ then $T$ becomes a contraction mapping. Hence it has a unique fixed point. But $Tf=f$ means that $f$ is a solution of the integral equation.