Solving nonlinear integral equation for unique solution

I am working on the following exercise in Kolmogorov's 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.

It seems to me that I need to use the definition of a contraction mapping and the fixed point theorem and to find a function, $$T$$, such that $$|T(f(x) - T(f(y))| \leq \alpha |x-y|$$ where $$\alpha < 1$$. This implies that there exists a fixed point, $$h$$, such that $$T(h) = h$$. I've gotten started on this, but do not know how to finish:

Define the function $$T: C_{[a,b]} \to C_{[a,b]}.$$ by the rule that $$T (f(x)) = \lambda \int_a^b K(x,y, f(y)) \ dy + \varphi(x)$$. For $$f, g \in C_{[a,b]}$$ and $$x \in [a,b]$$, we have: \begin{align*} |T(f(x)) - T((g(x))| & = \left \lvert \left[\lambda \int_a^b K (x,y, f(y)) \ dy + \varphi(x) \right] - \left[\lambda \int_a^b K(x,y, g(y)) \ dy + \varphi(x) \right] \right \rvert \\ & = \left \lvert \lambda \left[\int_a^b K(x,y, f(y)) \ dy + - \int_a^b K(x, y, g(y)) \ dy \right] \right \rvert \\ & = \left \lvert \lambda \int_a^b \left(K(x,y, f(y)) - K(x,y, g(y)) \right) \ dy \right \rvert \\ & = |\lambda| \left \lvert \int_a^b (K(x,y, f(y)) - K(x,y,g(y)) \ dy\right \rvert \\ & < \frac{1}{M(b-a)} \left \lvert \int_a^b (K(x,y, f(y)) - K(x,y,g(y)) \ dy\right \rvert \\ & < \frac{1}{M(b-a)} \left \lvert \int_a^b M |f(y) - g(y)| \right \rvert \end{align*}

I would greatly appreciate any help on this problem. I know I need to at some point introduce the $$L^{\infty}$$ norm, but I do not understand how to do so or the justification for doing this. I am also a bit confused on how to write the successive approximations.

For $$f=T(f),g=T(g)$$, \begin{align*} |f(x)-g(x)|&=|T(f)(x)-T(g)(x)|\\ &\leq\int_{a}^{b}|K(x,y,f(y))-K(x,y,g(y))|dy\\ &\leq\lambda M\int_{a}^{b}|f(y)-g(y)|dy\\ &\leq(\lambda M)^{2}\int_{a}^{b}\int_{a}^{b}|f(z)-g(z)|dzdy\\ &=(\lambda M)^{2}(b-a)\int_{a}^{b}|f(z)-g(z)|dz\\ &\leq(\lambda M)^{2}(b-a)\int_{a}^{b}\lambda M\int_{a}^{b}|f(w)-g(w)|dwdz\\ &=(\lambda M)^{3}(b-a)^{2}\int_{a}^{b}|f(w)-g(w)|dw. \end{align*} Denoting that \begin{align*} L=\dfrac{1}{b-a}\int_{a}^{b}|f(w)-g(w)|dw \end{align*} and proceeding in this way, we have \begin{align*} |f(x)-g(x)|\leq(\lambda M(b-a))^{n+1}L. \end{align*} Now $$\lambda M(b-a)<1$$ so $$(\lambda M(b-a))^{n+1}\rightarrow 0$$ and hence $$f(x)=g(x)$$.
• Use once again that $|f(y)-f(y)|\leq\lambda M\displaystyle\int_{a}^{b}|f(z)-g(z)|dz$ Commented Oct 26, 2019 at 19:03
• Pick any $f$, then consider the sequence $(T^{n}f)$, $T^{n}f=T\circ T\circ\cdots\circ T(f)$, this sequence will converge to a solution to that integral equation, not necessary $f$ itself, but some function anyway. Commented Oct 26, 2019 at 19:29