Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 22I / c / i

Set $y_k(0)=x_0$ and let $M={\sup }_{z\in {\mathbb{R}}^n}\Vert F(z)\Vert <\infty$. On the first interval, $y_k$ is the known constant $x_0$, so the integral equation defines $x_k$. Its endpoint value then defines the constant $y_k$ on the next interval, and induction over the $k$ subintervals defines the pair on all of $[0,1]$. An integral of the bounded, piecewise constant function $F\circ y_k$ is continuous, so every $x_k$ is continuous. In fact,so the approximations are uniformly Lipschitz continuous.