 Solution

For $0<\eta \le h$, work on the closed subsetIt is a nonempty complete metric space because it is closed in the Banach space of continuous functions with the supremum norm. Define the Picard iteration operatorChoose $\eta$ so that $M\eta \le k$ and $L\eta <1$, omitting a bound whose constant is zero. Such a positive $\eta$ exists. The first inequality makes $T$ a self-map, since $|(Ty)(x)-b|\le M|x-a|\le k$. The Lipschitz condition givesso the contraction mapping theorem yields a unique fixed point. The fundamental theorem of calculus makes it a differentiable solution of $y^{\prime }=f(x,y)$ with $y(a)=b$; conversely every solution staying in the rectangle satisfies the integral equation and is a fixed point. This is the local Picard-Lindelöf theorem construction. The interval-restricted space $C_{\eta }$ avoids treating functions defined on a shorter interval as elements of $C_h$.