How to use the contraction mapping theorem to prove the following result: Let $X$ and $Y$ be Banach spaces, let $a>0$, and let $$B\_a=B\_a\left(z\_0\right)=\left\{z \in X:\left\|z-z\_0\right\| \leq a\right\}.$$ Suppose that $F$ is a $C^1$ map of $B\_a$ into $Y$, with $F^{\prime}\left(z\_0\right)$ invertible, and satisfying, for some $0<\theta<1$,
$$
\left\|F^{\prime}\left(z\_0\right)^{-1} F\left(z\_0\right)\right\| \leq(1-\theta) a,
$$
and
$$
\left\|F^{\prime}\left(z\_0\right)^{-1}\right\|\left\|F^{\prime}(z)-F^{\prime}\left(z\_0\right)\right\| \leq \theta \quad \text { for all } z \in B\_a .
$$
Then there is a unique solution in $B\_a$ of $F(z)=0$.



A contraction mapping theorem
With $G(z)=F'(z\_0)^{-1}(F(z))$ you reduce your problem to the following assertion:



> 
> $\|G'(z)-id\|\le \theta$ for $\|z-z\_0\|\le a$ and $\|G(z\_0)\|\le (1-\theta)a$ imply that $G$ has a zero in $B\_a(z\_0)$.
> 
> 
> 


Indeed, we are looking for a fixed point of $H(z)=z-G(z)$ in $B\_a(z\_0)$. From $\|H'(z)\|=\|id-G'(z)\|\le \theta$ in $B\_a(z\_0)$ and the mean value inequality we get $\|H(z)-H(y)\|\le \theta \|z-y\|$ for all $z,y\in B\_a(z\_0)$ and to conclude with the fixed point theorem we need $H(z)\in B\_a(z\_0)$ for $z\in B\_a(z\_0)$. For $z\in B\_a(z\_0)$ this follows from $$\|H(z)-z\_0\|\le\|H(z)-H(z\_0)\|+\|H(z\_0)-z\_0\| $$ $$\le \theta \|z-z\_0\|+\|G(z\_0)\| \le \theta a +(1-\theta)a=a.$$


