How to use the contraction mapping theorem to prove the following result: Let $X$ and $Y$ be Banach spaces, let $a>0$, and let $$B_a=B_a\left(z_0\right)=\left\{z \in X:\left\zz_0\right\ \leq a\right\}.$$ 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 $F(z)=0$.

1$\begingroup$ Looks like an exercise. Where does the problem come from? $\endgroup$– Jochen WengenrothCommented Aug 6, 2023 at 11:46

1$\begingroup$ Actually, It can be seen in the paper: Y. Y. Li and L. Nirenberg, The Dirichlet problem for singularly perturbed elliptic equations, Comm. Partial Differential Equations 23 (1998), 487–545. but I don't know how to prove it. $\endgroup$– Davidi ConeCommented Aug 6, 2023 at 11:54
1 Answer
With $G(z)=F'(z_0)^{1}(F(z))$ you reduce your problem to the following assertion:
$\G'(z)id\\le \theta$ for $\zz_0\\le a$ 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 $H(z)=zG(z)$ in $B_a(z_0)$. From $\H'(z)\=\idG'(z)\\le \theta$ in $B_a(z_0)$ and the mean value inequality we get $\H(z)H(y)\\le \theta \zy\$ 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 $$\H(z)z_0\\le\H(z)H(z_0)\+\H(z_0)z_0\ $$ $$\le \theta \zz_0\+\G(z_0)\ \le \theta a +(1\theta)a=a.$$

$\begingroup$ I guess for any $z\in B_a(z_0)$, we have $ H(z)\in B_{\theta a}(H(z_0))$，right? $\endgroup$ Commented Aug 6, 2023 at 14:45