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 $F(z)=0$.
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1$\begingroup$ Looks like an exercise. Where does the problem come from? $\endgroup$– Jochen WengenrothCommented Aug 6, 2023 at 11:46
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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
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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 $$\|H(z)-z_0\|\le\|H(z)-H(z_0)\|+\|H(z_0)-z_0\| $$ $$\le \theta \|z-z_0\|+\|G(z_0)\| \le \theta a +(1-\theta)a=a.$$
answered Aug 6, 2023 at 12:41
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$\begingroup$ 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