A contraction mapping theorem

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 $$F(z)=0$$.

• Looks like an exercise. Where does the problem come from? Commented Aug 6, 2023 at 11:46
• 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. Commented Aug 6, 2023 at 11:54

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 $$\|H(z)-z_0\|\le\|H(z)-H(z_0)\|+\|H(z_0)-z_0\|$$ $$\le \theta \|z-z_0\|+\|G(z_0)\| \le \theta a +(1-\theta)a=a.$$
• I guess for any $z\in B_a(z_0)$, we have $H(z)\in B_{\theta a}(H(z_0))$，right? Commented Aug 6, 2023 at 14:45