Let $f : \mathcal{I} \subset \mathbb{R}^n \rightarrow \mathcal{I}$ is continuous, and $\mathcal{I}$ is a closed $\ell_{\infty}$ norm ball. From Brouwer fixed point theorem (p. 240, Real Mathematical Analysis by C. Pugh), a fixed point of $f$ exists in $\mathcal{I}$. To give a sufficient condition for $f$ to be a contraction on $\mathcal{I}$, can I then simply require $\big\Vert \frac{\partial f}{\partial x}\big\Vert_{\infty} < 1$? It seems from this post Mean value theorem for vector valued multivariable function such an approach is valid for $\ell_2$ norm, but I'm not sure if that's the case for $\ell_{\infty}$ norm.
This condition is from a proof in the appendix of a paper I'm trying to understand, but not much detail is provided. I'd appreciate any suggestions/counterexamples.
Best Answer
If I understand your question correctly:
Let $f: \mathcal{I} \to \mathbb{R}^n$ be continuously differentiable. Then the Frechet derivative at $x$ is the same, whether we use the $\ell^\infty$ or $\ell^2$. The reason is the following. Write $x=(x_1,\dots,x_n)$. Then $|x_k|=\sqrt{|x_k|^2}\leq \sqrt{|x_1|^2+\dots+|x_n|^2}=\|x\|_2$ for all $k$. Hence, $\|x\|_\infty \leq \|x\|_2$.
Similarly, $\|x\|_2=\sqrt{|x_1|^2+\dots+|x_n|^2}\leq \sqrt{n\|x\|_\infty^2}=\sqrt{n}\|x\|_\infty.$
Denote the Frechet derivative in the $\ell_2$ norm of $f$ at $x$ by $Df(x)$. Then $Df(x)$ is also the Frechet derivative with respect to the $\ell^\infty$ norm since $$ \frac{\|f(x+h)-f(x)-Df(x)h\|_\infty}{\|h\|_\infty}\leq \sqrt{n}\frac{\|f(x+h)-f(x)-Df(x)h\|_2}{\|h\|_2} $$ As $\|h\|_\infty \to 0$, we also have $\|h\|_2 \to 0$. The right hand side goes to zero. Therefore, the left hand side does as well by squeeze theorem so $Df(x)$ is the same in both norms.
In order to be a contraction, we use the fact that $$ |f(x)-f(y)|_\infty\leq \sup_{x\in \mathcal{I}}\|Df(x)\|\|x-y\|_\infty $$ If for all $x$, we have $\|Df(x)\|\leq \alpha <1$, then $f$ will be a contraction.