We have the system \begin{align*}&x_1=\left (5+x_1^2+x_2^2\right )^{-1} \\ &x_2=\left (x_1+x_2\right )^{\frac{1}{4}}\end{align*} and the set $G=\{\vec{x}\in \mathbb{R}^2: \|\vec{x}-\vec{c}\|_{\infty}\leq 0.2\}$ where $\vec{c}=(0.2,1)^T$.
I want to show with the Banach fixed-point theorem that the system has a solution in $G$.
I have done the following:
Let \begin{equation*}\Phi (x_1, x_2)=\begin{pmatrix}\left (5+x_1^2+x_2^2\right )^{-1} \\ \left (x_1+x_2\right )^{\frac{1}{4}}\end{pmatrix}\end{equation*} The jacobi matrix is \begin{equation*}\nabla \Phi =\begin{pmatrix}-\frac{2x_1}{(x_1^2+x_2^2+5)^2} & -\frac{2x_2}{(x_1^2+x_2^2+5)^2} \\ \frac{1}{4(x_1+x_2)^{\frac{3}{4}}} & \frac{1}{4(x_1+x_2)^{\frac{3}{4}}} \end{pmatrix}\end{equation*}
Then we get \begin{equation*}\|\nabla \Phi \|_{\infty}=\max \left \{\frac{2x_1+2x_2}{(x_1^2+x_2^2+5)^2}, \frac{1}{2(x_1+x_2)^{\frac{3}{4}}}\right \}\end{equation*}
We want to get an upper bound for that, don't we?
How can we do that? Could you give me a hint?
First note that $$G = \{(x_1,x_2)\mid \|(x_1,x_2)-(0.2,1)\|_{\infty}\leq 0.2\} =\{(x_1,x_2)\mid |x_1-0.2|\leq 0.2 \text{ and } |x_2-1|\leq 0.2\} \\= [0,0.4]\times [0.8,1.2].$$ Now, let $$\phi_1(x_1,x_2)=(5+x_1^2+x_2^2)^{-1}\quad\text{ and }\quad \phi_2(x_1,x_2)=(x_1+x_2)^{1/4},$$ so that $\Phi(x_1,x_2)=(\phi_1(x_1,x_2),\phi_2(x_1,x_2))$.
For $(x_1,x_2)\in G$ it holds $$0\leq \phi_1(0.2,1.2)\leq \phi_1(x_1,x_2)\leq \phi_1(0,0.8)\approx 0.1773\leq 0.4$$ and $$0.8 \leq 0.9457\approx \phi_2(0,0.8)\leq \phi_2(x_1,x_2)\leq \phi_2(0.2,1.2)\approx 1.0878 \leq 1.2$$ It follows that $\Phi(G)\subset G$.
Now, in order to apply the Banach fixed point theorem, you want to show that $\Phi$ is a strict contraction on $(G,\|\cdot\|_{\infty})$, i.e. there exists $\alpha <1$ such that $$\|\Phi(x)-\Phi(y)\|_{\infty}\leq \alpha \|x-y\|_{\infty},\qquad \forall x=(x_1,x_2),y=(y_1,y_2)\in G.\tag{1}$$ Let $x,y\in G$. As $\Phi$ is differentiable on $G$, by MVT, there exists $s,t\in [0,1]$ such that $u=s x+(1-s)y$ and $v=t x+(1-t)y$ satisfy $$(x-y)\cdot\nabla \phi_1(u) = \phi_1(x)-\phi_1(y)\quad\text{and}\quad (x-y)\cdot\nabla \phi_2(v) = \phi_2(x)-\phi_2(y).$$ Hence, if we can show that there exists $\alpha<1$ independent of $x,y$ such that
$$|(x-y)\cdot\nabla \phi_1(u)|\leq \alpha \|x-y\|_{\infty}\quad\text{and}\quad |(x-y)\cdot\nabla \phi_2(v)|\leq \alpha \|x-y\|_{\infty},$$ then, $\alpha$ satisfies $(1)$ and we are done. Now, note that $$\max\{|(x-y)\cdot\nabla \phi_1(u)|,|(x-y)\cdot\nabla \phi_2(v)|\}\leq \max\{\|\nabla\Phi(u)(x-y)\|_{\infty},\|\nabla\Phi(v)(x-y)\|_{\infty}\}\\ \leq \max\{\|\nabla \Phi(u)\|_{\infty,\infty},\|\nabla \Phi(v)\|_{\infty,\infty}\} \|x-y\|_{\infty}$$ where $$\|\nabla \Phi(z)\|_{\infty,\infty}=\max_{\|w\|_{\infty}\leq 1}\|\nabla\Phi(z)w\|_{\infty}.$$ Furthermore, as $u,v\in G$, combining the above arguments, we deduce that we can set $$\alpha = \max_{z\in G}\|\nabla \Phi(z)\|_{\infty,\infty}.$$ Now, for $z\in G$, we have, $$\|\nabla \Phi(z)\|_{\infty,\infty}=\max\{\|\nabla \phi_1(z)\|_{1},\|\nabla \phi_1(z)\|_{1}\},$$ and you have computed that $$\|\nabla \phi_1(z)\|_{1}=\frac{2z_1+2z_2}{(z_1^2+z_2^2+5)^2}, \qquad\text{and}\qquad\|\nabla \phi_2(z)\|_{1}=\frac{1}{2(z_1+z_2)^{3/4}}.$$ As $z\in G=[0,0.4]\times [0.8,1.2]$, we find that $$\frac{2z_1+2z_2}{(z_1^2+z_2^2+5)^2}\leq\frac{2(0.4+1.2)}{(0^2+(0.8)^2+5)^2}\approx 0.1006 \leq 0.2$$ $$\frac{1}{2(z_1+z_2)^{3/4}}\leq \frac{1}{2(0+0.8)^{3/4}}\approx 0.5911 \leq 0.6$$ Therefore, we conclude that with $\alpha=0.6$, $$\|\Phi(x)-\Phi(y)\|_{\infty}\leq 0.6 \|x-y\|_{\infty}$$ which implies that $\Phi\colon G\to G$ is a strict contraction with respect to $\|\cdot\|_{\infty}$ and thus, as $(G,\|\cdot\|_{\infty})$ is a complete metric space, the Banach fixed point theorem implies that $\Phi$ has a unique fixed point $p\in G$. Furthermore, for every $x\in G$, the iterative sequence $\Phi^k(x)$ converges towards $p$. A simple numerical experiments shows that $$p = (0.163190947349524, 1.049361520947913)$$