using fixed point theorem Hi I want to use the fixed point theorem to show that for 
$G: \mathbb{R}^n \rightarrow \mathbb{R}^n$
$G(x)= \epsilon M x + \max(x,y)$, here $y$ is given and $\max(x,y)$ is the vector with component $\max(x_i,y_i)$, $M$ is negative definite $n \times n$ matrix, $G(x)=x$ has a solution. (Note this is for any $\epsilon >0 $).
First, I showed that $G$ is a non expansive map for $\epsilon < \dfrac{1}{\|M\|}$. So the fixed point for contraction map wouldn't work so I changed direction.
I know that $G$ is a continuous map, so if I can find a closed set $B$ (ball) in $\mathbb R^n$ such that $G$ maps from $B$ to $B$ then the Brower's fixed point theorem says that $G$ has a fixed point. I can not figure out $B$. I had difficulty to deal with $|\max(x,y)|$.
Any thought on this would be very much appreciated!
 A: Alright, here's a solution (I actually had fun solving this problem, thanks)!
Without loss of generality, $\varepsilon = 1$. So you start with the original problem 
$$
G(x) = Mx + \max \{ x,y\}
$$
which under the substitution $x \mapsto x+y$ becomes
$$
x+y = M(x+y) + \max \{ x+y,y \} = M(x+y) + y + \max \{x,0\}. 
$$
or in other words, 
$$
x = Mx + My + \max \{x,0\}.
$$
Since $M$ is negative definite, $M$ cannot have $1$ as an eigenvalue, so $I-M$ is invertible ($I$ denotes the identity matrix). A solution to this equation therefore solves
$$
x = (I-M)^{-1} (My + \max \{x,0\}). 
$$
Now define $H(x) = (I-M)^{-1} (My + \max \{x,0\})$. Note that all my changes of variables were fixed-point preserving in both directions, so all we need to do now is find a fixed point of $H$. If we want to apply Banach's fixed point theorem on $H$, for $x_1, x_2 \in \mathbb R^n$, we compute
$$
\begin{aligned}
\| H(x_1) - H(x_2) \| & = \| (I-M)^{-1} (\max \{ x_1, 0\} - \max \{ x_2, 0\} ) \| \\
                      & \le \| (I-M)^{-1} \| \| \max \{ x_1, 0 \} - \max \{ x_2, 0 \} \| \\
                      & \le \| (I-M)^{-1} \| \| x_1 - x_2 \|
\end{aligned}
$$
The last inequality follows because for real numbers $t_1, t_2$, we have
$$
    \max \{ t_1, 0 \} \le \max\{ t_1 - t_2, 0 \} + \max \{ t_2, 0 \}
$$
hence by interchanging roles, you get
$$
\begin{aligned} 
|\max \{ t_1, 0 \} - \max \{ t_2, 0 \}| & \le \max \{ \max \{ t_1 - t_2 , 0 \} , \max \{ t_2 - t_1, 0 \} \} \\
 & \le \max \{ | t_1 - t_2 | , 0 \} \\
 & = |t_1 - t_2|
\end{aligned}
$$
so that if $x_j = (x_j^1 , \dots, x_j^n)$, $j=1,2$, we have
$$
\| \max \{ x_1, 0 \} - \max \{ x_2, 0 \} \| \le \sqrt{ \sum_{i=1}^n |x_1^i - x_2^i|^2 } = \| x_1 - x_2 \|. 
$$
All we need to show now is that $\|(I-M)^{-1}\| < 1$ and we can apply Banach's fixed point theorem. So we want to show that
$$
\begin{aligned}
1 > \|(I-M)^{-1}\| & = \sup_{x \neq 0} \frac{ \|(I - M)^{-1}x \|}{\|x\|} \\
& = \sup_{x \neq 0} \frac{\|x\|}{\|(I-M)x\|} \\
& = \sup_{\|x\| = 1} \frac 1{\|(I-M)x\|} \\
& = \frac 1{\inf_{\|x\| =1} \|(I-M)x\|}, \\
&  \\
&  \\
&  \\
\Longleftrightarrow & \quad \inf_{\|x\| = 1} \|(I-M)x\| > 1.
\end{aligned}
$$
The unit sphere is compact and linear applications/norms are continuous functions, hence this infimum is a minimum say at $\hat x$ with $\|\hat x\| = 1$. Fix an orthogonal basis $\{ \hat x, e_2, \dots, e_n \}$ of $\mathbb R^n$, so that since $\hat x^{\top} M \hat x < 0$, by Pythagoreas's theorem, we have
$$
\|(I-M)\hat x\|^2 = (\hat x^{\top}(I-M)\hat x)^2 + \sum_{i=2}^n (e_i^{\top} M \hat x)^2 > \| \hat x\|^2 = 1.
$$
Hope you liked it!
