I'm trying to prove that the map $\mathcal F$ below is well-defined and is a contraction. I'm able to complete the first part, but stuck at the second one.
Could you please leave me some hints to finish the second one? Thank you so much!
Let $\rho >0$ and $f,g \in \mathbb R^n$ in which $\mathbb R^n$ is endowed with the inner product $$\forall x,y \in \mathbb R^n:\langle x, y \rangle = \sum_{i=1}^n x_i y_i$$
Assume that $\textbf{M}$ is a squared matrix satisfying there is $\alpha >0$ such that $$\forall x \in \mathbb R^n: \langle x, \textbf{M}x \rangle \ge \alpha \|x\|^2$$
We denote $x \ge y$ if $x_i \ge y_i$ for all $i = \overline{1,n}$. We denote by $\mathcal F (x)$ the solution $y \ge f$ (if exists) of $$\forall v \ge f: \langle y-x + \rho(\textbf{M}x -g), v-y \rangle \ge 0$$
Prove that $\mathcal F$ is well defined and is a contraction.
My attempt:
Let $S := \{v \in \mathbb R^n \mid v \ge f\}$. It follows that $S$ is closed and convex.
The condition $$\forall v \ge f: \langle y-x + \rho(\textbf{M}x -g), v-y \rangle \ge 0$$ is then equivalent to $$\forall v \in S: \left \langle \big(x - \rho(\textbf{M}x -g)\big) - y, v-y \right\rangle \le 0$$
It follows that $y$ is the orthogonal projection of $x - \rho(\textbf{M}x -g)$ on the set $S$. Hence $\mathcal F:\mathbb R^n \to \mathbb R^n$ is well defined.
My professor replied that there is an error in the exercise. In fact, it is find $ρ>0$ such that $\mathcal F$ is a contraction.
I post it as an answer to peacefully close this question.