# Prove that if $x$ is a fixed point for $g$ then its projection to $S$ minimizes $f$ over $S$.

Discussion in 'Mathematics' started by ex.nihil, Oct 8, 2018.

1. ### ex.nihilGuest

Let $f:\mathbb{R}^n \longrightarrow \mathbb{R}$ be a convex and differentiable function and let $S$ be a nonempty, closed, and convex subset of $\mathbb{R}^n$.

The function $g:\mathbb{R}^n \longrightarrow \mathbb{R}^n$ is defined by: $$g(x) := -\big[ \nabla f\big(P_S(x)\big) - P_S(x)\big], \;\; \forall x \in \mathbb{R^n} \tag{1}$$ Where $P_S(x)$ is the projection of $x$ onto the set $S$.

Prove that if $x$ is a fixed point for $g$ then its projection to $S$ minimizes $f$ over $S$.

ATTEMPT

Since $x$ is a fixed point for $g$ then we have $g(x)=x$.

Therefore rearrange $(1)$ to obtain: $$\nabla f\big(P_S(x)\big) = P_S(x) - x \tag{2}$$ Case 1: $(x \in S)$
We have that $P_S(x) = x$ and thus $\nabla f(x)=x-x=0$.

Since $f$ is convex then $x=\text{argmin}_{x \in S} f(x)$, as desired.

Case 2: $(x \notin S)$
It is possible to write $(2)$ as: $$\nabla f\big(P_S(x)\big) = P_S(x) - x = d_S(x) = \inf_{s\in S} \Vert x-s\Vert \tag{3}$$ Where $d_S$ is the distance relative to $S$.

I am frankly stuck, and unsure what to do. Moreover, it is not very clear to me how to use the closedness and convexity of $S$.

Any help is greatly appreciated.