Proving a mathematical statement with projections

Question

Let $a, x \in \mathbb{R}$, $S \subset \mathbb{R}^n$ a convex and closed set and $L = S + \{x\}$. Prove that $L$ is convex and closed and that $proj_L(a) = x + proj_S(a - x)$. Guys, im stuck with this question, can someone give any hint or advice on how to prove it?

Answer 1

I am assuming that you are using the Euclidean metric here.

Note that the map $T_x y = x+y$ is affine (linear plus a constant) and so is its inverse $T_{-x}$. Hence $T_x$ maps closed convex sets to closed convex sets. Note that $L = T_x(S)$.

Note that $\operatorname{proj}_S$ is defined by the relationship $\| \operatorname{proj}_S a -a \| \le \| s -a \|$ for all $s \in S$.

Then $\| \operatorname{proj}_S (a-x) -(a-x) \| \le \| s -(a-x) \|$ for all $s \in S$.

Rearranging gives: $\| \operatorname{proj}_S (a-x)+x -a \| \le \| s+x -a \|$ for all $s \in S$ or $\| \operatorname{proj}_S (a-x)+x -a \| \le \| l -a \|$ for all $l \in L$.

Hence we have the desired result.