closed convex set, unique point that minimizes distance Let $E \subset \mathbb{R}^k$ be a closed convex set. How would I go about showing that for each $x \in \mathbb{R}^k$ there is a unique $p \in E$ such that $|x-p| = \inf_{y \in E} |x - y|$?
 A: It suffices to show that in any closed convex set there is a unique closest point to $0$. if $0 \in E$ this is obvious so assume $0 \notin E$. If $d = \inf_{y \in E} |y|$, then $E \cap \{y \text{ }|\text{ }|y| \le d + 1\}$ is closed and bounded, hence compact. The norm on $\mathbb{R}^n$ is continuous, and so the norm attains a minimum is attained at some point $y$, and $|y| = d$, since $d \ge \min\{|y|, d+1\}$.
Let $y_1$, $y_2$ be two closest points to $0$; then the fact that $y_1$ and $y_2$ both have minimal norm and Cauchy-Schwarz imply the string of inequalities $$\left|{{y_1 + y_2}\over2}\right|^2 \ge |y_1|^2 = (|y_1|^2 + 2|y_1| \cdot |y_2| + |y_2|^2)/4 \ge (|y_1|^2 + 2y_1 \cdot y_2 + |y_2|^2)/4 = \left|{{y_1+y_2}\over2}\right|^2.$$Thus all inequalities above are equalities, and equality holds in Cauchy-Schwarz. Thus $y_1$ is a scalar multiple of $y_2$. We have$${{y_1 + y_2}\over2} \neq 0,$$ so $y_1 \neq -y_2$, whence $y_1 = y_2$.
A: Suppose for some $x\in\mathbb{R}^k$, there are two distinct $p_1$ and $p_2$ s.t. $|x-p_1|=|x-p_2|=\min_{y\in E}|x-y|$.
But the function $f(t)=|x-(tp_1+(1-t)p_2)|$ is a strictly convex hyperbola (except if $x\in E$), and $f(0)=f(1)$. By Rolle's theorem, we know that $\exists c\in (0,1)$ s.t. $f'(c)=0$ and that must be the unique minimum point of the hyperbola and hence the unique minimum for the domain $[0,1]$. Contradiction.
