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 $|x-p| = \inf_{y \in E} |x – y|$?
Convex Analysis – Unique Point Minimizing Distance in Closed Convex Set
convex-analysisreal-analysis
Related Solutions
Thanks to those that have already responded, you were very helpful. Here I will give the solution I have come up with, with more exposition than is provided in some of the other responses.
First we need the following lemma:
Lemma: $\lim_{\|x\| \to \infty} f(x) = \infty$. Some authors refer to this as $f$ being coercive.
Proof: Let $x_0 \in \mathbb{R}^n$ and let $v$ be a subgradient of $g$ at $x_0$, i.e. $x_0 \in \partial g(x_0)$. By equivalence of norms in finite-dimensional vector spaces, there exists a constant $c > 0$ such that $\|x\|_2 \leq c \|x\|$ for all $x \in \mathbb{R}^n$. By Cauchy-Schwarz and the trinagle inequality, for $\|x\| > 0$ we have
$$ \begin{align*} \frac{| v^T(x - x_0) |}{\frac{m}{2}\|x\|^2} &\leq \frac{\|v\|_2 \|x - x_0\|_2}{\frac{m}{2}\|x\|^2} \\ &\leq \frac{\|v\|_2 \|x\|_2 + \|v\|_2 \|x_0\|_2}{\frac{m}{2}\|x\|^2} \\ &\leq \frac{c\|v\|_2 \|x\| + \|v\|_2 \|x_0\|_2}{\frac{m}{2}\|x\|^2} \\ &= \frac{2c\|v\|_2 \|x\|}{m} \frac{1}{\|x\|} + \frac{2\|v\|_2 \|x_0\|_2}{m} \frac{1}{\|x\|^2} \end{align*} $$
The far right hand side of this inequality $\to 0$ as $\|x\| \to \infty$, which implies that $v^T(x - x_0) + \frac{m}{2} \|x\|^2 \to \infty$ as $\|x\| \to \infty$. Now we use the definition of subgradient:
$$ \begin{align*} v^T(x - x_0) &\leq g(x) - g(x_0) \\ v^T(x - x_0) + \frac{m}{2}\|x\|^2 &\leq g(x) + \frac{m}{2}\|x\|^2 - g(x_0) \\ v^T(x - x_0) + \frac{m}{2}\|x\|^2 + g(x_0) &\leq f(x) \end{align*} $$
The left hand side of this $\to \infty$ as $\|x\| \to \infty$, so we conclude that $f(x) \to \infty$ as $\|x\| \to \infty$. $\square$
On to the main result. First, assume that $A$ is unbounded. If it is bounded, then it is compact, and the result follows immediately. There are 2 mutually exclusive possibilities:
Case 1: $f$ has a minimizer on $A$, in which case it is unique (see this thread).
Case 2: $f$ does not have a minimizer on $A$.
Assume we have case 2. Let $f^\star := \inf_{x \in A} f(x)$. $f^\star < \infty$ by assumption. Let $(x_k)$ be a sequence in $A$ such that $f(x_k) \to f^\star$. We now have two mutually exclusive subcases:
Subcase 2.1: $\sup_k \|x_k\| = d < \infty$. Define $B_d := \{ x \in \mathbb{R}^n \ : \ \|x\| \leq d\}$. Then for all $k$, $x_k \in \{ A \cap B_d \}$ which is closed and bounded and hence compact. Therefore $(x_k)$ converges in $A$, i.e. $x_k \to x^\star$ for some $x^\star \in A$. Continuity of $f$ then implies $f(x^\star) = f^\star$, which is a contradiction.
Subcase 2.2: $\sup_k \|x_k\| = \infty$. This implies $\|x_k\| \to \infty$, and by the Lemma this implies $f(x_k) \to \infty$ which implies $f^\star = \infty$ which contradicts $f^\star < \infty$.
Thus we conclude that Case 2 cannot occur, and therefore Case 1 must occur.
EDIT: After writing all of this out, it is clear that $f$ strongly convex is a stronger assumption than we require. $f$ strictly convex and coercive is sufficient for $f$ to have a unique global minimum on the convex set $A$.
How about $\mathbb R^2$ with $\|(x,y)\|=|x|+|y|,\ A=\{(1-t)(1,0)+t(0,1): 0\le t\le 1\}$ and $\vec x=(0,0).$
Best Answer
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 $y_1 = y_2$.