The projection of a point onto a convex set is unique with respect to any norm

Question

Given a convex set $C$, the projection of a point $z$ onto $C$ is a point $x$ in $C$ that minimizes $\|z- x\|$. Say the minimum is achieved at $x^*\in C$. My textbook shows such $x^*$ is unique under the euclidean norm, as shown below. My guess is the uniqueness should hold regardless what norm is chosen, but I have trouble proving it. Thanks!

Answer 1

Your guess is wrong, when dealing with a non-euclidean norm, a (closed) convex set does not necessarily have a unique element with minimal norm as it is shown in the following example:

Let endowed $\mathbb{R}^2$ with $\|\cdot\|_{\infty}$ defined by: $$\|(x,y)\|_{\infty}=\max(|x|,|y|).$$ Furthermore, let $D:=\{(1,t);t\in\mathbb{R}\}$, it is a (closed) convex subset of $\mathbb{R}^2$. For all $\mathbf{x}\in D$, one has: $$\|\mathbf{x}\|_{\infty}\geqslant 1.$$ However, for all $t\in [-1,1]$, $(1,t)\in D$ and $\|(1,t)\|_{\infty}=1$, so that there is an infinity number of elements with minimal norm in $D$.

Answer 2

Counterexample: Take $\mathbb R^2$ with $\|\cdot\|_1$. Be $C=\{x\in\mathbb R^2:\|x\|_1\le1\}$. Be $P=(2,2)$. Then for any point $x_\lambda=(\lambda,1-\lambda)$ with $\lambda\in[0,1]$ you have $\|P-x_\lambda\|_1 = |2-\lambda|+|2-1+\lambda| = 3$, which is also the minimum distance.

Note that $\|\cdot\|_1$ is not a differentiable function, therefore the proof doesn't apply.