If $y\in C$ then $||x-y||>||y-z||$.

Question

I'm trying to prove the Separating Hyperplane Theorem.

Let $C\subset \mathbb{R}^n$ be a closed and convex set, and $x\not \in C$. Then there exists $d\in \mathbb{R}^n$ and $\delta\in \mathbb{R}$ such that $$\langle d,x\rangle <\delta<\langle d,y\rangle, \quad \forall y\in C.$$

I proved that there is a unique point $z\in C$ such that $||x-y||> ||x-z||$ for all $y\in C$ and $y\neq z$. Let $d=z-x$ and $\delta = \frac{1}{2}\langle z-x, z+x \rangle$. I will prove that the hyperplane $\{t\in \mathbb{R}^n: \langle d,t\rangle =\delta\}$ separates $x$ and $C$. The inequality $\langle d, x\rangle<\delta$ is easy to prove. However, I was stuck with the inequality $\langle d, y\rangle >\delta$, where $y$ is an arbitrary point of $C$.The inequality $\langle d, y\rangle>\delta$ is equiavalent to $||x-y||>||y-z||$, and I don't know how to prove this.

Thank you very much.

Answer 1

Sometimes analytics may be guided by a geometric imagination. So, it seems the following. Since the set $C$ is convex, the segment $[y,z]$ is contained in $C$. If $\|x-y\|\le\|y-z\|$ then $[y,z]$ a longest side of a triangle $[x,y,z]$. Then a base $t$ of a height $[x,t]$ of the triangle $[x,y,z]$ is an inner point of the segment $[y,z]$. But then the point $t$ belongs to the set $C$ and $\|x-t\|<\|x-z\|$, a contradiction.