When I try to prove this problem, in $\mathbb{R}^n$, we know that a set $K$ which is closed, convex, then every point $x_0\in \partial K$ admits a supporting hyperplane, i.e there exists a vector $\alpha\in \mathbb{R}^n$ such that $$ \langle y, \alpha\rangle \leq \langle x,\alpha\rangle $$ for every $y\in K$. Where $\langle.,.\rangle$ denote the Euclid scalar product on $\mathbb{R^n}$. This result can be proved by using Hahn--Banach theorem in $\mathbb{R}^n$, ...etc. But such a vector like $\alpha$ for $x_0$ is not unique, incase $K$ is a triangle in $\mathbb{R}^2$ there are too many vector like that at every vertex. But if $K$ is strictly convex, i.e if $a,b\in K$ then for every $t\in (0,1)$ $$tx + (1-t)y \in \text{the interior of}\; K$$ In this case, I think the normal vector like that is unique at each point $x_0\in \partial K$, but I cannot prove this. Is this true?
I observe that if the supporting hyperplane $H =\{x: \langle \alpha, x\rangle = \langle \alpha, x_0 \rangle\}$ much have $$ H\cap K = \{x_0\}$$ incase $K$ is strictly convex.
