# The Existence of Support Hyperplane

Proposition 1. Let $K$ be a compact convex set in $\mathbb{R}^n$. For any $P\in \partial K$, there exists a support hyperplane of $K$ containing $P$.

Proof . For every $P\in \partial K$, take $f:~S^{n-1}\rightarrow \mathbb{R}$,
$f(u)=h(K,u)-P\cdot u, \quad u\in S^{n-1},$
where $h(K,\cdot)$ is support function of K. Thus, $f(u)\geqslant0$ for any $u\in S^{n-1}$. Since $h(K,\cdot)$ is continuous on $S^{n-1}$ which is a compact subset of $\mathbb{R}^{n-1}$, $f(\cdot)$ is continuous and there exists $u_1\in S^{n-1}$ such that

$f(u_1)=\min_{u\in S^{n-1}}f(u).$
Let $\rho=f(u_1)$. Thus, $\rho\geqslant0$.

Suppose $\rho>0$. Denote $B_P(\frac{1}{2}\rho)$ for the closed ball with center $P$ and radius $\frac{1}{2}\rho$, then, obviously, the support function of $B_P(\frac{1}{2}\rho)$ is
$h(B_P(\frac{1}{2}\rho), u)=P\cdot u+\frac{1}{2}\rho, \quad u\in S^{n-1}.$
Thus we have
$h(k,u)-h(B_P(\frac{1}{2}\rho), u)=h(K,u)-P\cdot u-\frac{1}{2}\rho\geqslant\frac{1}{2}\rho>0, \quad u\in S^{n-1}.$
i.e.,
$h(k,u)\geqslant h(B_P(\frac{1}{2}\rho), u), \quad\text{for any }u\in S_{n-1},$
which implies
$B_P(\frac{1}{2}\rho)\subset K.$
This contradicts the condition $P\in \partial K$. Therefore, we must have $\rho=0$, i.e.,  $f(u_1)=h(K,u_1)-P\cdot u_1=0$, which means $P\in H(K,u_1)$  and the desired results obtained.

### One Comment

1. Good, may be there need a figure and some introduction words about what you will show in this post, and may be add the definition of support plane will make the proof more readable.

At last, I want to emphasis that you should not press Enter key except that you want to have a new paragraph, or you will make the content have a new line, which in fact is not a new paragraph. Please compare with the content at $f(\cdot)$…