Supporting hyperplane of a convex set Let $\Omega$ be a bounded convex set in $\mathbb{R}^n$, and let $\partial \Omega$ denote its boundary.
Fix a point $p$ in $\Omega$, and let $c$ denote the point on $\partial \Omega$ that is closest to $p$.
Then, intuitively it seems that a hyperplane which goes through $c$ with the noraml vector parallel to the vector from $p$ to $c$ is a supporting hyperplane, but how can one prove or disprove this?
 A: I am assuming that $p \neq c$, that is, $p \in \Omega^\circ$, otherwise any
vector passes through $c-p$.
The essential idea is that a supporting hyperplace to $\Omega$ at $c$ is also a supporting hyperplane to $B(p,\|c-p\|)$ at $c$, and the direction of this
hyperplane is unique.
We use the following technical results:
If $x \in \Omega^\circ$ and $y \in \overline{\Omega}$, then $(1-t)x+ty \in \Omega^\circ$ for all $t \in [0,1)$ (see Theorem 6.1 in Rockafellar's
"Convex Analysis", for example).
Note that since $\Omega$ is convex, we have $\partial \Omega = \overline{\Omega} \setminus \Omega^\circ$.
If $y \in \overline{\Omega}^c$ and $y_t = (1-t)p+t y$, then there is
some $t^* \in (0,1)$ such that $y_{t^*} \in \partial \Omega$.
To see this,
let $I = \{ t \in [0,1] | y_t \in \Omega^\circ \}$. Since $\Omega^\circ$ is open and convex, we have $I=[0,t^*)$ for some $t^*$, and we must have
$t^* <1$ since $\overline{\Omega}^c$ is open. We see that $y_{t^*} \notin \Omega^\circ$, and $y_{t^*} \in \overline{\Omega}$, hence $y_{t^*} \in \partial \Omega$.
Let $r = \|p-c\|$. By choice of $c$, if
$y \in \partial \Omega$, then $\|p-y\| \ge r$.
If $y \in \overline{\Omega}^c$, then (from above) $y_{t^*} \in \partial \Omega$, and so $\|p-y_{t^*}\| = t^* \|p-y\| \ge r$, and so $\|p-y\| \ge r$.
Since $(\Omega^\circ)^c = \partial \Omega \cup \overline{\Omega}^c$, we see 
 that if $y \in (\Omega^\circ)^c$, then $\|p-y\| \ge r$. Hence
we have $B(p,r) \subset \Omega^\circ$.
Now suppose $H = \{x | \langle h, x \rangle = \alpha\}$ is a supporting hyperplane to $\Omega^\circ$ passing through $c$. 
We may assume $\langle h, p \rangle < \alpha$ and
hence $\langle h, x \rangle < \alpha$ for all $x \in \Omega^\circ$.
In particular, $\Omega^\circ$ does not intersect $H$,
and so $H$ is contained in $(\Omega^\circ)^c$. Hence $\|p-x\| \ge r$ for
all $x \in H$. Hence $c$ is the closest point to $p$ in $H$.
That is,
$c$ solves
$\min \{ {1 \over 2} \|x-p\|^2 |  \langle h, x \rangle = \alpha \}$, and
Lagrange gives
$c-p + \lambda h = 0$, so we see that $c-p$ and $h$ are collinear.
As an aside, this shows that this is the only separating hyperplane that passes through $c$.
A: Here's another way to see it. Assume that $c$ and $p$ are distinct, otherwise the vector from $p$ to $c$ doesn't determine a particular hyperplane. You'd like to show that, for $x$ in $\Omega$,
$$(x - c)\cdot (c-p) \le 0.$$
Suppose not, and consider an $x$ in $\Omega$ for which $(x - c)\cdot (c-p) > 0.$ Take a positive $\varepsilon$ and consider the point $c_\varepsilon\equiv c + \varepsilon(c-x)$. 
$$\begin{align}\|p-c_\varepsilon\|^2 &= \|p-c\|^2 + \varepsilon^2\|c-x\|^2 - 2\varepsilon (x-c)\cdot(c-p)\\
&< \|p-c\|^2,
\end{align}$$
for $\varepsilon$ sufficiently small. Fix this small $\varepsilon$. Then $c_\varepsilon$ is strictly closer to $c$ than $p$. If a neighbourhood of $c_\varepsilon$ were in $\Omega$, then $c$ would be an interior point of $\Omega$. So we may take a point $d$ in a neighbourhood of $c_\varepsilon$ but outside $\Omega$ such that $\|d-p\|<\|c-p\|$.  By a trivial topological argument, there is a boundary point of $\Omega$ in the line segment between $d$ and $p$. Such a boundary point is closer to $p$ than $d$ and hence also closer than $c$, contradicting the assumption that $c$ is the closest point on the boundary to $p$.
Interestingly, a small modification to the above argument shows that the same result holds for $p\notin\Omega$. 
