Separation theorem: Let $P, Q⊆\mathbb{R}^d$ be disjoint compact convex sets. Then there exist $v∈ \mathbb{R}^d$ and $c_1, c_2∈\mathbb{R}$ with $c_1<c_2$ such that
$x.v≤c_1$ for every $x∈P$
$x.v≥c_2$ for every $x∈Q$
Proof sketch:
There exist $p∈P, q∈Q$ with minimal distance.
Take hyperplanes through $p$ and $q$ perpendicular to the segment $pq.$
$\square$
The above proof is given by my professor.
But My question is how can I show that neither P nor Q can contain a point in the gap?
From the proof we have $v = q - p$, $c_1 = \left< p, v \right>$, $c_2 = \left< q, v \right>$.
Assume for contradiction that there exists $p' \in P$ such that $\left< p', v \right> > c_1$. By convexity, the segment $pp'$ is contained in $P$. Let $x(t) = (1-t)p + tp'$ for $t \in [0, 1]$. Then
$$\frac{\text{d}}{\text{d}t} \| x(t) - q \|^2 = \frac{\text{d}}{\text{d}t} \left< x(t) - q, x(t) - q \right> = 2 \left< x'(t), x(t) - q\right> = 2 \left< p'-p, x(t) - q \right>.$$
When $t = 0$,
$$\left< p'-p, x(t) - q \right> = \left< p' - p, -v \right> = c_1 - \left< p', v \right> < 0.$$
Hence for sufficiently small $t > 0$ we have
$$\| x(t) - q \|^2 < \| x(0) - q \|^2 = \| p - q \|^2,$$
contradicting the fact that $p$ was the closest to $q$ among all points in $P$.
