Let $f:\mathbb R^n\to\mathbb R$ be a concave continuous function and $g:\mathbb R^n\to\mathbb R$ be a convex continuous function such that $f\leq g$.
Then there exists an affine continuous function $h:\mathbb R^n\to\mathbb R$ such that $f\leq h\leq g$.
The proof for this just says: Consider the sets $\mathcal M_f:=\{(x,r)|x\in\mathbb R^n, r<f(x)\}\subset\mathbb R^{n+1}$ and $\mathcal M_g:=\{(x,r)|x\in\mathbb R^n, r>g(x)\}\subset\mathbb R^{n+1}$, then $h$ is the hyperplane separating these sets.
But how do I know that I can separate these sets? Isn't one of them concave and one of them convex? I only know separation theorems for convex sets.
Both of these sets are convex, because we can write the first as $\{-r>-f(x)\}$, and $-f(x)$ is convex.