uniform bound on derivative of convex functions Let $f$ and $g$ be smooth convex functions, let $h_{\alpha}$ be family of smooth convex functions satisfying the inequality $$f(u)\leq h_{\alpha}(u) \leq g(u) \text{ for all }\alpha \in \textit{F},u \in \mathbb{R}$$ then can we show that $h'_{\alpha}$ are uniformly bounded on compact sets.  
 A: Yes, that is true. The essential reason is that the graph of a (differentiable) convex function lies above all of its tangents. For $a < u < b$ we have (writing $h $ instead of $ h_\alpha$ for brevity):
$$
 h'(u) \le \frac{h(b)-h(u)}{b-u} \le h'(b) \le h(b+1)-h(b) \le g(b+1)-f(b)
$$
and
$$
 h'(u) \ge \frac{h(u)-h(a)}{u-a} \ge h'(a) \ge h(a)-h(a-1) \ge f(a) - g(a-1) \, .
$$
For a given compact set $K \subset \Bbb R$ we can choose $a < \inf K$ and $b > \sup K$, then
$$
 f(a)-g(a-1) \le h'(u) \le g(b+1)-f(b)
$$
for all $u \in K$, so that $h'(u)$ is uniformly bounded on $K$ for all convex functions satisfying $f \le h \le g$.
Note that the convexity and smoothness of $f$ and $g$ is not needed for this conclusion.
Even if $h$ is not differentiable: a convex function has one-sided derivatives at every point, and similar as above one can show that these are uniformly bounded on $K$:
$$
f(a) - g(a-1) \le h_-'(u) \le h_+'(u) \le g(b+1) - f(b)
$$
where $h_-'$, $h_+'$ denotes the left (resp. right) derivative $h$.
