Differentiability of a convex function Let $f,g\colon \mathbb{R}\rightarrow \mathbb{R}$ be convex functions such that $f\ge g$ and $f(0)=g(0)$. Show that if $f$ is differentiable in 0, then $g$ is too and 
$$ f'(0)=g'(0)$$
I have no idea what should I do with this exercise. Any help is appreciated.
 A: Since $f$, $g$ are convex, their "slope functions" (*) $F_a$, $G_a$ at any fixed point $a$ is non-decreasing—we'll use $a=0$—and therefore have both a left limit and a right limit at any point. You are basically asked to prove that $\lim_{0^-}G_0=\lim_{0^+}G_0=\lim_{0}F_0(=f^{\prime}(0))$. 
To see why, observe that


*

*for $x > 0$
$$
G_0(x) \stackrel{\rm def}{=} \frac{g(x)-g(0)}{x-0} = \frac{g(x)}{x} \leq \frac{f(x)}{x} = \frac{f(x)-f(0)}{x-0} \stackrel{\rm def}{=} F_0(x) \xrightarrow[x\to0^+]{} f^{\prime}(0)
$$
so $\lim_{0^+}G_0 \leq f^{\prime}(0)$.

*for $x < 0$, the inequality is reversed because of the sign
$$
G_0(x) = \frac{g(x)}{x} \geq \frac{f(x)}{x} = F_0(x) \xrightarrow[x\to0^-]{} f^{\prime}(0)
$$
so $\lim_{0^-}G_0 \geq f^{\prime}(0)$.
By monotonicity, one also has $\lim_{0^-}G_0\leq \lim_{0^+}G_0$, and thus
$$ f^{\prime}(0) \leq \lim_{0^-}G_0\leq \lim_{0^+}G_0 \leq f^{\prime}(0)$$
and $f^{\prime}(0) = \lim_{0^-}G_0 = \lim_{0^+}G_0$.
$\square$

(*) Slope function of $f$ at $a\in\mathbb{R}$:
$$F_a\colon x\in\mathbb{R}\setminus\{a\}\mapsto\frac{f(x)-f(a)}{x-a}.$$
