I am reading an article and need to prove the following proposition, which is used in the paper:
If a differentiable function $f \colon \mathbf{R} \to \mathbf{R}$ is convex, the derivative $f'$ is monotonically increasing and continuous.
I could prove the monotonicity like this.
It holds from the definition of convexity,
$$ f(rx_1+(1-r)x_3) \le rf(x_1)+(1-r)f(x_3) $$
for $x_1, x_3 \in \mathbf{R}$ and $r \in (0, 1)$ (and we assume $x_1 < x_3$ here).
Let $x_2 = rx_1+(1-r)x_3$, then
$$ \begin{align} \frac{f(x_2)-f(x_1)}{x_2-x_1} &\le \frac{(1-r))(-f(x_1)+f(x_3))}{(1-r)(-x_1+x_3)} \\ &=\frac{f(x_3)-f(x_1)}{x_3-x_1} \end{align} $$
As $x_2 \to x_1+0$, we obtain
$$ f'(x_1) \le \frac{f(x_3)-f(x_1)}{x_3-x_1} $$
because the right-hand derivative at $x_1$ is identical to the derivative due to differentiabllity.
We also get
$$ f'(x_3) \ge \frac{f(x_3)-f(x_1)}{x_3-x_1}. $$
These two inequalities result in $f'(x_1) \le f'(x_3)$, which proves the monotonicity.
However, I cannot prove the continuity of $f'$ by myself. Can anyone prove this? Correcting my idea above is also appreciated if I have some misunderstanding there.
Note that if $a<x$ then $\frac{f(x)-f(a)}{x-a}\ge f'(a)$ because the left hand side is equal to $f'(c)$ for some $a<c<x$, and since $f'$ is monotonic.
Differentiablity of $f$ implies that $\lim_{x\to a^{+}}\frac{f(x)-f(a)}{x-a}= f'(a)$
Let $\epsilon>0$ be fixed. By an elementary property of limits, there exists $\eta>0$ such that
$$y\in (a,a+\eta)\Rightarrow f'(a)\ge \frac{f(y)-f(a)}{y-a}-\epsilon.$$ By choosing $y=a+\frac \eta2$ and using the mean value theorem, there exists $c\in (a,a+\frac \eta2)$ such that $f'(a)\geq f'(c)-\epsilon$.
Since $f'$ is non-decreasing, we obtain for any $x\in (a,c)$
$f'(a)\ge f'(c)-\epsilon \ge f'(x)-\epsilon\Rightarrow f'(x)-f'(a)<\epsilon $.
A similar argument applies if $x<a$.