TITLE: $f : (0,1) \rightarrow \mathbb{R}$ countinous with non-negative right-hand derivative is non-decreasing
QUESTION [5 upvotes]: Let $f : (0,1) \rightarrow \mathbb{R}$ continous, right-hand derivable, such that $f'_+ \geqslant 0$ $\forall t \in (0,1)$. Show that $f$ is non-decreasing.
First I thought of Lagrange's theorem but I don't think that's useful here. Then, for every $x, y$, I tried to cover $[x, y]$ with a finite number of open subsets $(x_i, x_i+\delta_i)$ such that f was non-decreasing in every one of them, so that I could write $$f(y)-f(x)=f(y)-f(x_n)+f(x_n)-f(x_{n-1})+...-f(x) \geqslant 0$$ but I had no luck there.

REPLY [3 votes]: For differentiable functions this is a direct consequence of the mean-value theorem, which in turn is a consequence of Rolle's theorem.
In our case, where $f$ is only assumed to have a (nonnegative) right derivative at every point in the interval, one can mimic the proofs of these theorems as follows:
For $0 < a < b < 1$ consider the function
$$
 g(x) = f(x) - \frac{f(b)-f(a)}{b-a}(x-a) 
$$
which is right-differentiable with
$$
g_+'(x) = f_+'(x) - \frac{f(b)-f(a)}{b-a} \, .
$$
$g$ attains its maximum on the interval $[a, b]$ at some point $x_0$, and since $g(a) = g(b)$ we can assume that $a \le x_0 < b$. Then $g_+'(x_0) \le 0$, so that
$$
 \frac{f(b)-f(a)}{b-a} = f_+'(x_0)- g_+'(x_0) \ge 0
$$
and therefore $f(b) \ge f(a)$, i.e. $f$ is non-decreasing.
Remark: If $f_+'$ is strictly positive on the interval then the same proof shows that $f$ is strictly increasing.