# Increasing and differentiable implies nonnegative derivative that is not identically zero on any interval

## Statement

### On an open interval

Suppose $f$ is a function on an open interval $I$ that may be infinite in one or both directions (i..e, $I$ is of the form $\! (a,b)$, $(a,\infty)$, $(-\infty,b)$, or $(-\infty,\infty)$). Suppose the derivative of $f$ exists everywhere on $I$. Suppose further that $f$ is an increasing function on $I$, i.e.:

$x_1,x_2 \in I, x_1 < x_2 \implies f(x_1) < f(x_2)$

Then, $\! f'(x) \ge 0$ for all $x \in I$. Further, there is no sub-interval of $I$ such that $\! f'(x) = 0$ for all $x$ in the sub-interval.

### On a general interval

Suppose $f$ is a function on an interval $I$ that may be infinite in one or both directions and may be open or closed at either end. Suppose $f$ is a continuous function on all of $I$ and that the derivative of $f$ exists everywhere on the interior of $I$. Further, suppose $f$ is an increasing function on $I$, i.e.:

$x_1,x_2 \in I, x_1 < x_2 \implies f(x_1) < f(x_2)$

Then, $\! f'(x) \ge 0$ for all $x$ in the interior of $I$. Further, there is no sub-interval of $I$ such that $\! f'(x) = 0$ for all $x$ in the sub-interval.

## Proof

If $\! f$ is increasing on $I$, then every point in the interior of $I$ is a point of local maximum from the left and local minimum from the right. Thus, by Facts (1) and (2), both the left hand derivative and the right hand derivative of $f$, if they exist, are nonnegative at any point in the interior of $I$. In particular, if the derivative itself exists at a point in the interior of $I$, then it must be nonnegative at that point.

It remains to show that the derivative is not zero on any sub-interval of $I$. For this, note that by Fact (3), a derivative of zero forces the function to be constant on that sub-interval. This, however, contradicts the definition of an increasing function. We thus have the desired contradiction and we are done.