I'm familiar with the proof that if the first derivative is always positive then the function is increasing which uses the Mean Value Theorem. How can it be shown that the function not being differentiable at some point does not affect this property? For example a function may have a jump discontinuity with the derivative not being defined at a point but still be increasing.

  • $\begingroup$ Of course not. According to Wikipedia, increasing means monotonically increasing, which is definitely false if the function jumps down. $\endgroup$
    – mathlander
    Commented Dec 25, 2022 at 1:28
  • 1
    $\begingroup$ I'm not sure I completely understand. If $x$ is a point in the domain of a function $f$ such that $f'(x)$ is not defined, the increasing property may certainly be affected! If the function jumps down at $x$, then $f$ certainly won't be increasing. $\endgroup$
    – Michael
    Commented Dec 25, 2022 at 1:29
  • $\begingroup$ The result is true if $f$ absolutely continuous (hence differentiable ae) and $f'\ge 0$ where it exists, but the proof is fairly involved as it boils down to proving that the fundamental theorem of calculus holds with the integral of $f'$ being the Lebesgue one and then it's obvious; the counterexamples below are discontinuous $\endgroup$
    – Conrad
    Commented Dec 25, 2022 at 2:22
  • $\begingroup$ Actually even for $f$ continuous bounded variation the result holds since then $f'$ exists ae but $f(b)-f(a) \ge \int_a^b f'(x)dx$ so we do not necessarily have equality like in the absolutely continuous case but $f'\ge 0$ still implies $f(b)\ge f(a)$ when $b \ge a$ $\endgroup$
    – Conrad
    Commented Dec 25, 2022 at 2:51
  • $\begingroup$ The Weierstrass nowhere differentiable function $W(x)$ is a continuous function such that at each point where $W'(x)$ exists we have $W'(x) > 0$ (since there are no such points, this is vacuously true), but $W(x)$ is not increasing (or even nondecreasing) at any point (increasing at some point is a much weaker property than increasing in some interval). Of course, more interesting is to consider these properties when the set of points of differentiability is large in various ways. $\endgroup$ Commented Dec 25, 2022 at 7:34

2 Answers 2


It may not always be the case. Consider the "fractional part" function:

$$f(x) := \{x\} := x - \lfloor x \rfloor$$

This function, in some sense, returns the "decimal part" of $x$. Hence

  • $f(1.5) = 0.5$
  • $f(\pi) = 0.14159...$

...though, for negative numbers, it returns the decimal part in the "other direction":

  • $f(-1.3) = 0.7$
  • $f(-2.1) = 0.9$

...and so on.

The graph itself looks like $y=x$ for $0 \le x < 1$, repeated over and over again:

enter image description here

Where it is differentiable, $f'$ is always positive, but clearly it is not always increasing (because of the times it jumps down).


Define the function $f$ as $f(x) = x$, if $x<0$ and $f(x)=x-2$ otherwise. Wherever $f$ is differentiable we have $f'(x) = 1 > 0$, but $f$ is not differentiable at $x=0$ (the 'jump discontinuity' you talked about). We have that $f(-1)=-1$ and $f(0)=-2$, so $f(0)<f(-1)$, even though $-1<0$. So this is not true that the function is increasing ---- I suggest you to draw the graph.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .