Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Somebody told me that there exists a continuous function with a derivative of zero everywhere that is not constant.

I cannot imagine how that is possible and I am starting to doubt whether it's actually true. If it is true, could you show me an example? If it is not true, how would you go about disproving it?

share|cite|improve this question
If we are working on $\mathbb{Q}_p$ or $\mathbb{C}_p$, then nontrivial locally constant functions can exist. Also, if the domain $\Omega$ of the function $f : \Omega \stackrel{\mathbb{open}}{\subset} \mathbb{R} \to \mathbb{R}$ is disconnected, then $f$ can have many values while satisfying $f' \equiv 0$. But surely this kind of answer is not the one you are seeking for. – Sangchul Lee Feb 22 '12 at 17:42
up vote 15 down vote accepted

If it's differentiable at every point, then this can't happen. This follows from the mean value theorem:

If $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then for at least one point $c$ between $a$ and $b$, we have $$f'(c) = \dfrac{f(b)-f(a)}{b-a}.$$

If your $f(x)$ is not constant, but is differentiable everywhere, pick an $a$ and $b$ with $f(a)\neq f(b)$. By the MVT, we have $$f'(c) = \dfrac{f(b)-f(a)}{b-a} \neq 0$$ since $f(b) \neq f(a)$.

On the other hand, if you assert your function is differentiable only "almost everywhere" instead of "everywhere" (in a sense which can be made precise) and that the derivative "almost everywhere" is equal to $0$, then this can happen. The standard example is Cantor's function (also known as the Devil's Staircase). See

share|cite|improve this answer
Thanks, I think he probably was referring to the Cantor function. – Peter Olson Feb 22 '12 at 15:08
FYI, differentiable everywhere can be replaced with "differentiable on a set whose complement is countable". Some references are given at Also, it is not possible to improve this to some uncountable sets, but I don't know of any specific references for this right now. (I believe it's been proved that any exceptional set must have the property of containing no perfect subsets, and since the exceptional set in question is Borel, this forces the exceptional set to be countable.) – Dave L. Renfro Feb 22 '12 at 19:14

Since there are no restrictions on the domain, it is actually possible. Let $f:(0,1)\cup(2,3)\to \mathbb R$ be defined by $f(x)=\left\{ \begin{array}{ll} 0 & \mbox{if } x \in (0,1) \\ 1 & \mbox{if } x\in (2,3) \end{array} \right.$

share|cite|improve this answer

In short, no. Your friend may be misremembering examples of continuous nowhere-differentiable functions, e.g., the Weierstrass function or the boundary of the Koch snowflake.

I was gong to cite the mean value theorem, but Jason DeVito beat me to the punch. To add to his answer, you might also consider the Picard-Lindelof theorem. What would it mean for $f$ to adopt two different values, since the theorem asserts that its behavior is locally that of a constant function?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.