Can there exist a non-constant continuous function that has a derivative of zero everywhere? 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?
 A: 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 $f(x)=\left\{
 \begin{array}{ll}
  0  & \mbox{if } x \in (0,1) \\
  1 & \mbox{if } x\in (2,3)
 \end{array}
\right.$
A: 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 http://en.wikipedia.org/wiki/Cantor_function.
A: 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?
