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One of false beliefs in this question on Math Overflow is "If f is a smooth function with df=0, then f is constant". What is a counterexample to this statement? Can it be made correct by adding some restriction, e.g. that f is a function from reals to reals?

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This is described in the comments; f need only be locally constant, so as a counterexample take f : (0, 1) cup (2, 3) to R which is equal to 0 on the first component and 1 on the second. It is certainly true for functions from R to R, for example by the mean value theorem. –  Qiaochu Yuan Dec 9 '10 at 9:46
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@Qiaochu - meh, I was thinking that I missed something fundamental, and that the counter example will be more interesting. You should have posted this as an answer. –  ripper234 Dec 9 '10 at 10:00
    
If you're working in the complex numbers, there are many examples of non-holomorphic functions which are nonconstant but have zero derivative (at least in some sense). –  Marra Sep 15 '13 at 3:38
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This is described in the comments; $f$ need only be locally constant, so as a counterexample take $f : (0, 1) \cup (2, 3) \to \mathbb{R}$ which is equal to $0$ on the first component and $1$ on the second. It is certainly true for functions from $\mathbb{R}$ to $\mathbb{R}$, for example by the mean value theorem.

This seems trite, but it's still an important point to keep in mind. If you ever feel like you're missing something, then of course you should attempt to carefully prove the result that you thought was true and see where it fails (if it does).

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