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Is it true or false that if $$F'(x) = G'(x)$$ then $$F(x) = G(x) + C.$$ No more details are given (continuity, etc...)

I think this is false but I can't think of an example.

Chill2Macht
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Mor
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  • Do you want the equality to hold at every point of the domain? Or just almost everywhere? – Chill2Macht Jun 18 '16 at 11:20
  • They don't say... I just need a counter example – Mor Jun 18 '16 at 11:21
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    The example I gave in this answer (based on the stuff in this post) might be of interest. – J. M. ain't a mathematician Jun 18 '16 at 11:29
  • @J.M. that is a really good example. I am a little confused now though; the function given by the OP to that question has points of discontinuity, i.e. points where the derivative does not exist, so if we want the equality $F'(x)=G'(x)$ to hold everywhere, and not just almost everywhere, then it seems it would not be relevant (unless I am misunderstanding something). My gut feeling is that the statement might actually be true if we demand the equality holds everywhere, not just a.e., but I can't think of a proof or counterexample. Maybe something about Darboux functions? – Chill2Macht Jun 18 '16 at 11:36
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    I suppose a clarification is certainly in order, @William. You are right that the example I linked to is only "almost". (And that's part of why I did not write an answer.) – J. M. ain't a mathematician Jun 18 '16 at 11:40
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    @mor: As existing comments note, the (mutual) domain of $F'$ and $G'$ is a crucial detail. If $F'(x) = G'(x)$ for all $x$ in some interval $I$ of real numbers, then $F$ and $G$ differ by a constant on $I$. In general, you can have "different constants on different intervals", e.g., $\operatorname{sgn}(x) = x/|x|$ on the set of non-zero real numbers. (The function is non-constant, but its derivative exists and is identically zero throughout the set of non-zero reals.) – Andrew D. Hwang Jun 18 '16 at 12:09

2 Answers2

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Here's a simpler counterexample, but it may not be the correct solution, depending on subtle details of what is wanted.

Let $f(x) $ be the absolute value function and let $g(x)$ be the function which is $|x|$ for $x\le0$ and $x+1$ for $x>0$.

Then $f$ and $g$ have identical derivatives (both of which are defined if and only if $x≠0$) but do not differ by a constant function.

I would not normally hand over a solution like this, but I judge that it will be just as instructive, in this case, to decide whether the example is a valid answer to the question.

MJD
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This is too long for a comment, and is not a valid counterexample if we want the equality $$F'(x)=G'(x)$$ to hold everywhere and not just almost everywhere (i.e. everywhere except a set of measure zero).

With those caveats in mind, consider the Cantor function:

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(The definition is related to the Cantor set, although long enough that I don't care to post it again here.)

Except for a set of measure zero, the derivative of the Cantor function is zero, the same as any constant function. However, the Cantor function is clearly not constant, hence if we allow the derivative to be undefined except on a set of measure zero, then this is a valid counterexample.

Chill2Macht
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