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Let $f$ be a function such that

  1. $f:[0,1]\cap\mathbb Q\to\mathbb R$,
  2. $f(0)=0$, and
  3. $f'(x)=0$ for all $x$ in the domain.

Surprisingly, these conditions do not determine a unique function. The zero function satisfies them but so does Minkowski's question mark function. I'm interested in what can be added as a fourth condition to ensure that only the "natural option", i.e. in this case the zero function, is picked out. The condition should also work when condition 3 is replaced by $f'$ being any other uniformly continuous function.

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I will start out the list myself:

  1. $f$ has the Fundamental Theorem of Calculus-Property

By this I mean that for all $x$ in the domain it is the case that $f(x)=\int_0^xf'(t)dt$, where the integral is the Riemann integral.

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    That can hardly be called a condition. Or one might suggest a just as useless 4. $|f(x)|<\epsilon$ for all $\epsilon>0$ ... – 2017-01-22
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    @HagenvonEitzen: The useless 4 does not work when "condition 3 is replaced by $f'$ being any other uniformly continuous function". – 2017-01-22
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    Yes one might as well just assume the function is absolutely continuous. Or something along the line of restricting the total variation. – 2017-01-22
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    @jimh: Thanks! That's very useful! – 2017-01-22