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Let $f$ be a real-valued function satisfying the functional equation $$f(x)=f(x+y)+f(x+z)-f(x+y+z)$$ for all $x,y,z\in\mathbb{R}$. Is it true that $f$ must be the equation of a line, with no additional assumptions? Can one use calculus to see this without any a priori constraints on $f$ (that it be continuous, differentiable, etc.)?

Mohsen Shahriari
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Lepidopterist
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    You need some restriction on $f$. In fact, any solution to the Cauchy functional equation satisfies your equation as well. – Srivatsan Nov 17 '11 at 05:19
  • I originally found this question here: http://math.stackexchange.com/questions/9958/interesting-calculus-problems-of-medium-difficulty, where it was implied there was a solution without any conditions. – Lepidopterist Nov 17 '11 at 05:37
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    That question is slightly different. (Stare at the signs. =)) EDIT: But reading it again, it does not seem like any linear function satisfies the equation in the other question... – Srivatsan Nov 17 '11 at 05:39
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    I confess I am not at my best at the moment, but isn't the formulation on the page impossible? If you take the line $f(x)=x$, it doesn't satisfy $f(x)=f(x-a)+f(x-b)-f(x+a+b)$... – Lepidopterist Nov 17 '11 at 05:44
  • You are right. (See my previous comment.) – Srivatsan Nov 17 '11 at 05:44
  • A comment on your last sentence: to even contemplate using (differential) calculus is to assume that the conditions under which calculus can be applied is satisfied. And that requires your functions to be differentiable. – Willie Wong Nov 25 '11 at 12:11
  • It is enough to assume that $f$ is continuous. Or even Borel measurable. Or Lebesgue measurable. Then we must have $f(x) = ax+b$ for some $a,b$. – GEdgar Nov 25 '11 at 15:37
  • $ f $ is a solution of the given equation iff it's of the form $ f ( x ) = A ( x ) + a $ for some additive function $ A $ and some constant $ a $. The fact that these are indeed solution is straightforward to verify. For the converse, just let $ a = f ( 0 ) $, $ A ( x ) = f ( x ) - a $, and put $ x = 0 $ in the functional equation. All the above comments and the given answer will then be easy consequences of properties of additive functions. See Overview of basic facts about Cauchy functional equation. – Mohsen Shahriari Sep 24 '21 at 11:15

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No, it is not true. If you define $f$ arbitrarily on a basis of the vector space of real numbers over the rational numbers, you always get a linear function that is a solution of your equation.

Phira
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