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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.)?

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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 5:19
I originally found this question here:…, where it was implied there was a solution without any conditions. – Lepidopterist Nov 17 '11 at 5:37
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 5:39
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 5:44
You are right. (See my previous comment.) – Srivatsan Nov 17 '11 at 5:44
up vote 1 down vote accepted

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.

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