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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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    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 $a,b$.2011-11-25

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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.