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.)?
Is this a functional equation of a line?
1
$\begingroup$
calculus
functional-equations
-
3You need some restriction on $f$. In fact, any solution to the [Cauchy functional equation](http://en.wikipedia.org/wiki/Cauchy's_functional_equation) satisfies your equation as well. – 2011-11-17
-
0I 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. – 2011-11-17
-
1That 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... – 2011-11-17
-
1I 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 $f(x)=f(x-a)+f(x-b)-f(x+a+b)$... – 2011-11-17
-
0You are right. (See my previous comment.) – 2011-11-17
-
0A 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. – 2011-11-25
-
0It 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
1 Answers
1
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.