Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)?

share|improve this question
3  
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: math.stackexchange.com/questions/9958/…, where it was implied there was a solution without any conditions. –  jerome d Nov 17 '11 at 5:37
1  
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
1  
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 $f(x)=f(x-a)+f(x-b)-f(x+a+b)$... –  jerome d Nov 17 '11 at 5:44
    
You are right. (See my previous comment.) –  Srivatsan Nov 17 '11 at 5:44

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.