I am trying to solve the following problem:
Find all the continuous functions $f:\mathbb{R} \to \mathbb{R}$ that satisfy if $a$, $b$ and $c$ are nonzero real numbers such that $a+b+c=0$ then $f(a)+f(b)+f(c)=0$.
I have only proved that $f(0)=0$: Let $f$ be a function that holds the condition in the statement, and let $x$ be a nonzero real number. Because of the hypothesis on $f$, we have that $$f(2x) + f(-x) + f(-x) = 0.$$ Thus, $f(2x) = -2f(-x)$. Since $f$ is continuous, we get that $f(0) = -2f(0)$ as $x\to 0$. It follows that $f(0)=0$.
I guess that the only function that fulfills the statement is $f(x)=mx$ for any real $m$, but I don't know how to show it.
Best Answer
You have already proved that $f(0)=0$ and (by @JensSchwaiger's comment) $f(-x)=-f(x).$ Therefore, your equation boils down to $$f(a+b)=f(a)+f(b)$$ (whenever $a,b,a+b\ne0,$ but also when at least one of them is $0,$ by the two previous properties). This is Cauchy's functional equation, whose continuous solutions are indeed the linear functions $f(x)=mx,$ as you guessed.