# $f(a)+f(-b)=0$ implies $f(-a)+f(b)=0$. Then $f$ is an odd function?

If $$f(a)+f(-b)=0$$ for some $$(a,b)$$, then, $$f(-a)+f(b)=0$$.

Also, for all $$a$$, there exists $$b$$ such that $$f(a)+f(-b)=0$$.

$$f:\mathbb R\to\mathbb R$$ is continuous.

Based on the above conditions, can we prove or disprove that $$f$$ is an odd function: $$f(x)=-f(-x)$$?

I think we cannot prove this. Being odd seems to be a sufficient but not necessary. Yet I failed to find any counterexamples against the claim.

Update: Thanks to user WhatsUp, even functions $$f(x)=f(-x)$$ can also satisfy these conditions.

User Kavi suggests that if we add $$f(x)>0$$ for all $$x>0$$, then $$f$$ is odd. (please correct me if I am wrong) I will make a new question based on his intuition.

• What are domain and range of $f$? Jan 7, 2022 at 10:08
• @Mastrem Ok I will update. But your example seems violate the second condition.
– dodo
Jan 7, 2022 at 10:10
• Good point, I misread Jan 7, 2022 at 10:10
• The first sentence seems to mean that there is one pair such that $f(a)+f(-b)=0$ and $f(-a)+f(b)=0$.
– user1010241
Jan 7, 2022 at 10:12
• This implication is (vacuously) true for any strictly positive function. Jan 7, 2022 at 10:14

A quite simple counterexample is $$f(x) = \cos(x)$$.

• Ok it seems like any even function also satisfies these conditions. Good point!
– dodo
Jan 7, 2022 at 10:23

$$\forall a ~ \exists b: f(a)+f(-b)=0$$ Choose $$a=0$$. There exists $$b\in \mathbb{R}$$ such that $$f(0)+f(-b)=0$$

From the first condition, $$f(0)+f(b)=0$$ $$\therefore \exists b \in \mathbb{R} :f(b) = f(-b)$$ Thus, your function is certainly not odd.

• What about $b=0$ and $f(0)=f(0)$? If $f$ is an odd function then $f(0)=0$
– dodo
Jan 7, 2022 at 10:18