How to show that an odd function always goes through zero?

Question

I have the standard definition of an odd-function from wikipedia:

Again, let f(x) be a real-valued function of a real variable. Then f is odd if the following equation holds for all x and -x in the domain of $-f(x) = f(-x)$

Can anyone help me how to do this? Do I have to show, that it converges to zero?

Answer 1

The way I think of this property is through the intermediate value theorem. Assume that the odd function $f$, is continuous. Then choose some point $ a > 0 $. Suppose without loss of generality that $ f(a) > 0 $. Then since $ f(-a) = -f(a) <0 $, by the IVT, there must be a point $ b $ in $ (-a,a) $ such that $ f(b) = 0 $. Note that this only works if $ f $ is continuous.

Answer 2

HINT: Substitute $x=0$ into the equation defining oddness of $f$: $-f(0)=f(-0)$.

Answer 3

Nothing can be concluded about convergence. But to answer the question in the title, if $0$ is in the domain then $-f(0)=f(0)$ so it must be $0.$ If $0$ is not in the domain then the answer is that it doesn't necessarily do so.

Answer 4

If $-f(x)=f(-x)$ for all $x$ in the domain for which $-x$ is in the domain, then if $0$ is in the domain $$-f(0)=f(-0)=f(0),$$ which means $2f(0)=0$, and so $f(0)=0$.