Originally Posted by

**issacnewton** Hi

I have to prove the following theorem from my analysis book.

If $\displaystyle f$ and $\displaystyle f^{-1}$ are both defined on $\displaystyle \mathbb{R}$

and $\displaystyle f$ is odd, show that $\displaystyle f^{-1}$ is also odd. Is the statement

still true if $\displaystyle \mathbb{R}$ is replaced by any symmetric interval

$\displaystyle (-a,a)$ , with $\displaystyle a\neq 0 $ ?

Following is my attempt. Since both f and its inverse are defined on R ,

the function f is onto and off course one-to-one.

Let $\displaystyle y\in R$ be arbitrary. $\displaystyle \exists x \in R \backepsilon$

$\displaystyle f(x)=y$ and since $\displaystyle f^{-1}$ is also defined, we have

$\displaystyle f^{-1}(y)=x$. But since $\displaystyle \mathbb{R}$ is the domain of

$\displaystyle f^{-1}$ , $\displaystyle -y \in \mathbb{R}$.

$\displaystyle \Rightarrow \exists x_1 \in R \backepsilon f^{-1}(-y)=x_1$

or $\displaystyle f(x_1) =-y \, \Rightarrow -f(x_1)=y$

$\displaystyle \Rightarrow f(-x_1)=y$

The last step follows since f is an odd function.

$\displaystyle \Rightarrow -x_1=f^{-1}(y) \Rightarrow x_1=-f^{-1}(y)$

But we had assumed that $\displaystyle f^{-1}(-y)=x_1$

So it follows that $\displaystyle f^{-1}(-y)=-f^{-1}(y)$

So this proves the first part. For the second part, I will argue that since its

a symmetric interval , whenever $\displaystyle y\in (-a,a)\Rightarrow -y\in (-a,a)$

and the same reasoning will follow.

Can you comment on the proof ?

thanks

newton