# Thread: f and its inverse on symmetric interval

1. ## f and its inverse on symmetric interval

Hi

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

If $f$ and $f^{-1}$ are both defined on $\mathbb{R}$
and $f$ is odd, show that $f^{-1}$ is also odd. Is the statement
still true if $\mathbb{R}$ is replaced by any symmetric interval
$(-a,a)$ , with $a\neq 0$ ?

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

$f(x)=y$ and since $f^{-1}$ is also defined, we have
$f^{-1}(y)=x$. But since $\mathbb{R}$ is the domain of
$f^{-1}$ , $-y \in \mathbb{R}$.

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

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

$\Rightarrow f(-x_1)=y$

The last step follows since f is an odd function.
$\Rightarrow -x_1=f^{-1}(y) \Rightarrow x_1=-f^{-1}(y)$

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

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

So this proves the first part. For the second part, I will argue that since its
a symmetric interval , whenever $y\in (-a,a)\Rightarrow -y\in (-a,a)$
and the same reasoning will follow.

Can you comment on the proof ?

thanks
newton

2. Originally Posted by issacnewton
Hi

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

If $f$ and $f^{-1}$ are both defined on $\mathbb{R}$
and $f$ is odd, show that $f^{-1}$ is also odd. Is the statement
still true if $\mathbb{R}$ is replaced by any symmetric interval
$(-a,a)$ , with $a\neq 0$ ?

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

$f(x)=y$ and since $f^{-1}$ is also defined, we have
$f^{-1}(y)=x$. But since $\mathbb{R}$ is the domain of
$f^{-1}$ , $-y \in \mathbb{R}$.

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

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

$\Rightarrow f(-x_1)=y$

The last step follows since f is an odd function.
$\Rightarrow -x_1=f^{-1}(y) \Rightarrow x_1=-f^{-1}(y)$

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

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

So this proves the first part. For the second part, I will argue that since its
a symmetric interval , whenever $y\in (-a,a)\Rightarrow -y\in (-a,a)$
and the same reasoning will follow.

Can you comment on the proof ?

thanks
newton
looks right.

3. Thanks abhishek......... looks right to me too.......

4. ## Re: f and its inverse on symmetric interval

Originally Posted by issacnewton
Thanks abhishek......... looks right to me too.......