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Math Help - f and its inverse on symmetric interval

  1. #1
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    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
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by issacnewton View Post
    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.
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  3. #3
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    Thanks abhishek......... looks right to me too.......
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  4. #4
    Senior Member abhishekkgp's Avatar
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    Re: f and its inverse on symmetric interval

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