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If a function $f$ from $\mathbb{R}$ to $\mathbb{R}$ is one-to-one and bounded is it true that $f^{-1}$ is also one-to-one and bounded? I believe the answer is no but I'm not sure.

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The inverse is one-to-one (otherwise, $f$ wouldn't be a function). –  David Mitra Jan 14 '13 at 3:39
    
Technically, a bounded function from R to R can't have inverse –  Max Jan 14 '13 at 9:40
    
I think that $ f^{-1} $ here is defined as $ f^{-1} \stackrel{\text{def}}{=} \{ (b,a) \in \mathbb{R}^{2} ~|~ (a,b) \in f \} $. Therefore, $ \text{Dom}(f^{-1}) = \text{Range}(f) $ and $ \text{Range}(f^{-1}) = \text{Dom}(f) $. –  Haskell Curry Jan 16 '13 at 5:56

4 Answers 4

up vote 10 down vote accepted

The answer is always false. The range of $f^{-1}$ is the domain of $f$, which from the way the problem is stated is $\mathbb R$.

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+1: Nice general solution. :) –  Haskell Curry Jan 14 '13 at 3:29

A counter example: Logistic function.

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No. Consider $\arctan x$ which is bounded, but $\tan \theta$ isn't bounded.

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+1. Interesting to note that both of us had similar kind of sketch of the function in our mind. –  user17762 Jan 14 '13 at 3:33
    
@Marvis Seems the easiest to use. I wanted a 'simple' example, though OP should have seen Logistic functions under exponentials too. Just wondering, are you an analyst? Your 'calculus/analysis' proofs are very beautiful. –  Calvin Lin Jan 14 '13 at 4:34
    
Thanks and the answer to your question is "I don't know" :-). I am a doctoral student in computational mathematics working in numerical linear algebra/ algorithms. I just dabble in different areas of math with courses I have taken. –  user17762 Jan 14 '13 at 4:49
    
And brilliant.org looks like a great initiative! Good one. Is there a way I can contribute to it? –  user17762 Jan 14 '13 at 7:45
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Sure. I am also interested in Olympiad/Putnam type problem solving in general. Anyway, in case you guys plan to expand and need someone to contribute some time down the line, do keep me posted. :-) –  user17762 Jan 18 '13 at 0:52

Inverse function itself may not exist. For example the function $f(x) = \frac{1}{1+e^x}$ is defined from $\mathbb{R}$ to $\mathbb{R}$ which is one to one and bounded. But its inverse $\log\left(\frac{1}{y} -1\right)$ is not defined for all real numbers.

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