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Let $I\subset \mathbb R$ be an open interval and $f:I\to \mathbb R$ be a differentiable function. I would like to prove the following equivalence:

$$f:I\to\mathbb R\ \text{a diffeomorphism over $f(I)$}\Leftrightarrow f'(x)\cdot f'(y)>0\ \text{for every $x,y\in I$}$$

I'm trying to prove it using the inverse function theorem.

$\Leftarrow$

Since $f'(x)\neq 0$, take an arbitrary $x_0\in I$, there is an open subset $U\in I$ with $x_0\in U$ such that $f_{\restriction U}:U\to f(U)$ is a diffeomorphism. I don't know how to prove the function $f$ is a diffeomorphism on the entire $I$.

$\Rightarrow$

I have no idea, I need some hints.

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    "$f'(x)f'(y)>0$ for every $x, y \in I$" is a fancy way of saying that $f'$ never changes sign; it is either always positive or always negative. – 2017-01-10
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    @Arthur What can I do with this information? thank you for your comment. – 2017-01-10

1 Answers 1

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$\Leftarrow$: As commented by Arthur, $f'$ never changes sign on $I$. That is, there does not exist $x,y \in I$ with $f'(x)f'(y) < 0$. Moreover, this condition implies $f'(x) \neq 0$ for all $x \in I$, so $\mathrm Df$ is invertible on $I$. If you can also show that $f'$ is continuous, you can invoke the Inverse Function Theorem which asserts $f^{-1}:f[I] \to I$ is continuously differentiable, and hence $f$ is a $\mathrm C^1$-diffeomorphism.

$\Rightarrow:$ $f$ is a diffeomorphism implies $f' \neq 0$. Moreover, $f'$ is continuous (since $f$ is continuously differentiable), so $f'$ may never cross $0$. Hence $f'(x)f'(y) > 0$.

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    can we have $f'(x)<0$ for every $x\in I$? in this case we would have $f'(x)\cdot f'(y)>0$ too. – 2017-01-10
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    @user42912 You can. – 2017-01-10
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    So in the last line of your answer we could have $f'(x)\cdot f'(y)<0$ and it's not what we want. Thank you for your answer. – 2017-01-10
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    @user42912 That would be invalid, since $f'(x) \cdot f'(y) < 0$ iff $f'(x)$ and $f'(y)$ has different signs, which by continuous differentiability and Intermediate Value Theorem implies $f'(x) = 0$ somewhere. – 2017-01-10
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    You're right I'm sorry. – 2017-01-10
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    I'm using this version of inverse function theorem: http://www.math.tamu.edu/~tvogel/410/sect116.pdf. Thus I can use it without proving $f'$ is continuous, right? – 2017-01-10
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    @user42912 If the theorem allows you to do it then you can do it. – 2017-01-10
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    My only problem to use it is it asserts there is an open subset $U\in I$ such that $f:U\to f(U)$ is a diffeomorphism. We don't necessarily have $U=I$. – 2017-01-10
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    @user42912 The theorem you showed is only needed for the $\Leftarrow$ direction which you are proving diffeomorphism. – 2017-01-10
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    Yes I know. The only thing it misses to me to understand your answer is why you can take the open set to be the whole $I$. Thank you again – 2017-01-10
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    @user42912 The theorem says "Then there is some open set...". The existence of the open set is what the theorem asserts, not what you need to prove from other results. – 2017-01-10
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    Yes I know, the problem is in your answer you said $f^{-1}:f(I)\to I$ is continuously differentiable, but I think you can't say that, the only thing you can say is there is some open set $U\in I$ such that $f_{\restriction f(U)}^{-1}:f(U)\to U$ is continuously differentiable. Did you understand my point? – 2017-01-10