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If I have a monotone non-decreasing $f:[a,b]\rightarrow [c,d]$ which is also a homeomorphism, is it necessarily Lipschitz? If yes, what would be a good candidate for the Lipschitz constant?

I'm doing some path reparameterizations and this would make things a lot easier. Thanks in advance.

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    The question can be asked with fewer words, since every homeomorphism between closed intervals must be strictly monotone and it clearly has no bearing on the hypothesis whether $f$ is strictly increasing or strictly decreasing.2012-02-12

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The answer is no, since \begin{align} f:[0,1]&\rightarrow [0,1]\newline x&\mapsto \sqrt{x} \end{align} is obviously a monotone homeomorphism, but given $x\in (0,1]$ $d(f(x),f(0))=|\sqrt{x}-\sqrt{0}|=\frac{|x-0|}{\sqrt{x}}=\frac{d(x,0)}{\sqrt{x}} $ show us, that there is now possible $L>0$ such that $d(f(x),f(0))\leq L\cdot d(x,0)$ due to the fact that $\lim_{x\to 0^+}\frac{1}{\sqrt{x}}=\infty$

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    Thanks for the counter-example. You posted slightly earlier that Davide, so I'm 'accepting' your answer. Both explain the situation well.2012-02-12
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With $a=c=0$, $b=d=1$ and $f(x)=\sqrt x$ we get a counter-example since $|f(n^{-1})-f(0)|=\sqrt n^{-1}$ and $\frac{|f(n^{—1})-f(0)|}{n^{—1}-0}=\frac{\sqrt n^{-1}}{n^{-1}}=\sqrt n$. ($f$ is continuous, bijective since it's inverse is $f^{-1}(x)=x^2$ which is continuous)