Assume that $f$ is a homeomorphism of the unit circle onto itself. If $$1/M \le \frac{|f(e^{i(t+s)})-f(e^{i(t)})|}{|f(e^{i(t)})-f(e^{i(t-s)})|}\le M,$$ then we say that $f$ is $M-$quasi-symmetric selfmapping of the unit circle onto itself. Is every smooth quasi-simmetric mapping bi-Lipschitz?
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$\begingroup$ It seems that this is trivial, so you can close the question. $\endgroup$– user36162Sep 17, 2013 at 22:06
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$\begingroup$ What does it exactly mean "smooth"? $\endgroup$– Alexandre EremenkoSep 17, 2013 at 22:33
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$\begingroup$ The map $f(x)=x^3$ is quasi-symmetric and smooth on the unit interval. However, it is clearly not bi-Lipschitz. In the same fashion one gets examples for the circle. $\endgroup$– MishaSep 18, 2013 at 7:51
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