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Assume that $f(re^{it})$ is a bi-Lipschitz of the closed unit disk onto itself with $f(0)=0$. Is the function $h(r)=\max_{0\le t \le 2\pi}|f(re^{it})|$ bi-Lipschitz on $[0,1]$?

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It is easy to prove that $h$ is Lipschitz whenever $f$ is. Indeed, we simply take the supremum of the uniformly Lipschitz family of functions $\{f_t\}$, where $f_t(r)=|f(re^{it})|$.

Also, $h$ is bi-Lipschitz whenever $f$ is. Let $D_r$ be the closed disk of radius $r$. Let $L$ be the Lipschitz constant of the inverse $f^{-1}$. The preimage under $f$ of the $\epsilon/L$-neighborhood of $f(D_r)$ is contained in $D_{r+\epsilon}$. Therefore, $h(r+\epsilon)\ge h(r)+\epsilon/L$, which means the inverse of $h$ is also Lipschitz.


Answer to the original question: is $h$ smooth?

No, it's no better than Lipschitz. I don't feel like drawing an elaborate picture, but imagine the concentric circles being mapped onto circles with two smooth "horns" on opposite sides. For some values of $r$ the left horn is longer, for others the right horn is longer. Your function $h(r)$ ends up being the maximum of two smooth functions (lengths of horns). This makes it non-differentiable at the values of $r$ where one horn overtakes the other.

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    What about the question: Assume that $f(re^{it})$ is a diffeomorphism of the unit disk onto itself. Is the function $h(r)=\max_{0\le t \le 2\pi}|f(re^{it})|^2$ smooth on $(0,1)$?2012-09-16
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    @Malota That would not change anything. Squaring a positive function does not change its smoothness.2012-09-16
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    You mention Lipschitz, is this true for bi-Lipschitz setting?2012-09-16
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    Thanks, nice and short answer!2012-09-17
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    Assume that $f(re^{it})$ is Locally bi-Lipschitz of the closed unit disk onto itself with $f(0)=0$. Is the function $h(r)=\max_{0\le t \le 2\pi}|f(re^{it})|$ bi-Lipschitz on $[0,1]$?2012-09-29
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    @Malota "locally" implies "globally" on a compact set. And I already showed that $h$ is bi-Lipschitz (onto its image) without the assumption $f(0)=0$. With this additional assumption we have $h(0)=0$, hence $h$ is onto $[0,1]$.2012-09-29
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    $f:z\to e^z-1$ is locally bi-Lipschitz but is not globally bi-Lipschitz (neither on compacts), since it is not injective.2012-09-29
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    $Malota You are right. But the proof I indicated can be adapted to this case as well since it only needs local Lipschitz inverses.2012-09-29
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    Somehow the disk with center at $0$ is important. Your proof uses the following fact: $U(f(D_r),\epsilon/L)\subset f(D_{r+\epsilon})$ (this implies that $h(r+\epsilon)\ge h(r)+\epsilon/L$), however if instead of $D_r$ we take another set $A$ (which does not contain $0$ inside) then maybe we can have a problem?2012-09-29
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    @Malota The function $h$ measures the size of the smallest disk that contains $f(D_r)$. We want to prove that this disk will always expand at some rate. To this end, pick $z$ with $|z|=r$ such that $|f(z)|=h(r)$. For small $\epsilon$ the image of $\epsilon$-neighborhood of $z$ contains the $\epsilon /L$ neighborhood of $f(z)$. The result follows.2012-09-29