Bi-Lipschitzity of maximum function 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 $[0,1]$?
 A: 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. 
