Is this function Lipschitzian

Question

Let $h :S^1=\{x\in\mathbb R^d: \left \| x \right \|_2=1\} \to \mathbb R^+ $ be a Lipschitz function.

Let $\Omega\subset\mathbb R^d$ be a bounded open set and \begin{align} \Psi:B\setminus\{0\}\quad&\to\quad\Omega\setminus\{0\}\\[1ex] y\quad&\mapsto\quad y\cdot h\left(\frac y{\|y\|}\right). \end{align}

My question is: Is $\Psi$ a Lipschitz function?

Answer 1

Let $C$ be an upper bound for $h$ and $L$ be its Lipschitz constant. Then, \begin{align*} \|\psi(x) - \psi(y)\| &\le \|x-y\| h(\frac x{\|x\|}) + \|y\| \, \Big\| h(\frac x {\|x\|}) - h(\frac y{\|y\|}) \Big\| \\&\le C \|x-y\| + \| y \| \, L \, \Big| \frac x{\|x\|} - \frac{y}{\|y\|}\Big| \\&\le C \|x-y\| + L \, \Big| \frac {x\|y\|}{\|x\|} - y\Big| \\&\le C \|x-y\| + L \, \|x-y\| + L \, \Big| \frac {x\|y\|}{\|x\|} - x\Big| \\&\le C \|x-y\| + L \, \|x-y\| + L \, | \|x\| - \|y\| | \\&\le C \|x-y\| + 2 \, L \, \|x-y\|. \end{align*}