Let $f$ be a lipschitz continuous complex valued function in the complex plane, compactly supported in the closed disk of radius $\delta > 0$ with lipschitz constant $M$. Define $h(r,\theta) = f(r \exp(i \theta))$ on $\mathbb{R^2}$. Then $h$ is lipschitz continuous with lipschitz constant $M(1+ \delta)$.
I can sort of see why $h$ would be lipschitz, but have no idea how to establish the lipschitz constant in the statement above. Any help is greatly appreciated.
Edit: The norm everything is lipschitz with respect to is always the euclidean norm on the plane
Let $r,r',\theta,\theta'$ be four reals.
By the symmetry between the roles of $(r, \theta)$ and $(r', \theta')$, we only have to check the following two cases WLOG: $1)$ $|r'|$ is less than $\delta$, and $2)$ both $|r|$ and $|r'|$ are strictly greater than $\delta$.
In the latter case, both $f\left(re^{i\theta}\right)$ and $f\left(r'e^{i\theta'}\right)$ are zero, so in particular: $$|h(r,\theta) - h(r', \theta')| = 0 \leq M(1 + \delta) \|(r, \theta)\|_2$$
Thus there is only the former case left: assume that $|r'| \leq \delta$.
Then, we have: $$\begin{split}\left|f\left(re^{i\theta}\right) - f\left(r'e^{i\theta'}\right)\right| &\leq M\left|r e^{i\theta} - r' e^{i\theta'}\right|\\ &\leq M\left(\left|re^{i\theta} - r'e^{i\theta}\right| + \left|r'e^{i\theta} - r'e^{i\theta'}\right|\right)\\ &\leq M\left(|r - r'| + |r'|\left|e^{i\theta} - e^{i\theta'}\right|\right)\\ &\leq M\left(\sqrt{|r - r'|^2 + |\theta - \theta'|^2} + \delta\left|e^{i\theta} - e^{i\theta'}\right|\right)\\ &\leq M \|(r,\theta) - (r', \theta')\|_2 + M\delta \left|e^{i\theta} - e^{i\theta'}\right|\end{split}$$
Now, the derivative of $a : x \mapsto e^{ix}$ exists, and $a' : x \mapsto ie^{ix} (=i a(x))$, which is bounded by $1$ on $\mathbb{R}$, therefore $a$ is $1$-Lipschitz, hence: $$\left|e^{i\theta} - e^{i\theta'}\right| \leq |\theta - \theta'| \leq \|(r,\theta) - (r',\theta')\|_2$$ which lets us conclude.
One thing we can note is that we could have chosen the sup norm or whichever $\|\cdot\|_p$ on $\mathbb{R}^2$ due to how the calculations turned out.