# What is the condition for Continuously differentiable function to be Lipschitz (Holder)

I have function $f\in C^1(\overline{\Omega})$ for $\Omega\in\mathbb{R}^d$ bounded. What are the conditions on $\Omega$ for $f$ to be Lipschitz (or at least Holder)? Or at least locally if I have $f\in C^1(\Omega)$ with bounded derivative.

Note that $\Omega\in C^0$ is not sufficient as Gilbarg & Trudinger: Elliptic partial differential equations of second order showed that, $$u(x,y)=sgn(x)|y|^\beta\quad \text{for }y\geq0,\quad =0\quad \text{for }y<0,$$ on $$\Omega=\{ (x,y)\in\mathbb{R}^2,x^2+y^2<1,\ y<|x|^{1/2} \},$$ is not Holder continuous for $1\geq\alpha>\beta/2$.

I think that the answer might be $\Omega\in C^{0,\alpha}$ for $f$ to be $C^{0,\alpha}$ or $\Omega\in C^{0,1}$.

If $\Omega\in C^{0,1}$ then $\Omega$ is a $W^{1,\infty}$ extension domain, which means that you can extend $f$ to a function in $W^{1,\infty}(\mathbb{R}^d)$. This function would be Lipschitz. The idea is actually simple. Locally you can assume that $\Omega=\{(y,t): t>h(y)\}$ for some Lipschitz function $h$. To extend $f$ you would first flatten the boundary and then reflect $f$, that is, define $g(y,t):=f(y,2h(y)-t)$ if $t<h(y)$. Using this formula and the chain rule, you see that if $h$ is Lispchitz then so is $g$. You can play and see what happens if $h$ is only Holder. I did not try.