Suppose $D$ is a bounded, open, connected subset of $\mathbb{R}^n$.
Suppose that $u \in C^{1,\alpha}$ for $\alpha \in (0,1]$, i.e. $u$ has a Holder continuous derivative with exponent $\alpha$. Is it true that $u$ is Lipschitz?
My thoughts are that $u \in C^{1,\alpha}$ implies $\nabla u$ is uniformly continuous, and hence bounded since $D$ is bounded.
Then $\nabla u$ bounded implies $u$ is Lipschitz. I don't know why but I feel like I may be forgetting some hypotheses. It doesn't feel quite right.
Best Answer
And rightly so. Consider for example (identifying $\mathbb{C}$ with $\mathbb{R}^2$) a slit annulus,
$$D = \{ z \in \mathbb{C} : 1 < \lvert z\rvert < 2\} \setminus (1,2),$$
and $u(z) = \arg z$. Then the derivative of $u$ is Hölder-continuous, but $u$ is not Lipschitz continuous.
You can of course construct similar examples in higher dimensions.
What you need for the conclusion is that the distance of two points in $D$ doesn't differ by too much from the length of the shortest paths in $D$ between the points.