I got stuck with the following exercise.
Suppose $f$ is continuous and has compact support on the complex plane $\mathbb{C}$.
Let $$u(z) = \frac{1}{\pi} \int_{\mathbb{C}} \frac{f(\zeta)}{z-\zeta} dm(\zeta) = \frac{1}{\pi} \int_{\mathbb{C}} \frac{f(z-\zeta)}{\zeta} dm(\zeta),$$ where $dm(\zeta)$ is the Lebesgue measure in $\mathbb{C}$. (Basically we are treating $\mathbb{C}$ as $\mathbb{R}^2$.)
Show that $u$ belongs to $\mathrm{Lip}(\alpha)$, for every $\alpha < 1$.
It is easy to show that $u$ is continuous, by the uniform continuity of $f$ and the fact that the function $1/\zeta$ is integrable over compact sets in $\mathbb{C}$. I do not see how to prove $\frac{u(z+h)-u(z)}{h^\alpha}$ is bounded.