Denote by $\mathbb{D}$ the open unit disk in $\mathbb{R}^2$. Is it possible to find a bounded harmonic function $u : \mathbb{D} \to \mathbb{R}$ that is not uniformly continuous?
I tried using functions that oscillate near $\partial \mathbb{D}$ but was unable to get anything substantial.
On
Start with $$ \phi(z)=\frac{1+z}{1-z},\;\;\; z\in\mathbb{C},\; |z| < 1. $$ This function is holomorphic with \begin{align} \Re \phi(z) &= \Re\frac{1+z}{1-z}\frac{1-\overline{z}}{1-\overline{z}} \\ &=\Re\frac{1+z-\overline{z}-|z|^2}{|1-z|^2} \\ &= \frac{1-|z|^2}{|1-z|^2} > 0. \end{align} $\psi(z)=e^{-\phi(z)}$ is bounded and holomorphic in $|z| < 1$, but $\psi$ is not uniformly continuous in the open disk.
On
Let $\log$ denote the principal value $\log.$ Then $\log(1+z)$ is holomorphic in $\mathbb D.$ Hence its imaginary part, $u(z)=\arg (1+z),$ is harmonic in $\mathbb D.$ We have $|u|<\pi/2$ in the disc, so $u$ is bounded there. For small $r>0,$ $-1+re^{i\pi/4}\in \mathbb D.$ For such $r,$
$$u(-1+re^{i\pi/4})- u(-1+r) = \pi/4-0 = \pi/4.$$
But $(-1+re^{i\pi/4})-(-1+r) \to 0$ as $r\to 0.$ This shows $u$ cannot be uniformly continuous in $\mathbb D.$
For any bounded measurable function $f\in L^\infty(S^1)$, the Poisson integral $P[f]$ gives a bounded harmonic functions in $B_1$. This function is continuous (equivalently, uniformly continuous) if and only if $f$ itself is continuous.
Here you can find what the Poisson kernel is and how it's used to build harmonic functions in the ball.
Said otherwise, $f\mapsto P[f]$ is a linear isometry of $L^p(S^1)$ onto $h^p(B_1)$. We used the case $p=\infty$, where $h^\infty(B_1)$ stands for the the bounded harmonic functions in the ball. You can read more about it here.