# Bounded Harmonic Functions on the Disk

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.

• But aren't bounded harmonic functions constant ? Dec 5, 2018 at 6:07
• Wouldn't you need $u$ to be harmonic in all $\mathbb{R}^2$ for this?
– user596383
Dec 5, 2018 at 6:13
• For one we can use the inclusion map to make it harmonic on all of $\mathbb{R}^2$. I am not quite sure whether all the bounded harmonic functions on open disk are constant. Dec 5, 2018 at 6:26
• @YadatiKiran: Bounded harmonic functions in $\Bbb C$ are constant. $u(x, y) = x$ is harmonic and bounded in $\Bbb D$, but not constant. Dec 5, 2018 at 6:28

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.

• If $f$ is discontinuous on $S^1$, then $P[f]$ is discontinuous when viewed as a function on $\bar{\mathbb{D}}$. But $P[f]$ is still harmonic and continuous in $\mathbb{D}$? I'm confused as to how I can conclude from this that $P[f]$ is not uniformly continuous on $\mathbb{D}$.
– user596383
Dec 7, 2018 at 18:00
• If $P[f]$ were uniformly continuous in $\mathbb D$, it would extend continuously to the closure, but it can't, because its extension is discontinuous on $S^1$ Dec 7, 2018 at 18:03
• Yes you are right in saying that $P[f]$ is discontinuous in $\bar{\mathbb D}$ and continuous in $\mathbb D$. The point is that it cannot be uniformly continuous in $\mathbb D$, otherwise it would be in $\bar{\mathbb D}$ too. Dec 7, 2018 at 18:05
– user596383
Dec 7, 2018 at 18:07
• I should probably have written it better :) Dec 7, 2018 at 18:08

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.

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 $$\mathbb D.$$

• Are you considering the principal value $\log$ where $\theta\in\left(\dfrac{-\pi}{2},\dfrac{\pi}{2}\right]$ ? Dec 10, 2018 at 17:19
• @YadatiKiran Yes, but that should be $(-\pi,\pi).$
– zhw.
Dec 10, 2018 at 17:25
• If its $(-\pi,\pi)$ then how do you say $|u|<\pi/2$ in the disc? Am I missing something here? Dec 10, 2018 at 17:29
• @YadatiKiran Because $1+z$ is in the right half plane
– zhw.
Dec 10, 2018 at 17:41
• Oh Yes ! I get it. Dec 10, 2018 at 17:43