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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.

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  • $\begingroup$ But aren't bounded harmonic functions constant ? $\endgroup$ Dec 5 '18 at 6:07
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    $\begingroup$ Wouldn't you need $u$ to be harmonic in all $\mathbb{R}^2$ for this? $\endgroup$
    – user596383
    Dec 5 '18 at 6:13
  • $\begingroup$ 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. $\endgroup$ Dec 5 '18 at 6:26
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    $\begingroup$ @YadatiKiran: Bounded harmonic functions in $\Bbb C$ are constant. $u(x, y) = x$ is harmonic and bounded in $\Bbb D$, but not constant. $\endgroup$
    – Martin R
    Dec 5 '18 at 6:28
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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.

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  • $\begingroup$ 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 $\mathbb{D}$. $\endgroup$
    – user596383
    Dec 7 '18 at 18:00
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    $\begingroup$ 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 $S^1$ $\endgroup$
    – Federico
    Dec 7 '18 at 18:03
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    $\begingroup$ 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. $\endgroup$
    – Federico
    Dec 7 '18 at 18:05
  • $\begingroup$ Thanks, I had forgotten about this property. $\endgroup$
    – user596383
    Dec 7 '18 at 18:07
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    $\begingroup$ I should probably have written it better :) $\endgroup$
    – Federico
    Dec 7 '18 at 18:08
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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.

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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 $\mathbb D.$

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  • $\begingroup$ Are you considering the principal value $\log$ where $\theta\in\left(\dfrac{-\pi}{2},\dfrac{\pi}{2}\right]$ ? $\endgroup$ Dec 10 '18 at 17:19
  • $\begingroup$ @YadatiKiran Yes, but that should be $(-\pi,\pi).$ $\endgroup$
    – zhw.
    Dec 10 '18 at 17:25
  • $\begingroup$ If its $(-\pi,\pi)$ then how do you say $|u|<\pi/2$ in the disc? Am I missing something here? $\endgroup$ Dec 10 '18 at 17:29
  • $\begingroup$ @YadatiKiran Because $1+z$ is in the right half plane $\endgroup$
    – zhw.
    Dec 10 '18 at 17:41
  • $\begingroup$ Oh Yes ! I get it. $\endgroup$ Dec 10 '18 at 17:43

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