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I am trying to prove that if $u$ is bounded harmonic function on the upper half plane, such that $u$ limits to $1$ on the set $\mathbb{R}_{>0}$ and to $-1$ on the set $\mathbb{R}_{<0}$ then $u$ must equal $\frac{2}{\pi}Arg(-iz)$ where $Arg=Im(Log)$ and $Log$ is the main branch of the logarithm. This is my idea, suppose $u_{1},u_{2}$ are two functions satisfying this requirement. Then by Shwzartz reflection principle, $u=u_{1}-u_{2}$ can be extended to a bounded harmonic function on $\mathbb{C}\backslash\{0\}$, that is zero on the real axis. I want to prove that this means $u$ is always zero. Now, I know that if $f:\mathbb{C}\backslash\{0\}\to\mathbb{C}$ is holomorphic and bounded then $f$ is constant, but I can't seem to construct a bounded holomorphic function from this $u$, since $\partial_{z}u$ need not be bounded, even if $u$ is (I think). Is it possible to have a bounded harmonic function $u:\mathbb{C}\backslash\{0\}\to\mathbb{C}$ which is not constant?

Thank you!

p.s I couldn't use liouville argument here, because $u(x)$ is not the avarage of $u$ on large disks around $x$, since these disks will contain $0$.

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  • At an isolated singularity a bounded harmonic functions extends (isolated singularities can be logarithmic too but that is still unbounded so near an isolated singularity $u=c\log |z-z_0| + \Re f$ with holomorphic $f$ ) – Conrad Jan 08 '22 at 15:03
  • Check this: https://math.stackexchange.com/q/448443/42969, or this: https://math.stackexchange.com/q/3593085/42969, or this: https://math.stackexchange.com/q/459134/42969. – Martin R Jan 08 '22 at 15:04

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Answer - there are many proofs. Some one pointed out in a similar post that $h(e^{z})$ would then also be harmonic and bounded, but now it is defined on $\mathbb{C}$ and thus by Liouvill's theorem it is constant.

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