# Existence of a limit implies that a function can be harmonic extended

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• MathLearner123
MathLearner123
##\textbf{Theorem}##
If ##u: \mathbb{D'} = \mathbb{D} \setminus \{0\} \to \mathbb{R}## is harmonic and bounded, then ##u## extends to a function harmonic in ##\mathbb{D}##.

In the next proof ##\Pi^+## is the upper half-plane.

##\textbf{Proof}##: Define a function ##U: \Pi^{+} \rightarrow \mathbb{R}## by ##U(z)=u\left(e^{2 \pi i z}\right)##. Then ##U## is harmonic, being the composition of a harmonic function and a holomorphic function. Since ##\Pi^{+}## is simply connected, there exists ##F \in H\left(\Pi^{+}\right)## such that ##U=\operatorname{R e}(F)##. Now, although ##U(z+1)=U(z)## we need not have ##F(z+1)=F(z)##. But ##\operatorname{R e}(F(z+1)-F(z))=U(z+1)-U(z)=0## and so ##F(z+1)-F(z)## must be an imaginary constant: There exists ##c \in \mathbb{R}## such that ##F(z+1)-F(z)= ic## for all ##z \in \Pi^{+}##. Choose a real number ##\alpha \neq 0## so that ##c \alpha## is a multiple of ##2 \pi## (you can take ##\alpha=2 \pi / c## unless ##c=0## ). Let
$$E(z)=e^{\alpha F(z)}$$
It follows that ##E(z+1)=E(z)##, and so there exists ##f \in H\left(\mathbb{D}^{\prime}\right)## with
$$E(z)=f\left(e^{2 \pi i z}\right) .$$
Now the fact that ##u## is bounded shows that ##f## is bounded, so that ##f## has a removable singularity at the origin. The fact that ##u## is bounded also shows that ##f## is bounded away from ##0## , so in particular ##f(0) \neq 0##. Hence ##u## has a limit at ##0## , since ##u=\log (|f|) / \alpha##. (Since ##f\left(e^{2 \pi i z}\right)=e^{\alpha F(z)}## it follows that
$$\left.\log \left(\left|f\left(e^{2 \pi i z}\right)\right|\right)=\alpha \operatorname{R e}(F(z))=\alpha U(z)=\alpha u\left(e^{2 \pi i z}\right) .\right)$$

##\textbf{Question}##
Why the fact that ##u## has limit at ##0## implies that ##u## can be extended to a harmonic function ##u : \mathbb{D} \to \mathbb{R}##?

Last edited by a moderator:
It's an annoying quirk that you need double # instead of single \$ to make in line latex, if you could edit your post to find and replace all instances it would be easier to read

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