TheoremIf $u: \mathbb{D'} = \mathbb{D} \setminus \{0\} \to \mathbb{R}$ is harmonic and bounded, then $u$ extends to a function harmonic in $\mathbb{D}$.
In the next proof $\Pi^+$ is the upper half-plane.
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$$H\left(\Pi^{+}\right)$such that $U=\mathbb{R e}(F)$. Now, although $U(z+1)=U(z)$ we need not have $F(z+1)=F(z)$. But $\mathbb{R e}(F(z+1)-F(z))=U(z+1)-U(z)=0$. and so $F(z+1)-F(z)$ must be an imaginary constant: There exists $c \in \mathbb{R}$ such that $F(z+1)-F(z)=$ ic for all $z \in \Pi^{+}$. Choose a real number $\alpha \neq 0$ so that $c \alpha$ is a multiple of $2 \pi$ (you can take $\alpha=2 \pi / c$ unless $c=0$ ). Let$$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 \mathbb{R e}(F(z))=\alpha U(z)=\alpha u\left(e^{2 \pi i z}\right) .\right)$$
I don't understand why we can't have $F(z) = F(z+1)$ and why from $E(z) = E(z+1)$ results that "there exists $f \in H(\mathbb{D'})$ with $E(z) = f(e^{2 \pi i z})$