Proving continuous function $ u_{0}:\mathbb{R}\to\mathbb{R} $ can be extended to harmonic function Assume $ u_{0}:\mathbb{R}\to\mathbb{R} $ is continuous and such that $ \lim_{t\to-\infty}u_{0}\left(t\right)=\lim_{t\to\infty}u\left(t\right)=L $.
Prove that there exists unique $ u:\left\{ Im\left(z\right)\geq0\right\} \to\mathbb{R} $ such that $ u\upharpoonright_{\left\{ Im\left(z\right)>0\right\} } $ is harmonic, and $ u\upharpoonright_{\mathbb{R}}=u_{0} $.
Now, I do know that given $ v_{0}:C\left(0,1\right)\to\mathbb{R} $, there exists unique $ v:\overline{D\left(0,1\right)}\to\mathbb{R} $ such that $ v\upharpoonright_{D\left(0,1\right)} $ is harmonic, and $ v\upharpoonright_{C\left(0,1\right)}=v_{0} $, and $ v $ is given by $$ v\left(z\right)=\frac{1}{2\pi}\intop_{0}^{2\pi}v\left(e^{i\theta}\right)\text{Re}\left[\frac{e^{i\theta}-z}{e^{i\theta}+z}\right]d\theta $$
So I want to use it in order to prove my original question.
Here are my thoughts:
Note that given $ v $ as described above, we can use the biholomorphism $ g:\left\{ Im\left(z\right)\geq0\right\} \to\overline{D\left(0,1\right)} $ given by $ g\left(z\right)=\frac{z-i}{z+i} $.
This biholomorphism maps the real axis into the unit circle, if we consider $ h\left(z\right)=v\circ g:\left\{ Im\left(z\right)\geq0\right\} \to\mathbb{R} $
I think that we can prove that it would also be harmonic and satisfy the details we want, but even if it would, Im not sure how to connect it to a function from $\mathbb{R} \to \mathbb{R}$.
Thanks in advance.
 A: $g(z)=\frac{z-i}{z+i}$ maps the upper half-plane onto the unit disk $\Bbb D$, and $\Bbb R \cup \{ \infty \}$ to its boundary $\partial \Bbb D$, with $g(\infty) = 1$.
Therefore we can define $v_0: \partial \Bbb D \to \Bbb R$ as $v_0(\zeta) = u_0(g^{-1}(\zeta))$ for $\zeta \ne 1$, and $v_0(1) = L$. The condition $\lim_{t\to-\infty}u_{0}(t)=\lim_{t\to\infty}u_0(t)=L$ ensures that $v_0$ is continuous.
Let $v$ be the (unique) continuous function in $\overline{\Bbb D}$ which is harmonic in $\Bbb D$ and equal to $v$ on the boundary. Then $u = v \circ g$ is harmonic in the upper half-plane and has the boundary values $u_0$.
The solution can also be expressed using the Poisson kernel on the upper half-plane as
$$
 u(x+iy) = \frac 1\pi \int_{-\infty}^{\infty} \frac{y}{(x-t)^2 + y^2} u_0(t)\, dt \, .
$$
A: Let me add to Martin Rs answer: Your $g$ is actually a bit better. It is a Moebius transform, hence it preserves harmonicity, see e.g. https://www.uni-ulm.de/fileadmin/website_uni_ulm/mawi.inst.010/DallAcqua/DaSwChile.PDF Corollary 2, for a more general statement.
On the other hand, I do not think, your solution is unique without assuming any kind of growth conditions. Take $u_0(x)=0$. Then $u(x,y)=0$ is harmonic and satisfies $u(x,0)=u_0(x)=0$. On the other hand $v(x,y):=y$ is also harmonic, since $\Delta v=\partial_x^2v +\partial_y^2v=0$, and $v(x,0)=0=u_0(x)$. Hence you have two different solutions.
