Bounded harmonic function on upper half plane vanishing at real axis is constant 
I want to show that if $u(x,y)$ is bounded and harmonic on $\mathbb{H}$ and limits to zero as we approach any point on the real axis, then $u\equiv 0$.

I think one way to do this is to compose $u$ with an automorphism from $\mathbb{D} \to \mathbb{H}$ and use the maximum modulus principle for harmonic functions.
However, I was wondering if there is an easy way to do this using only the maximum modulus principle for holomorphic functions. For example, since $\mathbb{H}$ is simply connected, we know there is a function $f(z)$ holomorphic on $\mathbb{H}$ having real part $u$. How might we proceed from here?
 A: You need some growth condition. Consider $u(x,y)=y$.
A: You can extend $u$ to the lower half plane by Schwarz reflection, thus you get a bounded harmonic function in the whole space. Therefore by Liouville principle $u$ is constant.  
A: First, we prove that the real harmonic function $u(x,y)$ has a harmonic continuation $U(z)$:
Let \begin{align}
U(z)=\begin{cases}u(z),& z\in\mathbb H,\\ 0, & z\in\mathbb R,\\ -u(\overline{z}), & z\in -\mathbb H.\end{cases}
\end{align}
where $-\mathbb H$ denotes the lower half-plane. We claim that $U(z)$ is harmonic in $\mathbb C$. Since $u(\overline{z})=u(x,-y)$, by direct computation we know that $u(\overline z)$ is harmonic. Now it suffices to show that $U(z)$ has the mean value property on the real line:
\begin{align}
\frac{1}{2\pi}\int_{-\pi}^\pi U(x_0+re^{i\theta})d\theta &=\frac{1}{2\pi}\int_{-\pi}^0 -u(x_0+re^{-i\theta})d\theta+\frac{1}{2\pi}\int_0^{\pi}u(x_0+re^{i\theta})d\theta\\
&=\frac{1}{2\pi}\int_\pi^{0}u(x_0+re^{i\theta})d\theta+\frac{1}{2\pi}\int_0^{\pi}u(x_0+re^{i\theta})d\theta\\
&=0\\
&=U(x_0)
\end{align}
So $U(z)$ is a bounded harmonic function on $\mathbb C$, thus a constant. Hence $u(z)$ is also a constant. Since $U(x_0)=0$, we have $u(x,y)\equiv 0$.
