I encountered the following 2-part problem on a practice exam:
(a) Show that if $f:\Bbb C\to\Bbb C$ is entire and the real part of $f$ is always positive, then $f$ is constant. (b) Show that if $u:\Bbb R^2\to\Bbb R$ is a harmonic function with $u(x,y)>0$ for all $x,y\in\Bbb R$, then $u$ is constant.
Now, (a) was fairly simple. Putting $f=u+iv$, I took $g(z)=e^{-iz}$, so $|g(f(z))|=e^{-u(x,y)}<e^0=1$, so $g\circ f$ is a bounded entire function, so constant by Liouville's Theorem, from which it is readily seen that $f$ is also constant.
For (b), I wasn't certain what to do. According to the wikipedia article on harmonic conjugates, if the domain of a harmonic function is simply connected, then it admits a harmonic conjugate, and so (b) follows from (a), since the plane is of course simply connected. I had never seen this result before, so (obviously) didn't think to use it.
My question is this: Aside from proving that $u$ has a harmonic conjugate, I wonder if there are other ways that we can approach a proof of (b). My experience with harmonic analysis has been almost completely in the context of analytic functions. Any ideas?
Liouville's theorem holds also in the real case. (One of) the proof(s) is based on another result, known as Harnack's inequality, which in its general form reads
$$\Delta u=0\text{ in }\Omega \Rightarrow\underset{\Omega}{\sup}u\leq C \underset{\Omega}{\inf}u$$ where $C$ depends only on the domain. For the case $\Omega=\mathcal{B}(\underline{0},R)$ and $u$ non-negative we can use Poisson's formula for the ball and write a more useful version of the Harnack's inequality, namely
$$\frac{R^{n-2}(R-|\underline{x}|)}{(R+|\underline{x}|)^{n-1}}u(\underline{0})\leq u(\underline{x})\leq \frac{R^{n-2}(R+|\underline{x}|)}{(R-|\underline{x}|)^{n-1}}u(\underline{0})\tag{1}$$
Now suppose $\Delta u=0,\ u\geq M,\ \forall \underline{x}\in\mathbb{R}^n$. Then $w:=u-M$ is non-negative and we can use Harnack's inequality in $\mathcal{B}(\underline{0},R)$, with $R$ arbitrary. If in $(1)$ we take the limit as $R$ goes to infinity, we obtain
$$w(\underline{0})\leq w(\underline{x})\leq w(\underline{0})$$ that is, $w$ is constant.
Alternatively, if you don't want to invoke the Harnack's inequality, you can prove the Liouville's theorem applying the following result about harmonic functions:
If $u$ is harmonic in $\Omega$ and $\mathcal{B}(\underline{x},R)\subset\subset\Omega$ (meaning that the closure of $\mathcal{B}(\underline{x},R)$ is contained in $\Omega$), then
$$|u_{x_j}(\underline{x})|\leq \frac{n}{R}\underset{\partial \mathcal{B}(\underline{x},R)}{\max}|u|$$
This result can be actually generalized to derivatives of any order (the constant in front of the max will change depending on the order of differentiation).
Now, if $u$ is harmonic on $\mathbb{R}^n$, then the above result holds for every $R>0$ and every $\underline{x}$. Taking the limit as $R$ goes to infinity, you get the Liouville's theorem.