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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?

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You could just apply Liouville's theorem for harmonic functions, but it really is the same thing. –  tomasz Jun 19 '12 at 0:13
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Look at Nelson's proof of Liouville's theorem here: en.wikipedia.org/wiki/Harmonic_function –  Byron Schmuland Jun 19 '12 at 0:13
    
You can prove by using properties of harmonic functions, but I guess your method is the one you should use on the exam since it is complex analysis. –  timur Jun 19 '12 at 0:36
    
I previously thought Nelson's proof only worked for uniformly bounded functions, but now I see it works for the present case where $u$ is only bounded below. This essentially yields a Harnack-type inequality as in @bartgol's answer. –  Erick Wong Jun 19 '12 at 3:54

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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.

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