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?

  • $\begingroup$ You could just apply Liouville's theorem for harmonic functions, but it really is the same thing. $\endgroup$ – tomasz Jun 19 '12 at 0:13
  • 1
    $\begingroup$ Look at Nelson's proof of Liouville's theorem here: en.wikipedia.org/wiki/Harmonic_function $\endgroup$ – user940 Jun 19 '12 at 0:13
  • $\begingroup$ 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. $\endgroup$ – timur Jun 19 '12 at 0:36
  • $\begingroup$ 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. $\endgroup$ – Erick Wong Jun 19 '12 at 3:54

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.