# Confusion about Liouville's theorem for Harmonic functions

The excercises in my textbook ask me to prove Liouville's Theorem for harmonic functions. Their formulations are

Lioville's Theorem: Every bounded, harmonic function on the whole complex plane must be constant

and

A real valued function on a domain $$D$$ of the complex plane is Harmonic if $$u(x,y)\in C^2$$ and $$\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = 0$$ for all $$z=x+iy \in D$$

Although I don't even know how to interpret this, because harmonic functions are real valued. But the theorem says "harmonic on the whole complex plane"...

By the way, I should point out that in my book thus far, I have only learned up to Cauchy's integral formula. I have not learned residues or any more complicated integration methods. Thanks!

• Do you know about the existence of harmonic conjugates? – cmk May 27 at 15:39
• Do you want a proof? – Chinnapparaj R May 27 at 15:39
• @cmk I dont know the existence of conjugates – NazimJ May 27 at 15:45
• @ChinnapparajR A proof would be good as long as it clarifies the definitions at the same time :) – NazimJ May 27 at 15:48
• @NazimJ In response to your first paragraph after the yellow space: There is no problem there; "harmonic on the complex plane" has to do with the domain, while "real valued" has to do with the range. I think clasically, harmonic functions are from $\mathbb{R}^2 \to \mathbb{R}$, but here it seems they are looking at them as $\mathbb{C} \to \mathbb{R}$. – Ovi May 27 at 17:10

There are quite a few ways to do this, but I'll give a sketch of one. Recall the mean-value property

$$u(z_0)=\frac{1}{\pi r^2}\int\limits_{D_r(z_0)} u(z)dydx=\int\limits_{0}^{2\pi} \int\limits_0^r u( z_0+se^{i\theta})s ds d\theta.$$ Now, for any $$p,q\in \mathbb{C},$$ we have that $$|u(p)-u(q)|\leq \frac{1}{\pi r^2}\iint\limits_{\Delta (p,q,r)}|u(z)|dxdy,$$ where $$\Delta(p,q,r)=(D_r(p)\setminus D_r(q))\cup (D_r(q)\setminus D_r(p))$$ is the symmetric difference. Now, if $$d=|p-q|,$$ the $$\Delta (p,q,r)\subseteq D_{r+d}(p)\setminus D_{r-d}(p),$$ which implies that $$A(\Delta (p,q,r))\leq \pi ((r+d)^2-(r-d)^2)=4\pi dr.$$ So, if $$u$$ is bounded by $$M$$ over $$\mathbb{C},$$ then $$|u(p)-u(q)|\leq\frac{4Md}{r}$$ for all $$r>0$$. Taking the limit as $$r\rightarrow\infty$$ yields the result.

This proof comes from Taylor's notes on complex analysis. For alternative proof, still without harmonic conjugates, you can reference Evans, pages 29-30.

Cauchy's Integral Formula states that, for f(z) a holomorphic function (except possibly at finitely many points)

$$f(w) = \int_{\gamma} \frac{f(z)}{z-w} dz$$ for $$\gamma$$ a circular contour within which $$w$$ resides.

Now assume f is bounded, say $$|f(z)|\leq M$$. Then we, for a sufficiently large radius of contour

$$|f(w_1)-f(w_2)| \\= | \int_{\gamma} \frac{f(z)}{z-w_1} - \frac{f(z)}{z-w_2} dz | \\= | \int_{\gamma} \frac{(w_1-w_2)f(z)}{(z-w_1)(z-w_2)} dz| \\ \leq \text{sup}_{\gamma} |\frac{(w_1-w_2)f(z)}{(z-w_1)(z-w_2)}| \text{length}(\gamma) \\ \leq \frac{M2\pi R |w_1-w_2|}{(R-|w_1|)(R-|w_2|)}$$

Now taking R to infinity gives the desired result, for holomorphic functions. I'll leave it to you to show any harmonic function may be written as a holomorphic one.