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There is another post with this exact same prompt which got several down-votes for not showing their work. So I'll show what work I've got. I know being a harmonic function implies satisfying the Mean Value Property, thus what I thought I'd do is consider two arbitrary points in $\mathbb{C}$ and prove that:

$$|u(z) - u(w)|=|\frac{1}{2\pi}\int_0^{2\pi}u(z - re^{i\theta})d\theta - \frac{1}{2\pi}\int_0^{2\pi}u(w - te^{i\theta})d\theta| $$ $$= |\frac{1}{2\pi}\int_0^{2\pi}u(z - re^{i\theta}) - u(w - te^{i\theta})d\theta|$$

Now what I wan't to do is take r and t to infinity and prove this integral goes to zero, but I can't figure out how to do that. I've used the fact that my harmonic function u is defined on all of $\mathbb{C}$ in taking t and r to infinity but I still need to use the fact that it's bounded.

I also read through the proof of Liouville's Theorem, given as an answer in that other post, which seems intuitively correct but rather un-rigorous. It actually says that the value in the center of a ball is equal to the average over the ball's volume. I imagine this can be proven from the average over the boundary since the average over the boundary will not change as you continuously shrink the boundary down to its center. This however would require the difference of two surface integrals, or is there a simpler way?

I was also considering using the Maximum Modulus Principle somehow but I haven't found a way yet. I've probably made this way more complicated than it has to be, hopefully someone can help me, thanks! =].

Edit: Ok after thinking some more about the accepted answer in that other post, I've decided what I need to do is look at the canonical holomorphic function whose real part is my harmonic function u, and prove that holomorphic function is bounded. From there I can prove that since it's entire and bounded that it's constant, and from there prove that then its real part must be constant. So could someone talk a bit about the construction of this canonical holomorphic function?

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You want $f= u + iv $ to be complex analytic, and $u$ being your harmonic function. Construct $v$ such that the Cauchy-Riemann equations are satisfied. –  Ragib Zaman Apr 21 '12 at 4:54
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2 Answers

up vote 3 down vote accepted

I'll expand on Nelson's proof a bit.

You know the Mean Value Property that the value of a harmonic function $u$ at $z$ equals the average of its values on any circle centered at $z$. We can turn this into a fact about averages over balls (which in this case are disks) just by using polar coordinates: $$\int_{B(z,R)} u(w)\,dw = \int_0^R \int_0^{2\pi} u(z + r e^{i \theta}) r \,d\theta\,dr = \int_0^R 2 \pi u(z) r\,dr = u(z) \pi R^2.$$ Dividing both sides by $\pi R^2$, we see that $u(z)$ equals the average value of $u$ over the ball $B(z,R)$.

Now fix $z,w \in \mathbb{C}$, and consider disks $B(z,R)$, $B(w,R)$ centered at each of them. Let $L = B(z,R) \cap B(w,R)$ be the lens-shaped region lying in both disks. Let $L_z = B(z,r) \backslash L$ be the region in the $z$ disk but not the $w$ disk. (Draw a picture.) One can work out the area $A(L)$ of $L$ by a bit of multivariable calculus (or look it up on Mathworld). A bit more calculus shows that the ratio $\frac{A(L)}{\pi R^2} \to 1$ as $R \to \infty$, so that the lens occupies "most" of the disk. Since $A(L) + A(L_z) = \pi R^2$, we must therefore have $\frac{A(L_z)}{\pi R^2} \to 0$.

Now by the mean value property, for any $R$ we have $$u(z) = \frac{1}{\pi R^2} \int_{B(z,R)} u(\zeta)\,d\zeta = \frac{1}{\pi R^2} \int_L u(\zeta)\,d\zeta + \frac{1}{\pi R^2} \int_{L_z} u(\zeta)\,d\zeta.$$ For the second term, if $|u|$ is bounded by some constant $M$, we have as $R \to \infty$, $$\left|\frac{1}{\pi R^2} \int_{L_z} u(\zeta)\,d\zeta \right| \le \frac{M A(L_z)}{\pi R^2} \to 0.$$ Thus $$u(z) = \lim_{R \to \infty} \frac{1}{\pi R^2} \int_L u(\zeta)\,d\zeta.$$ But if we replace $z$ with $w$ we can make exactly the same argument (replacing $L_z$ by $L_w$ whose area is the same, by symmetry). So we also have $$u(w) = \lim_{R \to \infty} \frac{1}{\pi R^2} \int_L u(\zeta)\,d\zeta.$$ Thus $u(z) = u(w)$.

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You could consider Cauchy's integral formula used in this way: since $f$ is entire it can be written $f(z)=f(0)+f'(0)z+...$, this being valid for all complex $z$. Then Cauchy's formula gives $$f^{(k)}(0)=\frac{k!}{2\pi i}\int_{|z|=r} \frac{f(z)}{z^{k+1}}dz$$ for all $z$. Passing to the modulus you get the estimate $$\frac{|f^{(k)}(0)|}{k!}\leq \frac{\max_{|z|=r}|f(z)|}{r^{k}} \leq \frac{C}{r^{k}}$$ (C is the constant that limits the modulus of the function $f$). Letting r tend to infinity you find that all the coefficients of the above series are zero except for the term $f(0)$, so $f$ is constant.

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Are harmonic functions automatically analytic in $\mathbb{C}$? I was under the impression that they only necessarily have continuous partial derivatives satisfying the Laplacian. –  heat death Apr 20 '12 at 21:51
    
They aren't but you can always make an analytic function out of them, see the accepted answer in the thread you posted. –  Ragib Zaman Apr 21 '12 at 4:52
    
but how do we know that f is bounded? Only its real part is necessarily bounded. –  heat death Apr 22 '12 at 17:24
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