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Nelson's proof of Liouville's theorem (in the case $n=2$) is as follows:

Consider a bounded harmonic function on Euclidean space. Since it is harmonic, its value at any point is its average over any sphere, and hence over any ball, with the point as center. Given two points, choose two balls with the given points as centers and of equal radius. If the radius is large enough, the two balls will coincide except for an arbitrarily small proportion of their volume. Since the function is bounded, the averages of it over the two balls are arbitrarily close, and so the function assumes the same value at any two points. Thus a bounded harmonic function on Euclidean space is a constant.

I tried to formalize it.

Let $f:\mathbb{C}\to\mathbb{C}$ be a holomorphic bounded function. If $z,w\in\mathbb{C}$ we have that \begin{align*} |f(z)-f(w)| &= \frac{1}{\pi r^2}\left|\int_{D(z,r)}f(x+iy)\:\mathrm{d}x\:\mathrm{d}y - \int_{D(w,r)}f(x+iy)\:\mathrm{d}x\:\mathrm{d}y\right|\\ &= \frac{1}{\pi r^2}\left|\int_{A}f(x+iy)\:\mathrm{d}x\:\mathrm{d}y - \int_{B}f(x+iy)\:\mathrm{d}x\:\mathrm{d}y\right| \\ &\leq \frac{2}{\pi r^2}(\sup |f|)\int_A 1\:\mathrm{d}x\:\mathrm{d}y \\ &= \frac{2}{\pi r^2}(\sup |f|) \left(2r^2\cos^{-1}\left(\frac{d}{2r}\right)-\frac{d}{2}\sqrt{4r^2-d^2}, \right) \end{align*}

where $d=|z-w|$. Observe what are $A$ and $B$ in the following drawing: I would expect that the right hand side tends to $0$ when $r\to\infty$. However, that is not the case. How should I formalize Nelson's proof?

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  • $\begingroup$ I get that $|A| = \mathcal O(r)$. $\endgroup$ – amsmath Sep 10 '18 at 19:56
  • $\begingroup$ Perhaps you could explain how you computed $\int_A 1\,dx\,dy$, because I agree with amsmath that the answer looks suspicious. $\endgroup$ – Nate Eldredge Sep 10 '18 at 20:03
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Your computation of the area of $A$ doesn't look correct.

It would be easier to bound the integral as follows: the width of $A$ is never more than $d$, and its $y$-coordinates are bounded between $-r$ and $r$. Thus $\int_A 1\,dx\,dy \le \int_{-r}^r d\,dy = 2rd$.

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  • $\begingroup$ I don't agree with this argumentation. Think of two parallel lines between the lines $y=-r$ and $y=r$ with "x-distance" $d$ whose slope is very small. Making the slope smaller and smaller, the area between the lines tends to $\infty$. $\endgroup$ – amsmath Sep 10 '18 at 20:21
  • $\begingroup$ @amsmath: No, it doesn't! The area of such a parallelogram is exactly $2rd$ regardless of the slope. $\endgroup$ – Nate Eldredge Sep 10 '18 at 20:22
  • $\begingroup$ Nate, you are right, of course. $\endgroup$ – amsmath Sep 10 '18 at 20:27
  • $\begingroup$ I upvoted for your answer because I think it's better. $\endgroup$ – amsmath Sep 10 '18 at 20:39
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Ok, let's calculate the volume $|A|$ of $A$. WLOG I let $z=0$ and $w=d > 0$. Then $$ \frac{|A|}2 = \int_{d/2}^{r+d}\sqrt{r^2-(x-d)^2}\,dx - \int_{d/2}^{r}\sqrt{r^2-x^2}\,dx = \int_{-d/2}^{d/2}\sqrt{r^2-x^2}\,dx. $$ Substituting $x = r\sin t$ gives $$ \frac{|A|}2 = r^2\int_{-\arcsin(d/2r)}^{\arcsin(d/2r)}\cos^2t\,dt = \frac{r^2}2\left[\cos t\sin t + t\right]_{-\arcsin(d/2r)}^{\arcsin(d/2r)}. $$ As $\cos(\arcsin(x)) = \sqrt{1-x^2}$, $$ \frac{|A|}2 = r^2\left(\frac{d}{2r}\sqrt{1-\frac{d^2}{4r^2}}+\arcsin\frac{d}{2r}\right)\,\le\,r^2\left(\frac d{2r} + \arcsin\frac d{2r}\right). $$ Now, for small positive $x$ we have $\arcsin x\le 2x$, hence $|A|\le 3dr$ for large $r$.

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