July 7th – 9th: Application of Removable Singularity Theorem. Liouville’s Theorem in Euclidean space. Application of barrier method.

Jul 7—2025

Proof of Liouville’s Theorem using Removable Singularities Theorem

Let f(z) = u(z) + iv(z) be bounded and holomorphic, i.e. there exists $M \geq 0$ such that $|f(z)| \leq M$ for all $z \in \mathbb{C}$ .

Then, $|f(z)|^2 \leq M^2$

$\Leftrightarrow u(z)^2 + v^2(z) \leq M^2$

and so, $u^2, v^2 \leq M^2$ . u and v are bounded harmonic functions on the plane.

Key step: we show that u and v are bounded harmonic functions on the Riemann sphere (i.e. the plane with one point added). Since the Riemann sphere is a compact set, and by the maximum value property, u and v cannot achieve maximum value on the surface, it follows that u and v must be constants.

This key step follows from the removable singularity theorem. The Riemann sphere is described by two charts z and w such that $w = \frac{1}{z}$ . The point at infinity for z is given by w = 0. From the above $u(z)$ is a function defined on the z chart. Suppose we look at $u(z)$ on the w chart. Then, it is defined everywhere (and bounded and harmonic) on this chart except at w = 0. It follows that we can continuously extend u to w = 0 and this resulting new function is harmonic.

Claim: if a harmonic function defined on the plane is bounded on one side, then it is a constant function.

This is quite an interesting claim as it tells us that a harmonic function cannot just approach infinity in one direction, it must approach it in both the negative and positive directions.

For instance, take the real and imaginary parts of $\frac{1}{z}$ : $\frac{x}{x^2 + y^2}$ , $\frac{-y}{x^2 + y^2}$ .

For $\frac{x}{x^2 + y^2}$ , when we approach the origin along the positive x-axis, the function tends towards positive infinity. When we approach the origin along the negative x-axis, the function tends towards negative infinity. Similar things could be said for $\frac{-y}{x^2 + y^2}$ .

We now verify the above claim. Suppose there exists $M \in \mathbb{R}$ such that $u(z) \leq M$ . Then, there exists a holomorphic function f such that $f = u + iv$ .

Consider $g(z) = e^{f}$ , which is a holomorphic function.

$|g(z)| = |e^{f}| \leq |e^{u+iv}| \leq e^u \leq e^{M}$ .

Since $g(z)$ is bounded and holomorphic, it follows that $g(z) = C$ , a constant. Hence, $e^u = |C|$ , and so $u$ is a constant.

Generalization: We now prove Liouville’s in $\mathbb{R}^n$ (this is my understanding of the proof by E. Nelson; I find it very short and elegant)

Suppose u to be a bounded harmonic function in $\mathbb{R}^n$ . We apply the mean value property of harmonic functions, i.e. for any point, its value is equal to the average of the values over an arbitrary sphere with the point as its center (by taking limit, this value would also be equal to the average over the corresponding ball; since on each ‘layer’ inside the ball, the average is the same). For any two points, say $P_1, P_2$ , we construct two spheres of equal radius (why?) with centers at the two respective points. For radius large enough, the two balls coincide except for a small proportion of their volumes (this is the most important part of the observation. Why does this make sense? It’s a generalization of the overlapping of two circles on the plane. May need to make this part more rigorous. (*)) Now, because the function is bounded, the averages of the function over the two balls can get arbitrarily close. This is because the proportion of the non-overlapping part goes to 0 (and f is bounded over the non-overlapping part). Hence, the value of the function at those two points are the same.

Miscellaneous

July 7th – 9th: Application of Removable Singularity Theorem. Liouville’s Theorem in Euclidean space. Application of barrier method.

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