Is a Positive Subharmonic Function on R^2 Always Constant?

[Combinatorics]

Homework Statement

Given a positive subharmonic function [itex] u[/itex] , defined on [itex]R^2 [/itex], how can I prove that [itex] u[ /itex] must be constant?

Homework Equations

[itex] \Delta u \leq 0 [/itex] is the definition of subharmonic function !

The Attempt at a Solution

I've tried solving this by using a new function [itex] v_\epsilon = u+ \epsilon log(\sqrt{x^2+y^2 } ) [/itex] , and then messing out with this [itex] /epsilon [/itex], but without any success.


Hope you'll be able to help !

Thanks !

 

[mighty2000]

Thank you for your question. Proving that a positive subharmonic function must be constant is a well-known result in harmonic analysis. Here is one approach you can take to prove this statement:

1. Recall that a function u:R^2 \to R is subharmonic if it satisfies the inequality \Delta u \leq 0 , where \Delta is the Laplace operator. This means that the function u is "less than harmonic", or in other words, it has a tendency to be "less curved" than a harmonic function.

2. Consider the function v = e^u . Note that since u is positive, v is also positive. Moreover, we have \Delta v = e^u \Delta u \leq 0 , since \Delta u \leq 0 by definition. This means that v is also subharmonic.

3. Now, let's assume that u is not constant. This means that there exist two points x_1 and x_2 in R^2 such that u(x_1) < u(x_2) . Without loss of generality, we can assume that u(x_1) = 0 and u(x_2) = 1 .

4. Consider the line segment L connecting x_1 and x_2 . Since u is continuous, there exists a point x_0 on this line segment where u(x_0) = 1/2 .

5. Now, consider the circle C centered at x_0 with radius r = |x_1 - x_0| . Since u is subharmonic, we have u(x_0) \leq \frac{1}{2\pi r}\iint_C u(x) dS = \frac{1}{2\pi r}\int_0^{2\pi} u(x_0 + r(\cos t, \sin t)) r dt .

6. Since u is positive and u(x_0) = 1/2 , we have that the integral on the right-hand side is strictly greater than 1/2 . This contradicts the fact that u(x_0