Thread: A positive harmonic function on C is constant.

Hello,

I would love to see direction on a proof of the following claim:

If u:C->R is a harmonic function and u(z) >= 0 for all z, then u is constant.

I am studying for a qualifying exam in single variable complex analysis. I can prove this claim using the mean value property, but would like to avoid this fact as this problem is presented in the theory before the mean value integral is taught.

I was developing the following line of reasoning to no avail: since the domain is simply connected, we are furnished with a harmonic conjugate v, and consequently an entire function f = u + iv. We may bear upon this the theorem of Liouville, but I cannot determine a bound for f given the hypotheses. Nor am I certain this is even a viable logical thread pursuant to my goal.

Random fact given that u >= 0:
f maps the whole plane into the right half plane, and hence |f|>0 (to avoid defiance of the open mapping theorem). Seeking a bounded entire function for the application of Liouville, this fact enables a reasonable g = 1/f function should it be required.

Insights appreciated.

It is curious when one will finally achieve a moment of insight.

Only after submitting my post a moment ago and reviewing it for typographical error, did the solution come to mind.

I have established that f maps C to the right half plane. I can now map the right half plane to the unit disk by a Mobius transformation, say g(z) = (1-z)/(1+z). The composition map gf is entire, and bounded by 1, so gf is constant by Liouville's Theorem. Supposing gf(z) = C,

[1 - f(z)] / [1 + f(z)] = C implies f(z) = (1-c)/(1+c), for which u being constant is requisite.

I am glad I was able to do this myself, but thank you anyway.

(As an interesting corollary, there must be no entire function reducing to a half plane)