Harmonic function bounded below also bounded above

[Grothard]

I am trying to prove that a nonnegative (so bounded below) harmonic function is constant. I know that Liouville's Theorem states that if a function is bounded both above and below then it is constant.
This means it would be sufficient to prove that a harmonic function bounded below is always bounded above. How can one show that?

[snipez90]

Consider the harmonic conjugate of your harmonic function u, which exists assuming u is harmonic on R^2. Then you have an analytic function and it boils down to complex analytic methods. Giving away this next part is giving away the entire problem, but if you've seen enough applications of Liouville you'll know what to do next.

[Grothard]

I'm still a bit confused. I now have an analytic function whose real part is bounded below, but Liouville requires it to be bounded from both sides. I tried playing around with Cauchy-Riemann, but didn't get anywhere. Could I get some more help?

[snipez90]

The idea is that if you have two entire functions g and f, then we can try to apply Liouville to g \circ f, which is also entire. If this last function is bounded, hence constant, we try to deduce that f is constant. Here we take f to be the analytic function whose real part is bounded below. We need to find the g just described. There is a well known function g with the property that |g(z)| is simply g(Re(z)).

[Grothard]

Here is the solution I came up with. I'm not quite sure if it's valid, though.

Let x(z) >= 0 be the given harmonic function and let y(z) be its harmonic conjugate. Then f(z) = x(z) + iy(z) is analytic and entire.
Let g(z) = 2/(z + conjugate(z) + 2). Then g(f(z)) = 1/(x(z) + 1). Since x(z) >= 0, |g(f(z))| <= 1, so g(f(z)) = k.
This means 1/(x(z) + 1) = k, so x(z) = (1 - k)/k.
This proves that x(z) is a constant and we are done.

The problem I have here is that I don't think g(z) is entire or analytic, so Liouville doesn't necessarily hold... or does it? How could I fix it?

[snipez90]

Right, g is not entire since it has a singularity at z = -1. You need your functions to be entire to apply Liouville. When you think of a specific example of an entire function, the exponential function should come to mind. Furthermore, the modulus of the exponential is the exponential of the real part.