A positive harmonic function on the punctured plane is constant

Question

Let $f(z)$ be a positive harmonic function on $\mathbb{C}\backslash \{0\}$. Prove that $f(z)$ is constant.

I have no idea to prove this statement.

Answer 1

Add a constant so it is bounded away from zero. It is still harmonic. But now the reciprocal is bounded. ... and sub-harmonic.

Answer 2

In RGB's answer, a justification is asked as to why the reciprocal of a harmonic function is subharmonic. This follows from the following :

Proposition

Let $u$ be a real harmonic function on some open set $\Omega$, and let $\phi$ be convex on $\mathbb{R}$. Then $\phi \circ u$ is subharmonic on $\Omega$.

proof: If $\overline{\mathbb{D}(a,r)} \subset \Omega$, then $$\phi(u(a)) = \phi \left( \frac{1}{2\pi}\int_{0}^{2\pi} u(a+re^{i\theta})d\theta \right) \leq \frac{1}{2\pi}\int_{0}^{2\pi} \phi(u(a+re^{i\theta}))$$ by harmonicity of $u$ and Jensen's inequality.