Invariance of subharmonicity under a conformal map

I want to prove the following (exercise from Ahlfors’ text):

Prove that a subharmonic function remains subharmonic if the independent variable is subjected to a conformal mapping.

Here is my attempt, please tell me if it’s correct.

Let $v:\Omega \subset \mathbb C \to \mathbb R$ be a subharmonic function, and let $f: \Omega \to \Omega$ be some conformal map. Suppose that $v \circ f$ is not subharmonic in $\Omega$. Then, there exist a point $z_0 \in \Omega$, a harmonic function $u_0: \Omega_0 \subset \Omega \to \mathbb R$ where $z_0 \in \Omega_0$, such that $v \circ f-u_0$ takes its maximal value in $\overline{\Omega_0}$ at $z_0$ without reducing to a constant. Since $f'(z_0) \neq 0$ we may restrict the domain $\Omega_0$ into some disk $\Delta_0=\Delta_0(z_0,r)$ such that the restricted $f$ results in a biholomorphism.

For $z \in \Delta_0$ we may write $$v \circ f-u_0=v \circ f-u_0 \circ f^{-1} \circ f, $$
and since $\Delta_0$ is simply connected, $u_0$ is the real part of some holomorphic function $f_0$, and thus $u:=u_0 \circ f^{-1}= \Re (f_0 \circ f^{-1})$ is harmonic as well.

Plugging in $f(z)=w$ we see that $$v(w)-u(w) $$ has its maximal value in $\overline{f(\Delta_0)}$ at the (interior) point $w_0=f(z_0)$ without reducing to a constant.

This is in contradiction with the subharmonicity of $v$.

QED (?)

Solutions Collecting From Web of "Invariance of subharmonicity under a conformal map"

Your proof is basically correct. Once you have the characterization of subharmonic functions in terms of harmonic, it follows quite abstractly that subharmonicity is preserved by whatever transformations preserve harmonicity.

That said, I don’t think that Ahlfors had in mind only conformal automorphisms of $\Omega$. In general $f$ can be a conformal map between different domains, and you should account for that in your proof.

It’s worth pointing out that the composition $v\circ f$ is subharmonic as long as $v$ is subharmonic and $f$ is holomorphic (not necessarily invertible). The proof goes along the same lines, because $h\circ f$ is harmonic whenever $h$ is. Alternatively, one can reduce to smooth $v$ by approximation and compute the Laplacian of composition explicitly.