I am trying to prove the following conjecture (or find a counterexample):
Claim: Let $f(z)$ and $g(z)$ be holomorphic functions defined on a simply connected bounded domain $\Omega \subset \mathbb C$ in the complex plane.
We know that at any point in the domain the following inequality holds
$$ |f(z)| > |g(z)|, $$
then the function $\zeta(z) = \frac{1}{|f(z)|-|g(z)|}$ is subharmonic in $\Omega$.
Some properties that (might) be useful:
1) From the above inequality we also know that $f$ does not vanish in the domain, i.e. $\forall z\in \Omega$, $f(z)\neq 0$.
2) $log|f(z)|$ and $log|g(z)|$ are subharmonic and since $f$ does not vanish in the domain $log|f(z)|$ is also harmonic.
3) The function $\zeta(z)$ is strictly positive.
4) The modulus of a holomorphic function is subharmonic. In particular, $|f(z)|$ and $|g(z)|$ are subharmonic.
5) Linear combinations of subharmonic functions with positive weights is subharmonic. In particular the sum of two subharmonic functions is subharmonic (but the difference of two subharmonic functions is not in general subharmonic).
Best Answer
It is correct. First, note that $$\frac{1}{|f(z)|-|g(z)|}=\frac{1}{|f(z)|}\frac{1}{1-\frac{|g(z)|}{|f(z)|}}=\frac{1}{|f(z)|}\sum_{k=0}^{\infty}\frac{|g(z)|^k}{|f(z)|^k}=\sum_{k=0}^{\infty}\frac{|g(z)|^k}{|f(z)|^{k+1}}.$$ For any $k\in\mathbb N$, since $f$ does not vanish and $\Omega$ is simply connected, there exists a holomorphic branch $F_k(z)=f^{-\frac{k+1}{k}}(z)$. Then, write $$\frac{|g(z)|^k}{|f(z)|^{k+1}}=|g(z)F_k(z)|^k,$$ and since $gF_k$ is holomorphic, we obtain that the summands are subharmonic functions.
Edit: To show subharmonicity of the sum, consider a ball $B$ in the domain and let $u$ be a harmonic function in $B$, with $\frac{1}{|f|-|g|}\leq u$ on $\partial B$. Set $$v_n(z)=\sum_{k=0}^n\frac{|g(z)|^k}{|f(z)|^{k+1}},$$ then $v_n\to\frac{1}{|f|-|g|}$ pointwise, as $n\to\infty$, and $v_n\leq\frac{1}{|f|-|g|}$ for all $n$. Therefore, $v_n\leq u$ on $\partial B$, and since $v_n$ is subharmonic, we obtain that $v_n\leq u$ in $B$. By letting $n\to\infty$ we get that $\frac{1}{|f|-|g|}\leq u$, therefore $\frac{1}{|f|-|g|}$ is subharmonic in $B$.