# Prove that |f|-|g| attains minimum on the boundary

I am trying to prove the following conjecture:

Let $f(z)$ and $g(z)$ be holomorphic functions defined on a simplify connected subregion $\Omega$ of the complex plane, where $\forall z\in \Omega$, $f(z)\neq 0$.

Prove that the function $|f|-|g|$ attains its minimum on the boundary of $\Omega$.

Some insights:

Since $f$ and $g$ are holomorphic, their modulus $|f|$ and $|g|$ are subharmonic and attains maximum on the boundary. $-|g|$ is superharmonic and attains minimum on the boundary. Since $f$ does not vanish in the domain $log|f|$ is harmonic and attains minimum on the boundary. So in fact since $log$ is a monotonic function, $|f|$ also attains minimum on the boundary.

In total we have a sum of two functions, $|f|$ and $-|g|$, both attain their minimum on the boundary. For general functions, this does not imply that the sum should attain minimum on the boundary. However, I believe that it is true in this particular case.

Any help proving this (or finding a counter example) would be appreciated!

• I've been playing around with the problem and it appears to true. I can't seem to find a counterexample. Two observations: 1) $|f|$ and $-|g|$ aren't necessarily minimal at the same $z$. 2) $|f|-|g|$ reaches a minimum on the boundary in a certain $z_0$. (Weierstrass) If you could prove that forall $z\in \mathring{\Omega}$ $|f(z)|-|g(z)|$ would be larger then $|f(z_0)|-|g(z_0)|$ the statement would be proven. I haven't succeeded so far... Dec 30 '14 at 16:34

The problem of your question can be solved by the Dirichlet principle. First I add some missing prerequisites concerning the boundary:

Claim: Let $\Omega \subset \mathbb C$ be a bounded domain where the Dirichlet principle has a solution, e.g., each component of the complement $\mathbb C - \Omega$ comprises more than one point. Consider two functions $f,g$ which are continuous on $\overline{\Omega}$ and holomorphic on $\Omega$. Assume $f$ without zeros in $\overline{\Omega}$. Then $|f|-|g|$ attends its minimum on the boundary.

Proof: The claim is equivalent to the claim that $log|f|-|g|$ attends its minimum on the boundary, because the function $log$ is strictly monotonic increasing. The latter claim is equivalent to the claim that $|g|-log|f|$ attends its maximum on the boundary.

By the Dirichlet principle a function $h$ on $\overline{\Omega}$ exists which is harmonic on $\Omega$ and equals $|g|-log|f|$ on $\partial \Omega$. By the maximum principle the harmonic function $h$ attends its maximum on $\partial \Omega$. Furthermore, the function $|g|$ is subharmonic and the function $log|f|$ is harmonic. Hence also $-log|f|$ is harmonic, specifically subharmonic, and the sum

$$|g|+(-log|f|) = |g|-log|f|$$

is subharmonic. Hence for all $z \in \Omega$:

$$|g(z)|-log|f(z)| \leq h(z) \leq max\{|g(w)|-log|f(w)|: w \in \partial \Omega\}, q.e.d.$$