# Maximum Modulus Principle and Open Mapping Theorem

The Maximum Modulus Principal states: let $$f$$ be a function holomorphic on some connected open subset $$D$$ of the complex plane $$\mathbb{C}$$ and taking complex values. If $$z_{0}$$ is a point in $$D$$ such that $$|f(z_{0})|\geq |f(z)|$$ for all $$z$$ in a neighborhood of $$z_{0}$$, then the function $$f$$ is constant on $$D$$.

The notes further state that: “alternatively, the maximum modulus principle can be viewed as a special case of the open mapping theorem, which states that a nonconstant holomorphic function maps open sets to open sets. If $$|f|$$ attains a local maximum at $$z$$, then the image of a sufficiently small open neighborhood of $$z$$ cannot be open. Therefore, $$f$$ is constant.

Can some explain why intuitively, if $$|f|$$ attains a local maximum at $$z$$, then the image of a sufficiently small open neighborhood of $$z$$ cannot be open?

• I don't understand why they don't say the maximum modulus and open mapping are easy consequences of the holomorphic $\implies$ analytic theorem (the latter means $f(z) = f(z_0) + C(z-z_0)^n+O((z-z_0)^{n+1})$) – reuns Dec 8 '18 at 20:56

## 1 Answer

If $$z$$ is a local maximum of $$\vert f \vert$$, then by definition for a sufficiently small open neighborhood of $$U$$ of $$z$$,

$$y \in U \Longrightarrow \vert f(y) \vert \le \vert f(z) \vert; \tag 1$$

this in turn implies

$$\vert f \vert (U) \subset (\vert f(x) \vert - \epsilon, \vert f(z) \vert] \tag 2$$

for some $$\epsilon > 0$$, which is not open since it does not contain an open set containing $$\vert f(z) \vert \in \Bbb R$$.

It is worth observing that only the continuity of $$f$$ is used here, not fact that $$f$$ is holomorphic.