# Deriving Inequality of Holomorphic Function Using Maximum modulus principle

Let $$f$$ be a holomorphic on an open set containing $$\overline{\Bbb{D}}$$. Use the maximum modulus principle to prove that there exists $$z_0 \in \partial \Bbb{D}$$ such that $$\left| \frac{1}{z_0} - f(z_0) \right| \ge 1$$

Here is my attempt.

Let $$f : \overline{\Bbb{D}} \subset U \to \Bbb{C}$$ be holomorphic. First, if $$f$$ is identically $$0$$ on $$U$$, then

$$\left| \frac{1}{z_0} - f(z_0) \right| = \left|\frac{1}{z_0} \right| = 1$$ for every $$z_0 \in \partial \Bbb{D}$$, and so we're finished. Now, suppose that $$f$$ is not identically $$0$$ and consider the function $$h : U \to \Bbb{C}$$ defined by $$h(z) = 1-zf(z)$$. If $$h$$ were constant, say equal to $$c$$, then $$h(z) = c$$ would imply $$f(z) = \frac{1-c}{z}$$ for all $$z \in U \setminus \{0\}$$. First, if $$c=1$$, then continuity would tell us $$f$$ is identically $$0$$, which is contrary to our assumption, so $$c \neq 1$$. Now, note that

$$\lim_{z \to 0 } z f(z) = \lim_{z \to 0} z \frac{1-c}{z} = 1-c \neq 0$$ which shows that $$f$$ has a non-removable singularity at $$z=0$$, which contradicts the fact that $$f$$ is holomorphic everywhere on $$U$$. Hence, $$h$$ cannot be constant. So, the maximum modulus principle tells us that there exists a point $$z_0 \in \partial D$$ such that

$$|h(z)| \le |h(z_0)|$$ for every $$z \in \overline{\Bbb{D}}$$ or $$|1-zf(z)| \le |1-z_0f(z_0)|$$ for every $$z \in \overline{\Bbb{D}}$$. In particular, when $$z=0$$ we obtain

$$1 \le |1-z_0f(z_0)|$$ or $$1 = \frac{1}{|z_0|} \le \left|\frac{1}{z_0} - f(z_0) \right|$$

Does this seem okay, or is it a bit round about?

• Why so complicated? $h$ is holomorphic in an open set containing $\overline{\Bbb{D}}$. Therefore $|h(z_0)| \ge |h(0)| = 1$ for some $z_0 \in \partial \Bbb D$. Done. Considering all these different cases is not necessary. Oct 13, 2021 at 13:42
• @MartinR Oh, whoops. I thought that the maximum modulus principle was applicable only in the case that $h$ is constant. Oct 13, 2021 at 13:47
• There are different ways to formulate the MMP. One is that $|f(z)| \le \max \{ |f(w)| : w \in \partial D \}$. That holds for non-constant functions and (trivially) also for constant functions. Oct 13, 2021 at 13:51

The function $$h(z) = 1-zf(z)$$ is holomorphic in $$\Bbb D$$ and continuous in $$\overline{\Bbb D}$$. The maximum modulus principle implies that $$1 = |h(0)| \le \max_{z \in \partial \Bbb D} |h(z)| = |h(z_0)|$$ for some $$z_0 \in \partial \Bbb D$$. But $$|h(z_0)| = |1-z_0 f(z_0)| = \left| \frac{1}{z_0} - f(z_0) \right| \, ,$$ so this is the desired conclusion.