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?

  • $\begingroup$ 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. $\endgroup$
    – Martin R
    Oct 13, 2021 at 13:42
  • $\begingroup$ @MartinR Oh, whoops. I thought that the maximum modulus principle was applicable only in the case that $h$ is constant. $\endgroup$
    – user193319
    Oct 13, 2021 at 13:47
  • $\begingroup$ 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. $\endgroup$
    – Martin R
    Oct 13, 2021 at 13:51

1 Answer 1


Your proof is correct and uses the right idea, but it is too complicated. Considering all these different cases is not necessary.

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.


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