Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $f$ is a non-constant entire function. Define $$ d_f(a,b) = \inf_{\gamma} \ell(f\circ \gamma), $$ where $a,b \in \mathbb{C}$, $\ell$ is the euclidean length, and $\gamma$ is a path connecting $a$ and $b$. I'm trying to prove $d_f$ is a metric. I'm having trouble showing that $$ a\ne b \, \Longrightarrow \, d_f(a,b) \ne 0 .$$ Now if $f(a)\ne f(b)$, this is clear. If, however, $f(a)=f(b)$, is it not possible that there are paths, $\gamma_n$, connecting $a$ and $b$, such that $\ell(f\circ \gamma_n)$ approches $0$ ?

share|cite|improve this question
up vote 2 down vote accepted

Let $a \neq b$ such that $f(a) = f(b)$. Suppose we can find a sequence of paths $\{\gamma_n\}$, all with endpoints $a$ and $b$, such that

$$ \ell(f\circ \gamma_n) \to 0 $$

as $n \to \infty$. Then it must be true that every point on the paths $f \circ \gamma_n$ approaches $f(a)$, and thus every point on $\gamma_n$ must approach a zero of $f(z) - f(a)$. But $f(z) - f(a)$ is entire, so its zeros are isolated. Since $a \in \gamma_n$ for all $n$, we must have $\gamma_n \to a$ (all points of $\gamma_n$ must converge to $a$). But we must similarly conclude that $\gamma_n \to b$. Since $a \neq b$, we arrive at a contradiction.

share|cite|improve this answer
Technically this is just a contrapositive argument, not contradiction, but it felt more natural this way considering the statement of the question. – Antonio Vargas Aug 27 '12 at 14:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.