# Showing that the pull-back of the Euclidean metric by an entire function is a metric

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 $0$ ?

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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.

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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