# Entire function, Liouville and zeroes

Suppose $f:\mathbb{C}\rightarrow\mathbb{C}$ is an entire function. Let $g:\mathbb{C}\rightarrow\mathbb{C}$ be an entire function, which has no zeros.

I have shown that $\vert f(z) \vert \leq \vert g(z) \vert$ for all $z\in\mathbb{C}$ implies $f(z)=Cg(z)$ for some constant $C\in\mathbb{C}.$

I have to decide if this also holds if $g$ is allowed to have zeros. Help?

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Did you mean to write $|f(z)|\leq |g(z)|$ ? (In this case, yes, it holds even if $g$ has zeroes because they're isolated and hence singularities for $f/g$ are removable). – Jose27 Nov 3 '12 at 23:56
@Jose27 Perhaps you should add that as an answer. – EuYu Nov 4 '12 at 0:07
Perhaps he should wait for the OP to explain complex inequalities. – GEdgar Nov 4 '12 at 0:10
Yes. I meant $\vert f(z) \vert \leq \vert g(z) \vert$ – whoisitnow Nov 4 '12 at 7:25

If $f \equiv 0$, then choose $C = 0$. Also, if $g \equiv 0$, then this forces $f \equiv 0$, so we can choose any $C \in \mathbb{C}$. Henceforth, let us assume that $f,g \not\equiv 0$.

Suppose that $g$ never vanishes. Then $\dfrac{f}{g}$ is entire. As $\left| \dfrac{f}{g} \right| \leq 1$, it is also bounded. Hence, by Liouville's Theorem, $\dfrac{f}{g} = C$ for some constant $C$, which yields $f = Cg$.

Now, suppose that $g$ has zeroes. Denote the zero sets of $f$ and $g$ by $\mathcal{Z}_{f}$ and $\mathcal{Z}_{g}$ respectively. Clearly, $\mathcal{Z}_{g} \subseteq \mathcal{Z}_{f}$. By the Quotient Rule, $\dfrac{f}{g}$ is complex-differentiable, thus holomorphic, on $\mathbb{C} \setminus \mathcal{Z}_{g}$. Pick $z_{0} \in \mathcal{Z}_{g}$. As $\mathcal{Z}_{g}$ is a discrete subset of $\mathbb{C}$, there exists an $r_{0} > 0$ such that $g(z) \neq 0$ for all $z \in D(z_{0},r_{0}) \setminus \{ z_{0} \}$. Let the complex power series representations of $f$ and $g$ on $D(z_{0},r_{0})$ be given by $\displaystyle \sum_{k=0}^{\infty} a_{k} (z - z_{0})^{k}$ and $\displaystyle \sum_{k=0}^{\infty} b_{k} (z - z_{0})^{k}$ respectively. Clearly, $a_{0} = b_{0} = 0$. However, there must exist smallest $M,N \in \mathbb{N}$ such that $a_{M},b_{N} \neq 0$, otherwise $f = g \equiv 0$, contrary to our earlier assumption (at the end of the first paragraph). Hence, $$\forall z \in D(z_{0},r_{0}) \setminus \{ z_{0} \}: \quad \left( \frac{f}{g} \right)(z) = \frac{\displaystyle \sum_{k=M}^{\infty} a_{k} (z - z_{0})^{k}}{\displaystyle \sum_{k=N}^{\infty} b_{k} (z - z_{0})^{k}} = (z - z_{0})^{M - N} \cdot \frac{\displaystyle \sum_{k=0}^{\infty} a_{k + M} (z - z_{0})^{k}}{\displaystyle \sum_{k=0}^{\infty} b_{k + N} (z - z_{0})^{k}}.$$ It is easily seen from this that $$\lim_{\substack{z \rightarrow z_{0}; \\ z \in D(z_{0},r_{0}) \setminus \{ z_{0} \}}} \left| \left( \frac{f}{g} \right)(z) \right|$$ (i) converges when $M \geq N$ and (ii) diverges to $\infty$ when $M < N$. The second scenario cannot happen because $\left| \left( \dfrac{f}{g} \right)(z) \right| \leq 1$ for all $z \in \mathbb{C} \setminus \mathcal{Z}_{g}$. Therefore, $M \geq N$, which allows us to define a holomorphic $F_{0}: D(z_{0},r_{0}) \rightarrow \mathbb{C}$ by $$\forall z \in D(z_{0},r_{0}): \quad {F_{0}}(z) = (z - z_{0})^{M - N} \cdot \frac{\displaystyle \sum_{k=0}^{\infty} a_{k + M} (z - z_{0})^{k}}{\displaystyle \sum_{k=0}^{\infty} b_{k + N} (z - z_{0})^{k}}.$$ Clearly, $F_{0}$ agrees with $\dfrac{f}{g}$ on the punctured disk $D(z_{0},r_{0}) \setminus \{ z_{0} \}$, and $|F_{0}| \leq 1$. Enumerating the elements of $\mathcal{Z}_{g}$ as $z_{i}$, starting with the $z_{0}$ just considered, we can find corresponding disks $D(z_{i},r_{i})$ and holomorphic functions $F_{i}: D(z_{i},r_{i}) \rightarrow \mathbb{C}$ that agree with $\dfrac{f}{g}$ on those disks. Then $$F := \left( \frac{f}{g} \right) \cup \left( \bigcup_{i} F_{i} \right)$$ is an entire function satisfying $|F| \leq 1$. By Liouville's Theorem, $F \equiv C$ for some constant $C$, and so $\dfrac{f}{g} = C|_{\mathbb{C} \setminus \mathcal{Z}_{g}}$. This implies that $f|_{\mathbb{C} \setminus \mathcal{Z}_{g}} = C \cdot g|_{\mathbb{C} \setminus \mathcal{Z}_{g}}$. By the discreteness of $\mathcal{Z}_{g}$ and by the continuity of both $f$ and $g$, we can conclude that $f = Cg$.

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Thanks! I have a little trouble with this part: "The second scenario cannot happen because $\mathcal{Z}_{g}$ is a discrete subset of C and $\vert (\frac{f}{g})(z) \vert\leq1$ for all $z\in \mathbb{C}\setminus\mathcal{Z}_{g}$" Can you explain in details, why you can deduce that? Which theorems, lemmas ect.? – whoisitnow Nov 4 '12 at 7:32
Oh :) I think I got it. It's because when a singularity is removable, there exist a number $r>0$ such that $\frac{f}{g}$ is bounded in $K'(a,r)$, which in this case is true, because every zero is isolated and $\vert \frac{f}{g}(z) \vert \leq 1$. – whoisitnow Nov 4 '12 at 7:43
Yep! That's what I meant! :) – Haskell Curry Nov 4 '12 at 9:16

The zeros of $g(z)$ are isolated (otherwise it is identically zero and so is $f$). Then use Riemann's removable singularity Theorem: An isolated singularity $z_0$ is removable iff $f$ is bounded near $z_0$.

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