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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
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2 Answers 2

up vote 3 down vote accepted

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
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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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