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I am asked to prove this:

Let $f$ be an entire function such that $f(1)=2f(0)$. Prove that $\forall\epsilon>0, \exists z\in\mathbb{C}$ such that $|f(z)|<\epsilon$

I considered a function $g(z)=f(z+1)-2f(z)$, which is also entire and has a zero at $z=0$, but I am not sure this is going to help me solve the problem.

NFC
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3 Answers3

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Hint Assume by contradiction that this is not true. Show that $g(z)=\frac{1}{f(z)}$ is entire and bounded.

N. S.
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  • I see. So, by Liouville's theorem we obtain that $g(z)$ is constant, which implies that $f(z)$ is constant. But that would mean that $f(0)=f(1)$ and therefore $f(0)=f(1)=0$, arriving at a contradiction. – NFC Jun 04 '19 at 16:34
  • @NFC Exactly... Note that the given relation is only used to deduce that $C=0$. – N. S. Jun 04 '19 at 16:36
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I was just giving my answer a final polish edit when N.S. posted; our proofs are essentially the same, though I have fleshed out a few details. Perhaps someone will find it useful.

If

$\forall \epsilon > 0 \exists z \in \Bbb C, \; \vert f(z) \vert < \epsilon \tag 1$

is not true, then

$\exists \epsilon > 0 \forall z \in \Bbb C, \; \vert f(z) \vert \ge \epsilon \tag 2$

is true.

Thus

$g(z) = \dfrac{1}{f(z)} \tag 3$

is a well-defined entire function, and

$\forall z \in \Bbb C, \; \vert g(z) \vert = \dfrac{1}{\vert f(z) \vert} \le \dfrac{1}{\epsilon}; \tag 4$

thus $g(z)$ is a bounded entire function, hence constant by Liouville's theorem; hence $f(z)$ must be constant as well; in fact via (1),

$\forall \epsilon > 0 \forall z \in \Bbb C, \; \vert f(z) \vert < \epsilon \Longrightarrow \forall z \in \Bbb C, \; f(z) = 0; \tag 5$

that is, $f(z)$ is identically zero.

The hypothesis $f(1) = 2f(0)$ is not necessary to attain this result.

Robert Lewis
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    in (4) the RHS should be $\frac{1}{\epsilon}$. – N. S. Jun 04 '19 at 16:41
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    Also, Step (5) is wrong. You start by assuming that (1) is not true, and in Step (5) you USE (1). – N. S. Jun 04 '19 at 16:42
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    And the given hypothesis is necessary, otherwise $f(z)=2019$ for example is a counterexample.... The given hypotthesis is how you fix step (5) ;) – N. S. Jun 04 '19 at 16:44
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If $f(0)=0$ the result is immediate. If $f(0)\ne 0$, we have that $f(1)\ne f(0)$ and so $f$ is not constant. According to Picard's little theorem, a non-constant entire function takes every complex value, with one possible exception. Even if that possible exception turns out to be zero, all other complex values are attained by $f$ and the result follows.

PierreCarre
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