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Let $f$ and $g$ be entire functions and $g(z)\neq 0$ for all $z\in \mathbb{C}$. If $|f(z)|\le |g(z)|$, can we say $f$ is constant?

Liouville theorem says that an entire bounded function is constant, but g is not given bounded here. So I think it should be false. What else can we say about $f$?

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    $\begingroup$ If you set $g = f$ then this also holds. $\endgroup$ Dec 20 '16 at 19:18
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    $\begingroup$ What you have is that $f/g$ is bounded. Thus $f/g$ is constant. That is, $g=cf$ for some real constant $c$ with $|c|\ge 1.$ $\endgroup$
    – mfl
    Dec 20 '16 at 19:20
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    $\begingroup$ In any case, you have $\phi = {f \over g}$ is entire and $|\phi(z)| \le 1$ for all $z$, hence $\phi$ is constant. Bingo. $\endgroup$
    – copper.hat
    Dec 20 '16 at 19:20
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    $\begingroup$ ohh yes. this was the use of $g(z)\neq 0$ $\endgroup$
    – Departed
    Dec 20 '16 at 19:20
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    $\begingroup$ You are right in supposing $f$ need not be constant: Let $g$ be any unbounded entire function and $f(z) = g(z)/2$. Then $f(z)$ is entire but non-constant. $\endgroup$ Dec 20 '16 at 19:21
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$\frac{f(z)}{g(z)}$ is entire and bounded => $\frac{f(z)}{g(z)}$ is constant => f(z) = c*g(z)

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No, we cannot conclude that $f$ is constant. Here's a counter-example:

If $f(z) = e^z$ and $g(z) = 2f(z) = 2e^z$. Clearly $f$ and $g$ are both entire and non-constant. We also see that $|f(z)| \leq |g(z)|$ for all $z$ and $g(z) \neq 0$.

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If this holds then $f = e^{\phi(z)}$ for some entire $\phi$.

Using Weierstrass Factorisation Theorem it can be shown that $g = e^{\psi(z)}$ for some entire $\psi$. (See this)

As copper.hat pointed out, $f/g$ must be constant, so $f$ must also be in the form $e^{\phi(z)}$.

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  • $\begingroup$ That's a big weapon to use here. $\endgroup$
    – zhw.
    Dec 23 '16 at 18:40

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