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The Liouville theorem state:

Let $f$ be an entire function for which there exists a positive number M such that $|f(z)|\leq M$ for all z in $\mathbb{C}$, the $f$ must be a constant.

more general form of this theorem is :

Let $f$ be an entire function for which there exists a positive number M and a polynomial $P$ such that $|f(z)|\leq M |P(z)|$ for all z in $\mathbb{C}$ then $f(z)=k P(z)$ for $k$ a constant.

who know a good proof of this generalized form ?

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I don't think your generalized form is correct. Consider $f=z^2$, $P=z^2+1$, $M=1$. Then surely $|f(z)|=|z^2|\leq 1\cdot(|z^2|+1)$, but $z^2$ is not a constant multiple of $z^2+1$. I believe the correct statement should be $|f(z)\leq M|P(z)|$ implies $f(z)=kP(z)$. Then the proof would consist of showing that $\frac fP$ is entire, and using the usual form of Liouville's theorem. –  Vladimir Sotirov Mar 17 '12 at 19:17
    
yes, maybe $|f(z)|\leq M|P(z)|$ for all $z\in\mathbb{C}$. thans for the remarks :) –  Abdelmajid Khadari Mar 17 '12 at 19:42

3 Answers 3

up vote 2 down vote accepted

Claim $\frac{f(z)}{P(z)}$ is an entire function.

Proof: $\frac{f(z)}{P(z)}$ is analytic on $\{ z| P(z) \neq 0 \}$. To prove the claim, we need to show that if $P(z_0)=0$ then $\frac{f(z)}{P(z)}$ is analytic at $z_0$.

Since $P(z_0)=0$ we have $f(z_0)=0$. Let $m$ be the order of $z_0$ for $f(z)$ and $n$ be the order of $z_0$ for $P(z)$ and. Then, there exists analytic functions $h(z)$ and polynomial $Q(z)$ so that

$$f(z)=(z-z_0)^mh(z) \,;\, P(z)=(z-z_0)^nQ(z) \,,$$

and $Q(z_0) \neq 0 \,;\, h(z_0)\neq 0$$

We have

$$\frac{f(z)}{P(z)}=(z-z_0)^{n-m} \frac{h(z)}{Q(z)}$$

Since $Q(z_0) \neq 0$, if $n -m \geq 0$ we are done. We finish the proof by showing that this is the case.

Assume by contradiction that $n<m$. Then, the inequality above yields:

$$|(z-z_0)^n h(z)| \leq M|(z-z_0)^m Q(z)| \Rightarrow |h(z)| \leq M|(z-z_0)^{m-n} Q(z)| \forall z \neq z_0 $$

Since $n-m >0$ we get that

$$|h(z_0)|=\lim_{z \to z_0}|h(z)| \leq \lim_{z \to z_0}M|(z-z_0)^{m-n} Q(z)| \leq 0 \,.$$

Thus $h(z_0)=0$ Contradiction.

This proves the claim.

The Strong Liouville Theorem follows by applying the Liouville Theorem to $\frac{f(z)}{P(z)}$.

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Generalized Liouville's theorem. Suppose $f(z)$ is an entire function such that $|f(z)|\le c|z|^n$ for some constant $c$ and some nonnegative integer $n$ for sufficiently large $|z|$. Then $f$ is a polynomial of degree at most $n$.

Proof. By Liouville's theorem, the statement is true for $n=0$. Suppose true for $n-1$. Consider $$g(z)=\left\{\begin{array}{ll} \frac{f(z)-f(0)}{z}, & z\neq 0\\ f'(0), & z=0\end{array}\right.$$ Then $g$ is entire, and for sufficiently large $|z|$, $$|g(z)|\le c_1|z|^{n-1}$$ for some constant $c_1$. (This is because $|f(0)|/|z|, |f'(0)|$ are bounded for large $|z|$, and $a+b|z|^{n-1}\le c_1|z|^{n-1}$ for large $|z|$.)

Hence by induction hypothesis, $g$ is a polynomial of degree at most $n-1$. Then it follows easily by the definition of $g$ that $f$ is a polynomial of degree at most $n$.

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Why $g$ is entire? –  Idonknow Nov 5 '13 at 16:46
    
Because 0 is a removable singularity. Or just use the series expansion of f. –  TCL Nov 10 '13 at 0:16

If f is dominated by P like in your statement, then it is dominated by many other polynomials. If your generalization were correct, it would imply that all these polynomials are collinear. This can not be true, as soon as you consider nonconstant polynomials P. You might want to prove that f is a polynomial, not necessarily of the type you mention. Maybe you can start by looking at the case when $|f(z)\leq M|z|^n$. Then you can see that the general case boils down to the latter.

Reference: see exercises 3 and 4 p.227 in Rudin's Real and Complex Analysis.

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That is true only over reals, in complex numbers by the Fundamental Theorem of Algebra, a polynomial cannot dominate a different polynomial. –  N. S. May 10 '13 at 16:54

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