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I had this question on last semesters qualifying exam in complex analysis, and I've attempted it several times since to little result.

Let $f$ be an entire function with $|f(z)|\geq 1$ for all $|z|\geq 1$. Prove that $f$ is a polynomial.

I was trying to use something about $f$ being uniformly convergent to a power series, but I can't get it to go anywhere.

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I assume you mean for all $z$ with $|z|\geq 1$, since $z\geq 1$ has no meaning for complex numbers $z$. –  Eric Naslund Dec 18 '12 at 0:00

2 Answers 2

up vote 2 down vote accepted

By Casorati-Weierstrass theorem (or whatever name you want to give to the following fact), if a point $z\in\mathbb{C}\cup\{\infty\}$ is a singularity for $f$, but it is not a pole, then every neighborhood of $z$ has dense image in $\mathbb{C}$.

If $|f(z)|\ge1$ whenever $|z|\ge1$, then $z=\infty$ is a pole for $f$, because the image of any neighborhood ( $|z|\ge M$ is a basis of neighborhoods for $\infty$ as $M\to\infty$) is not dense (it misses the whole open set $|z|<1$).

This means that there exists $n$ such that $z^{-n}f(z)$ is holomorphic at $\infty$, which means bounded, hence $|f(z)|\le C|z|^n$ for some $C$ and $n$, as $z\to\infty$. This forces an entire function to be a polynomial (a generalization of Liouville theorem).

An alternative way could be to note that the negative part of Laurent series of $f$ around $\infty$ corresponds to the power series for $f$ centered in $0$, so if the former is finite (because $\infty$ is a pole), then also the latter has to be finite, so $f$ is a polynomial. (But here I am avoiding the computation of the Laurent series of $f$ around $z=\infty$, which may anyway be a useful exercise!)

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And with this proof you get actually the following: take two open bounded subsets of $\mathbb{C}$, $A$ and $B$; if $f(z)$ is an entire function such that $f(z)\not\in A$ for every $z\not\in B$, then it is a polynomial. –  wisefool Dec 18 '12 at 0:28

Picard's Theorem proves this instantly; which states:

Let $f$ be a transcendental (non-polynomial) entire function. Then $f-a$ must have infinitely many zeros for every $a$ (except for possibly one exception, called the lacunary value).

For example, $e^z-a$ will have infinitely many zeros except for $a=0$ and so the lacunary value of $e^z$ is zero.

Your inequality implies that $f$ and $f-\frac{1}{2}$ have only a finite number of zeros. Thus $f$ cannot be transcendental.

Of course this is what we call hitting a tac with a sludge hammer. A more realistic approach might be the following:

Certainly $f$ has a finite number of zeros say $a_1,\ldots,a_n$, so write $f=(z-a_1)\cdots(z-a_n)\cdot h$, where $h$ is some non-zero entire function. Then the inequalities above give us

$|\frac{1}{h}|\le \max\left\{\max_{z\in D(0,2)}|\frac{1}{h}|, |(z-a_1)\cdots (z-a_n)|\right\}$ on the entire complex plane.

Said more simply $|\frac{1}{h}|<|p(z)|$ for every $z\in C$ for some polynomial $p$. That implies $\frac{1}{h}$ is a polynomial. But remember that $\frac{1}{h}$ is nonzero, so $h$ is a constant and $f$ must therefore be a polynomial.

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Welcome to Math.SE. Thank you for your answer. Since this question was asked 5 months ago, the asker might not get as much value out as the hard work you put in. –  vadim123 May 10 '13 at 1:08

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