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Let $f$ be an entire function such for some $N \in \mathbb{N}$ and $R >0$, the following property holds:

$|f(z)| \geq |z|^N$ $\forall z \in \mathbb{C}$ with $|z| \geq R$.

Show that $f$ is a polynomial of degree greater than or equal to $N$.

Progress thus far:

Clearly, $f$ is not identically zero.

Case 1) If $f$ is also nowhere 0, it's not hard to show $f$ is bounded and therefore, by Louisville's Theorem, a constant.

Case 2) If $f$ is neither identically zero nor nowhere 0, then we can show that it has a finite number of zeros in the ball $B(0,R)$. From these zeros we can get a polynomial $p$, whose zeros are exactly those of $f$, and an entire function $g$ such that $f = pg$. Consequently, $g$ is nowhere zero.

At this point, I'd like to show $g \equiv c$, where $c$ is some complex constant.

A more general problem has been solved here, but I'm trying to avoid using big hammers like Casorati-Weierstrass. (When this problem was assigned, we hadn't covered it yet.)

Community
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artificial_moonlet
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2 Answers2

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You're almost there, and your idea in case 2 is the right one. Though we can argue more simply by just using $z^N$ instead of introduction a new polynomial $p$.

Consider the function $g(z)=\frac{z^N}{f(z)}$. This function is holomorphic except possibly at the points where $f(z)$ equals zero. But we know that $|g(z)| \leq 1$, so each of these points is actually a removable singularity of $g(z)$ so $g$ extends to a bounded holomorphic function, and is thus constant.

It's interesting to note here that there was nothing special about $|f(z)| \geq |z|^N$. The inequality could have been the other way and the result would still hold. More generally, no entire function can dominate another entire function without being a constant multiple of one another.

Ryan Hunter
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  • Ah, thank you, that's so simple. In fact, the general result you've mentioned should follow in a similar fashion, correct? – artificial_moonlet Aug 06 '14 at 20:37
  • Yes, definitely! Though to be very careful, in all these situations you might want to mention why the zeros of the denominator $f(z)$ can't accumulate so that each singularity is in fact isolated. – Ryan Hunter Aug 06 '14 at 20:39
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    Actually, now that I've taken a longer look, your solution doesn't work. $g$ would only be bounded for $|z| \geq R$. – artificial_moonlet Aug 06 '14 at 20:47
  • The estimate holds only for $\lvert z\rvert \geqslant R$. The global result holds only for a global estimate. In the setting of the question, $f$ generally has zeros other than $0$. – Daniel Fischer Aug 06 '14 at 21:32
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You are almost there. The idea to show that $g$ must be constant is correct, all that remains is to play with the given constraints until it falls out. Like in the proof of the fundamental theorem of algebra, we look at the reciprocal $1/g$ to obtain the desired conclusion:

If you take $d = \deg p$, then you have

$$\lvert g(z)\rvert = \frac{\lvert f(z)\rvert}{\lvert p(z)\rvert} \geqslant \frac{\lvert z\rvert^N}{\lvert p(z)\rvert} \geqslant C\cdot \lvert z\rvert^{N-d}\tag{1}$$

for some $C > 0$ and all $z$ with $\lvert z\rvert \geqslant \tilde{R} \geqslant R$. Now, $g$ is an entire function without zeros, so we can take the reciprocal of $(1)$:

$$\left\lvert \frac{1}{g(z)}\right\rvert \leqslant C^{-1}\cdot \lvert z\rvert^{d-N}\tag{2}$$

for $\lvert z\rvert \geqslant \tilde{R}$. If $d \leqslant N$, $(2)$ shows that $1/g$ is bounded, hence constant. If $d > N$, the Cauchy inequalities show that $1/g$ is a polynomial of degree at most $d-N$. But $1/g$ has no zeros, and a polynomial without zeros is constant by the fundamental theorem of algebra.

Daniel Fischer
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