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Suppose $f(z)$ is holomorphic in the whole plane, and that $f(z)$ does not have an essential singularity at $\infty$. Prove that $f(z)$ is a polynomial.

I've tried following the hint given in this question. Since $f(z)$ has a nonessential singularity at $\infty$, so $g(z)=f(1/z)$ has a nonessential singularity at $0$. There are two cases:

1) $g(z)$ has a removable singularity at $0$. This means $\lim_{z\rightarrow 0}zg(z)=0$.

2) $g(z)$ has a pole at $0$. This means $g(z)=h(z)/z^k$ for some analytic function $h(z)$ such that $h(0)\neq 0$.

How can I finish each of these cases?

Community
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PJ Miller
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  • In both cases, $z^k g(z)$ is analytic in some neighborhood of $0$. So $z^kg(z)$ has a power series expansion valid in some neighborhood of $0$. $f$ is holomorphic, so $f$ also has a power series around $0$. How are these two power series related? – Gyu Eun Lee Oct 19 '13 at 04:16
  • We can write $f(z)=f(0)+f'(0)z+f''(0)z^2/2!+\ldots$. For $z^kg(z)$, the $i$th derivatives at $0$ vanishes for $i=1,2,\ldots,k-1$, and the $k$th derivative should be $k!g(0)$. How is this helpful? – PJ Miller Oct 19 '13 at 04:27
  • $z^k g(1/z)=z^k f(z)$, and since the right side is holomorphic the left side is too. But we already know that $z^k g(z)$ is holomorphic at $0$, so $z^k g(z)$ has a power series expansion around $0$. So write $z^k g(z)$ has a power series, and then write $z^k g(1/z)$ in terms of this power series. This power series has terms of the form $z^{-n}$, yet it is holomorphic at $0$. How can this be? – Gyu Eun Lee Oct 19 '13 at 04:35

3 Answers3

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1) Since $g$ is continuous, we can bound $g(z)$ inside some interval of $z$. Since $g(z)$ is defined as $f(1/z)$, it turns out that from the bound on $z$ we deduce that $f$ is a constant function:

There exists $M, \varepsilon >0$ such that $|g(z)|\leq M$ for all $|z| \in (0,\varepsilon)$. Hence $|f(z)|\leq M$ for all $|z| > 1/\varepsilon$. Letting $\varepsilon \to \infty$ $f$ is bounded and entire. It follows by Liouville's Theorem that $f$ is a constant function.

2) We use that $g$ has a Laurent expansion at $0$ and $f$ has a Taylor expansion since $f$ is entire. We can 'invert' the Laurent expansion so to say and by uniqueness of the Laurent expansion see that $f$ must be a polynomial.

Since the pole is of order $m$, the Laurent expansion of $g$ at $0$ is $$g(z) = \sum_{k=-m}^{\infty} a_k z^k$$ for $|z|\in (0,\varepsilon)$. We can invert this to get $f$: $$f(z) = \sum_{k = -\infty}^{m} a_{-k}z^k$$ for $|z|>1/\varepsilon$. The Taylor expansion of $f$ around $0$ given by $$f(z)=\sum_{k=0}^{\infty}b_kz^k$$ and the Laurent expansion must be equal by uniqueness, so $$f(z)=\sum_{k=0}^{m} b_k z^k$$ where $a_{-k}=b_k$ for all $k$. So $f$ is a polynomial.

See the following pdf: math.berkeley.edu/~mjv/Math185hw8.pdf

Numbersandsoon
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Since we have a non-essential singularity at $\infty$ we have an $h$ such that $z^{-h}f(z)$ has a limit as $z$ tends to $\infty$ that is neither $0$ nor $\infty$. This means that $f$ cannot have more than $h$ zeroes at $0$ because otherwise $f(z) = z^{h+k}f_k(z)$ and hence limit of $z^{-h}f(z)$ would not be finite. This also means that $f^{(h+1)}(0) \neq 0$ (the $h+1$st derivative is not $0$). Now $$f(z) = f(0) + f'(0)z + ... + z^{h+1}g(z)$$ where $g(z)$ is analytic in all of complex plane (including $0$). Hence $g(0)$ is finite and $\neq 0$ (because $g(0) = f^{h+1}(0)$) Also observe that we can write: $$z^{-h}f(z) = z^{-h}(f(0) + f'(0)z + ... + z^{h+1}g(z))$$

Now since $\lim_{z \to \infty} z^{-h}f(z)$ exists we see that $\lim_{z \to \infty} zg(z)$ also exists.

Also observe that $g(z)$ cannot be $0$ at infinity because otherwise $g(1/t) = t^{k}g_2(t) \implies g(z) = \frac{g_2(1/z)}{z^{k}}$ and would mean $g(0)$ is not defined.

All this implies that $g(z) = 0$

Ankit Sahu
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I think I have a found a proof for the second part that is shorter than the ones already given here. Either it is correct or it is horribly flawed. Please don't downvote if it is latter, and I am sorry in that case.

If $f$ has a pole at $\infty$, then $\lim_{|z|\to\infty}f(z)=\infty$, then there is an $R>0$ such that if $R<|z|$ then $|f(z)|>0$. Let $S=f^{-1}(\{0\})\cap\overline{D(0,R)}$ which is a compact set. If $S$ contains an accumulation point, then by the identity theorem $f=0$, which is a contradiction. So $S$ is a discrete set (because we are in a metric space), by the compactness property, $S$ is finite.

IJM98
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