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?

  • $\begingroup$ 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? $\endgroup$ – Gyu Eun Lee Oct 19 '13 at 4:16
  • $\begingroup$ 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? $\endgroup$ – PJ Miller Oct 19 '13 at 4:27
  • $\begingroup$ $z^k g(1/z)=z^k f(z)$, 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? $\endgroup$ – Gyu Eun Lee Oct 19 '13 at 4:35

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

  • 3
    $\begingroup$ The above link is not active $\endgroup$ – Tutankhamun Aug 9 '18 at 14:54

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 $g(z) = 0$


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