I've been stumbling across a couple of questions in Ahlfors that fall along the lines of "If $f$ is an analytic function with property $X$, show that $f$ reduces to a polynomial."
One concrete example would be:
Show that a function which is analytic in the whole plane and has a nonessential singularity at $\infty$ reduces to a polynomial.
I was wondering what the general approach to these problems is. Is it to use some sort of factorization of the zeros/poles of the function and then to use Cauchy's estimate?
My thought process for this particular problem has been as follows:
Consider $g(z) = \frac{1}{f(\frac{1}{z})}$. Then $h(z)=0$ for $z=0$. $h$ cannot vanish identically or else $f$ is identically $\infty$ and then, $f$ does not have a nonessential singularity at $\infty$. Thus, we know that $g(z) = z^h g_h(z)$ where $g_h$ is an analytic function with $g_h(0) \neq 0$. Then we get $\frac{1}{f(\frac{1}{z})} = z^h \frac{1}{f_h(\frac{1}{z})}$ (where $\frac{1}{f_h(\frac{1}{z})} = g_h(z)$), which implies that $f(\frac{1}{z}) = z^{-h} f_h(\frac{1}{z})$. This last expression can also be equivalently written as $f(z) = z^h f_h(z)$. Then, since $f_h(z)$ is bounded as $|z| \to \infty$, we can use Cauchy's estimate to show that the $(h+1)$-th derivative of $f$ is equal to $0$ for all $z$, which implies that the function must reduce to a polynomial.
I'd appreciate it if any of you could tell me if that's correct and/or if there is a better/different way to go about answering the question.
As an extra note, would it be wrong to assume that any approach toward these problems with polynomials is also going to extend in some sense toward problems about rational functions?
Best Answer
When I look at this problem, Cauchy's Integral Formula definitely jumps out. Other than that, I am not too sure of any "general" approaches.
Solution: Recall the singularity at $\infty$ of $f(z)$ is the same as the singularity at $0$ of $f\left(\frac{1}{z}\right)$. Say $f$ has a pole of order $m$ at $\infty$. Then $$z^m\cdot f\left(\frac{1}{z}\right)$$ can be made is an analytic function at $z=0$. Thus, $f(z)=O(z^m)$ as $z\rightarrow \infty$.
Now apply Cauchy's Integral Formula to a sufficiently high derivative of $f$. Using the contour $C_R$, the circle of radius $R$, we can find that this derivative will be bounded on the complex plane. (Show it is bounded in the circle of radius $r$ for every $r$ by taking $R$ large enough, and using $f(z)=O(z^m)$) Hence by Liouvilles Theorem it is a constant.
Thus we conclude $f$ is a polynomial.
Approaches: It looks like your solution is basically the same as what I wrote, so this answer may not be too much help. (Although, I don't fully see the use of defining the function $g$?) I am curious to see what other solutions people conceive. In short, I think you are thinking about it in a fairly good way.
Alternative: Suppose $f$ is analytic but not a polynomial. Then in the Taylor expansion around $0$, there must be infinitely many terms. Hence, the Taylor expansion of $f(1/z)$ at $0$ will have arbitrarily large negative powers, so that $0$ cannot be a pole of finite order.
Alternative 2: Recall that if $h(z)$ has a pole of order $n$ at $z=a$, then $h'(z)$ has a pole of order $n + 1$ at $z = a$ ($n\geq 1$). Consequently, if $f(1/z)$ has a pole of order $m$ at $0$, then $$f'(1/z)\cdot \frac{-1}{z^2}$$ has a pole order $n+1$, and hence $f'(1/z)$ has a pole order $n-1$. Thus, $f^{(n)}(z)$ is analytic everywhere on the Riemann sphere, and hence a constant.
Edit: I added two alternative solutions. I view them as the same, since the key ideas employed in each are identical.