Entire function is polynomial of degree $n$ iff $z^{n}f(1/z) \to \alpha$ as $z \to 0$.

Question

I am working through old qualifier questions in analysis over break, and I was hoping someone would be willing to help me verify (and correct) my proof for the following statement. I feel fairly confident in the forward side, but less so in the converse. In particular, am I on the right track in reaching for uniform convergence to move the limit inside the series? If not, what tools should I be reaching for instead? Thank you in advance for any help!

Theorem: Let $f$ be an entire function. Prove carefully that $f$ is a polynomial of degree $n$ if and only if there exists $\alpha \in \mathbb{C}$ such that $\lim_{z \to 0} z^{n} f(1/z) = \alpha$.

Proof: Suppose first $f$ is a polynomial of degree $n$. Then: $$f(z) = \sum_{k=0}^{n} a_{k}z^{k}$$

And:

$$z^{n}f(1/z) = \sum_{k=0}^{n} a_{k} z^{n-k}$$

And $z^{n}f(1/z) \to a_{n}$ as $z \to 0$. So $a_{n}$ is our choice of $\alpha$.

Conversely, let $\alpha \in \mathbb{C}$ and suppose $\lim_{z \to 0} z^{n}f(1/z) = \alpha$. As $f$ is entire, we have that $f$ is equal to its power series representation for all $z \in \mathbb{C}$: $$f(z) = \sum_{k=0}^{\infty} \dfrac{f^{(k)}(0)}{k!} z^{k}$$

As $f$ is entire and the Cauchy root test, this power series converges uniformly on $\mathbb{C}$. Now consider: $$z^{n}f(1/z) = \sum_{k=0}^{\infty} \dfrac{f^{(k)}(0)}{k!} z^{n-k}$$

I show that the power series representation of $z^{n}f(1/z)$ converges uniformly using the Weierstrass $M$-test. Let $r > 0$. We have that for any $z \in B_{r}(0)$ that: $$\left|\dfrac{f^{(k)}(0)}{k!} z^{n-k} \right| \leq \left|\dfrac{f^{(k)}(0)}{k!} \right| r^{k}$$

Let $M_{k} := \left|\dfrac{f^{(k)}(0)}{k!} \right| r^{k}$. By the Ratio Test, we see that $\sum M_{k}$ converges absolutely. So by the Weierstrass $M$-test, the power series representation of $z^{n}f(1/z)$ converges uniformly in $B_{r}(0)$. As $r$ was arbitrary, the power series representation of $z^{n}f(1/z)$ converges uniformly everywhere. Thus, we can move the limit inside the series:

$$\alpha = \lim_{z \to 0} z^{n} f(1/z) = \lim_{z \to 0} \sum_{k=0}^{\infty} \dfrac{f^{(k)}(0)}{k!} z^{n-k} = \sum_{k=0}^{\infty} \left(\lim_{z \to 0} \dfrac{f^{(k)}(0)}{k!} z^{n-k} \right) $$

It follows that $\lim_{z \to 0} \dfrac{f^{(k)}(0)}{k!} z^{n-k} = 0$ whenever $|n-k| < 0$, which implies that $f^{(k)}(0) = 0$ whenever $n-k < 0$. Now when $k = n$, $\lim_{z \to 0} \dfrac{f^{(n)}(0)}{k!} z^{0} = \alpha$. So $f$ is a polynomial of degree $n$ (provided $\alpha \neq 0$). QED.

Answer 1

Since $$ \begin{align} \alpha &=\lim_{z\to0}z^nf\left(\frac1z\right)\\ &=\lim_{z\to\infty}\frac1{z^n}f(z) \end{align} $$ as $z\to\infty$ $$ f(z)=\alpha z^n+o\!\left(z^n\right) $$ Thus, for all $k\gt n$, we have $$ \begin{align} f^{(k)}(0) &=\frac{k!}{2\pi i}\oint_{|z|=R}\frac{f(z)}{(z-0)^{k+1}}\,\mathrm{d}z\\ &=\lim_{R\to\infty}\frac{k!}{2\pi i}\oint_{|z|=R}\frac{f(z)}{z^{k+1}}\,\mathrm{d}z\\ &=\lim_{R\to\infty}\frac{k!}{2\pi i}\oint_{|z|=R}\frac{\alpha z^n+o\!\left(z^n\right)}{z^{k+1}}\,\mathrm{d}z\\[6pt] &=0 \end{align} $$ Therefore, the Taylor series has degree at most $n$.

Furthermore, $$ \begin{align} f^{(n)}(0) &=\frac{n!}{2\pi i}\oint_{|z|=R}\frac{f(z)}{(z-0)^{n+1}}\,\mathrm{d}z\\ &=\lim_{R\to\infty}\frac{n!}{2\pi i}\oint_{|z|=R}\frac{f(z)}{z^{n+1}}\,\mathrm{d}z\\ &=\lim_{R\to\infty}\frac{n!}{2\pi i}\oint_{|z|=R}\frac{az^n+o\!\left(z^n\right)}{z^{n+1}}\,\mathrm{d}z\\[6pt] &=\alpha n! \end{align} $$ Thus, the coefficient of $z^n$ in $f(z)$ is $\alpha$. Therefore, $f$ has degree $n$.

Answer 2

Consider the holomorphic function $g(z)=f(z)-\sum_{k=0}^{n-1}\frac{f^{(k)}(0)}{k!}z^k$ on $\mathbb{C}$. It too satisfies $$\lim_{z\to\infty}\frac1{z^n}g(z)=\alpha$$ and since the power series of $g$ starts in degree $n$ we get that $$\lim_{z\to 0}\frac1{z^n}g(z)=\frac{f^{(n)}(0)}{n!}$$ so that $z\mapsto\frac1{z^n}g(z)$ is bounded near zero. By the Riemann extension theorem it extends to an entire function. Since that function has a limit at infinity by the first point, it is bounded, hence constant by Liouville's theorem, equal to $\alpha$ at infinity and equal to $\frac{f^{(n)}(0)}{n!}$ at zero, so that $\alpha=\frac{f^{(n)}(0)}{n!}$.

Hence $f$ is polynomial of degree $\leq n$.

Answer 3

Your use of the Weierstrass $M$-test does not work. In particular, the inequality $$ \left| \frac{f^{(k)}(0)}{k!}z^{n-k}\right| \leq \left| \frac{f^{(k)}(0)}{k!}\right|r^k $$ is false if $f^{(k)}(0) \neq 0$ and $k>n$, because the LHS blows up as $z\to 0$ and the RHS goes to $0$ as $r\to 0$, a contradiction. In particular, notice that by this point you have not used anything about the behavior of $f$ other than that it is entire: by your logic, you would conclude that $z^ng(1/z)$ is analytic in a neighborhood of $0$ for every entire function $g$, which is not true, as exemplified by the standard example $g(z) = e^z$, $g(1/z) = e^{1/z}$. (The first is entire, the second has an essential singularity at $0$.)

I think there are a few ways that work for this result. The simplest I can think of is to replace $z$ by $1/z$ to see that $|f(z)| \leq |\alpha| R^n$ for $|z|=R$ sufficiently large. What happens if you apply this estimate with Cauchy's estimates for $f^{(m)}(0)$, $m>n$?