Suppose $f(z)$ is analytic on $\mathbb{C}\setminus\{0\}$ and $|f(z)| \leq |z|^{5/2} + |z|^{-1/2}$ for all $z\in \mathbb{C}\setminus\{0\}$...

Question

This is a question from a previous qualifying exam that I'm working through to study for my own upcoming qual. I can't seem to figure out where to go with it, so if someone could walk me through this proof, it would be very helpful.

Problem:

Suppose $f(z)$ is analytic on $\mathbb{C}\setminus\{0\}$ and $|f(z)| \leq |z|^{5/2} + |z|^{-1/2}$ for all $z\in \mathbb{C}\setminus\{0\}$. Prove that $f(z)$ is a polynomial of degree at most two.

Thoughts:

I don't think I can use that there exists a power series $f(z) = \sum_{k=0}^\infty a_kz^k$, because zero is excluded from the domain. (Is this right?)

I was thinking of instead using that there exists a Laurent series (since it can be in an annulus) such that $$ f(z) = \sum_{k=0}^\infty a_kz^k + \sum_{k=1}^\infty a_{-k}z^{-k}, $$

and then try to bound this somehow? Can someone please walk me through this or a better method for this problem?

Update (based on answer from @Martin R):

Continuing with the Laurent series $$ f(z) = \sum_{-\infty}^{\infty} a_kz^k, $$

a generalization of the Cauchy Integral Theorem yields the coefficients $$ a_k = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z^{k+1}}\; dz $$

for a positively oriented curve $\gamma: \{|z|=r,\; r>0\}$. We now consider the magnitude of these coefficients, such that $$ |a_k| = \left\vert\frac{1}{2\pi i}\right\vert\cdot\left\vert \oint_\gamma \frac{f(z)}{z^{k+1}}\; dz\right\vert = \frac{1}{2\pi }\cdot\left\vert \oint_\gamma \frac{f(z)}{z^{k+1}}\; dz\right\vert $$

Since we have a bound for $f(z)$, we can use the ML estimation lemma such that $$ |a_k| \leq \frac{1}{2\pi}\cdot ML, $$

where $M = \max_{|z|=r} \left\vert\frac{f(z)}{z^{k+1}}\right\vert = \frac{\max_{|z|=r} |f(z)|}{r^{k+1}}$ and $L = 2\pi r$ is the length of the curve, which yields $$ |a_k| \leq \frac{1}{2\pi}\cdot \left(\frac{\max_{|z|=r} |f(z)|}{r^{k+1}}\right)\cdot 2\pi r = \frac{\max_{|z|=r} |f(z)|}{r^{k}}. $$

The given bound on $f(z)$ is $|f(z)| \leq |z|^{5/2} + |z|^{-1/2}$, thus we have $$ |a_k| \leq \frac{r^{5/2} + r^{-1/2}}{r^{k}} = r^{5/2-k} + r^{-1/2-k}. $$

Now, since we want this result to hold for all $z\in \mathbb{C}\setminus \{0\}$, we must consider what happens as $r$ gets very large.

For $k>\frac{5}{2}$, we find that $|a_k|\to 0$ as $r\to\infty$. Similarly, for $k<-\frac{1}{2}$. Thus, for this bound to hold, we must have $-\frac{1}{2} < k < \frac{5}{2}$.

Returning to our original Laurent series then, we must have $$ f(z) = \sum_{k>-1/2}^{k<5/2} a_kz^k = \sum_{k=0}^{2} a_kz^k, $$ since $k$ must be an integer. This means that $f(z)$ must be a polynomial of degree 2 or less, as desired. $\blacksquare$

Answer 1

If $f$ is analytic in $\Bbb C \setminus \{ 0 \}$ then – as you said – it can be developed into a Laurent series $$ f(z) = \sum_{n=-\infty}^\infty a_n z^n \, , $$ and the coefficients of the Laurent series are given by $$ a_n = \frac{1}{2 \pi i}\int_\gamma \frac{f(z)}{z^{n+1}} \, dz $$ where $\gamma$ is any closed curve surrounding the origin exactly once in counter-clockwise (mathematical positive) direction.

Choosing a circle of radius $r$ as the integration path and taking absolute values then shows that $$ |a_n| \le \frac{\max_{|z|=r} |f(z)|}{r^n} $$ for all $r > 0$.

Now assume that $f$ satisfies a growth condition $$ \tag{$*$} |f(z)| \le A |z|^\alpha + B |z|^{-\beta} $$ with non-negative real constants $A, B, \alpha, \beta$. Then $$ \tag{$**$} |a_n| \le A r^{\alpha - n} + B r^{-\beta -n} $$ for all $r > 0$. If $n > \alpha$ then the right-hand side of $(**)$ tends to zero for $r \to \infty$, so that necessarily $a_n = 0$. And if $n < - \beta$ then the right-hand side of $(**)$ tends to zero for $r \to 0$, so that also $a_n = 0$.

This shows that if $f$ satisfies the growth condition $(*)$ then $a_n \ne 0$ is only possible for $-\beta \le n \le \alpha$, i.e. $f$ is a rational function with the finite Laurent series $$ f(z) = \sum_{n = \lceil - \beta \rceil}^{\lfloor \alpha \rfloor} a_n z^n \, . $$

In your case, with $\alpha = 5/2$ and $\beta = -1/2$, the only non-zero Laurent coefficients are in the range $-1/2 \le n \le 5/2$, and $f(z) = a_0 + a_1 z + a_2 z^2$ is a polynomial of degree at most two.