Non-polynomial entire function with finitely many zeros tends to zero on circles

Question

Suppose $f$ is entire with finitely many zeros. Assume $f$ is not a polynomial. Let $m(r) = \inf_{|z| = r} |f(z)|$. I want to show that $$\lim_{r \rightarrow \infty} m(r) =0.$$ I want to note that a previous problem required me to prove that the function $M_f(r) = \sup_{|z|=r} |f(z)|$ is a non-decreasing function of $r$ (should be true for any entire function).

A hint I was given for this problem was to factor out the zeros to obtain a non-vanishing entire function $g$. Then supposing that $m(r)$ does not go to zero, I should try to estimate $m(r)$ using $M_{1/g}(r)$ and show that $1/g$ is a polynomial.

My current progress so far is that I can write $f(z) = g(z) \prod_{i=1}^k (z-z_i)^k$ where $z_1, \ldots, z_k$ are the finitely many zeros of $f$, the $k_i$ are the respective orders of the zeros, and $g(z)$ is entire and non-vanishing. Then analyzing $M_{1/g}(r)$, we have for large enough $r$ so that $f(z) \neq 0$, $$ M_{1/g}(r) = \sup_{|z|=r} \frac{1}{|g(z)|} = \sup_{|z|=r} \frac{\prod_{i=1}^k |z-z_i|^k}{|f(z)|} \leq \frac{\sup_{|z|=r}\prod_{i=1}^k |z-z_i|^k}{\inf_{|z|=r}|f(z)|} = \frac{\sup_{|z|=r}\prod_{i=1}^k |z-z_i|^k}{m(r)}. $$ So rearranging gives us $$ m(r) \leq \frac{\sup_{|z|=r}\prod_{i=1}^k |z-z_i|^k}{M_{1/g}(r)}. $$ From here, I'm not sure how to make use the assumption that $m(r)$ does not go to zero and how to show $1/g$ is a polynomial. Furthermore, I'm not sure how that leads to contradiction. I would appreciate any hints.

Answer 1

Choose $R > 0$ so large that all zeros of $f$ are contained in $|z| < R$, so that $1/f$ is holomorphic for $|z| > R$. If $m(r)$ does not converge to zero then there is a sequence $$ R < r_1 < r_2 < r_3 < \cdots $$ with $r_n \to \infty$ and a constant $\epsilon > 0$ such that $m(r_n) \ge \epsilon$. Then $$ M_{1/f} (r_n) \le \frac{1}{\epsilon} $$ for all $n$, and since $1/f$ is holomorphic for $|z| > R$ it follows by the maximum modulus principle that $$ \frac{1}{|f(z)|} \le \frac{1}{\epsilon} \text{ for } |z| \ge r_1 \, . $$ So $1/f$ can not have a pole or an essential singularity at $\infty$ (the latter is excluded by the Casorati–Weierstrass theorem, one can also use Riemann's theorem on removable singularities here).

It follows that $1/f$ has a removable singularity at $\infty$, so that $f$ has a removable singularity or a pole at $\infty$, and that means that $f$ is a polynomial.

Answer 2

(1) You have $f(z) = p(z)\cdot g(z)$ where $p$ is a degree $d$ polynomial function and $g$ is non-zero and entire. Suppose $\vert f(z)\vert \geq M \gt 0$ for all $\vert z \vert \geq R$ for some positive number $R$, then the entire function given by $\frac{1}{g(z)}$ observes for all large enough $\vert z \vert$
$\big \vert \frac{1}{g(z)}\big \vert =\big \vert \frac{p(z)}{f(z)}\big \vert\leq \frac{1}{M}\vert p(z)\vert \leq \vert z\vert^{d+1}$
$\implies \frac{1}{g(z)}$ is a polynomial [e.g. by Cauchy's Inequalities], albeit has no zeros hence is constant$\implies f$ is a polynomial function.

(2) suppose WLOG that all zeros of (non-constant) $f$ occur in the unit disc. Then restrict the domain to the closed Annulus $A_r$ given by $1\leq \vert z\vert \leq r$. For reasons of compactness $\vert f\vert$ attains a minimum on $A_r$ but since it is non-zero, the open mapping theorem implies the minimum modulus cannot be in the interior hence it must be on the boundary. Then $m_r := \min_{z \in A_r}\vert f(z)\vert$ and for any strictly increasing sequence $r_i\geq 1$ where $r_i\to \infty$ gives $m_{r_1}\geq m_{r_2}\geq m_{r_3}\geq\dots$ which is a monotone non-increasing sequence bounded below by zero, hence has a limit $L$. We conclude $L=0$, since otherwise $(1)$ would tell us that $f$ is a polynomial function. [And for avoidance of doubt, notice this implies the minimum occurs on the outer boundary for all but finitely many of the $r_i$ in the sequence.]