The convergence of a sequence of polynomials and the distribution of their zeros

TL;DR

This paper investigates the conditions under which sequences of polynomials with zeros constrained to the upper half-plane or with bounded non-real zeros converge uniformly, extending classical theorems and conjectures in complex analysis.

Contribution

It establishes new convergence criteria for polynomial sequences with zeros in specific regions, confirming a theorem of Benz and a conjecture of Pólya.

Findings

Sequences with zeros in the upper half-plane converge uniformly if derivatives at zero stabilize.

Real polynomial sequences with bounded non-real zeros also exhibit uniform convergence.

Results extend classical theorems and confirm longstanding conjectures in polynomial zero distribution.

Abstract

Suppose that $\langle f_n \rangle$ is a sequence of polynomials, $\langle f_n^{(k)}(0)\rangle$ converges for every non-negative integer $k$, and that the limit is not $0$ for some $k$. It is shown that if all the zeros of $f_1, f_2, \dots$ lie in the closed upper half plane $\rm{Im}\ z\geq 0$, or if $f_1, f_2, \dots$ are real polynomials and the numbers of their non-real zeros are uniformly bounded, then the sequence converges uniformly on compact sets in the complex plane. The results imply a theorem of Benz and a conjecture of P\'olya.

Equations

$$f(z)=c{z}^m{e}^{-\alpha z^2+\beta z}\prod _j\left(1-\frac{z}{a_j}\right)e^{z/a_j}(z\in \mathbb{C}),$$

$$\frac{1}{k!}\underset{n\to \infty }{\lim }{f}_n^{(k)}(0)$$

$${\tilde{s}}_k(f;X)=\sum _{a_j\in X}|a_j{|}^{-k};$$

$$s_k(f;X)=\sum _{a_j\in X}{a}_j^{-k}.$$

$$f(z)=f(0)\exp \left(-\sum _{k=1}^{\infty }\frac{1}{k}{s}_k(f)z^k\right)(z\in D_r).$$

$$s_k(f)=\frac{-1}{(k-1)!}{\left(\frac{f^{\prime }}{f}\right)}^{(k-1)}(0)(k=1,2,\dots ),$$

$$\underset{n\to \infty }{\lim }{s}_k(f_n)=\frac{-1}{(k-1)!}{\left(\frac{f^{\prime }}{f}\right)}^{(k-1)}(0)(k=1,2,\dots ).$$

$$|1-z|\le \exp \left(-\text{Re}\sum _{k=1}^{p-1}\frac{z^k}{k}+c_p|z{|}^p\right)(z\in \mathbb{C}).$$

$$|f(z)|\le e^{{\tilde{s}}_1(f)|z|}(z\in \mathbb{C}),$$

$$|f(z)|\le \exp \left(-\text{Re}\sum _{k=1}^{p-1}\frac{1}{k}{s}_k(f)z^k+c_p{\tilde{s}}_p(f)|z{|}^p\right)(z\in \mathbb{C}).$$

$$f(z)=e^{\alpha z^p}g(z)(z\in \mathbb{C}),$$

$$|f_n(z)|\le a{e}^{b|z{|}^p}(n=1,2,\dots ;z\in \mathbb{C}),$$

$$|f(z)|\le a{e}^{b|z{|}^p}(z\in \mathbb{C}),$$

$${\tilde{s}}_p(f)=\underset{R\to \infty }{\lim }{\tilde{s}}_p(f;D_R)=\underset{R\to \infty }{\lim }\underset{n\to \infty }{\lim }{\tilde{s}}_p(f_n;D_R)\le M;$$

$$S_c=\left\{z\in \mathbb{C}:|z|\le \sqrt{1+c^2}\text{Re}z\right\}.$$

$$S_c^{1/p}=\{z\in \mathbb{C}:z^p\in S_c\}.$$

$$z\in S_c^{1/p}\setminus \{0\}\iff z^{-1}\in S_c^{1/p}\setminus \{0\}.$$

$$f(z)=e^{\alpha z^{2p}}g(z)(z\in \mathbb{C}),$$

$$z\in S_{\infty }^{1/p}\setminus S_c^{1/p}\Rightarrow -z^{2p}\in S_{c_1}.$$

$${\tilde{s}}_p(g_n)\le \sqrt{1+c^2}\text{Re}{s}_p(g_n)\le \sqrt{1+c^2}\text{Re}{s}_p(f_n)$$

$${\tilde{s}}_{2p}(h_n)\le -\sqrt{1+c_1^2}\text{Re}{s}_{2p}(h_n)\le \sqrt{1+c_1^2}\left|s_{2p}(h_n)\right|$$

$$f(z)=f(0)e^{\alpha z^2-\beta z}\prod _j\left(1-\frac{z}{a_j}\right)e^{z/a_j}(z\in \mathbb{C}),$$

$$\sum _j\text{Re}\left(a_j^{-1}\right)\le \text{Re}\beta$$

$$\beta =-\left(\frac{f^{\prime }}{f}\right)(0)=\underset{n\to \infty }{\lim }{s}_1(f_n)$$

$$2\alpha -s_2(f)={\left(\frac{f^{\prime }}{f}\right)}^{\prime }(0)=-\underset{n\to \infty }{\lim }{s}_2(f_n).$$

$$\sum _j\text{Re}\left(a_j^{-1}\right)=\underset{R>0}{\sup }\text{Re}{s}_1(f;D_R)$$

$$\text{Re}{s}_1(f;D_R)$$

$$=\underset{n\to \infty }{\lim }\text{Re}{s}_1(f_n;D_R)$$

$$\le \underset{n\to \infty }{\mathrm{liminf}}\text{Re}{s}_1(f_n)$$

$$=\text{Re}\beta$$

$$\left|s_2(f_n;S_c\setminus D_R)\right|$$

$$\le R^{-1}{\tilde{s}}_1(f_n;S_c)$$

Full text

††dedicatory: Dedicated to the memory of the late Professor Jehpill Kim (1930–2016). ††support: This work was supported by the SNU Mathematical Sciences Division for Creative Human Resources Development.

1. Introduction

This paper is concerned with the distribution of zeros of entire functions and the convergence of polynomial sequences. Let $f$ be an entire function which is not identically equal to [math]. We denote the zero set of $f$ by $\Cal{Z}(f)$, that is, $\Cal{Z}(f)=\{z\in\mathbb{C}:f(z)=0\}$. If $X$ is a subset of the complex plane, the number of zeros of $f$ that lie in $X$ counted according to their multiplicities will be denoted by $N(f;X)$. In the case where $f$ is identically equal to [math], we set $\Cal{Z}(f)=\emptyset$ and $N(f;X)=0$. The closed upper half plane $\{z\in\mathbb{C}:\text{\rm{Im}}\,z\geq 0\}$ is denoted by $\mathbb{H}$, and for $r>0$ the open disk $\{z\in\mathbb{C}:|z|<r\}$ is denoted by $D_{r}$.

The Pólya-Obrechkoff class $\Cal{PO}$ is the collection of entire functions $f$ which can be represented in the form

[TABLE]

where $c$ is a constant, $m$ is a nonnegative integer, $\alpha\geq 0$, $a_{1},a_{2},\dots\in{\mathbb{H}}\setminus\{0\}$, $\sum_{j}|a_{j}|^{-2}<\infty$ and $\sum_{j}\text{\rm{Im}}\,\left(-a_{j}^{-1}\right)\leq\text{\rm{Im}}\,\beta$. If $f\in\Cal{PO}$, then it is easy to see that there is a sequence $\langle f_{n}\rangle$ of polynomials such that $f_{n}\to f$ uniformly on compact sets in the complex plane and $\Cal{Z}(f_{n})\subset{\mathbb{H}}$ for all $n$. In 1914 Lindwart and Pólya proved a strong version of the converse [4, Satz II B].

Theorem (Lindwart-P'olya)

Suppose that $\langle f_{n}\rangle$ is a sequence of polynomials, $\langle f_{n}\rangle$ converges uniformly on a disk $D_{r}$, the limit is not identically equal to [math], and that $\Cal{Z}(f_{n})\subset{\mathbb{H}}$ for all $n$. Then it converges to an entire function $f\in\Cal{PO}$ uniformly on compact sets in the complex plane.

Since then “a disk $D_{r}$” in the theorem has been replaced with considerably smaller sets by several authors: Korevaar and Loewner replaced it with a certain arc [3], Levin replaced it with a certain countable set, and later Clunie and Kuijlaars proved Levin’s theorem in a new and simple way, and extended it [2].

In this paper, we approach the problem in a different direction. If a sequence $\langle f_{n}\rangle$ of analytic functions converges uniformly on a disk $D_{r}$, then for every nonnegative integer $k$ the sequence $\langle f_{n}^{(k)}(0)\rangle$ of complex numbers converges, but the converse does not hold in general. Let us say that $\langle f_{n}\rangle$ converges weakly if $\langle f_{n}^{(k)}(0)\rangle$ converges for every nonnegative integer $k$. In this case, the limit is the formal power series whose $k$-th coefficient is given by

[TABLE]

for every $k$. For notational simplicity, we say that a sequence of entire functions converges strongly if it converges uniformly on compact sets in the complex plane.

In 1934, Benz proved that a formal power series $f$ represents an entire function (the radius of convergence is $\infty$) in $\Cal{PO}$ if there is a sequence $\langle f_{n}\rangle$ of polynomials such that $f_{n}\to f$ weakly and $\Cal{Z}(f_{n})\subset{\mathbb{H}}$ for all $n$ [1, Satz 2]. However, the result implies nothing about uniform convergence of the sequence, except in the case of the Jensen sequence introduced in Section 3 below.

As our first result, we improve the Lindwart-Pólya theorem as well as Benz’s result.

Theorem 1.1

Suppose that $\langle f_{n}\rangle$ is a sequence of polynomials, $f_{n}\to f$ weakly, some coefficient of $f$ is not [math], and that $\Cal{Z}(f_{n})\subset{\mathbb{H}}$ for all $n$. Then $f$ represents an entire function in the Pólya-Obrechkoff class and $f_{n}\to f$ strongly.

Remark Remarks

(i) As the example $f_{n}(z)=z^{n}$ shows, the condition that some coefficient of $f$ is not [math] is necessary. (ii) It may be remarked that in the general case any of the convergence conditions considered in [2] and [3] does not imply weak convergence and vice versa.

If $f$ is of the form (1.1) with $c,\beta,a_{1},a_{2},\dots\in\mathbb{R}$, then $f$ is said to be in the Laguerre-Pólya class $\Cal{LP}$. It is clear that $\Cal{LP}\subset\Cal{PO}$, $\Cal{Z}(f)\subset\mathbb{R}$ for all $f\in\Cal{LP}$, and that $f\in\Cal{LP}$ if and only if $f\in\Cal{PO}$ and $f$ is a real entire function, that is, $f(\mathbb{R})\subset\mathbb{R}$. In [5], Pólya introduced another extension of $\Cal{LP}$, namely the class $\Cal{LP}^{*}$ of real entire functions $f$ which are of the form $f=Pg$ where $P$ is a real polynomial and $g\in\Cal{LP}$. If $f\in\Cal{LP}^{*}$, then $N(f;\mathbb{C}\setminus\mathbb{R})<\infty$ and there is a sequence $\langle f_{n}\rangle$ of real polynomials such that $f_{n}\to f$ strongly and $N(f_{n};\mathbb{C}\setminus\mathbb{R})=N(f;\mathbb{C}\setminus\mathbb{R})$ for all $n$. As Pólya stated in the same paper, if a sequence $\langle f_{n}\rangle$ of real polynomials converges uniformly on a disk $D_{r}$, the limit is not identically equal to [math] and $\langle N(f_{n};\mathbb{C}\setminus\mathbb{R})\rangle$ is bounded above, then it converges strongly to an entire function in $\Cal{LP}^{*}$ [5, §4 Satz II]. It is not clear whether the results of [2] and [3] extend to the case of $\Cal{LP}^{*}$, but weak convergence implies strong convergence in this case too.

Theorem 1.2

Suppose that $\langle f_{n}\rangle$ is a sequence of real polynomials, $f_{n}\to f$ weakly, some coefficient of $f$ is not [math], and that $\langle N(f_{n};\mathbb{C}\setminus\mathbb{R})\rangle$ is bounded. Then $f$ represents an entire function in $\Cal{LP}^{*}$ and $f_{n}\to f$ strongly.

As we shall see in the sequel, this theorem implies a conjecture of Pólya which has remained open since 1915.

In Section 2, we prove Theorems 1.1 and 1.2 by generalizing a theorem of Lindwart and Pólya. Finally, we apply the results to give a simple proof of the original version of Benz’s theorem mentioned above and to improve some classical theorems of Pólya (Section 3).

2. Proofs of Theorems 1.1 and 1.2

In this section, we obtain some generalizations (Theorems 2.3 and 2.4 below) of results in [4], and prove Theorems 1.1 and 1.2.

Let $f$ be an entire function such that $f(0)\neq 0$, and let $a_{1},a_{2},\dots$ be the zeros of $f$ listed according to their multiplicities. If $X\subset\mathbb{C}$ and $k$ is a positive integer, we put

[TABLE]

and in the case where $\tilde{s}_{k}(f;X)<\infty$, we put

[TABLE]

We also write $s_{k}(f)=s_{k}(f;\mathbb{C})$ and $\tilde{s}_{k}(f)=\tilde{s}_{k}(f;\mathbb{C})$.

Suppose that $f$ is a polynomial and $f(0)\neq 0$. Then there is a positive constant $r$ such that

[TABLE]

Thus we have

[TABLE]

and it follows that $s_{k}(f)$ is a rational function of $f(0),f^{\prime}(0),\dots,f^{(k)}(0)$ for $k=1,2,\dots$. For instance, $s_{1}(f)=-f^{\prime}(0)/f(0)$, $s_{2}(f)=\left(f^{\prime}(0)^{2}-f(0)f^{\prime\prime}(0)\right)/f(0)^{2}$, and so on. This observation leads to the following:

Proposition 2.1

Suppose that $\langle f_{n}\rangle$ is a sequence of polynomials, $f_{n}(0)\neq 0$ for all $n$, $\lim_{n\to\infty}f_{n}(0)\neq 0$ and $f_{n}\to f$ weakly. Then the sequences $\langle s_{k}(f_{n})\rangle$, $k=1,2,\dots$, are all convergent. If $f$ represents an analytic function in a neighborhood of [math], then we have

[TABLE]

It is obvious that $|1-z|\leq e^{|z|}$ for all $z\in\mathbb{C}$. More generally, we have the following, which is also trivially proved.

Proposition 2.2

Let $p\geq 2$ be an integer. Then there is a positive constant $c_{p}$ such that

[TABLE]

Corollary

Suppose that $f$ is a polynomial and $f(0)=1$. Then

[TABLE]

and for $p=2,3,\dots$ we have

[TABLE]

In [4], Lindwart and Pólya proved the following theorem under the assumption that the sequence converges uniformly on a disk $D_{r}$. Fortunately, their proof works in the case of weak convergence as well.

Theorem 2.3

Suppose that $p$ is a positive integer, $M>0$, $\langle f_{n}\rangle$ is a sequence of polynomials, $f_{n}(0)\neq 0$ for all $n$, $\lim_{n\to\infty}f_{n}(0)\neq 0$, $f_{n}\to f$ weakly, and that $\tilde{s}_{p}(f_{n})\leq M$ for all $n$. Then $f_{n}\to f$ strongly, and $f$ is of the form

[TABLE]

where $\alpha$ is a constant and $g$ is an entire function of genus at most $p-1$.

Demonstration Proof

By replacing $\langle f_{n}\rangle$ with $\langle f_{n}(0)^{-1}f_{n}\rangle$, we may assume that $f_{n}(0)=1$ for all $n$. Since $\langle f_{n}\rangle$ converges weakly and $\tilde{s}_{p}(f_{n})\leq M$ for all $n$, there are positive constants $a$ and $b$ such that

[TABLE]

by Proposition 2.1 and the corollary to Proposition 2.2. In particular, the polynomials $f_{1},f_{2},\dots$ are uniformly bounded on compact sets in the complex plane, hence Montel’s theorem implies that $\langle f_{n}\rangle$ has a subsequence which converges strongly to an entire function; and since $f_{n}\to f$ weakly, the Maclaurin series of the entire function and the formal power series $f$ coincide. Thus $f$ represents an entire function. Furthermore, the same argument shows that every subsequence of $\langle f_{n}\rangle$ has a subsequence which converges to $f$ strongly. Therefore $f_{n}\to f$ strongly.

Finally, (2.1) implies that

[TABLE]

and we have

[TABLE]

hence the last assertion follows from Hadamard’s factorization theorem. ∎

For $0\leq c<\infty$ we denote the sector $\left\{re^{i\theta}:r\geq 0,\ |\theta|\leq\arctan c\right\}$ by $S_{c}$, which may be expressed as

[TABLE]

We denote the closed right half plane $\{z\in\mathbb{C}:\text{\rm{Re}}\,z\geq 0\}$ by $S_{\infty}$. It is clear that $S_{c}$ is closed under addition for $0\leq c\leq\infty$. For $p=1,2,\dots$ and for $0\leq c\leq\infty$ we put

[TABLE]

Each $S_{c}^{1/p}$ is a closed subset of the complex plane, and we have

[TABLE]

The following theorem plays a crucial role in our proofs of Theorems 1.1 and 1.2.

Theorem 2.4

Suppose that $p$ is a positive integer, $\langle f_{n}\rangle$ is a sequence of polynomials, $f_{n}\to f$ weakly, $\lim_{n\to\infty}f_{n}(0)\neq 0$, and that $\Cal{Z}(f_{n})\subset S_{\infty}^{1/p}$ for all $n$. Then $f_{n}\to f$ strongly, $\Cal{Z}(f)\subset S_{\infty}^{1/p}$, and $f$ is of the form

[TABLE]

where $\alpha$ is a constant and $g$ is an entire function of genus at most $2p-1$.

Demonstration Proof

First of all, we may assume that $f_{n}(0)\neq 0$ for all $n$. To show that $\langle\tilde{s}_{2p}(f_{n})\rangle$ is bounded, let $c$ be a real number $>1$ and $c_{1}=2c/(c^{2}-1)$. We have $0<c_{1}<\infty$ and

[TABLE]

Let $\langle g_{n}\rangle$ and $\langle h_{n}\rangle$ be sequences of polynomials such that $f_{n}=g_{n}h_{n}$, $\Cal{Z}(g_{n})\subset S_{c}^{1/p}$ and $\Cal{Z}(h_{n})\subset S_{\infty}^{1/p}\setminus S_{c}^{1/p}$ for all $n$. Then we have

[TABLE]

and

[TABLE]

for all $n$. Since $\langle f_{n}\rangle$ converges weakly, (2.3) and Proposition 2.1 imply that $\langle\tilde{s}_{p}(g_{n})\rangle$ is bounded, and we have $\tilde{s}_{2p}(g_{n})\leq\left(\tilde{s}_{p}(g_{n})\right)^{2}$ for all $n$. Thus $\langle\tilde{s}_{2p}(g_{n})\rangle$ is bounded. Since $s_{2p}(f_{n})=s_{2p}(g_{n})+s_{2p}(h_{n})$ and $|s_{2p}(g_{n})|\leq\tilde{s}_{2p}(g_{n})$ for all $n$, it follows that $\langle s_{2p}(h_{n})\rangle$ is bounded, hence (2.4) implies that $\langle\tilde{s}_{2p}(h_{n})\rangle$ is bounded. Therefore $\langle\tilde{s}_{2p}(f_{n})\rangle$ is bounded, and Theorem 2.3 implies that $f_{n}\to f$ strongly and $f$ is of the form (2.2). Finally, it is clear that $\Cal{Z}(f)\subset S_{\infty}^{1/p}$, because $S_{\infty}^{1/p}$ is closed and $\Cal{Z}(f_{n})\subset S_{\infty}^{1/p}$ for all $n$. ∎

Remark Remark

This theorem is also proved in [4] under the assumption that the sequence converges uniformly on some $D_{r}$. Unlike in the case of Theorem 2.3, our proof is different from theirs.

A large part of the following theorem is known, but we provide a detailed proof for completeness as well as for the reader’s convenience.

Theorem 2.5

Suppose that $\langle f_{n}\rangle$ is a weakly convergent sequence of polynomials, $\lim_{n\to\infty}f_{n}(0)\neq 0$, and that $\Cal{Z}(f_{n})\subset S_{\infty}$ for all $n$. Then $\langle f_{n}\rangle$ converges strongly, and the limit $f$ is of the form

[TABLE]

where $\alpha,\beta$ are constants, $a_{j}\in S_{\infty}\setminus\{0\}$ for all $j$ and $\sum|a_{j}|^{-2}<\infty$. Furthermore,

[TABLE]

and $\alpha\geq 0$.

Demonstration Proof

We need only to prove the last two inequalities, because the remaining assertions are immediate consequences of the case $p=1$ of Theorem 2.4.

First of all, we may assume that $f_{n}(0)\neq 0$ for all $n$. Then Proposition 2.1 implies that

[TABLE]

and

[TABLE]

Since $\text{\rm{Re}}\,\left(a^{-1}\right)\geq 0$ for all $a\in S_{\infty}\setminus\{0\}$, we have

[TABLE]

and

[TABLE]

for every $R>0$. Hence (2.5) holds.

To prove $\alpha\geq 0$, let $c$ be a real number $>1$ and $c_{1}=2c/(c^{2}-1)$. Then we have

[TABLE]

and it follows that

[TABLE]

If $a\in S_{\infty}\setminus S_{c}$ and $a\neq 0$, then $-a^{-2}\in S_{c_{1}}$, hence $2\alpha\in S_{c_{1}}$. Since $c>1$ was arbitrary and $c_{1}\to 0$ as $c\to\infty$, we conclude that $\alpha\in S_{0}=[0,\infty)$. ∎

Remark Remark

If there is a $c$ such that $0\leq c<\infty$ and $\Cal{Z}(f_{n})\subset S_{c}$ for all $n$, then (2.5) and the argument given in the last paragraph of the proof show that $\tilde{s}_{1}(f)<\infty$ and $\alpha=0$; hence we have

[TABLE]

where $\gamma=\beta-s_{1}(f)$. In this case, we have

[TABLE]

Corollary

Suppose that $\langle f_{n}\rangle$ is a weakly convergent sequence of polynomials, $\lim_{n\to\infty}f_{n}(0)\neq 0$, and that $\Cal{Z}(f_{n})\subset\mathbb{H}$ for all $n$. Then $\langle f_{n}\rangle$ converges strongly, and the limit is in the Pólya-Obrechkoff class. If the polynomials $f_{1},f_{2},\dots$ are real, then the limit is in the Laguerre-Pólya class.

Demonstration Proof

If $f$ and $g$ are entire functions and $f(z)=g(iz)$, then $\Cal{Z}(f)\subset S_{\infty}$ if and only if $\Cal{Z}(g)\subset\mathbb{H}$, and $f$ is as in Theorem 2.5 if and only if $g\in\Cal{PO}$. This proves the first assertion. The second assertion is obvious. ∎

Now, we are ready to prove Theorem 1.1.

Demonstration Proof of Theorem 1.1

First of all, it is trivial to see that if $\langle f_{n}\rangle$ converges weakly and $\langle f_{n}^{(k)}\rangle$ converges strongly for some $k$, then the original sequence converges strongly.

Let $k$ be such that $\lim_{n\to\infty}f_{n}^{(k)}(0)\neq 0$. Since $f_{1},f_{2},\dots$ are polynomials, and since $\Cal{Z}(f_{n})\subset\mathbb{H}$ for all $n$, it follows from the Gauss-Lucas theorem that $\Cal{Z}(f_{n}^{(k)})\subset\mathbb{H}$ for all $n$. We have $\lim_{n\to\infty}f_{n}^{(k)}(0)\neq 0$ and it is obvious that $\langle f_{n}^{(k)}\rangle$ converges weakly. Hence the corollary to Theorem 2.5 implies that $\langle f_{n}^{(k)}\rangle$ converges strongly, and we conclude that $f_{n}\to f$ strongly.

It remains to show that $f\in\Cal{PO}$. If $t\in\mathbb{R}$ and $g(z)=f(z+t)$, then it is easy to see that $f\in\Cal{PO}$ if and only if $g\in\Cal{PO}$. Since $f^{(k)}(0)\neq 0$, there is a $t\in\mathbb{R}$ such that $f(t)\neq 0$. If we put $g(z)=f(z+t)$ and $g_{n}(z)=f_{n}(z+t)$, then $g_{n}\to g$ strongly, $g(0)\neq 0$ and $\Cal{Z}(g_{n})\subset\mathbb{H}$ for all $n$, hence the corollary to Theorem 2.5 implies that $g\in\Cal{PO}$, as desired. ∎

In order to prove Theorem 1.2, we need a technical result.

Proposition 2.6

Let $N$ be a positive integer. Then there is a finite set $\Cal{Q}$ of positive even integers having the following property: If $z_{1},\dots,z_{N}\in\mathbb{C}$, then there is some $q\in\Cal{Q}$ such that $z_{1}^{q},\dots,z_{N}^{q}\in S_{\infty}$.

Demonstration Proof

The unit circle $\mathbb{T}=\{\zeta\in\mathbb{C}:|\zeta|=1\}$ is a compact set in $\mathbb{C}$ and is a multiplicative group. For $q=2,4,\dots$ we put

[TABLE]

It is clear that $U_{2},U_{4},\dots$ are open subsets of the compact metric space $\mathbb{T}^{N}$. If $(\zeta_{1},\dots,\zeta_{N})\in\mathbb{T}^{N}$, then $\zeta_{1}^{q},\dots,\zeta_{N}^{q}\in\text{int}\,S_{\infty}$ for some positive even integer $q$. In fact, if $g$ is an element of a sequentially compact topological group, any neighborhood of the unit element contains $g^{n}$ for infinitely many positive integers $n$. We have shown that $\{U_{q}:q=2,4,\dots\}$ is an open cover of $\mathbb{T}^{N}$. Hence there is a finite set $\Cal{Q}$ of positive even integers having the property. ∎

If $t\in\mathbb{R}$ and $g(z)=f(z+t)$, then (as in the proof of Theorem 1.1) $f\in\Cal{LP}^{*}$ if and only if $g\in\Cal{LP}^{*}$.

Demonstration Proof of Theorem 1.2

We first consider the case where $\lim_{n\to\infty}f_{n}(0)\neq 0$. In this case, we may assume that $f_{n}(0)=1$ for all $n$. Let $N$ be a positive integer such that $N(f_{n};\mathbb{C}\setminus\mathbb{R})\leq N$ for all $n$. Then there are sequences $\langle P_{n}\rangle$ and $\langle g_{n}\rangle$ of real polynomials such that $f_{n}=P_{n}g_{n}$, $P_{n}(0)=g_{n}(0)=1$, $\Cal{Z}(P_{n})\cap\mathbb{R}=\emptyset$, $\deg P_{n}\leq N$ and $\Cal{Z}(g_{n})\subset\mathbb{R}$ for all $n$. Let $\Cal{Q}$ be as in Proposition 2.6. Since each $P_{n}$ has at most $N$ zeros, it follows that for every $n$ there is a $q\in\Cal{Q}$ such that $\Cal{Z}(P_{n})\subset S_{\infty}^{1/q}$. Since every $q\in\Cal{Q}$ is even and $\Cal{Z}(g_{n})\subset\mathbb{R}$ for all $n$, we have $\Cal{Z}(g_{n})\subset S_{\infty}^{1/q}$ for all $n$ and for all $q\in\Cal{Q}$. Consequently, for every $n$ there is a $q\in\Cal{Q}$ such that $\Cal{Z}(f_{n})\subset S_{\infty}^{1/q}$; and it follows from Theorem 2.4 that $\langle f_{n}\rangle$ is the union of a finite number ($\leq|\Cal{Q}|$) of strongly convergent subsequences. Therefore $f_{n}\to f$ strongly.

Since $f_{n}\to f$ strongly and $f_{n}(0)=1$ for all $n$, there is a positive constant $r$ such that $\Cal{Z}(f_{n})\cap D_{r}=\emptyset$ for all $n$, and it follows that $\Cal{Z}(P_{n})\cap D_{r}=\emptyset$ for all $n$. Since $\deg P_{n}\leq N$ and $P_{n}(0)=1$ for all $n$, the coefficients of the polynomials $P_{1},P_{2},\dots$ are uniformly bounded. Hence there is a strictly increasing sequence $\langle n(l)\rangle$ of positive integers such that $\langle P_{n(l)}\rangle$ converges strongly to a real polynomial $P$, and it follows that $\langle g_{n(l)}\rangle$ converges uniformly on a disk centered at the origin. Since $g_{n}(0)=1$ and $\Cal{Z}(g_{n})\subset\mathbb{R}$ ($\subset\mathbb{H}$) for all $n$, and since $g_{1},g_{2},\dots$ are real polynomials, the corollary to Theorem 2.5 implies that $\langle g_{n(l)}\rangle$ converges strongly to an entire function $g\in\Cal{LP}$. Therefore $f=Pg\in\Cal{LP}^{*}$.

In the general case, there is an integer $k$ such that $\lim_{n\to\infty}f_{n}^{(k)}(0)\neq 0$. Since $f_{1},f_{2},\dots$ are real polynomials, Rolle’s theorem implies that $N(f_{n}^{(k)};\mathbb{C}\setminus\mathbb{R})\leq N(f_{n};\mathbb{C}\setminus\mathbb{R})$ for all $n$. Hence $\langle f_{n}^{(k)}\rangle$ converges strongly, and it follows that $\langle f_{n}\rangle$ converges strongly. Since $f^{(k)}(0)\neq 0$, there is some $t\in\mathbb{R}$ such that $f(t)\neq 0$, and the same argument as in the proof of Theorem 1.1 shows that $f\in\Cal{LP}^{*}$. ∎

3. Zeros of Appell and Jensen Polynomials

Let $f$ be a formal power series given by

[TABLE]

For $n=0,1,2,\dots$ the $n$-th Appell and Jensen polynomials $A{(f,n)}$ and $J{(f,n)}$ of $f$ are defined by

[TABLE]

respectively. We have

[TABLE]

and it follows that $N(A{(f,n)};\mathbb{C}\setminus\mathbb{R})=N(J{(f,n)};\mathbb{C}\setminus\mathbb{R})$ and $N(A{(f,n)};\mathbb{C}\setminus S_{0})=N(J{(f,n)};\mathbb{C}\setminus S_{0})$ for all $n$. As in [1], we define the sequence $\langle f^{*}_{n}\rangle$ of polynomials by

[TABLE]

which may be called the Jensen sequence of $f$. It is easy to see that $f^{*}_{n}\to f$ weakly, in the general case; that $f^{*}_{n}\to f$ uniformly on a disk $D_{r}$ if and only if the radius of convergence is positive; and that $f^{*}_{n}\to f$ strongly if and only if $f$ represents an entire function. It is also easy to see that if $f_{k}\to f$ weakly as $k\to\infty$, then $A(f_{k},n)\to A(f,n)$ and $J(f_{k},n)\to J(f,n)$ strongly as $k\to\infty$ for every $n$. The following two propositions are known, and will be proved at the end of this section.

Proposition 3.1

If $f$ is a real polynomial, then

[TABLE]

and

[TABLE]

for all $n$.

Remark Remark

Since $N(A{(f,n)};\mathbb{C}\setminus\mathbb{R})=N(J{(f,n)};\mathbb{C}\setminus\mathbb{R})$ and $N(A{(f,n)};\mathbb{C}\setminus S_{0})=N(J{(f,n)};\mathbb{C}\setminus S_{0})$ for all $n$, the term $J(f,n)$ may be replaced with $A(f,n)$.

Proposition 3.2

If $f$ is a polynomial and $\Cal{Z}(f)\subset\mathbb{H}$, then $\Cal{Z}(J(f,n))\subset\mathbb{H}$ for all $n$.

In 1914, Pólya and Schur proved that a formal power series $f$ with real coefficients represents an entire function in $\Cal{LP}$ if and only if $\Cal{Z}(J(f,n))\subset\mathbb{R}$ for all $n$ [7, §6 Satz IV], and later Benz generalized the result [1, Satz 2]. Theorem 1.1 immediately implies Benz’s theorem.

Theorem 3.3

Suppose that $f$ is a formal power series. Then the following are equivalent:

Demonstration Proof

If $f$ represents an entire function in $\Cal{PO}$, then there is a sequence $\langle f_{k}\rangle$ of polynomials such that $f_{k}\to f$ strongly (hence weakly) and $\Cal{Z}(f_{k})\subset\mathbb{H}$ for all $k$.

Next, suppose that $\langle f_{k}\rangle$ is a sequence of polynomials such that $f_{k}\to f$ weakly and $\Cal{Z}(f_{k})\subset\mathbb{H}$ for all $k$. Let $n$ be arbitrary. Then $J(f_{k},n)\to J(f,n)$ strongly as $k\to\infty$, and Proposition 3.2 implies that $\Cal{Z}(J(f_{k},n))\subset\mathbb{H}$ for all $k$. Therefore we have $\Cal{Z}(J(f,n))\subset\mathbb{H}$.

Finally, suppose that $\Cal{Z}(J(f,n))\subset\mathbb{H}$ for all $n$. If every coefficient of $f$ is [math], then it is obvious that $f\in\Cal{PO}$, otherwise the Jensen sequence $\langle f^{*}_{n}\rangle$ satisfies the conditions of Theorem 1.1, and we conclude that $f$ represents an entire function in $\Cal{PO}$. ∎

Remark Remarks

(i) If (2) holds, then Theorem 1.1 implies that the sequence converges strongly, unless $f$ is identically equal to [math]. (ii) Benz proved the implication (3) $\Rightarrow$ (1) by showing that the $f_{n}^{*}\to f$ strongly, but the argument does not imply Theorem 1.1.

The Laguerre-Pólya class has a subclass which plays an important role in the theory of multiplier sequences [7]. Let $\Cal{LP}(S_{0})$ be the class of real entire functions $f$ which are of the form

[TABLE]

where $c\in\mathbb{R}$, $m$ is a non-negative integer, $\alpha\geq 0$, $a_{1},a_{2},\dots>0$ and $\sum_{j}a_{j}^{-1}<\infty$. Likewise, $\Cal{LP}(S_{0})^{*}$ will denote the class of real entire functions of the form $Pg$ where $P$ is a real polynomial and $g\in\Cal{LP}(S_{0})$. It is clear that

[TABLE]

and it is known that $f\in\Cal{LP}(S_{0})^{*}$ if and only if there is a sequence $\langle f_{n}\rangle$ of real polynomials such that $f_{n}\to f$ strongly and $\langle N(f_{n};\mathbb{C}\setminus S_{0})\rangle$ is bounded [5, §4]. Since $N(f;\mathbb{C}\setminus\mathbb{R})\leq N(f;\mathbb{C}\setminus S_{0})$ for every entire function $f$, Theorem 1.2 implies the following:

Theorem 3.4

Suppose that $\langle f_{n}\rangle$ is a sequence of real polynomials, $f_{n}\to f$ weakly, some coefficient of $f$ is not [math], and that $\langle N(f_{n};\mathbb{C}\setminus S_{0})\rangle$ is bounded. Then $f$ represents an entire function in $\Cal{LP}(S_{0})^{*}$ and $f_{n}\to f$ strongly.

Remark Remark

This theorem may be proved directly by the same argument (with $S_{0}$ instead of $\mathbb{R}$) as in the proof of Theorem 1.2. See the remark after the proof of Theorem 2.5.

In the following two theorems, Pólya proved (1) and (2), proved (3) under the additional assumption that the radius of convergence of $f$ is positive [5, §4; 6, Satz II], and conjectured that the assumption could be unnecessary [5, p.246 footnote $^*)$]. It seems that no progress has been made about the conjecture.

Theorem 3.5

Let $f$ be a formal power series with real coefficients, and for $n=0,1,2,\dots$ let $N_{n}=N(A(f,n);\mathbb{C}\setminus S_{0})$. Then

Theorem 3.6

Let $f$ be a formal power series with real coefficients, and for $n=0,1,2,\dots$ let $N_{n}=N(A{(f,n)};\mathbb{C}\setminus\mathbb{R})$. Then

Since the proofs of these theorems are almost identical, we prove the first one only.

Demonstration Proof of Theorem 3.5

The equality in (1) is obvious, and the remaining inequalities follow from Rolle’s theorem, because we have $A(f,n)^{\prime}=nA(f,n-1)$ for every positive integer $n$.

To prove (2), suppose that $f\in\Cal{LP}(S_{0})^{*}$ and $N=N(f;\mathbb{C}\setminus S_{0})$. Then there is a sequence $\langle f_{k}\rangle$ of real polynomials such that $f_{k}\to f$ strongly as $k\to\infty$ and $N(f_{k};\mathbb{C}\setminus S_{0})=N$ for all $k$. Let $n$ be an arbitrary positive integer. Then $J(f_{k},n)\to J(f,n)$ strongly as $k\to\infty$, hence

[TABLE]

for all sufficiently large $k$. On the other hand, Proposition 3.1 implies that

[TABLE]

for all $k$. Hence $N_{n}\leq N$.

Since $f$ represents an entire function, it follows that the Jensen sequence $\langle f^{*}_{n}\rangle$ converges to $f$ strongly, and we have $N(f^{*}_{n};\mathbb{C}\setminus S_{0})=N_{n}$ for all $n$. Hence we have

[TABLE]

for all sufficiently large $n$, and (2) is proved.

Finally, (3) is proved by the same argument as in the proof of the implication (3) $\Rightarrow$ (1) in Theorem 3.3, except that Theorem 3.4 is applied. ∎

Demonstration Proof of Proposition 3.1

Suppose that $f$ is a real polynomial and $N=N(f;\mathbb{C}\setminus\mathbb{R})$. If $f$ has no real zeros, so that $N=\deg f$, then

[TABLE]

for every non-negative integer $n$. On the other hand, it is trivial to see that if $a$ is a constant and $f(z)=(z-a)g(z)$, then

[TABLE]

for all $n$. Since $N(J{(f,n)};\mathbb{C}\setminus\mathbb{R})=N(A{(f,n)};\mathbb{C}\setminus\mathbb{R})$ for all $n$, the first inequality follows from an induction (on the number of real zeros of $f$) based on the Hermite-Poulain theorem, which states that if $h$ is a real polynomial and $b$ is a real constant, then

[TABLE]

The second inequality is proved in the same way except that the induction is on the number of non-negative real zeros of $f$ and the corresponding Hermite-Poulain theorem states that if $h$ is a real polynomial and $b\geq 0$, then

[TABLE]

Demonstration Proof of Proposition 3.2

Suppose that $a\in\mathbb{H}$, $f(z)=(z-a)g(z)$, $n$ is a non-negative integer and $\Cal{Z}(J(g,n))\subset\mathbb{H}$. Then $A(g,n)$ is a polynomial of degree $\leq n$, $\Cal{Z}(A(g,n))\subset-\mathbb{H}$ and

[TABLE]

Let $d=\deg A(g,n)$ and let $a_{1},\dots,a_{d}$ be the zeros of $A(g,n)$. Then $\text{\rm{Im}}\,a_{j}\leq 0$ for all $j$, and we have

[TABLE]

If $\text{\rm{Im}}\,z>0$, then $A(g,n)(z)\neq 0$, and we have

[TABLE]

Thus $\Cal{Z}(A(f,n))\subset-\mathbb{H}$, and we conclude that $\Cal{Z}(J(f,n))\subset\mathbb{H}$.

Now it is clear that the result follows from an induction on the degree of $f$. ∎