Entire bivariate functions of exponential type

Abstract

In this paper we will analysis the concepts of bivariate entire complex valued functions of exponential type. To accomplish this goal, we begin with the presentation of a notion of bounded index for bivariate complex functions. Using this notion we present a series of sufficient conditions that ensure that exponential type is preserved.

1 Introduction

An entire function $f(z)$ is said to be of exponential type if there is a constant $\alpha$ such that

$$\begin{aligned} |f(z)|< const.e^{\alpha |z|} \end{aligned}$$

everywhere. The infimum of the set of these $\alpha$ is called the type of $f(z)$.

If $f(z,w)$ is a bivariate entire function in the bicylinder

$$\begin{aligned} \{(z,w)\in \mathbb {C}^{2}:|z-a|<r_{1},|w-b|<r_{2}\} \end{aligned}$$

then at point $(a,b)$, $f(z,w)$ have a bivariate Taylor expansion

$$\begin{aligned} f(z,w)=\sum _{k=0}^{\infty }\sum _{l=0}^{\infty }c_{kl}(z-a)^{k}(w-b)^{l} \end{aligned}$$

where

$$\begin{aligned} c_{kl}=\frac{1}{k!l!}\left[ \frac{\partial ^{k+l}f(z,w)}{\partial w^{k}\partial z^{l}}\right] _{z=a;w=b}=\frac{1}{k!l!}f^{(k,l)}(a,b). \end{aligned}$$

Similar to Gross [5] we presented in [9] the following notion of bounded index of bivariate entire function.

Definition 1

A bivariate entire function $f$ is said to be of bounded index provided that there exist integers $M$ and $N$ independent of $z$ and $w$ such that

$$\begin{aligned} \max _{0\le k\le M;0\le l\le N} \left\{ \frac{\left| f^{(k,l)}(z,w)\right| }{k!l!} \right\} \ge \frac{\left| f^{(i,j)}(z,w)\right| }{i!j!} \end{aligned}$$

for all $i=0,1,2,\ldots$ and $j=0,1,2,\ldots$ and all $z$ and $w$.

We shall say that $f$ is of index $(M,N)$ if N ad M are the smallest integers for which above inequality holds. A bivariate entire function which is not of bounded index is said to be of unbounded index. One should observe that a f bivariate entire function is of bounded index then there exist integers $M\ge 0$, $N\ge 0$ and some $C>0$,

$$\begin{aligned} \sum _{k=0}^{M}\sum _{l=0}^{N}\frac{\left| f^{(k,l)}(z,w)\right| }{k!l!} \ge C \frac{\left| f^{(i,j)}(z,w)\right| }{i!j!} \end{aligned}$$

(1.1)

where $i=M+1,M+2,\ldots$ and $j= N+1,N+2,\ldots$.

In addition if the last inequality holds then

$$\begin{aligned} \max _{0\le k\le M;0\le l\le N} \left\{ \frac{\left| f^{(k,l)}(z,w)\right| }{k!l!} \right\} \ge \frac{1}{(M+1)(N+1)}\frac{\left| f^{(i,j)}(z,w)\right| }{i!j!} \end{aligned}$$

(1.2)

where $i, j= 0,1,2,3,\ldots$.

Slight variation of (1.1) is following

$$\begin{aligned} \sum _{i=0}^{M}\sum _{j=0}^{N}\frac{\left| f^{(i,j)}(z,w)\right| }{i!j!} > C \sum _{k=M+1}^{\infty }\sum _{l=N+1}^{\infty }\frac{\left| f^{(k,l)}(z,w)\right| }{k!l!} \end{aligned}$$

We also consider those variation of (1.2) obtained by replacing $|f^{(i,j)}(z,w)|$ by

$$\begin{aligned} \left( \int _{0}^{2\pi }\int _{0}^{2\pi }\left| f^{(i,j)}\left( r_{1}e^{\mathbf {i}\theta _{1}},r_{2}e^{\mathbf {i}\theta _{2}}\right) \right| ^{p}d\theta _{1}\theta _{2}\right) ^{\frac{1}{p}} \end{aligned}$$

where $p$ is any positive integer. We shall show in the sequel that entire bivariate functions satisfying inequality (1.2) or any of the conditions obtained from it by the substitutions suggested above are bivariate functions of exponential type.

2 Main result

Let $f(z,w)$ be a bivariate function. Suppose that $0<\rho <\infty$ and let as define

$$\begin{aligned} \tau =\limsup _{r_{1},r_{2}\rightarrow \infty }\frac{\log M_{f}(r_{1},r_{2})}{r_{1}^{\rho }r_{2}^{\rho }}. \end{aligned}$$

where

$$\begin{aligned} M_{f}(r_{1},r_{2})=\sup _{|z|\le r_{1};|w|\le r_{2}}|f(z,w)|. \end{aligned}$$

The functions which satisfy the above equality are said to be functions of exponential type $\tau$.

We can state the classical Borel [1] lemma for bivariate functions as follows:

Lemma 1

Let T be a continuous nondecreasing function on $[r_{0}, \infty )\times [s_{0}, \infty )$ for some $r_{0}$ and $s_{0}$ such that $T (r_{0},s_{0}) \ge 1$. Then

$$\begin{aligned} T(r_{1}+\frac{1}{T(r_{1},r_{2})},r_{2}+\frac{1}{T(r_{1},r_{2})})\le 4T(r_{1},r_{2}) \end{aligned}$$

for all $r_{1}$ and $r_{2}$ outside a possible exceptional set $E$ whose measure is at most 4, that is,

$$\begin{aligned} \int \int _{(s,t)\in E\cap [r_{0},\infty )\times [s_{0},\infty )}dsdt\le 4. \end{aligned}$$

Theorem 1

Let $f(z,w)$ be a bivariate entire function and $C$ be a positive constant. If f satisfies one of the following for $i=0,1,2,3,\ldots ,M; j=0,1,2,3,\ldots ,N$ and for all $(z,w)$ with $|z|$ and $|w|$ sufficiently large:

• (a)
  $$\begin{aligned} \sum _{i=0}^{M}\sum _{j=0}^{N}\frac{|f^{(i,j)}|}{i!lj!} > C \sum _{k=M+1}^{\infty }\sum _{l=N+1}^{\infty }\frac{|f^{(k,l)}|}{k!l!} \end{aligned}$$
• (b)
  $$\begin{aligned} \sum _{i=0}^{M}\sum _{j=0}^{N}\frac{\left( \int _{0}^{2\pi }\int _{0}^{2\pi }\left| f^{(i,j)}\left( r_{1}e^{\mathbf {i}\theta _{1}},r_{2}e^{\mathbf {i}\theta _{2}}\right) \right| ^{p}d\theta _{1}\theta _{2}\right) ^{\frac{1}{p}}}{i!j!}>C \end{aligned}$$
  $$\begin{aligned} \sum _{k=M+1}^{\infty }\sum _{l=N+1}^{\infty }\frac{\left( \int _{0}^{2\pi }\int _{0}^{2\pi } \left| f^{(k,l)}\left( r_{1}e^{\mathbf {i}\theta _{1}},r_{2}e^{\mathbf {i}\theta _{2}}\right) \right| ^{p}d\theta _{1}\theta _{2}\right) ^{\frac{1}{p}}}{k!l!} \end{aligned}$$

  for some integer $p$,

• (c)
  $$\begin{aligned} \sum _{i=0}^{M}\sum _{j=0}^{N}\frac{|M_{f^{(i,j)}}(r_{1},r_{2})|}{i!lj!}>C\sum _{i=M+1}^{\infty }\sum _{j=N+1}^{\infty }\frac{|M_{f^{(i,j)}}(r_{1},r_{2})|}{i!lj!} \end{aligned}$$

  where $M_{f^{(i,j)}}(r_{1},r_{2})$ is the maximum modulus of $f^{(i,j)}$ on $|z|=r_{1}$ and $|w|=r_{2}$, that is

  $$\begin{aligned} M_{f^{(i,j)}}(r_{1},r_{2})=\sup _{|z|\le r_{1};|w|\le r_{2}}|f^{(i,j)}(z,w)|,\quad \quad i=0,1,2,\ldots ;\quad j=0,1,2,\ldots \end{aligned}$$

then $f$ is of exponential type.

Proof

For convenience we choose $C=1$. The proof of other $C$ is similar. For any arbitrary bivariate entire function $F$ and any complex numbers $A$ and $B$, we have

$$\begin{aligned} F(z,w)=\sum _{i=0}^{\infty }\sum _{j=0}^{\infty }\frac{F^{(i,j)}(A,B)}{i!j!}(z-a)^{i}(w-b)^{j}. \end{aligned}$$

Let $m$ and $n$ any integers, $a,b,\xi ,\zeta$ complex numbers with $|\xi |=1$ and $|\zeta |=1$. Choosing $A=(m-1)\xi a$, $B=(n-1)\zeta b$, $z=m\xi +a$, $n\zeta +b$ and $F=f^{(i,j)}$ we obtain

$$\begin{aligned} |f^{(i,j)}(a+m\xi ,b+n\zeta )|\le \sum _{i=0}^{\infty }\sum _{j=0}^{\infty }|f^{(i,j)}(a+(m-1)\xi ,b+(n-1)\zeta )| \end{aligned}$$

(2.1)

for $m=0,1,2,\ldots ; n=0,1,2,\ldots$. Assume that (a) holds. Then (2.1) yields

$$\begin{aligned} |f^{(i,j)}(a+m\xi ,b+n\zeta )|\le 2\sum _{i=0}^{M}\sum _{j=0}^{N}|f^{(i,j)}(a+(m-1)\xi ,b+(n-1)\zeta )|\qquad \end{aligned}$$

(2.2)

for $m=0,1,2,\ldots ,M; n=0,1,2,\ldots ,N.$ One observe that for $i\le M$ and $j\le N$

$$\begin{aligned}&\sum _{i=0}^{M}\sum _{j=0}^{N}\frac{|f^{(i+k,j+l)}(z,w)|}{i!j!}\nonumber \\&\quad =\sum _{i=0}^{M}\sum _{j=0}^{N}\frac{|f^{(i+k,j+l)}(z,w)|}{i!j!}\frac{(i+k)!(j+l)!}{(i+k)!(j+l)!}\nonumber \\&\quad \le \sum _{u=0}^{2M}\sum _{v=0}^{2N}\frac{|f^{(u,v)}(z,w)|}{u!v!}(2M)^{M}(2N)^{N}\nonumber \\&\quad \le 2\sum _{u=0}^{M}\sum _{v=0}^{N}\frac{|f^{(u,v)}(z,w)|}{u!v!}(2M)^{M}(2N)^{N}\nonumber \\&\quad \le 2^{M+1}2^{N+1}M^{M}N^{N}\sum _{i=0}^{M}\sum _{j=0}^{N}\frac{|f^{(i,j)}(z,w)|}{i!j!}. \end{aligned}$$

(2.3)

Equations (2.2) and (2.3) yield

$$\begin{aligned}&\sum _{i=0}^{M}\sum _{j=0}^{N}\frac{|f^{(i,j)}(a+m\xi ,b+n\zeta )|}{i!j!}\le (M+1)(N+1)2^{M+1}2^{N+1}\nonumber \\&\quad \times \sum _{i=0}^{M}\sum _{j=0}^{N}\frac{|f^{(i,j)}(a+m\xi ,b+n\zeta )|}{i!j!}. \end{aligned}$$

(2.4)

Letting

$$\begin{aligned} \lambda =(M+1)2^{M+1}M^{M},\ \mu =(N+1)2^{N+1}N^{N} \end{aligned}$$

and using (2.4) recursively we have

$$\begin{aligned}&\sum _{i=0}^{M}\sum _{j=0}^{N}\frac{|f^{(i,j)}(a+m\xi ,b+n\zeta )|}{i!j!}<\lambda ^{m}\mu ^{n}\sum _{i=0}^{M}\sum _{j=0}^{N}\frac{|f^{(i,j)}(a,b)|}{i!j!}. \end{aligned}$$

(2.5)

For $|a|<1$, $|b|<1$, we get

$$\begin{aligned} \sum _{i=0}^{M}\sum _{j=0}^{N}\frac{|f^{(i,j)}(a+m\xi ,b+n\zeta )|}{i!j!}<C \lambda ^{m}\mu ^{n}. \end{aligned}$$

Letting $z=a+m\xi$, $w=b+n\zeta$, $r_{1}=|z|$, $r_{2}=|w|$ we get

$$\begin{aligned} |f(z,w)|<C\lambda ^{2r_{1}}\mu ^{2r_{2}}. \end{aligned}$$

(2.6)

Hence, f must be of exponential type.

If (c) is assumed instead of (a) the argument is almost identical to the one above. If (b) is assumed instead of (a), then letting $M_{p}(r_{1},r_{2})$ denote

$$\begin{aligned} \left( \frac{1}{4\pi ^{2}}\int _{0}^{2\pi }\int _{0}^{2\pi }\left| f^{(i,j)}\left( r_{1}e^{i\theta _{1}},r_{2}e^{i\theta _{2}}\right) \right| ^{p}\right) ^{\frac{1}{p}} \end{aligned}$$

we have as before

$$\begin{aligned} M_{p}(r_{1},r_{2})<C\lambda ^{2r_{1}}\mu ^{2r_{2}}. \end{aligned}$$

One readily sees, by means of Cauchy’s formula that $R_{1}>r_{1}>0$, $R_{2}>r_{2}>0$

$$\begin{aligned} M_{p}(r_{1},r_{2})\le \left( \frac{R_{1}}{R_{1}-r_{1}}\right) ^{\frac{1}{p}}\left( \frac{R_{2}}{R_{2}-r_{2}}\right) ^{\frac{1}{p}}M_{p}(R_{1},R_{2}). \end{aligned}$$

Furthermore it is easy to verify that, $M_{p}(R_{1},R_{2})$ is continuous and increases faster than any power of $r_{1}$ and $r_{2}$ whenever $f$ is transcendental.

Letting

$$\begin{aligned} R_{1}=\frac{1}{log M_{p}(r_{1},r_{2})}+r_{1}, \quad R_{2}=\frac{1}{log M_{p}(r_{1},r_{2})}+r_{2} \end{aligned}$$

and by Lemma 1 we get for some $\epsilon >0$ that

$$\begin{aligned} M_{f}(r_{1},r_{2})<M_{p}(r_{1},r_{2})^{1+\epsilon } \end{aligned}$$

outside a set of $r_{1}+r_{2}$ of finite measure. One can easly show that however that if there exist infinite sequences $(r_{1_{n}})$ and $(r_{2_{n}})$ such that

$$\begin{aligned} M_{f}(r_{1_{n}},r_{2_{n}})>e^{C_{n}(r_{1_{n}}+r_{2_{n}})}, \quad C_{n}\rightarrow \infty , \end{aligned}$$

then there exist a set of $r_{1}+r_{2}$ of infinite measure with the same property. Thus

$$\begin{aligned} M_{f}(r_{1},r_{2})<C\lambda ^{2r_{1}}\mu ^{2r_{2}} \end{aligned}$$

for sufficiently large $r_{1}$ and $r_{2}$, so proof is completed. $\square$