We consider vector-valued functions, with components which are entire functions, for growth problems. All such functions, when they satisfy a class of differential equations, are of bounded index and exponential type, and their components are also of bounded index

Roy, Ranjan

Shah, S.M.

Elsevier - Publisher Connector 

JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS 116, 349-362 (1986) 
Vector-Valued Entire Functions Satisfying 
a Differential Equation 
RAN JAN ROY 
Department of Mathematics, Beloit College, Wisconsin 53511 
AND 
S. M. SHAH 
Department of Mathematics, University of Kentucky, Lexington, Kentucky 40506 
Submitted by R. P. Boas 
We consider vector-valued functions, with components which are entire 
functions, for growth problems_ All such functions, when they satisfy a class of dif- 
ferential equations, are of bounded index and exponential type, and their com- 
ponents are also of bounded index_ © 1986 Academic Press, Inc. 
1. INTRODUCTION 
Let F: C IoC  m be a vector-valued function whose components  
fk: C' ~ C I are all entire functions. We write 
F(z)  = 
fro(z) 
which for convenience in printing we shall write 
F(z) = ( f  l(z),..., f,,(z) ). 
We now define two norms for F: (i) the sup norm 
IJF(z)ll, = max(]  f , (z)[ ;  1 <~i<~m} (1.1) 
and (ii) the euclidean orm 
J] f (z)  J]E = )fi(z)] 2 (12) 
i 1 
349 
0022-247X/86 $3.00 
Copyright © 1986 by Academic Press, Inc. 
All rights of reproduction ira any form reserved. 
350 ROY AND SHAH 
DEFINITION. A vector-valued entire function F is said to be of bounded 
index (BI) with sup norm if there exists an integer N= Ns such that 
II ~i)(z)II II F~k)(z)Iq~ max s >/ (1.3) 
o<i<u i! k! 
for all z e C and k = 0, 1,.... The least such integer N, is called the index of 
F, 
If we use in (1.3) the euclidean orm we will have a class of functions of 
BI with euclidean orm and index Ne. The following theorem shows that 
these two definitions of BI are equivalent. 
THEOREM 1. I f  F is of  B I  with supnorm then it is of  B I  with euclidean 
norm and vice versa. The two indices Ns and N e may possibly be different. 
Note that this definition of BI for vector-valued entire functions is 
similar to that for scalar functions. See Lepson [7], and Shah [9]. For 
vector-valued functions of BI with sup norm see Heath [5]. In the follow- 
ing we will use the sup norm definition (1.3) of BI and write N s = N. 
We now show that even if F is of BI, the components fk not be of BI. 
EXAMPLE 1.1. Let {kn}~ be a strictly increasing sequence of positive 
integers and let {an } 1" be a strictly increasing sequence of positive numbers. 
Then 
f ( z )  - 1 - - -  
1 an /  
where 32 k, /a ,  < oe is an entire function of unbounded index for all such 
choices of {kn} and {an}- By suitably choosing such {kn} and {an} we can 
show that (Shah [10] ) f ( z ) -  c, where e e C, c ~ 0, is of BI. Hence F(z )= 
(f(z) ,  f ( z ) -c )  is of BI but one component f i s  not of BI. (See also Heath 
[5].) 
If we assume that F satisfies a differential equation (DE) with coefficients 
which are matrices with entries in R (see Theorem 2 below) then we can 
show that each component is of BI. 
Let R denote the class of all rational functions r(z) bounded at infinity, 
and Qi(z) (1 <. i ~ n) denote an m x m matrix with entries in R. Write 
Q,(z)= (apq,i(z)), lim lapq,i(z)l  = Idpq,il (1.4) 
and 
sup(I Apq,il, 1 <~ p, q <. m) = L Ai[. 
VECTOR-VALUED ENTIRE FUNCTIONS 351 
THEOREM 2. Let F: C1~ C m be a vector-valued function whose com- 
ponents f l, f2,..., fm are all entire functions. Suppose that F satisfies the DE 
Ln(W,z , Q)= w(n)(z)+ QI(Z) W (n 1)(z)--[- --. + Qn(z) w(z)=0.  (1.5) 
Then each fj satisfies a DE of the form (1.5) (with poosibly different n and 
coefficients) and hence each fj is of BI. 
2. GROWTH BOUNDS 
Let 
M(r, F) = max II F(z)Pr- 
Izl ~r 
THEOREM 3. If F(z)is of BI with index N, then 
IJF(z)H ~A exp((N+ 1)[z[), (2.1) 
where 
A= max IJF(k)(O)ll/(N+ 1) k. 
O<~k<~N 
The result is sharp. 
THEOREM 4. I f  X(z) is a vector-valued entire solution of the DE 
L,(W, z, Q)= g(z) (2.2) 
where L,(W, z, Q) is as in (1.5) and g(z) is a vector-valued entire function of 
BI, then X(z) is of BI. 
We give later an example to show that if the entries in Qi are not in R, 
then F may not be of BI. 
Note that if g'(z) is of BI, then on differentiating (2.2) we see that X'(z) 
is also of BI. We give an example to show that in general F(z) may be of BI 
but F'(z) may not be. 
We next consider the DE (2.2) when g(z)= 0 and obtain (i) bound on 
the index and (ii) bounds on [I WII and M(r, W). 
THEOREM 5. Let W(z)~ 0 be a vector-valued entire function satisfying 
the DE 
r°(w, z, Q)=0 
352 ROY AND SHAH 
where Q i (1 ~ i <~ n) are all matrices with constant elements, and p >t 0 is any 
integer such that 
[IA,I + IA21 IA,,[ ] 
mkn+p (n+p) (n+p_ l )  + . . -  + (n+p~.- - - (p+l  i ~<1 (2.3) 
then the index N(W) of W(z), is less than or equal to n + p - 1. The bound 
on N is best possible. 
THEOREM 6. Let 
DE. 
Then 
W(z) be a vector-valued entire function satisfying the 
L , (W,z ,  Q)=0. 
l imsup l °gM(r 'W)  ~<max {1, m~ [A,[}. (2,4) 
r~0 r 1 
THEOREM 7. Let q be the least positive integer such that 
m{IA l l (n+q-1) !+ ' "+ lA , [q !}<(n+q) ! ,  where IAi[ are as in 
Theorem 6. I f  W(z) is a vector-valued entire function satisfying the DE. 
L.(W, z, Q)=0, 
II W(z)ll ~A exp{(n + q)]zb} (2.5) 
then 
where A & a constant. 
Then 
(i) 
3. PROOF OF THEOREM 1 
Suppose F is of BI with sup norm, BI(s), and index N,. Write 
11 F(k)II s = IJ F(k)Jl and N~ = N.  
Now for all k >t 0, 
max - -  >/ 
O~k~N k! (N+I)! 
II F (k) IL <~ It F~k)II e <~  II F (k) II- 
VECTOR-VALUED ENTIRE FUNCTIONS 353 
Hence 
IIF(N+')IIE _ 7-IIF(N+a)II <~X/f~ max IIF(k)ll 
(N+i )~ ~x/m (N+I ) !  O<k<N k! 
II F (k) II E 
<~ x//-~ max 
O<~k<~N k! ' 
that is, 
[fF(N+')/[E~<C 1 max--IIF(k)lle<~C max [IF(k)[I E, (3.1) 
o~<k<N k! o<k~<N 
Here C, and C are constants and we may suppose C > 1. Le t ,  e C, ] ~ I = i. 
Fix ~ and consider for x ~> 0, 
G(x) = max l[ F(k)(Zo + ax)lIE. 
O<~j<~N 
We use (3.1) and obtain (cf. Hayman [4, Theorem 2] )  
G(x) <~ G(O) exp(Cx). 
Writing z = Zo + ctx, we have 
max NF(k)(z)}le<~exp(CIz-zol)max ]t~k)(Zo)lbe. (3.2) 
O<~j<~N O<~j<~N 
Now for any component f j and R t> 1 we have 
1 
[f}')(Zo)[~<~-~ max [~(z) l  
pz-zol =R 
1 
~<~-~ max IPF(z) 0le, 
]z--zol=R 
and so by (3.2) 
1 
[f}')(zo)[ ~< ~ exp(CR) max [I F(k)(zo)I[E- 
O<~j<~N 
This holds for all n ~> 0. Adding these inequalities we get 
x//-m exp(CR) max [I Ftk)(Zo)II e- ll F~')(Zo)II e < -~-  
O<-.j<~N 
Choose R = 2 and no such that 
(x//--~ exp( 2C) )/2 "° <~ no !/ N~, 
354 ROY AND SHAH 
and we have 
IIr"(zo)llE IIt k)(Zo)ll  
n! ~< (2 "°-") max o~k~u k! 
II  k)(Zo)II 
~< max 
o~<k~N k! 
for all n >/n o. Hence F is of BI(E) with NE ~< no. 
(ii) Suppose now F is of BI(E). Then we have on writing N = Ne, 
[I F(N+ 1)l[ ~ C max I[ F~k)I[ 
O<.k<~N 
for some C> 1; and the above argument shows that F is of BI(S). 
We also get (cf. Hayman [4] for scalar functions). 
COROLLARY 1.1. F(z) is of BI if and only if there exists a number C and 
an integer N such that 
[I F(N+ 1)(Z) II ~ C max II F¢i)(z) II 
o<~i~N 
for all z. 
4. PROOF OF THEOREM 2 
We consider first the case when m---2 = n. Write 
and 
Q,  , fal(z),a2(z)) 
'tzI=~a3(z),a4(z)]' Q2(z)=~b3(z),b4(z)] 
F(z) = (f(z), g(z)), 
where a i, bi eR. From (1.4) and (4.1) we have 
f "  +a l f '  +azg' +bxf  +bzg=O, 
g" +a3f '  +a4g' +b3f  +b4g=O. 
Hence 
(4.1) 
(4.2) 
(4.3) 
f "  +aaf '  +b l f  = -a2g ' -b2g .  (4.4) 
VECTOR-VALUED ENTIRE FUNCTIONS 355 
Write the right side of (4.4) as A2 g' + B2 g and the left side as Dz(f). We 
differentiate (4.4) and use the equation for g"  and get 
D3( f )=(a2a4-a ; -bz )g '+(azb4-b 'z )g -A3g '+B3g.  (4.5) 
Similarly we get 
D, ( f )= Ang' + B~g (4.6) 
where the following recurrence formulae connecting An+~ and B.+~ hold. 
An + 1 = --Ana4 + A'n + B, 
B,+I =B'~-b4A,-  
We note that for each n i> 2, D~(f) is of the form 
f (m+Rl ,  f ( " - l )+  ".- + Rn, fwhere  each Rg, (1 <~i<~n)~R; that is, D, ( f )  
is a monic operator with coefficients in R. 
Further for each n ~> 2, A, and B, ~ R. Now we use the following lemma 
due to A. Sathaye. 
LEMMA 2.1. Let R denote the class of rational functions bounded at 
infinity. Let {Fn},~,,o be a sequence of monic operators with coefficients in R 
such that the degrees of F~ form a strictly increasing sequence and 
F , ( f )  = ~ 2}")vi, n ~> no (4.7) 
i= l  
where 2~") ~ R and vl,..., vr are all functions and f ~ C ~. Then there exists a 
sequence of monic operators {G,},~>,~ such that the degrees of G, form a 
strictly increasing sequence and 
r 1 
G,(U) = ~ ~_i,(")v.,, n >~ n 1 , (4.8) 
i= l  
where p}") ~ R. In particular if r = 1 then G,( f )  = 0 for n >i nl. 
Proof For a~R, ord~a=orda  denotes the order of a at infinity. 
Recall that ord a = oe e:~ a = 0. 
Choose m ~> no such that ord ~'~) -- min{ord 2~")ln ~> m }. 
If ord 2~ ") = oe then 2~ ") = 0 for all n ~> m, and if m = nl, G, = F, for n ~> nl 
and p}")=2~") for i=  1, 2,..., r -1  then (4.8) is satisfied. Now assume 
2! ") ~ 0. Since 
ord 2r (") >t- ord 2~ m) (4.9) 
409/116/2-4 
356 ROY AND SHAH 
we see that 
For n/> m consider 
Set n~ =m+ 1, 
and 
(n) ~m) 2~ /2~ ~R. 
Fn(f) - (2~n)12~m))  F,~(f) 
~''r I ' ' r  lt~t J ~l '  
i=1 
G. = F. (n) (m) - -  (2  r /2  r )Fro for n>~nl, 
/~In) ~In)__ (n) (m) (m)  = )2 ;  
Then it is evident hat 
(4.10) 
for i=1 ,2  ..... r -1  andn>~n~. (4.11) 
r - - I  
G.(f)  = ~, ~I" )V i  ~- {2~ " ) -  (2~") l~rn) )2~rn)}Vr  
i=1 
r--1 
= 
i= l  
Thus (4.8) holds. Since n>/nl ~m, it is evident from (4.10) that G~ is 
monic and the degrees of Gn are strictly increasing for n >~ nl. Also #I ") e R, 
because of (4.9) and (4.11). The remark about r= 1 is obvious; and the 
proof of the lemma is complete. 
COROLLARY 2.2. I f  F., f are as above then there exist monic operators 
H. with coefficients in R such that 
H , ( f )  = 0 for sufficiently large n. 
Proof Use induction on r. 
To complete the proof of the theorem, when m = 2 -- n, observe that r = 2 
and vl = g, v2 = g' and An, Bn eR. 
The proof for the general case when m>2 and F satisfies the DE 
Lp(W, z, Q)=0,  p~> 1, is similar. We follow the same process and obtain 
for any large n (n>(m-1)p)  a relation of the form (4.6), where f is 
replaced by one of the components, ay f l -  The right side will consist of 
(m - 1 ) terms involving f}k) (2 ~< j ~< m, 0 ~< k ~< p - 1 ) and the coefficients 
in R. Lemma 2.1 and Corollary 2.2 then give the required result. 
VECTOR-VALUED ENTIRE FUNCTIONS 357 
5. PROOF OF THEOREM 3 
Since F is of B IN  we have 
II F(N+ l)(z/N + 1)II 
(N+ 1)! 
~< max 
O<~k<~N 
fl F~k)(z/N + 1)II 
k~ 
Set Y(z) = F(z/N + 1). Then 
II y(~V+ 1)(z ) II (N+ 1) (N+ 17 
(N+ 1)r 
and hence 
II Y(N+ 1)(z) 1[ 
max 
O~k~N 
rl Y(k)(z)ll (N+ i) k 
k~ 
max JI Y(k)(z) II, 
O<~k<~N 
For a fixed real 0 and R > 0, we set 
g(t) = max II Y(k)( tei°)]l, 
O<~k<~n 
O<,t<R 
and obtain, as in the proof of Theorem 1, 
IL Y(tei°) II <~ max LI Y(k)(O) II e'. 
O<~k<~N 
This gives, on writing z = te ~°, 
II F(k)(0)II IIF(z/N+ 1)Pr ~< max e Izl. 
O<~k~<N (N--~- 1) k 
(5.1) 
We now replace z/N + 1 by z. The proof is complete. 
To show that the result is sharp we give: 
EXAMPLE 3.1. Let F(z): C 1 ~ C m have components which are all equal 
to f ( z )= exp((N+ 1)z). For this function F, there is equality sign in (2.1). 
Remark. For similar results for scalar functions see Hayman [-4], 
Fricke and Shah [2]. 
6. PROOF OF THEOREM 4 
We require two lemmas: 
LEMMA 4.1. I f  X (z )50  is an entire vector-valued function and T is a 
given positive number, then there exists an integer k > 0 such that for every z, 
JzJ<~T, 
max{I] X(z)]], 1[ X(1)(z) [I ,1X(k)(z) [1 } 1 
1 ! ' "  k! >~ 2 -~ 
358 ROY AND SHAH 
We omit the proof of this lemma and the next one. See for similar lem- 
mas for scalar functions, Shah [9]. 
LEMMA 4.2. I f  X(z) is an entire vector-valued function and T is a given 
positive number, then there is an integer L such that 
{ } II X<~)(z) I[ max II y(z)II II X~(z)II >i 
' "  L~ j~ 
for [ z l <<, T and j = l, 2,.... 
Proof of the Theorem. Since g is of BI, there exists by Corollary 1.1 an 
integer N and a number C such that 
Ilg(~V+l)(z)l[ ~<C max II g(J)(z)ll 
O<~j<~N 
for all zE C. Thus from Eq. (2.2) we have (cf. Fricke and Shah [1]) 
n dN+l -J)(z) 
II g<N+ 1)(z) II = iIj~=od-~--k-~Qj(z)X<" 
--~ ~0 N~I (N;1)Q}k)(z ) x(n_ j+N+I_k) (z  ) 
j k=0 
~< C max I[ g(Jl(z)I[ 
O~j<~N 
Hence by a simple transposition 
i[X(,+N+l)(z)ll ~ (k) (n--j+N+l 
j k=0 
O<~j<~N k=O t:0 
Now, since the functions in R are all bounded at infinity, we may choose 
Ro large enough so that (see (1.4)) for ]zl >Ro and O<<.k<<.N+n, 
a(k).(Z][ < M pq,t \ 
for some constant M. 
Thus, when P = (C+ 1) m(n + 1)(N+ 1)(N+ 2) 
IlX('+N+l)(z)]l <~PM max IlX(J)(z)ll, 
O<~j<~N+n 
VECTOR-VALUED ENTIRE FUNCTIONS 359 
for Izl >Ro. By Lemma 4,2, this inequality holds also for Izl ~<Ro, 
provided we replace PM by a suitable constant K. Hence by Corollary 1.1, 
X(z) is of BI. 
EXAMVLE 4.1. Consider the DE 
W' - 2z lW = 0 
where I is a unit matrix. This equation is satisfied by W= (fl,  f2,-.,, fro), 
where all components fj(z) are equal to exp(z2). The coefficient is not in R 
and W(z) is not of BI. 
EXA~VLE 4.2. Let F be as defined in Example I.i. Then F is of BI but 
F' is not. 
7. PROOF OF THEOREM 5 
We differentiate Eq. (1.5) p times and get 
W/"+p)= - (Q1 W ~"+p- ~)+ "" + Q.W~P)} - 
This implies 
I[ W("+P)[] mlAl[ I[ w(n+p-I)II m[A,,[ II (We')It <~--  +. . .+  
(n+p)! n+p (n+p-  1)r (n+p) . . . (p+l )  p! 
Choose p such that (2.3) holds. Then 
[[ w¢"+"~(z)[I [I w"~(z)II 
~< max 
(n+p)! o<~i<~.+p-1 i! 
for all z ~ C. It is clear from the above argument that p can be replaced by 
p+l ,p+2 .... and hence 
max 
o~i<~+p-1 i! j! 
~°(z )  ll II w~J~(z) 
for j--- 0, 1, 2 .... and all z ~ C. This, gives 
N(W)<~ n+ p-1 .  
To show that this bound is best possible, we consider: 
360 
EXAMPLE 5.1. Let 
ROY AND SHAH 
(o i) X'= X. 
- -1  - -  
This equation is satisfied by W(z) = (f(z), f'(z)), where 
f(z) = e -z/2 Ca Cos -7  + C2 Sin 
and C1, C2 are any two constants. Choose 0 < C1 < C2 x//3. Now 
[I W(0)II = max C1, 2 ' 
Ca + C2 ~f3 
II W'(0)II -- 2 > II w(o)II- 
Hence N(W) ~> 1. But since p is the smallest non-negative integer such that 
2/(1+p)~<1 we takep=l ,  andsoN(W)~l+l - l= l .  Hence N( W) = l_ 
8. PROOF OF THEOREM 6 
By Theorem 4, W(z) is of BI. From the equation (1.5) we have 
II W(n)(z)II ~< IIOa(z) W(n-a)(z)ll + " + II O,(z) W(z)lL. 
Given e>0,  we have, from (1.4), for z>~Ra > Ro(e), 
]l W(")(z) [l <~ {m ~ IA~l+e} max [LW(°(z)l[, 
i=1 O<~i<~n--1 
and consequently we have by the argument of Theorem 1, 
i=1 
for ]zl~> R 1. Since ~ can be chosen arbitrarily small, this gives (2.4). 
9. PROOF OF THEOREM 7 
We have from Eq. (1.5) 
II W~"+q)ll 1 IIQ1W~n+q-1)l] 1 Q,,w(q)ll 
~<--  + "+ [ I - -  (n + q)! (n + q) (n+q- -  1)! (n + q). " (q + l ) q! 
+ similar terms involving derivatives of Qi (i = 1, 2,..., n). 
VECTOR-VALUED ENTIRE FUNCTIONS 361 
Hence for [zl 1> R2 > R0(e) 
[I W~"+q)l[ <m(lAl]  +~) 
(n+q) !  (n+q)  
+ 
II w ~j> II
max - -  + ' "  
o~<j~<.+q-1  j !  
m([A . I  +e)  I1 W<'[I 
max 
(n+q) - - . (q+ l)o<_j<.,+q-~ j! 
+ e max 
O<~j<~n+q--1 j! 
[I W <j~ II
Since e can be chosen arbitrarily small we have for I zl~> R2 > Ro 
Ir m(n + q) II II w(J~l[  
~< max 
(n+q)!  o_<j~,+q 1 j! 
And this gives N(W) ~< n + q - 1. We now use Theorem 3 to get the result. 
Remark. For similar and other results on scalar functions of BI, see 
Shah [11, 12] and Fricke, Roy and Shah [3]; and for functions of several 
complex variables ee Krishna and Shah [-6]. 
ACKNOWLEDGMENTS 
We express our thanks to Dr. Avinash Sathaye for Lemma 2.1. 
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Vector-valued entire functions satisfying a differential equation 

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Vector-valued entire functions satisfying a differential equation

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