Sign regularity of Maclaurin coefficients of functions in the Laguerre-P\'olya class

0.16 MB·

by  Dimitar K. Dimitrov

SIGN REGULARITY OF MACLAURIN COEFFICIENTS OF
FUNCTIONS IN THE LAGUERRE-PO´LYA CLASS
6
1
DIMITARK.DIMITROVANDWILLIAN D.OLIVEIRA
0
2
n Abstract. WeprovethatthesignsoftheMaclaurincoefficientsofawideclassof
a entirefunctionsthatbelongtotheLaguerre-P´olyaclasspossesaregularbehaviour.
J

1
2
1. Introduction

]
V A real entire function ϕ is in the Laguerre-Po´lyaclass, written ϕ , if
∈LP
C ω
h. (1.1) ϕ(z)=czmeaz2+bz (1−z/xk)eλz/xk, 0≤ω ≤∞,
t kY=1
a forsomenonnegativeintegerm,c,a,b R,a 0,λ 0,1 andx R 0 suchthat
m k
∈ ≤ ∈{ } ∈ \{ }
x −λ−1 < . When ω =0 the product on the right-hand side of (1.1) is defined
[ k| k| ∞
to be identically one. Otherwise the terms of the product are arranged according to
P
1 the increasing order of x .
k
| |
v Observe that ϕ if and only if ϕ(z) = exp(az2)φ(z), where a 0 and φ is a
7 ∈ LP ≤
real entire function with real zeros of genus at most one.
8
Ifa=λ=0andbα 0foreveryk Nintherepresentationofϕ(z),thefunction
4 k ≤ ∈
5 ϕ(z) is said to belong to the Laguerre Polya Class of type I, denoted by ϕ I. In
∈LP
0 other words, ϕ I if and only if either ϕ(z) or ϕ( z) can be represented in the
∈ LP −
. form
1
ω
0 ϕ(z)=czmebz (1 z/x ),
6 − k
1 kY=1
: where c R, m N 0 , b 0, x >0 and x−1 < .
v ∈ ∈ ∪{ } ≤ k k k ∞
The functions in and only these are uniform limits on the compact sets of
i
X the complex plane (lLoPcally uniform limits) oPf hyperbolic polynomials, that is, real
r polynomials with only real zeros. Similarly, the functions in I are the locally
a LP
uniform limits of hyperbolic polynomials with zeros of the same sign.
The class offunctions wasstudiedfirstby Laguerre[4]andlaterthenbyP´olya,
LP
Jensen, Schur, Obrechkoff (see [5, 6] and the references therein) and other celebrated
mathematicians in the beginning of the twentieth century because of the attempts to
settle the Riemann hypothesis. The connection between the latter and the Laguerre-
Po´lya Class is straightforwardand we refer to [2, 3] for the details.
The properties of the functions in the Laguerre-Po´lya class are tightly connected
with the notion of multiplier sequence. A real sequence γ ∞ is called a multiplier
{ n}n=0
2010 Mathematics Subject Classification. 30D10,30D15.
Keywordsandphrases. Laguerre-P´olyaclassofentirefunctions,Maclaurincoefficients,multiplier
sequence, signregularity.
Research supported by by the Brazilian foundations CNPq under Grant 307183/2013-0 and
FAPESPunderGrant2013/14881-9.
1
2 DIMITARK.DIMITROVANDWILLIAND.OLIVEIRA
sequence if for any hyperbolic polynomial
(1.2) p(z)=a +a z+ +a zm
0 1 m
···
with zeros of the same sign, the polynomial
(1.3) q(z)=a γ +a γ z+ +a γ zm
0 0 1 1 n m
···
is a hyperbolic polynomial. Similarly, the sequence γ ∞ is said to be a multiplier
{ n}n=0
sequenceoftypeI ifforanyhyperbolicpolynomialpofthe form(1.2),thepolynomial
q given by (1.3) is a hyperbolic polynomial too. Po´lya and Schur [7] proved that the
sequence γ ∞ is a multiplier sequence if and only if the series
{ n}n=0
∞
γ
(1.4) nzn,
n!
n=0
X
represents an entire function in class. In particular the sequence γ ∞ is a
LP { n}n=0
multipliersequenceoftypeI ifandonlyifthelatterseriesrepresentsanentirefunction
in I class.
LP
Therefore, given a real entire function with Maclaurin expansion
∞
ϕ(z)= a zn,
n
n=0
X
it is of interest to provide necessary and/or sufficient conditions in terms of the be-
haviour of the sequence a in order that ϕ belongs to . In the present note we
n
{ } LP
establish some results of this nature. We are interested if the signs of the coefficients
a exhibit some regular patterns. For this purpose it is convenient to divide into
n
three subclasses. The firstone is simply I. The secondclass,denotedby LPa con-
LP LP
sists of function in for which the constant a in the exponent in the representation
LP
(1.1) is nonzero. The third class consists of entire function in for which a=0 but
LP
do not belong to I and is denoted by 0.
LP LP
It is straightforward that the Maclaurin coefficients of a nonpolynomial function
ϕ I are either all of the same sign or their signs alternate. In fact, Po´lya and
∈ LP
Schur [7] provedthat a realentire function in the Laguerre-Po´lyaclass which exhibits
this sign regularity of its coefficients is necessarily in I.
LP
However,Laguerre[4]providedabeautiful exampleofaparametricfamily ofentire
functionsintheLaguere-P´olyaclasswhosecoefficientsmayexhibitsignsformarather
irregular sequence for certain choices of the parameters. He proved that
∞
cos(ϑ+nθ)
ϕ(z)= zn
n!
n=0
X
is in . In fact, it is straightforwardto observe that
LP
ϕ(z)=ezcosθ cos(ϑ+zsinθ).
We emphasise that a = 0 and ϕ / I because its zeros are not of the same sign.
Therefore ϕ 0. Varying the∈chLoPices of ϑ and θ we observe a great variety of
∈ LP
possibilities for the distribution of the signs in the sequence cos(ϑ+nθ), especially
when θ/2π is an irrational number. Laguerre’s example shows that one can hardly
expect regularity in the distribution of the signs of the Maclaurin coefficients of a
function in 0, similar to the one in I.
ThesituaLtiPonchangesdramaticallywLhPenoneconsiderstheclass a. Thepresence
LP
ofthe factor exp(az2), a<0, seems to put order in the distribution ofthe signs ofa .
n
MACLAURIN COEFFICIENTS OF FUNCTIONS IN THE LAGUERRE-PO´LYA CLASS 3
Letusconsiderthreeillustrativeexamples. ThesignsofthefirstMaclaurincoefficients
of exp( z2)exp(z), exp( z2)(1/Γ(z) and exp( z2)cos√z are
− − −
1,1, 1, 1,1,1,1, 1, 1,1,1, 1, 1,1,1, 1, 1,1,1, 1, 1, 1,1,1, 1, 1,1,1 ,
{ − − − − − − − − − − − − − }
0,1,1, 1, 1,1,1, 1, 1,1,1, 1,1,1, 1, 1,1,1, 1, 1,1,1, 1, 1,1,1, 1, 1 ,
{ − − − − − − − − − − − − − }
1, 1, 1,1,1, 1, 1,1,1, 1, 1,1,1, 1, 1,1,1, 1, 1,1,1, 1, 1,1,1, 1, 1 ,
{ − − − − − − − − − − − − − − }
respectively. Observe that the common pattern 1,1, 1, 1 appears frequently. A
detailedanalysisoftheseandavastnumberofoth{erfun−ctio−ni}n a ledustoobserve
LP
that this phenomenon occurs infinitely many times. This pattern can be formalized
stating that the inequalities a a < 0 hold asymptotically. These observations
n−1 n+1
made us pose the following:
Question A. Let ϕ(z)= ∞ a zn be a real entire function with only real zeros. Is
n=0 n
it true that ϕ a if and only if 0<limsup na 2/n < and there exists n N,
n 0
∈LP P | | ∞ ∈
such that a a 0 for all n>n ?
n−1 n+1 0
≤
We have substituted the strict inequalities a a < 0 by a a 0 in
n−1 n+1 n−1 n+1
≤
order to cover the possibility of occurence of zeros in the sequence a which happens,
n
for instance, when the entire function is either even or odd. We establish a partial
results towards an affirmative answer to Question A, proving that the inequalities
a a 0, together with adequate asymptotic rate of a guarantees that a real
n−1 n+1 n
entire func≤tion with real zeros indeed belongs to a. Formally, one of our main
LP
results reads as:
Theorem 1. Suppose that the real function
∞
ϕ(z)= a zn
n
n=0
X
possesses only real zeros. If there exist n N, such that a a 0 for all n>n
0 n−1 n+1 0
∈ ≤
and 0<limsup na 2/n < , then ϕ a.
n→∞ | n| ∞ ∈LP
We state a relevant result concerning the necessity statement of the question too.
Theorem 2. If
∞
ϕ(z)= a zn a
n
∈LP
n=0
X
can be represented in the form
ϕ(z)=exp(az2)P(z),
where a < 0 and P is a hyperbolic polynomial, then 0 < limsup na 2/n <
and there exists n N, such that a a 0 for every n>nn.→∞ | n| ∞
0 n−1 n+1 0
∈ ≤
However,thepatternofthesignsofthecoefficientsoffunctionsin a weobserved
LP
beforestatingtheabovequestion,admitsexceptions. HencetheanswertoQuestionA
is negative. Moreover, exceptions occur rather frequently, for entire function in a
LP
of any possible order, as seen in the following:
Theorem 3. For every ρ with 0 ρ 2, there is an entire function Ψ(z) of order
≤ ≤
ρ =ρ, of the form
Ψ
∞
z λz
(1.5) ψ(z)= 1 exp ,
− x x
n=1(cid:18) n(cid:19) (cid:18) n(cid:19)
Y
4 DIMITARK.DIMITROVANDWILLIAND.OLIVEIRA
such that, for every a<0, the function
∞
ϕ(z)=exp(az2)ψ(z)= a zn
n
n=0
X
is in a and possesses the property that for every n N, there is n>n , such that
0 0
LP ∈
a a >0.
n−1 n+1
2. Preliminaries
Despitethatmostofthefactinthissectionarebasicinthetheoryofentirefunctions
and can be found in classical reading as [1, 5], we shall list them in the form we need
in the proofs in order to make the reading relatively self-contained.
Given a sequence of nonzero complex numbers zn n∈N, without accumulation
{ }
points, we consider:
(i) the associated canonic product
∞
z
f(z)= G ;p ,
z
n=1 (cid:18) n (cid:19)
Y
where p is the genus of the canonic product, that is, the smallest integer for
which
∞
1
z p+1
n
n=1| |
X
converges and
G(u;p)=(1 u)eu+u2/2+···+up/p;
−
(ii) the exponent of convergence λ of the sequence as the infimum of all positive
numbers t such that
∞
1
z t
n
n=1| |
X
converges;
(iii) the superior density ∆ of the sequence defined by
n(r)
∆=limsup ,
rλ
r→∞
where n(r)=# n: z r .
n
{ | |≤ }
As it is well known, an entire function
∞
f(z)= a zn
n
n=0
X
is of order ρ if
f
loglogM(r)
limsup =ρ ,
f
logr
r→∞
where M(r)=M (r)=max f(z). If ρ < , then f is said to be of type σ if
f |z|=r f f
| | ∞
logM(r)
limsup =σ .
rρ f
r→∞
MACLAURIN COEFFICIENTS OF FUNCTIONS IN THE LAGUERRE-PO´LYA CLASS 5
When0<ρ < theorderandthetypearedeterminedintermsofthecoefficients
f
∞
via
nlogn
ρ = limsup ,
f
− log a
n→∞ n
| |
(2.6)
nn an ρf
σ = limsup | | .
f
eρ
n→∞ p f
Borel’stheoremclaimsthat the orderofa canonic productis equalto the exponent of
convergence of its zeros.
The category κ of the entire function f is the pair of its order and type, that is
f
κ = (ρ ,σ ). The entire functions are partially ordered (denoted ) according to
f f f
(cid:22)
alphabetic order of their categories. Precisely, f and g belong to the same category
provided both their orders and types coincide. Otherwise, if ρ > ρ then f belongs
f g
to a higher category. Finally, if ρ = ρ but σ > σ then f is of a higher category.
f g f g
The so-called Theorem of categories (see [5, Theorem 12, p. 23]) states if the entire
functions f and g belong to distinct categories, then the product fg belongs to the
category of the factor with the higher category. In other words,
(2.7) κ =max κ ,κ .
fg f g
{ }
Hadamard’s theorem claims that every entire function f of finite order ρ can be
f
represented in the form
∞
z
(2.8) f(z)=zkexp(P(z)) G ;p ,
z
n=1 (cid:18) n (cid:19)
Y
where k N 0 , P is an algebraic polynomial of degree not exceeding ρ , and z
f n
∈ ∪{ }
are the zeros of f distinct from the origin.
We state the classical theorem of Lindelo¨f in its complete form, as in [5, Theorem
15, p. 28].
Theorem A. (Lindel¨of) Let f be an entire function with Hadamard product (2.8), of
finiteorder ρ , type σ , genus p of its canonic product and superior density of its zeros
f f
∆ . Then the following hold:
f
(i) If ρ is not integer, then ∆ and σ are simultaneously equal to either 0, a
f f f
finite number, or .
∞
(ii) If ρ is an integer and ρ > p, then f is of type σ = α , where α is the
f f f | ρf| ρf
coefficient of zρf in the expansion of the polynomial P(z).
(iii) If ρ =p and
f
1
δ (r)= α + z−ρf , δ¯ =limsupδ (r), γ =max(∆ ,δ¯ ),
f (cid:12) ρf ρ n (cid:12) f f f f f
(cid:12)(cid:12) f |zXn|<r (cid:12)(cid:12) r→∞
(cid:12) (cid:12)
then σ(cid:12)f and γf are simultan(cid:12)eously equal to either 0, a finite number, or .
(cid:12) (cid:12) ∞
Finally, the classical Hermite-Biehler theorem states that all zeros of the algebraic
polynomial
r(z)=u(z)+iv(z),
where u(z) and v(z) are polynomials with real coefficients, belong to one of the open
semi-planes determined by the real axis, if and only if both u and v are hyperbolic
polynomials and their zeros interlace.
6 DIMITARK.DIMITROVANDWILLIAND.OLIVEIRA
3. Proofs
We begin this section with a technical lemma which shows how to determine the
order and the type of an entire function from the asymptotic behaviour of its Taylor
coefficients. Theresultmightbeofindependentinterestandtheideaoftheproofgoes
back to [5, Theorem 2, p. 4].
Lemma 1. Let ρ > 0 and f(z) = ∞ a zn be an entire function, such that 0 <
n=0 n
limsup (nn a ρ)< . Then ρ =ρ and 0<σ < .
n→∞ | n| ∞ Pf f ∞
Proof. The propof goes by reductio ad absurdum. First we assume that ρ > ρ. It
f
follows from limsup (nn a ρ)< that there is K >0, such that
n→∞ | n| ∞
pnn a ρ
n
| | <K for n>n(K),
ρe
p
or equivalently,
n/ρ
ρeK
(3.9) a < for n>n(K).
n
| | n
(cid:18) (cid:19)
Let z =r,wherer >r(K)islargeenoughsothattheinequalityN(r)=[2ρρeKrρ]>
| |
n(K) holds. Here [] stands for the integer part. Then we employ the estimate (3.9)
·
to obtain
N(r) ∞
f(z) a rn+ a rn
n n
| | ≤ | | | |
nX=0 n=NX(r)+1
N(r)
a rn+2−N(r)
n
≤ | |
n=0
X
< (N(r)+1)max a rn +2−N(r).
n
n { }
Hence,
M (r)<(2ρρeKrρ+1) max a rn +1 for r >r(K).
f n
n { }
Observe that f is not a polynomial because we assumed that ρ > ρ > 0. Then the
f
Cauchy estimate yields that M (r) increases faster than any power of r. The latter
f
inequality implies that max a rn increases faster than any power of r too. This
n n
{ }
means that the index n, where the maximum is attained increases going to infinity as
r . Redefining r(K), if necessary, so that the index n where max a rn occurs
n n
→∞ { }
be such that n>n(K), we obtain
ρeK n/ρ
max a rn rn, r >r(K).
n
n { }≤ n
(cid:18) (cid:19)
Astraightforwardanalysisshowthatthemaximumontheright-handsideofthelatter
is attained at n=ρKrρ. Thus
max a rn eKrρ, r >r(K).
n
n { }≤
Then
M (r)<(2ρρeKrρ+2)eKrρ for r >r(K),
f
and, increasing r(K) if necessary,
(3.10) M (r)<e2Krρ, r >r(K).
f
MACLAURIN COEFFICIENTS OF FUNCTIONS IN THE LAGUERRE-PO´LYA CLASS 7
Ontheotherhand,thedefinitionoftheorderofanentirefunctionandtheassumption
that ρ > ρ show that we may chose ǫ > 0 in such a way that ρ ǫ > ρ and
f K f K
−
rρf−ǫK−ρ >2K for every r >r(ǫK) and
M (r)>exp(rρf−ǫk)=exp(rρf−ǫK−ρrρ)>exp(2Kr2), r >r(ǫ ).
f K
The latter contradicts (3.10).
Assume now that ρ <ρ. Then, given ǫ>0, there exists r(ǫ) such that
f
M (r) exp(ǫrρ), r >r(ǫ).
f
≤
By the Cauchy estimate
M (r) exp(ǫrρ)
f
a , r>r(ǫ).
| n|≤ rn ≤ rn
The minimum of r−nexp(ǫrρ), when r >0, is attained at r =(n/ǫρ)1/ρ. Choose now
n(ǫ) such that (n/ǫρ)1/ρ >r(ǫ) for all n>n(ǫ), to obtain
ρǫe n/ρ
a , n>n(ǫ),
n
| |≤ n
(cid:16) (cid:17)
which is equivalent to
nn a ρ <ρǫe, n>n(ǫ).
n
| |
Since the above reasonings hold for any ǫ>0, then
p
limsup(nn a ρ)=0.
n
| |
n→∞
p
This a contradiction to the hypothesis that limsup (nn a ρ)>0.
Therefore, ρ =ρ. The fact that 0<σ < folnlo→w∞s from|(n2.|6). (cid:3)
f f ∞ p
Proof. (of Theorem 1) By Lemma 1, ρ =2 and 0<σ < . Hadamard’s factorisa-
ϕ ϕ
∞
tion theorem implies that ϕ can be represented as
∞
z
(3.11) ϕ(z)=czkexp(az2+bz) G ;p ,
x
n=1 (cid:18) n (cid:19)
Y
where k N 0 , a,b,c R, and x are the zeros of ϕ distinct from 0. We shall
n
∈ ∪{ } ∈
prove that the series
∞
1
(3.12)
x2
n=1 n
X
converges. Let ρ and σ be the order and the type of the canonic product in (3.11).
Π Π
Since ρ =2, then (2.7) implies
ϕ
(3.13) κ (2,σ ).
Π ϕ
(cid:22)
Assume that the series (3.12) diverges. Then the exponent of convergence λ of the
Π
canonicproduct in (3.11) obeys λ 2. It follows fromthe abovementioned theorem
Π
≥
ofBoreland(3.13)thatλ =ρ =2. Since (3.12)diverges,then wemusthavep=2.
Π Π
Hence ρ =p and, by item (iii) of Teorema A,
Π
1 1
σ =limsup = .
Π 2 x2 ∞
r→∞ |xXn|<r n
 
Therefore, κ = (2, ), so that, by (3.13), κ = (2, ). This contradicts σ < .
Π ϕ ϕ
∞ ∞ ∞
Hence, the series (3.12) converges.
8 DIMITARK.DIMITROVANDWILLIAND.OLIVEIRA
The convergence of (3.12) implies that p 1 and that (3.11) can be written in the
≤
form
∞
z λz
(3.14) ϕ(z)=czkexp(az2+bz) 1 exp ,
− x x
n=1(cid:18) n(cid:19) (cid:18) n(cid:19)
Y
with k N 0 , a,b,c R, where x R for every n N, λ 0,1 and
n
∈ ∪ { } ∈ ∈ ∈ ∈ { }
(1/x )λ+1 < . It remainto proveonly thata<0 in(3.14). Assume the contrary,
n
∞
that a 0.
P ≥
However, there is n , such that a a 0 for all n > n . Then we can write
0 n−1 n+1 0
ϕ(z)= ∞ a zn in the form ≤
n=0 n
(3.15) P ϕ(z)=Pn0(z)+g(z),
with a polynomial P (z) of degree n and g an entire function of the form
n0 0
∞
g(z)= sgn(b ) b zn with sgn(b )sgn(b ) 0 for all n>0.
n n n−1 n+1
| | ≤
n=0
X
If t>0 then
∞ ∞ ∞
g(it) = sgn(b )b (it)n = sgn(b ) b t2n+isgn(b ) b t2n+1
n n 0 2n 1 2n+1
| | (cid:12) | | (cid:12) (cid:12) | | | | (cid:12)
(cid:12)nX=0 (cid:12) (cid:12) nX=0 nX=0 (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) ∞ 2 (cid:12) ∞(cid:12) 2 (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
= b t2n + b t2n+1
vu | 2n| ! | 2n+1| !
u nX=0 nX=0
t ∞
1 1 1
b tn max g(z) = M (t).
n g
≥ √2 | | ≥ √2 |z|=t| | √2
n=0
X
Since ρ =2>p, then item (ii) of TheoremA yields 0<σ =a. Now (2.6) allows us
ϕ ϕ
to conclude that ρ =2 and σ =a>0. This implies that there is t >0, such that
g g 0
√2P (it) 1
M (t)>exp((a/2)t2) and | n0 | < for every t>t .
g exp((a/2)t2) 2 0
Therefore
1
ϕ(it) g(it) P (it) M (t) P (it)
| | ≥ | |−| n0 |≥ √2 g −| n0 |
1 √2P (it)
exp((a/2)t2) 1 | n0 | .
≥ √2 − exp((a/2)t2)!
Finally we conclude that
1
(3.16) ϕ(it) > exp((a/2)t2) for every t>t .
0
| | 2√2
The convergence of the series (3.12) shows that there are two possibilities: either
ρ < 2 or ρ = 2 and, in the latter case, σ = 0 because of item (ii) of Theorem A.
Π Π Π
In both cases, there is t >0, such that
1
∞
it λit
1 exp <exp((a/2)t2) and ctkexp( (a/2)t2)<1 for t>t .
1
(cid:12)(cid:12)nY=1(cid:18) − an(cid:19) (cid:18)an(cid:19)(cid:12)(cid:12) | | −
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
MACLAURIN COEFFICIENTS OF FUNCTIONS IN THE LAGUERRE-PO´LYA CLASS 9
Using this inequality in (3.14) we obtain
∞
it λit
ϕ(it) = ctkexp( at2) 1 exp < ctkexp( (a/2)t2)<1, t>t .
1
| | | | − (cid:12)(cid:12)nY=1(cid:18) − αn(cid:19) (cid:18)αn(cid:19)(cid:12)(cid:12) | | −
This contradicts (3.16). (cid:12)(cid:12)Therefore a<0 in (3.14) a(cid:12)(cid:12)nd ϕ a. (cid:3)
(cid:12) (cid:12) ∈LP
Corollary 1. Let ϕ(z) = ∞ a zn be an even (odd) function with only real zeros.
n=0 n
Then ϕ a if and only if 0<limsup(nn a 2)< and a a 0 for every
n N. ∈LP P | n| ∞ n−1 n+1 ≤
∈ p
Proof. If 0 < limsup(nn a 2) < and a a 0 for every n N, then, by
n n−1 n+1
Theorem 1, ϕ a. | | ∞ ≤ ∈
Suppose tha∈tLϕP pa is even (odd). It can be written in the form
∈LP
∞ z2
ϕ(z)=czkexp(az2) 1 .
− x2
n=1(cid:18) n(cid:19)
Y
It suffices to apply Cauchy’s multiplication formula to see that a a 0 for all
n−1 n+1
n N. Since ∞ (1/x2)< , then item (ii) of TheoremA yields σ = ≤a. Finally,
∈ n=1 n ∞ ϕ −
(2.6) implies 0<limsup(nn a 2)< . (cid:3)
P | n| ∞
Proof. (of Theorem 2) Sincpe ρ =2 and σ = a, then (2.6) yields
ϕ ϕ
−
0<limsup(nn a 2)< .
n
| | ∞
Let p
2m
P(z)= b zn,
n
n=0
X
withtheobviousconventionthatifP isofodddegree,thenb =0. Consideranother
2m
polynomial,
P (z)=amb +am−1b z+ +b z(z 1)...(z m+1).
1 0 2 2m
··· − −
Thereexists k N, suchthat P (x) is strictly positive orstrictly negativefor x>k .
0 1 0
∈
Letn=2k+1 be anodd number with k >k andk m. The Cauchymultiplication
0
≥
formula yields
m ak−j ak−m m
a = b = am−jb k(k 1)...(k j+1)
n−1 2j 2j
(k j)! k! − −
j=0 − j=0
X X
ak−m
= P (k),
1
k!
m ak+1−j ak+1−m m
a = b = am−jb (k+1)(k)...(k j+2)
n+1 2j 2j
(k+1 j)! (k+1)! −
j=0 − j=0
X X
ak+1−m
= P (k+1).
1
(k+1)!
Having in mind that a<0 and P (k) and P (k+1) have equal signs when k >k , we
1 1 0
see that
a2(k−m)+1
a a = P (k)P (k+1) 0,
n−1 n+1 1 1
k!(k+1)! ≤
10 DIMITARK.DIMITROVANDWILLIAND.OLIVEIRA
for all odd indices n > 2k +1. When n is even, it suffices to apply an analogous
0
argument but employing the polynomial
P (x)=am−1b +am−2b (x 1)+ +b (x 1)(x 2)...(x m+1)
2 1 3 2m−1
− ··· − − −
instead of P . Again, there is n , such that a a 0 for all n>n . (cid:3)
1 0 n−1 n+1 0
≤
Now we establish a lemma whose proof might be of independent interest because
it is simpler than the proof of the corresponding general claim about functions in the
Hermite–Biehler space.
Lemma 2. Let ϕ(z) = ∞ a zn be an entire function with only real zeros, all of
n=0 n
the same sign and 0 ρ <1. Then all the zeros of both
ϕ
≤ P
∞
ϕo(z) = a z2n+1,
2n+1
n=0
X
∞
ϕe(z) = a z2n
2n
n=0
X
are purely imaginary.
Proof. Since 0 ρ <1, then the Hadamard product of ϕ is
ϕ
≤
∞
z
ϕ(z)=ϕ(0) 1 .
− x
n=1(cid:18) n(cid:19)
Y
Clearly, ϕ is a local uniform limit of hyperbolic polynomials
m ∞ ∞
z
P (z)=ϕ(0) 1 = a z2n+ a z2n+1,
m 2n,m 2n+1,m
− x
n=1(cid:18) n(cid:19) n=0 n=0
Y X X
with a R and a =a =0, when 2n>m. Then obviously
n,m 2n,m 2n+1,m
∈
m ∞ ∞
iz
P (iz)=ϕ(0) 1 = ( 1)na z2n+i ( 1)na z2n+1.
m 2n,m 2n+1,m
− x − −
n=1(cid:18) n(cid:19) n=0 n=0
Y X X
For each fixed m N, the zeros of the polynomial P (iz) are ix ,..., ix and
m 1 m
∈ − −
they posses imaginary parts of the same sign because the zeros x of ϕ are all of the
n
samesign. BytheTheoremofHermite-Biehler,thezerosofboth ∞ ( 1)na z2n
and ∞ ( 1)na z2n+1 are all real [6]. Therefore, the zerosno=f0 −∞ a2n,mz2n
and n∞=0a− 2nz+2n1,+m1 are all purely imaginary. Since these pPolynomina=ls0 c2onn,vmerge
Pn=0 2n+1,m P
locally uniformly to the functions ϕo(z) and ϕe(z), when m , it follows from the
TheoPrem of Hurwitz, that the zeros of ϕo(z) e ϕe(z) are pure→ly∞imaginary too. (cid:3)
Proof. (of Theorem 3). First we shall prove that if ψ is an entire function with the
property that at least one of the functions
∞ ∞
ψo(z)= b z2n+1 and ψe(z)= b z2n
2n+1 2n
n=0 n=0
X X
has an infinite number of purely imaginary zeros, then, for any a<0, the function
∞
ϕ(z)=exp(az2)ψ(z)= a zn
n
n=0
X