On the number of real critical points of logarithmic derivatives and the Hawaii conjecture

0.46 MB·English

by  Mikhail Tyaglov

On the number of real critical points of logarithmic derivatives and
the Hawaii conjecture
Mikhail Tyaglov
Institut fu¨r Mathematik, MA 4–5
9
0 Technische Universit¨at Berlin
0
D–10623 Berlin, Germany
2
[email protected]
b
e
F February 3, 2009

3

]
A
C
Dedicated with gratitude to Thomas Craven,
.
h George Csordas, and Wayne Smith.
t
a
m
[ Abstract

2 Foragivenrealentirefunctionϕwithfinitely manynonrealzeros,weestablishaconnection
v ′ ′ ′′ ′ ′
between the number of real zeros of the functions Q = (ϕ/ϕ) and Q1 = (ϕ /ϕ). This
3
connectionleadsto aproofoftheHawaiiconjecture[T.Craven,G.Csordas,andW.Smith, The
1
4 zeros of derivatives of entire functions and the Po´lya-Wiman conjecture, Ann. of Math. (2)
0 125 (1987), 405–431]stating that the number of real zeros of Q does not exceed the number of
2. nonreal zeros of ϕ.
0
9
0 Introduction
:
v
i Inthis paper,weinvestigate therealcritical pointsoflogarithmic derivatives ofrealentirefunction,
X
ϕ(z), in the class ∗ (see Definition 2). Our main result establishes bounds on the number
r L−P
a of real zeros of the derivative of the logarithmic derivative of the function ϕ(z). The idea behind
these bounds arose from attempts to prove the following twenty year-old conjecture of T. Craven,
G. Csordas and W. Smith [3] (see also [5]), which was nicknamed by A. Eremenko the Hawaii
conjecture.
The Hawaii conjecture. If a real polynomial p has 2m nonreal zeros, then the rational function
d p′(z) dp(z)
Q(z) = , where p′(z) = ,
dz p(z) dz
(cid:18) (cid:19)
has at most 2m real zeros, counting multiplicities.
Various attempts were made to resolve this conjecture by geometric and topologic methods [2,
6, 7]. In particular, J.Borcea and B.Shapiro [2] developed a general theory of level sets, which
1
may imply the validity of the Hawaii conjecture as a special case. In [13, Chapter 9], the Hawaii
conjecture was proved only for a few particular cases.
We remark that while the upper bound of the number of real zeros of Q was only conjectured
recently, the lower bound was known a long time ago, at least in the following special case:
Problem 133 ([9]). Let the real polynomial f(x) have only real zeros and suppose that the poly-
nomial f(x) + a, where a R 0 , has 2m nonreal zeros. Prove that the equation (f′(x))2
∈ \{ } −
f(x)f′′(x) af′′(x) = 0 has at least 2m real roots.
−
If p(z) = f(z)+a, then it follows that p has only simple zeros and exactly 2m nonreal zeros.
Moreover, p′ = f′ has only real zeros. By Problem 133, Q associated with the polynomial p(z) =
f(z)+a has at least 2m real zeros. In fact, Q has exactly 2m real zeros [3, Theorem 1]. The hint
provided for Problem 133 in [9] suggests using Rolle’s Theorem. This approach ultimately leads us
to a more precise result, namely, the following proposition.
Proposition 1. Suppose that the polynomial p has 2m nonreal zeros and its derivative p′ has 2m
1
nonreal zeros, then Q has at least 2m 2m real zeros.
1
−
Thus,thelower boundofthenumberofrealzeros ofQcanbeeasily determined. Unfortunately,
even this simple fact was not well known. We establish this lower bound in Theorem 2 (see (1.154)
and Remark 1.14).
In Section 1, we prove our main result, Theorem 2, which provides the bounds of the number
of critical points of the logarithmic derivative of a real entire function ϕ ∗ (Definition 2)
∈ L−P
possessing property A (Definition 4). In Section 2, we prove the Hawaii conjecture (Theorem 8) for
real entire functions in the class ∗ using Theorem 2 and one technical result, Theorem 7.
L−P
1 Bounds on the number of real critical points of the logarithmic
derivatives of functions in
∗
L − P
In this section, we need the following definitions.
Definition 1 ([10, 12]). The function ϕ is said to be in the Laguerre-Po´lya class, ϕ , if
∈ L−P
ω z z
ϕ(z) = czde−γz2+βz 1 eαj, (0 6 ω 6 ),
− α ∞
j
j=1(cid:18) (cid:19)
Y
where c,β,α R, γ > 0, d is a nonnegative integer and α−2 < .
j ∈ j ∞
Definition 2. The function ϕ is in the class ∗ if ϕP= pf where f and p is a real
L−P ∈ L−P
polynomial with no real zeros.
For ϕ ∗, by ZC(ϕ) we denote the number of nonreal zeros of ϕ, counting multiplicities.
∈ L−P
Iff is arealmeromorphicfunctionhavingonly afinitenumberof realzeros, then ZR(f)willdenote
the number of real zeros of f, counting multiplicities. In the sequel, we also denote the number
of zeros of the function f in an interval (a,b) and at a point α R by Z (f) and Z (f),
(a,b) {α}
∈
respectively1. Generally, the number of zeros of f on a set X will be denoted by Z (f).
X
1Thus, ZR(f)=Z(−∞,+∞)(f).
2
Let ϕ ∗. Between any two consecutive real zeros, say a and b, a < b, of ϕ, ϕ′ has an
∈ L−P
odd number of real zeros (and a fortiori at least one) by Rolle’s theorem. Counting all zeros with
multiplicities, suppose that ϕ′ has 2r+1 zeros between a and b. Then we will say that ϕ′ has 2r
extra zeros between a and b. If ϕ has the largest zero a (or the smallest zero a ), then any real
L S
zero ϕ′ in (a , ) (and in ( ,a )) is also called an extra zero of ϕ′. The total number of extra
L S
∞ −∞
zeros of ϕ′, counting multiplicities, will be denoted by E(ϕ′).
Remark 1.1. The multiple real zeros of ϕ are not counted as real extra zeros of ϕ′.
Remark 1.2. Let ϕ be a real polynomial of degree n > 1, and suppose that ϕ has exactly 2m
nonreal zeros. Then
ZC(ϕ) ZC(ϕ′), if n > 2m,
E(ϕ′)= − (1.1)
( ZC(ϕ) ZC(ϕ′) 1, if n = 2m.
− −
This fact will be used in the proof of Theorem 3.
For thereader’s convenience, werecall thefact proved in[3, Lemma3, p.411] that thefollowing
inequalities hold for ϕ ∗.
∈ L−P
ZC(ϕ) 6 E(ϕ′)+ZC(ϕ′) 6 ZC(ϕ)+2. (1.2)
However, if ϕ has an infinite number of real zeros, then a tighter upper bound can be established
(see [4, p. 325]):
ZC(ϕ) 6 E(ϕ′)+ZC(ϕ′) 6 ZC(ϕ′)+1. (1.3)
Let ϕ ∗ and let the function Q = Q[ϕ] associated with ϕ be defined as
∈ L−P
d ϕ′(z) ϕ(z)ϕ′′(z) (ϕ′(z))2
Q(z) = Q[ϕ](z) = = − . (1.4)
dz ϕ(z) (ϕ(z))2
(cid:18) (cid:19)
We note that if ϕ(z) = Ceβz, where C,β R, then Q(z) 0. Hence, we adopt the following
∈ ≡
convention throughout this paper.
Convention. If ϕ ∗, then ϕ is assumed not to be of the form ϕ(z) = Ceβz, C,β R.
∈ L−P ∈
With this convention, we set out to prove that Q has only finitely many real zeros. This fact is
essentially known from[3], butwe still includetheprooffor completeness. Note thatthis is obvious
if ϕ has a finite number of zeros. Here is a proof for the general case.
Theorem 1. Let ϕ ∗. Then the function Q has finitely many real zeros.
∈ L−P
Proof. By definition, ϕ is a product ϕ= pψ, where ψ and p is a real polynomial with no
∈ L−P
real zeros. Let degp = 2m >0.
If m = 0, then p(z) const. In this case, ϕ . But it is well known (see, for example, [3,
≡ ∈ L−P
4]) that the logarithmic derivative of a function from the Laguerre-Po´lya class is a decreasing
function on the intervals where it has no poles. Therefore, Q(z) < 0 for any real z, which is not a
pole of this function. Thus, if m = 0, then Q has no real zeros.
We assume now that m > 0. Observe that
Q = Q[ϕ] = Q[p]+Q[ψ].
3
From this equality it follows that all zeros of Q are roots of the equation
Q[p](z) = Q[ψ](z). (1.5)
−
It is easy to see that if p(z) = a z2m+... (a = 0), then
0 0
6
2ma2z4m−2+...
Q[p](z) = − 0 .
a2z4m+...
0
This formula shows that Q[p](z) 0 whenever z . Consequently, there exist two real
→ − → ±∞
numbers a and a (a < a ) such that Q[p](z) < 0 for z ( ,a ] [a ,+ ). But since
1 2 1 2 1 2
∈ −∞ ∪ ∞
ψ , we have Q[ψ](z) < 0 for z R as we mentioned above. Thus, the right hand side of the
∈ L−P ∈
equation(1.5)ispositiveforallz Rbutitslefthandsideisnegativeforallz ( ,a ] [a ,+ ).
1 2
∈ ∈ −∞ ∪ ∞
Therefore, all real roots of the equation (1.5) and, consequently, all real zeros of the function Q
belongtotheinterval (a ,a ), andQ(z) < 0outsidethisinterval. Sincerealzerosof ameromorphic
1 2
function are isolated, Q has only finitely many real zeros, as required.
Later (see Corollary 2) we show that the number of real zeros of the function Q is even.
Analogously to (1.4), we introduce the related function
d ϕ′′(z) ϕ′(z)ϕ′′′(z) (ϕ′′(z))2
Q (z) = Q[ϕ′](z) = = − . (1.6)
1 dz ϕ′(z) (ϕ′(z))2
(cid:18) (cid:19)
To expedite our presentation, we also introduce the following definition.
Definition 3. Let ϕ ∗ and let α be a real zero of ϕ. Suppose that β and β , β < α< β ,
1 2 1 2
∈ L−P
are real zeros of ϕ′ such that ϕ′(z) = 0 for z (β ,α) (α,β ). The function ϕ is said to possess
1 2
6 ∈ ∪
property A at its real zero α if Q has no real zeros in at least one of the intervals (β ,α) and
1
(α,β ). If α is the smallest zero of ϕ, then set β = , and if α is the largest zero of ϕ, then set
2 1
−∞
β = + .
2
∞
Definition 4. A function ϕ ∗ is said to possess property A if ϕ possesses property A at
∈ L−P
each of its real zeros. In particular, ϕ without real zeros possesses property A.
Attheendofthissection,weprovethefollowingtheorem,whichprovidesboundsforthenumber
of real zeros of Q.
Theorem 2. Let ϕ ∗ and suppose that ϕ has exactly 2m nonreal zeros. If the function ϕ
∈ L−P
possesses property A, then
2m 2m1 6 ZR(Q) 6 2m 2m1+ZR(Q1), (1.7)
− −
where 2m1 = ZC(ϕ′) and the functions Q and Q1 are defined in (1.4) and (1.6).
Before we establish Theorem 2, we prove a few lemmas and their corollaries. For a given
function ϕ, by F and F we will denote the following functions
1
F(z) = ϕ(z)ϕ′′(z) ϕ′(z) 2, F (z) = ϕ′(z)ϕ′′′(z) ϕ′′(z) 2. (1.8)
1
− −
At first, we estimate the parities o(cid:0)f the(cid:1)numbers of real zeros of Q i(cid:0)n cert(cid:1)ain intervals and on
the real axis.
4
Lemma 1. Let ϕ ∗ and let β and β be two real zeros of ϕ′.
1 2
∈ L−P
I. If β and β are consecutive real zeros of ϕ′, and ϕ(z) = 0 for z (β ,β ), then Q has an
1 2 1 2
6 ∈
odd number of real zeros in (β ,β ), counting multiplicities.
1 2
II. If β and β are two real zeros of ϕ′ such that ϕ has a unique real zero α in (β ,β ) and
1 2 1 2
ϕ′(z) = 0 for z (β ,α) (α,β ), then Q has an even number of real zeros, counting
1 2
6 ∈ ∪
multiplicities, in each of the intervals (β ,α) and (α,β ).
1 2
Proof. I. In fact, since ϕ′/ϕ equals zero at the points β and β , its derivative, the function Q,
1 2
has an odd number of zeros in (β ,β ) by Rolle’s theorem.
1 2
II. If α is a zero of ϕ of multiplicity M > 1, then ϕ(z) = (z α)Mψ(z), where ψ(α) = 0. Thus, we
− 6
have
M ψ(z)ψ′′(z) (ψ′(z))2
Q(z) = + − .
−(z α)2 (ψ(z))2
−
Consequently, for all sufficiently small ε> 0,
Q(α ε) < 0. (1.9)
±
Thus, ϕ′/ϕ is decreasing in a small left-sided vicinity of α, and ϕ′(z)/ϕ(z) whenever z α.
→ −∞ ր
Since ϕ′/ϕ equals zero at β , its derivative, the function Q, must have an even number of zeros in
1
(β ,α) by Rolle’s theorem.
1
By the same argumentation, Q has an even number of zeros in (α,β ).
2
Lemma 2. Let ϕ ∗. If ϕ has the largest zero α (or the smallest zero α ), then Q has an
L S
∈L−P
even number of real zeros in (α ,+ ) (or in ( ,α )), counting multiplicities.
L S
∞ −∞
Proof. The inequality (1.9) holds for any real zero of ϕ, consequently, Q is negative for z suf-
ficiently close to α (or to α ). But it was already proved in Theorem 1 (see also (3.11) in [3,
L S
p.415] and subsequent remark there) that Q(z) < 0 for all sufficiently large real z. Therefore,
Q has an even number of zeros in (α ,+ ) (and in ( ,a ) if ϕ has the smallest real zero α ),
L S S
∞ −∞
counting multiplicities, since Q(z) is negative for all real z sufficiently close to the ends of the
interval (α ,+ ) (or of the interval ( ,α )).
L S
∞ −∞
If additional information on the number of real zeros of ϕ′ is available, then Lemma 1 and
Lemma 2 can be used to derive the following sharper result.
Corollary 1. Let ϕ ∗ and suppose that ϕ has the largest zero α and ϕ′ has exactly r extra
L
∈ L−P
zeros, counting multiplicities, in the interval (α ,+ ). If β is the largest zero of ϕ′ in (α ,+ ),
L L L
∞ ∞
then Q has an odd (even) number of real zeros in (β ,+ ) whenever r is an even (odd) number.
L
∞
Proof. Letϕ′ havel 6r distinct zeros, say β < β < ... < β = β , intheinterval(α ,+ ). Ac-
1 2 l L L
∞
cording to Lemma 2, Q has an even number of real zeros in (α ,+ ), counting multiplicities. But
L
∞
fromLemma1itfollowsthatQhasanevennumberofrealzerosin(α ,β ),countingmultiplicities,
L 1
and an odd number of real zeros, say 2M +1, in each of the intervals (β ,β ) (i = 1,2,...,l 1).
i i i+1
Hence, Q has exactly l−1(2M +1) real zeros, counting multiplicities, in l−1(β ,β ). −
i=1 i i=1 i i+1
Moreover, from(1.4) itfollows that β is azero of Qof multiplicity M 1whenever β is azero of
P − S
ϕ′ ofmultiplicityM. Consequently,inourcase,Qhasexactlyr lrealzeros,countingmultiplicities,
−
5
atthepointsβ thataremultiplezerosofϕ′. Thus,Qhasr l+ l−1(2M +1) = r 1+ l−12M
i − i=1 i − i=1 i
real zeros, counting multiplicities, in the interval [β ,β ]. Therefore, if r is an even (odd) number,
1 L
P P
then Q has an odd (even) number of real zeros in (α ,β ]. Recall that Q has an even number of
L L
zerosin(α ,β )byLemma1. Consequently,Qhasanodd(even)numberofrealzerosin(β ,+ ),
L 1 L
∞
since Q has an even number of real zeros in (α ,+ ), according to Lemma 2.
L
∞
Remark 1.3. Corollary 1 is valid with respective modification in the case when ϕ has the smallest
zero a .
S
So far, we have considered semi-infinite intervals, our next statement addresses the entire real axis.
Corollary 2 (Craven–Csordas–Smith [3], p. 415). If ϕ ∗, then the function Q associated
∈ L−P
with ϕ has an even number of real zeros, counting multiplicity.
Proof. In fact, if ϕ has no real zeros, then Q has no real poles, and the number ZR(Q) is even,
since Q(z) < 0 for all sufficiently large real z.
If ϕ has only one real zero α, then, according to Lemma 2, Q has an even number of zeros in
each of the intervals ( ,α) and (α,+ ). Thus, ZR(Q) is also even in this case.
−∞ ∞
Letϕhaveatleast tworealzeros. Ifα andα aretwo consecutive zeros ofϕ, then, according
j j+1
to (1.9), Q has an even number of zeros, counting multiplicities, in the interval (α ,α ). If ϕ has
j j+1
the largest (or/and the smallest) real zero, say α (α ), then, by Lemma 2, Q has an even number
L S
of realzeros, counting multiplicities, in(aL,+ )(and in( ,αS)). Therefore, thenumberZR(Q)
∞ −∞
is even.
Remark 1.4. Analogously, the function Q associated with a function in the class ∗ has an
1
L−P
even number of real zeros, counting multiplicities, since the class ∗ is closed with respect to
L−P
differentiation [3].
The following lemma is the first in a series of lemmata that estimate the number of real zeros
of Q on a finite interval, given specific information on ϕ, ϕ′, ϕ′′ and Q .
1
Lemma 3. Let ϕ ∗ and let a and b be real and let ϕ(z) = 0, ϕ′(z) = 0, ϕ′′(z) = 0,
∈ L−P 6 6 6
Q (z) =0 in the interval (a,b).
1
6
I. If, for all sufficiently small δ > 0,
ϕ′(a+δ)ϕ′′(a+δ)Q(a+δ)Q (a+δ) > 0, (1.10)
1
then Q has no real zeros in (a,b].
II. If, for all sufficiently small δ > 0,
ϕ′(a+δ)ϕ′′(a+δ)Q(a+δ)Q (a+δ) < 0, (1.11)
1
then Q has at most one real zero in (a,b), counting multiplicities. Moreover, if Q(ζ) = 0 for
some ζ (a,b), then Q(b) = 0 (if Q is finite at b).
∈ 6
Proof. The condition ϕ(z) = 0 for z (a,b) means that Q is finite at every point of (a,b).
6 ∈
If ζ (a,b) and Q(ζ)= 0, then F(ζ) = 0 and (1.8) implies
∈
ϕ(ζ)ϕ′′(ζ)
ϕ′(ζ)= . (1.12)
ϕ′(ζ)
6
Now we consider F . From (1.8) and (1.12) it is easy to derive that
1
ϕ(ζ)ϕ′′(ζ)ϕ′′′(ζ)
F (ζ)= ϕ′(ζ)ϕ′′′(ζ) (ϕ′′(ζ))2 = (ϕ′′(ζ))2 =
1 − ϕ′(ζ) −
(1.13)
ϕ′′(ζ) ϕ′′(ζ)
= [ϕ(ζ)ϕ′′′(ζ) ϕ′(ζ)ϕ′′(ζ)] = F′(ζ).
ϕ′(ζ) − ϕ′(ζ)
Sinceϕ′(z) = 0,ϕ′′(z) = 0,Q (z) =0 (and therefore F (z) = 0) in (a,b) by assumption, from (1.13)
1 1
6 6 6 6
it follows that ζ is a simple zero of Q. That is, all zeros of Q in (a,b) are simple.
I. Let the inequality (1.10) hold. Assume that, for all sufficiently small δ > 0,
ϕ′(a+δ)ϕ′′(a+δ)Q (a+δ) > 0, (1.14)
1
then Q(a+δ) > 0, that is, F(a+δ) > 0. Therefore, if ζ is the leftmost zero of Q in (a,b), then
F′(ζ) < 0. This inequality contradicts (1.13), since
ϕ′(z)ϕ′′(z)Q (z) > 0
1
for z (a,b), which follows from (1.14) and from the assumption of the lemma. Consequently,
∈
Q cannot have zeros in the interval (a,b) if the inequalities (1.10) and (1.14) hold. In the same
way, one can prove that if ϕ′(a+δ)ϕ′′(a+δ)Q (a+δ) <0 for all sufficiently small δ > 0 and if the
1
inequality (1.10) hold, then Q(z) = 0 for z (a,b).
6 ∈
Thus, Q has no zeros in the interval (a,b) if the inequality (1.10) holds. Moreover, it is easy to
show that Q(b) = 0 as well. In fact, if F(b) = 0, then, for all sufficiently small ε > 0,
6
ϕ′(b ε)
sign − F (b ε) = sign(F′(b ε)) (1.15)
ϕ′′(b ε) 1 − −
(cid:18) − (cid:19)
according to (1.13). But if the inequality (1.10) holds, then
ϕ′(b ε)
sign − F (b ε) = sign(F(b ε)) (1.16)
ϕ′′(b ε) 1 − −
(cid:18) − (cid:19)
for all sufficiently small ε > 0, since ϕ′(z) = 0, ϕ′′(z) = 0, Q (z) = 0 in the interval (a,b) by
1
6 6 6
assumption and since Q(z) = 0 in (a,b), which was proved above. So, if the inequality (1.10) holds
6
and if F(b) = 0, then from (1.15) and (1.16) we obtain that
F(b ε)F′(b ε) > 0
− −
for all sufficiently small ε > 0. This inequality contradicts the analyticity2 of the function F.
Therefore, if the inequality (1.10) holds and if Q is finite at the point b, then Q(b) = 0. Thus, the
6
first part of the lemma is proved.
2If a function f is analytic at some neighborhood of a real point a and equals zero at this point, then, for all
sufficiently small ε>0,
f(a−ε)f′(a−ε)<0.
7
II. Let the inequality (1.11) hold, then Q can have zeros in (a,b). But it cannot have more than
one zero. In fact, if ζ is the leftmost zero of Q in (a,b), then this zero is simple as we proved above.
Therefore the following inequality holds for all sufficiently small ε > 0
ϕ′(ζ +ε)ϕ′′(ζ +ε)Q(ζ +ε)Q (ζ +ε) > 0.
1
Consequently, Q has no zeros in (ζ,b] according to Case I of the lemma.
Remark 1.5. Lemma 3 is also true if (a,b) is a half-infinite interval, that is, (a,+ ) or ( ,b).
∞ −∞
Lemma3addressesthecaseofafiniteintervalandRemark1.5thecaseofahalf-infiniteinterval.
Our next statement concerns the entire real line.
Corollary 3. Let ϕ ∗. If ϕ′(z) = 0, ϕ′′(z) = 0, Q1(z) = 0 for z R, then ZR(Q) = 0, i.e.
∈ L−P 6 6 6 ∈
Q has no real zeros.
Proof. The function ϕ may have real zeros. But by Rolle’s theorem, ϕ has at most one real zero,
counting multiplicity, since ϕ′(z) = 0 for z R by assumption.
If ϕ(z) = 0 for z R, then, by6 Lemma ∈3 applied on the real axis, Q has at most one real zero.
6 ∈
But by Corollary 2, the number of real zeros of Q is even. Consequently, ZR(Q) = 0.
If ϕ has one real zero, counting multiplicity, say α, then α is a unique pole of Q. Moreover,
since ϕ′′(z) = 0 for z R by assumption, the function ϕϕ′′ changes its sign at α and, therefore,
6 ∈
ϕ(z)ϕ′′(z) < 0 in one the intervals ( ,α) and (α,+ ). Consequently, by (1.4), Q(z) = 0
−∞ ∞ 6
in one of the intervals ( ,α) and (α,+ ). Without loss of generality, we may suppose that
−∞ ∞
Z (Q) = 0. Then according to Lemma 3 and Remark 1.5, we obtain that Q has at most one
(−∞,α]
zero in the interval (α,+ ). But the number of real zeros of Q is even by Corollary 2, and Q
∞
has no zeros in the interval ( ,α]. Therefore, Q cannot have zeros in the interval (α,+ ), so
−∞ ∞
ZR(Q) = 0, as required.
Thus, we have found out that Q has at most one real zero, counting multiplicity, in an interval
if the functions ϕ, ϕ′, ϕ′′ and Q have no real zeros in this interval. Now we study the multiple
1
zeros of Q and its zeros common with some other above-mentioned functions. From (1.4) it follows
that all zeros of ϕ′ of multiplicity at least 2 are also zeros of Q and all zeros of ϕ′ of multiplicity
at least 3 are multiple zeros of Q. The following lemma provides the information about common
zeros of Q and Q .
1
Lemma 4. Let ϕ ∗ and let a and b be real and let ϕ(z) = 0, ϕ′(z) = 0, ϕ′′(z) = 0 in the
∈ L−P 6 6 6
interval (a,b). Suppose that Q has a unique zero ξ (a,b) of multiplicity M in (a,b). If Q(ξ) = 0,
1
∈
then ξ is a zero of Q of multiplicity M +1, and Q(z) = 0 for z (a,ξ) (ξ,b].
6 ∈ ∪
Proof. The condition ϕ(z) = 0 for z (a,b) means that Q is finite at every point of (a,b).
6 ∈
By assumption, ξ is a zero of F of multiplicity M and F(ξ) = 0. First, we prove that ξ is
1
a zero of F of multiplicity M +1.
Since ϕ′(z) =0 for z (a,b) by assumption, from (1.8) it follows that
6 ∈
ϕ(z)ϕ′′(z) F(z)
ϕ′(z) = .
ϕ′(z) − ϕ′(z)
8
Substituting this expression into the formula (1.8), we obtain
ϕ(z)ϕ′′(z)ϕ′′′(z) ϕ′(z)ϕ′′(z)ϕ′′(z) F(z)ϕ′′′(z)
F (z) = =
1 ϕ′(z) − ϕ′(z) − ϕ′(z)
ϕ′′(z) F(z)ϕ′′′(z)
= F′(z) ,
ϕ′(z) · − ϕ′(z)
or, equivalently,
ϕ′(z) F(z) ′
F (z) = ,
[ϕ′′(z)]2 1 ϕ′′(z)
(cid:18) (cid:19)
since ϕ′′(z) = 0 for z (a,b) by assumption. Differentiating this equality j times with respect to z,
6 ∈
we get
ϕ′(z) (j) F(z) (j+1)
F (z) = . (1.17)
[ϕ′′(z)]2 1 ϕ′′(z)
(cid:18) (cid:19) (cid:18) (cid:19)
From (1.17) it follows that F(j+1)(ξ) = 0 if ϕ′(ξ) = 0, ϕ′′(ξ) = 0, F(i)(ξ) = 0 and F(i)(ξ) = 0,
6 6 1
i = 0,1,...,j. Consequently, ξ is a zero of F of multiplicity at least M +1. But by assumptions,
the formula (1.17) gives the following equality
ϕ′(ξ)F(M+1)(ξ) = ϕ′′(ξ)F(M+2)(ξ) = 0.
1 6
Hence, ξ is a zero of F of multiplicity exactly M +1. But ϕ(ξ) = 0 by assumption, therefore, ξ is
6
a zero of Q of multiplicity M +1.
It remains to prove that Q has no zeros in (a,b] except ξ. In fact, consider the interval (a,ξ).
According to Lemma 3, Q can have a zero at ξ only if the inequality (1.11) holds and Q(z) = 0
6
for z (a,ξ). Furthermore, the function ϕ′ϕ′′ does not change its sign at ξ but the function QQ
1
∈
does, since ξ is a zero of QQ of multiplicity 2M +1. Thus, for all sufficiently small δ > 0,
1
ϕ′(ξ+δ)ϕ′′(ξ +δ)Q(ξ +δ)Q (ξ+δ) > 0, (1.18)
1
since the inequality (1.11) must hold in the interval (a,ξ) by Lemma 3. From (1.18) it follows that
Case I of Lemma 3 holds in the interval (ξ,b), so Q(z) = 0 for z (ξ,b].
6 ∈
Remark 1.6. Lemma 4 remains valid if (a,b) is a half-infinite interval, that is, (a,+ ) or ( ,b).
∞ −∞
Corollary 4. Let ϕ ∗ and let ϕ′(z) = 0, ϕ′′(z) = 0 for z R. If Q has only one real zero,
1
∈ L−P 6 6 ∈
then Q has no real zeros, i.e. ZR(Q) = 0.
Proof. As in Corollary 3, we note that ϕ has at most one real zero by Rolle’s theorem.
CaseI.Letϕ(z) =0forz Randletξ beauniquerealzeroofQ . ByCorollary2(seeRemark1.4),
1
6 ∈
ξ is a zero of Q of even multiplicity 2M. In this case, the number ξ cannot be zero of Q. In fact,
1
if Q(ξ) =0, then, by Lemma 4 applied on the real axis, ξ is a unique real zero of Q of multiplicity
2M +1, that is, Z{ξ}(Q) =ZR(Q) =2M +1. This contradicts Corollary 2. So, Q(ξ) = 0.
6
Sinceξ isauniquerealzeroofQ ofevenmultiplicity, Q hasequalsignsintheintervals( ,ξ)
1 1
−∞
and (ξ,+ ). By Lemma 3 applied to these intervals, Q can have at most one zero in each of the
∞
intervals ( ,ξ) and (ξ,+ ). But if Q has a zero in the interval ( ,ξ), then Q(z) = 0
−∞ ∞ −∞ 6
for z (ξ,+ ). In fact, if ζ ( ,ξ) is a zero of Q, then ζ is simple zero of Q by Lemma 3.
∈ ∞ ∈ −∞
9
Moreover, Q(z) = 0 for z (ζ,ξ], since (see the proof of Lemma 3), for all sufficiently small δ > 0,
6 ∈
we have
ϕ′(ζ +δ)ϕ′′(ζ +δ)Q(ζ +δ)Q (ζ +δ) > 0. (1.19)
1
But the functions ϕ′, ϕ′′, Q and Q do not change their signs at ξ, therefore, we have the in-
1
equality (1.19) in a small right-sided neighborhood of ξ. Consequently, Q(z) = 0 in (ξ,+ ) by
6 ∞
Lemma 3.
Thus,ifQhasrealzeros,thenithasatmostonezeroinoneoftheintervals( ,ξ)and(ξ,+ ),
−∞ ∞
that is, ZR(Q) 6 1. Now Corollary 2 implies ZR(Q) = 0.
Case II. If ϕ has one real zero, counting multiplicity, say α, then α is a unique pole of Q. In this
case, as in the proof of Corollary 3, one can show that Q(z) = 0 in one of the intervals ( ,α)
6 −∞
and (α,+ ). Without loss of generality, we may assume that Q(z) = 0 for z ( ,α], that is,
∞ 6 ∈ −∞
Z (Q)= 0. Then by Corollary 2, the number Z (Q) is even.
(−∞,α] (α,+∞)
Let ξ be a unique real zero of Q and ξ ( ,α], then, by Lemma 3 (see Remark 1.5), Q has
1
∈ −∞
at most one zero in (α,+ ). Since Z(α,+∞)(Q) is an even number, ZR(Q)= 0.
∞
If ξ (α,+ ), then, by the same argument as in Case I, one can show that Z (Q) = 0.
(α,+∞)
∈ ∞
Therefore, ZR(Q)= 0, since Z(−∞,α](Q) = 0 by assumption.
We now provide a general bound on the number of real zeros of Q in terms of the number of
real zeros of Q in a given interval.
1
Lemma 5. Let ϕ ∗ and let a and b be real. If ϕ(z) = 0, ϕ′(z) = 0 and ϕ′′(z) = 0 for
∈ L−P 6 6 6
z (a,b), then
∈
Z (Q) 6 1+Z (Q ). (1.20)
(a,b) (a,b) 1
Proof. If ϕ(z)ϕ′′(z) < 0 in (a,b), then Q(z) < 0 for z (a,b) by (1.4), that is, Z (Q) = 0.
(a,b)
∈
Therefore, the inequality (1.20) holds automatically in this case.
Let now ϕ(z)ϕ′′(z) > 0 for z (a,b). If Q (z) = 0 in (a,b), then, by Lemma 3, Q has at most
1
∈ 6
one real zero, counting multiplicity, in (a,b). Therefore, (1.20) also holds in this case.
If Q has a unique zero ξ in (a,b) and Q(ξ) = 0, then, by Lemma 3, Q has at most one real
1
6
zero in each of the intervals (a,ξ) and (ξ,b):
Z (Q) 6 1+Z (Q ), (1.21)
(a,ξ) (a,ξ) 1
where Z (Q ) = 0, and
(a,ξ) 1
Z (Q) 6 1+Z (Q ), (1.22)
(ξ,b) (ξ,b) 1
where Z (Q ) = 0. But since Q(ξ) = 0 and Q (ξ)= 0, we have
(ξ,b) 1 1
6
0 = Z (Q) 6 1+Z (Q ). (1.23)
{ξ} {ξ} 1
−
Thus, summing the inequalities (1.21)–(1.23), we obtain (1.20).
If Q has a unique zero ξ in (a,b) and Q(ξ) =0, then, by Lemma 4, we have
1
Z (Q) = 1+Z (Q ),
{ξ} {ξ} 1
and Q(z) = 0 for z (a,ξ) (ξ,b). Therefore, the inequality (1.20) is also true in this case.
6 ∈ ∪
10