Remarks on the Ilieff-Sendov Problem

ANNALES

UNI VERSIT ATIS MARIAE C U RI E-S K LO D O W S K A LUBLIN-POLONIA

VOL. XLVIII, 9________________SECTIOA__________________________1994

Peter PFLUG (Vechta) Gerald SCHMIEDER (Oldenburg)

Remarks on the Ilieff-Sendov Problem

Abstract. Letp be a polynomial whose zeros are contained in theclosed unit disk E. According to the Ilieff-Sendov conjecturethere exist a, z* eE such that

(*) p(“)=0. j>'(**)=o> I«-»*I<1 •

Theauthors define certain polynomials 0t=0*(a,z2)... ,z„), k=O,...,n—1, and es­

tablish inequalities involving the values of fit, at ag[0,l] and z2l...,z„6E which imply the existence pf z* such that (*) holds.

The problem known for more than thirty years as the Ilieff- Sendov conjecture is as follows:

Let p E C[z] be a polynomial whose zeros belong to the closed unit disc E with E := {A E C : |A| < 1}. Let p(a) = 0. Then it is asked whether there exists z* E a+ E withp'(z*) = 0.

There is already along list of papers attacking this question (cf.

[l]-[27]). But so far it is only known that the conjecture is true if the degree of p is less or equal than 6 (cf. [11]). Proofs use very ad hoc methods. So the aim of our short note is to reduce the above problem to a more geometric question.

Fix a E [0,1] and let Z2,...,2„ € £(n > 3). We define Ok •= <7fc(a,Z2,...,Zn)

:= 52 (a“*h)-- (a “*>*), fc = l,...,n-l, 2<ji <---<>* <n

and

:=/3t(a,2r2,..-,-?n)

••= 7^ir^n-i-t(a,22,-• •,Zn), fc = 0,...,n - 1, v k )

with

a0 := 1.

We will prove the following:

Theorem . Let p(z) .= (z — a){J"=2(z ~ zi) a, z2> • • •»*„

as above, n > 3. Assume that at least one of the following conditions is fulfilled:

1) there exists some ( 6 E such that for some v E {3, ...,n} the inequality K/Sj + /30| < + /3„_2| holds

or

2)

IA>| < |A~ll(=

Then there exists z* € a + E with p'(z*) = 0.

Proof. Put q(z) := p(z +a) and

p=0 i/=0 ' '

For 1 < n — v <n — 3 we consider the polynomial

Note that L(z) has all its zerosin E if and only if Q(z) = zn~2 L(l/z) does not vanish in E. And this is obviously true in the case that

100 P. Pflug and G. Schmieder

Assume that (*) holds and take some £ € E.

Then all the zeros ofthe test-polynomial W = -L(z)(z -() -

< + (-1)1 (" 7 ’ 'J + (-1)"- (" ~

\ 1 / n — 1 \n — v) v

+ (_l)n-v+1/' «"I (.n_vzn~v+x

\n —i/ + l/ — 1)

are as well contained in E.

The (somewhat mysterious) theorem of Grace (cf. [26] Satz 1') can be stated as follows:

Let Ao,...,Ajv, ao,...,on be complex numbers with Xn / 0 whichfulfill the following “apolarity condition”

Aoayv + Ajajv-i + • • • + Ajva0 = 0 .

Then every closed disk D in C containing all the roots of the polyno­ mial

A(z)— Ao — i Ai z2----+•••• + (—l)WAjvzN

contains at least one zero of the polynomial

A(z) = a0 + Qi z -|---1- aNzN .

Hence in order to apply the theorem of Grace we have to make sure that there are and ¿n_2 with (*) and £n_2 0such that the

apolarity conditionholds (a separate discussion of the cases 1/ = n —1 and v = 3 (two exponents in Lq coincide) is not necessary):

This is equivalent to

4-2

= — (^ßn-\ + ßn-2 + ( 1/ •

From (*) we obtain

<

v

I4-

|^

(,ßn-l + ßn-2 + ^£n-v[(,ßv-\ + /?»/—2]

Division by |4-»| and considering the limit 4-p —» oo gives the assumption 1) in the theorem. Therefore, whenever 1) is fulfilled then the above apolarity condition for L^z) and q'(z') is fulfilled for a suitable choice of 4-f, 4-2- From (*) we obtain the desired zero 2* by the theorem of Grace.

The polynomial = — /30//?n—l + (—l)n_1 «n_1 is always ap­

olar to g'(z) and therefore the assumption 2) in the theorem comes out to be sufficient for the claimed zero z*. □

Remarks.

1. Note that the condition 1) is equivalent to the following state­

ment which is freeof the parameter £:

(la) I0ol<lftl

102 P. Pflug and G. Schmieder or, for some v G {3,...,n},

(lb) |/%|2 + |/3i |2 — |/3P-i|2 — |/?i,_212 < 2|/3p-i/3„-2 — Z?i /?o | Thisfollows fromthe fact that a Moebius transformation is an inner function of the unit disk iff the image of dE is part of E and there is nopole inside E.

Hence,in order to solvethe Hieff-Sendov conjectureit suffices to show that there are no points a, Z2,... ,zn € E (as above) satisfying the following inequalities:

|an_i(a,z2,... ,zn)| > n,

|a„_i(a,z2,...,zn)| > --- 72 |crn-2(a,22,...,zn)|

n — 1 and

|a„_i(a,22,... ,zn)|2 + --- —|crn_2(a,22,...,2n)|

> 2

(n - 1)^

i/!(n — i/)!

. (n-1)!

(i/ — l)!(n —v+ 1)!

j/!(p — l)!(n— i/)!(n — t/+1)1

CTn_„(a,z2,...,zn)|

|crn—p+i(a, ^2,• ••> ¿n)|

2

[(n - 1)!]2 ^{CTn —p ‘ Gn—v+1} («- 1)<^n—2*^n—1 if 1/ = 3,4, ... ,n.

2. Setting a = 0 ourTheorem shows thatthe Ilieff-Sendov conjecture is true for every polynomial p(z) = (z — a) f|”_2(z — 2«') = p(z + a) and g'(z) = f°r which the coefficients fulfill

|/?o| < |/3jt| for some k G {1,...n - 1}, i.e. < (t + 1)

for some k.

S2<h<...<jt<„(a-2ji) 1 •••(<* 1

3. From the above inequality for k = n — 1 we see that it suffices to have

< n .

>=2

Therefore the desired zero z* of q' can be found if

for all j = 2,...,n. Roughly speaking: the conjecture is true for a if theother zeros axepretty close to a. In particular,if a < —1+ n_^/n, then the conjecture is always true for a.

4. Observe that the inequality |/?o| > |^i| covers the often discussed case that 53”=2 1/rJ < (n — 1)/(1 + a), wherer; := \zj — a|.

Moreover, |/?o| > |/?i | implies that

n - 1 Ê— ä£Re—

2--' Zj —a Zj — a 2a 2a ' r2

j=2 1 J=2 1 j=2 i

which gives r; < + a for at least one j.

Hence we have the following Corollary:

Corollary . If p(z) = (z — a)n>=2(3 — ” as a^ove an^ lf l*j —a| > \/\z}l + a for a^j> then the Hieff-Sendov conjecture is true for a, i. e. there exists z* Ç. a + Ë with p'{z*j = 0.

We emphasize that this remark solves the Ilieff-Sendov conjec­

ture for all n if \zj — a| > \/l + a for j = 2,..., n, i. e. if one zero

°f p (here a) is strongly separated from the remaining n — 1 zeros.

Roughly speaking this means: the conjecture is true for a if theother zeros are pretty fax away from a.

Using the Corollary it is easy to see that already the assumption

r2 < (ti/a/I + a)1^" 2 (cf. Remark 3) implies that the Ilieff-Sendov conjecture is true for a.

104 P. Pflug and G. Schmieder

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