Some In tegral Mean Estim ates for Polynom ialswith Restricted Zeros

Abdu llah M ir，Q.M.Daw ood and BilalDar

Department ofM athematics，University ofKashm ir，Srinagar190006，India

Some In tegral Mean Estim ates for Polynom ialswith Restricted Zeros

Abdu llah M ir∗，Q.M.Daw ood and BilalDar

Department ofM athematics，University ofKashm ir，Srinagar190006，India

Received 19 January 2014；Accep ted（in revised version）11March 2015

.Let P（z）be a polynom ialof degree n having all its zeros in|z|≤k.For k=1，it is know n that for each r＞0 and|α|≥1， In this paper，we shall firstconsider the casew hen k≥1 and presentcertain generalizationsof this inequality.Also for k≤1，we shallprovean interesting resu lt for Lacunary type of polynom ials from which m any Results can be easily deduced.

Polynom ial，zeros，polar derivative.

AM SSub jectClassifications：30A 10，30C10，30D 15

1 In troduction and statem en tof Results

Let P（z）be a polynom ial of degree n and P′（z）be its derivative.Itwas show n by Turan［21］that if P（z）has all its zeros in|z|≤1，then

M ore generally，if the polynom ial P（z）has all its zeros in|z|≤k≤1，itwas proved by M alik［12］that the inequality（1.1）can be rep laced by

while as Govil［6］proved that if all the zerosof P（z）lie in|z|≤k where k≥1，then

As an im provem entof（1.3），Govil［7］proved that if P（z）hasall its zeros in|z|≤k where k≥1，then

Let DαP（z）denotes the polar derivative of the polynom ial P（z）of degree n with respect to the pointα.Then

The polynom ial DαP（z）is of degree atm ost n-1 and it generalizes the ord inary derivative in the sense that

Shah［18］extended（1.1）to the polar derivative of P（z）and proved that if all the zerosof the polynom ial P（z）lie in|z|≤1，then

Aziz and Rather［3］generalised（1.6）which also extends（1.2）to the polar derivative of a polynom ial.In fact，they proved that if all the zerosof P（z）lie in|z|≤k where k≤1，then for every realor com p lex num berαwith|α|≥k，

Further as a generalization of（1.3）to the polar derivative of a polynom ial，Aziz and Rather［3］proved that if all the zeros of P（z）lie in|z|≤k where k≥1，then for every real or com p lex num berαwith|α|≥k，

Recently Govil and M cTum e［8］sharpened（1.8）and proved that if all the zeros of P（z）lie in|z|≤k，k≥1，then for every realor com p lex num berαwith|α|≥1+k+kn，

On the other hand，M alik［13］obtained an Lranalogue of（1.1）by p roving that if P（z）has all its zeros in|z|≤1，then for each r＞0，

As an extension of（1.3），Aziz［1］proved that if P（z）has all its zeros in|z|≤k，k≥1，then for each r≥1，

M ore recently，Dew an，Singh，M ir and Bhat［5］generalized（1.6）by obtaining an Lranalogue of it.M ore p recisely，they proved that if P（z）has all its zeros in|z|≤1，then for every realor com p lex num berαwith|α|≥1 and for each r＞0，

If we let r→∞in（1.12）and m ake use of the well-know n fact from analysis（see for example［17，pp.73］or［20，pp.91］）that

we get（1.6）.

In this paper，we shall first present certain generalizations of the inequality（1.12）by considering polynom ials having allzeros in|z|≤k，k≥1.We shallalso prove a resu lt for Lacunary type of polynom ials having all zeros in|z|≤k，k≤1 from which m any Results can be easily deduced.

where

Rem ark 1.1.Ifwe let r→∞and p→∞（so that q→1）in（1.13）we get（1.8）.Ifwe d ivide both sidesof（1.13）by|α|and let|α|→∞，wegeta resu lt recently proved M irand Dar［15］. Ifwe take k=1 in（1.13）and note that Cr=1，we obtain a generalization of（1.12）in the sense that the righthand side of（1.12）is rep laced by a factor involving the in tegralMean of|DαP（z）|on|z|=1.

The follow ing corollary imm ed iately follows by letting p→∞（so that q→1）in Theorem 1.1.

Coro llary 1.1.If P（z）=∑nν=0aνzνisa polynom ialof degree n having all its zeros in|z|≤k where k≥1，then for every com p lex num berαwith|α|≥k and for each r＞0，

Rem ark 1.2.Divid ing both sides of（1.14）by|α|and let|α|→∞，we get（1.11）and also extends it to the values r∈（0，1）.For k=1，Corollary 1.1 reduces to inequality（1.12）.

Ou r next resu lt is a generalization of Theorem 1.1which in tu rn p rovides extensions and generalizations of Results of Aziz and Ahem ad［2］.Wewill see that as a special case Theorem 1.2 givesa resu ltof Goviland M cTum e［8，Theorem 3］.

Rem ark 1.3.A variety of interesting Results can be easily deduced from Theorem 1.2 in the sam e w ay aswe have deduced from Theorem 1.1.Here wem en tion a few of these. Divid ing the tw o sidesof（1.15）by|α|and let|α|→∞，wegeta resu lt recently proved M ir and Dar［15］.M oreover，ifwe take k=1 in（1.15）（noting that Cr=1）and then d ivide both sidesof itby|α|and let|α|→∞，we geta resu ltof Aziz and Ahem ad［2，Theorem 2］.

If in（1.15），we let p→∞（so that q→1），we get

Ifwe d ivide both sides of（1.16）by|α|and let|α|→∞，we get a resu lt of Aziz and Ahem ad［2，Theorem 4］and also extends it for 0＜r＜1 aswell.Forλ=0，（1.16）reduces to（1.14）.Fu rther，ifwe let r→∞in（1.16）and assum e|α|≥1+k+kn，we get

Let z0be a poin ton|z|=1 such that|P（z0）|=m ax|z|=1|P（z）|，then from（1.17），we get

Ifwe choose the argum entofλsuch that

then from（1.18），we get

which is equivalen t to

If in（1.19）wem ake|λ|→1，we get

which is exactly inequality（1.9）.

Rem ark 1.4.Inequality（1.20）sharpens inequality（1.8）.Also itgeneralise inequality（1.4）and to obtain（1.4）from（1.20）sim p ly d ivide both sidesof（1.20）by|α|and let|α|→∞.

Finally，we prove the follow ing resu lt from which a variety of interesting Results follows as special cases.

Theorem 1.3.IfP（z）=anzn+∑nν=µan-νzn-ν，1≤µ≤n，isapolynomialofdegree n having all itszeros in|z|≤k，k≤1，then for every complex numbersα，βwith|α|≥k and|β|≤1，wehave

Theresult isbestpossibleand equality holds in（1.21）for P（z）=γzn，γ∈C.

Rem ark 1.5.Forµ=k=1，Theorem 1.3 reduces to a resu lt of Lim an，M ohapatra and Shah［11，Lemma 3］.Ifwe d ivideboth sidesof inequality（1.21）by|α|and let|α|→∞，we get

Forµ=k=1，inequality（1.22）reduces to a resu lt of Jain［10，Lemma 3］and forµ=1，inequality（1.22）reduces to a resu ltof Soleim an etal.［19，Lemma 3］.

2 Lemmas

For the p roofof these theorem swe shallm ake use of the follow ing Lemmas.

The above Lemma is due to Aziz and Shah［4］.

Proof.If Q（z）=znP（1/z），then P（z）=znQ（1/z）and one can easily verify that for|z|=1，

which implies

By com bining（2.1）and（2.3），weobtain

Now for every com p lex num berαwith|α|≥k（≥kµ），

which implies that for|z|=1，

Inequality（2.5）w hen com bined with Lemma 2.1 gives

Inequality（2.6）in con junction with（2.4）gives

which proves Lemma 2.2 com p letely.

3 Proof of theorem s

Proof of Theorem 1.1.Since P（z）hasall its zeros in|z|≤k，k≥1，it follows that the polynom ial G（z）=P（kz）has all its zeros in|z|≤1.Hence the polynom ial H（z）=znG（1/z）has all itszeros in|z|≥1 and|G（z）|=|H（z）|for|z|=1.Also it iseasy to verify that for|z|=1，

and

Again since G（z）has all its zeros in|z|≤1，we have by Lemma 2.1（for k=µ=1），

Using（3.1）in（3.3），we get

Now for every com p lex num berαwith|α|≥k，we have

which gives by（3.2）and（3.3）for|z|=1，that

or

Also，by the Guass-Lucas theorem，all the zeros of G′（z）lie in|z|≤1.This implies that the polynom ial

doesnot vanish in|z|＜1.Therefore，it follows from（3.4）that the function

isanalytic for|z|≤1 and|W（z）|≤1 for|z|≤1.Fu rtherm ore，W（0）=0 and so the function 1+W（z）is subord inate to the function 1+z for|z|≤1.Hence by awell-know n p roperty of sub ord ination［9］，we have for each r＞0，

Now

which giveswith the help of（3.1）that for|z|=1，

From（3.5），（3.6）and（3.7），we deduce for each r＞0，

If F（z）is a polynom ial of degree n which does not vanish in|z|＜1，then accord ing to a resu ltofRahm an and Schem eisser［16］，we have for every R≥1 and r＞0，

where

Since H（z）isa polynom ialof degree n and H（z）/=0 in|z|＜1，weapp ly（3.9）with R=k≥1 to H（z）and obtain

Also，since H（z）=znG（1/z）=znP（k/z），therefore，for 0≤θ＜2π，we have

Hence，from（3.8），（3.10）and（3.11），it follows for each r＞0，

which giveswith thehelp of Holder's inequality for each r＞0，p＞1，q＞1with p-1+q-1= 1，

Equivalently，

Since

isa polynom ialofdegree n-1，therefore foreach t＞0and R≥1，wehaveby an inequality（see［16］）that

App lying this in（3.12）with R rep laced by k and t by pr，we obtain for each r＞0，

which proves Theorem 1.1.

□

Proof of Theorem 1.2.We assum e with ou t loss of generality that P（z）has all its zeros in|z|＜k，k≥1，for if P（z）has a zero on|z|=k，then m=0 and in view of Theorem 1.1，the theorem holds trivially.Since P（z）has all its zeros in|z|＜k where k≥1，so that m in|z|=k|P（z）|=m＞0 and for everyλ∈C with|λ|＜1，wehave|λm|＜m≤|P（z）|，for|z|=k. By Rouche's theorem the polynom ial P（z）+λm also hasall its zeros in|z|＜k where k≥1. App lying Theorem 1.1 to the polynom ial P（z）+λm and noting that Dα（P（z）+λm）= DαP（z）+λmn，Theorem 1.2 follows.□Proof of Theorem 1.3.If P（z）has a zero on|z|=k，then the theorem is trivial.So，we assum e that P（z）has all its zeros in|z|＜k，thereforeand hence for every com p lex num berγwith|γ|＜1，we haveIt follows by Rouche's theorem that the polynom ial P（z）-γmzn/knof degree n has all its zeros in |z|＜k，k≤1.On app lying Lemma 2.2 to P（z）-γmzn/kn，we have for every com p lex num berαwith|α|≥k，

Equivalently，

Since by Laguerre's theorem（see［14，pp.52］），the polynom ial

hasallzeros in|z|＜k for every com p lex num berαwith|α|≥k，therefore，for any com p lex βwith|β|＜1，the polynom ial

Since k≤1，we have T（z）/=0 for|z|≥1 also.

Now choosing theargum entofγin（3.14）suitably and letting|γ|→1，weget for|z|=1 and|β|＜1，

or

Forβ，with|β|=1，above inequality holds by continuity.□

Acknowledgments

The w ork of the first au thor is supported by UGC underm ajor research p roject schem e vide No.MRP-MAJOR-MATH-2013-29143.

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∗Correspond ing author.Emailaddresses：mabdul lah mir@yahoo.co.in（A.M ir），qdawood@gmai l.com（Q.M. Daw ood），darbi lal85@ymai l.com（B.Dar）