Probl. Anal. Issues Anal. Vol. 13 (31), No3, 2024, pp. 101-117
DOI: 10.15393/j3.art.2024.16050
101
UDC 517.53
N. A. Rather, N. Wani, A. Bhat
INTEGRAL MEAN ESTIMATE FOR POLYNOMIALS WITH
RESTRICTED ZEROS
Abstract. In this paper, we present certain sharp Lp-inequalities for polynomials with restricted zeros. Our results improve and generalize some known integral inequalities for polynomials in the complex domain.
Key words: polynomials, inequalities, complex domain
2020 Mathematical Subject Classification: 26D10, 41A17, 30C15.
1. Introduction and statements of the main results. Let Pn
denote the set of all complex polynomials P(z) = YTj=0 bjof degree n. The subset P°(p) consists of polynomials whose zeros all lie within the disk defined by \z\ ^ p. Specifically, = P°(1) represents polynomials with zeros inside the unit disk. The set P8 includes polynomials whose zeros are located in the region \z\ ^ 1.
For a polynomial P e Pn, the p-norm in the Hardy space is defined as
/ 1 f \ Vp
\\p||P = J \P(ez&)\p M) , 0 < p <8.
0
It is not hard to observe that lim \\PL = max \ P(z)\. For this reason,
the uniform norm max\P(z)\ of P(z) is denoted by \\P\\8. On the other
M=i
/ 27r \ hand, lim \\P\\p = exM 2L ^ ln\P(eid)\dO) (see [15, p. 139], [21]). This P^o+ V 2 0 /
is known as the Mahler measure of P(z) and is denoted by \\P\\0. As an
© Petrozavodsk State University, 2024
application of Jensen's inequality, the Mahler measure of the n-th degree
n
polynomial P(z) = b ^ (z ~ zv) can be explicitly given by
V=i
\P||o = \b\ nmax(1, \zv|).
V = i
If Pe P„, then
and
max\P'(z)\ ^ nmax\P(z)\
H=i M=i
max \P(z)\ ^ Rn max\P(z)\. (3)
!z!-R>i !z!=i
Inequality (2) is an immediate consequence of S. Bernstein's Theorem [8] on the derivative of a trigonometric polynomial. Inequality (3) is a simple deduction from the maximum modulus principle. The equality in (2) and (3) holds for P(z) = azn, a ^ 0.
If we restrict ourselves to the class of polynomials P e P8, then inequalities (2) and (3) can be, respectively, replaced by
max\P'(z)\ ^ -max\P(Z)\ (4)
M"1 2 !^!"1
and
Rn + 1
max \P(z)\ ^ —-— max\P(z)\. (5)
!z!-R>i 2 M=i
Inequality (4) was conjectured by P. Erdos and later verified by P. D. Lax [14]. Ankeny and Rivlin [1] used (4) to prove inequality (5). The equality in (4) and (5) holds for P(z) = azn + b, \a\ = \6\ ^ 0.
As an analogue of Bernstein's inequality in the Hardy space norm, Zygmund [23] proved that if P(z) is a polynomial of degree n, then
\\P'\\P < n\\P\\p, p ^ 1. (6)
De Bruijn and Springer [10] and later Mahler [21] proved that this inequality also holds for p = 0, but for the case 0 < p < 1 its validity remained an open question for quite a long time. Finally, Arestov [3] obtained an inequality concerning the Schur-Szego product of polynomials, which among other things also answered the question.
n
The Schur-Szego composition of a polynomial P(z) = bjzj e Pn
j-0
n n
with another polynomial Q(z) = (ra) jjzj is defined as P*Q = jjbjzj.
3=0 3 j=0
For Q * P, Arestov [2] (see also [3]) proved the following inequality, which also includes the case 0 ^ p < 1 of (6) as a special case:
| | Q * P||p ^ ||Q\\oIIPIIp, for p > 0. (7)
Inequality (6) follows at once from (7) by taking Q(z) = nz(z + 1)ra~1 =
n
= Zj (™) Jzj • For the class of polynomials P e Vf, inequality (6) can be
3=0 3
sharpened. In fact, in this case inequality (6) can be replaced by
IIP'IIp ^ n ,1|P|py , P > 0. (8)
11 1 ^ znp
This inequality is due to N. G. De Bruijn [10] for p ^ 1, whereas Rahman and Schmeisser [18] extended it for 0 ^ p < 1.
As a generalization of (8) and in the spirit of (7), Arestov [4] also
n
proved that if P(z) e V8 and Q(z) = ^ ljz3 e then
3=0
IIP * Q\\P ^ IIPII» P > 0. (9)
1 1 1 ' ^11 p
n
Inequality (8) follows from (9) by choosing Q(z) = nz(z+1)n~1 = (ra) jz^.
3=0 3
For polynomials P e Pf, Boas and Rahman [9] established an analogue of inequality (5) in the Lp-norm for p ^ 1:
I I * + Rn I I P
I I P (Rz) 11 p ^ 1 1 |U ■ HI1 1 p I I P (z) 11 p , R > 1. (10)
1 1 z ~r 1 11 p
Equality in (10) holds for P(z) = azn + b, with |a| = |6| ^ 0. By letting p in (10), one can recover inequality (5).
Rahman and Schmeisser [18] (see also [4]) later showed that inequality (9) also holds for 0 ^ p < 1. The above inequality has been generalized in several ways and a good number of papers are available (see, for example, [7], [6], [19], [20]). By applying inequality (10) to the polynomial znP (1),
we immediately deduce that for P e Pn and 0 < r ^ 1, the following inequality holds for each p > 0:
I rn 7 + 111
I I P M }} p ^ 11 P (z) }} p. (11)
I 1 Z + 1 11 p
This result is sharp and equality in (11) holds for P(z) = azn + b with \a\ = \b\ ^ 0.
By letting p ^ œ in (11), we obtain the following sharp inequality under the conditions of (11):
rn + i
+ 1 1 '12)
ML < IIP(z
In this paper, we present the following result, which is a generalization as well as a refinement of inequality (11). More precisely, we prove
n
Theorem 1. For any polynomial P (z) = bj ze P°, 0 ^ p < œ,
j-0
0 < r < 1, and 0 ^ t < 1, we have
.. _ r.n \\rn + zI \p(rz)\+ tm^—- ^ "y—f IIP(z)II„ , (13)
1 + pr p II^r + ZI
up
where m = min \p(z)\ and
\z\ = l
_ ( r\60\ + rtm + \bn\ \ ^r ^ \bo\ + tm + r\bn\ J
The result is sharp and equality in (13) holds for P(z) = azn + b, \a\ = \b\ ^ 0.
n
Remark 1. Since P(z) = £ b3zj e P0,
j-0
n
Q(z) = znnW) = 2 b:i^ e Pn .
3=0
By Lemma 4, we have \bn\ ^ \&0\+ m, wherem = min\Q(z)\ = min\P (z)\.
\z\ = l \z\ = l
This implies for 0 ^ t ^ 1 \bn\ ^ \b0\ + tm, which gives for 0 < r ^ 1
(1 - r)\bn\ ^ (1 - r)\b0\ + tm(1 - r),
or, equivalently,
|bA + Abo| + rtm > 1 |bo| + Ab,n\ + tm ^ ' That is, Tr ^ 1 for 0 < r ^ 1.
Since | | fir + z 11 p ^ | | 1 +z | | p , p ^ 0, inequality (13) refines inequality (11). For t = 0, inequality (13) reduces to the following refinement of inequality (11).
n
Corollary 1. If P(z) =2] bvzv e P0, then for each r < 1, 0 ^ p <8:
v=0
I I + z 11 p
11P M 11 p ^ lTT—r 11P (z) 11 p , (14)
1 1 °r + ^1 1 p
where
0 = A M +1 U r | b01 + A M.
The result is sharp and equality in (14) holds for P(z) = azn + b with |a| = ^ ^ 0.
By letting p ^ 8 in (13) and noting that tr ^ 1, we obtain the following refinement of inequality (12):
Corollary 2. For any polynomial P(z) = bjZJ e 0 < r < 1, and
3=0
0 ^ t < 1, we have
1 1 p ^ I-«()lP WI L -tm (T+w), (15)
where m = min ^(z)| and Tr is given by (13). The inequality is sharp
M"1
and equality in (15) holds for P(z) = azn + b, |a| = |6| ^ 0.
If p(z) = Y, bjZj e P-, then the polynomial P*(z) = znP(1/z) e
3=0
Applying Theorem 1 to the polynomial P*(z) with r = , we obtain the following refinement of inequality (10):
n
Corollary 3. IfP(z) = 2 bjZj e P-, then for each R > 1, 0 ^p <8,
3=0
and 0 ^ t ^ 1:
Rn5n- 1 lRn + z ||p
^ (RzK+ tm—^— ^ " + } I IP (z)I Ip , (16)
1 + O r p 11 Or + Z11 p p
where m = min \ P(z)\ and
\z\ = l
R\bo\ + \ bn\ + tm R " \bo\ + R\bn\ + Rtm U ).
The bound is sharp and equality in (16) holds for P(z) = azn + b, \a\ = \6\ ^ 0. Since \\6R + z\\p ^ \ \ 1 + z\ \ p, p ^ 0, inequality (16) refines inequality (10). Fort = 0, inequality (16) also refines inequality (10).
A. Aziz [5] proved that if P(z) = YTj=0 e Pn(p) where p ^ 1, then for 1 ^ p <8 and 0 ^ t ^ 1:
\ \ P'(Z) \\ 8 ^ J^-^ \\ P(*) \\P . (17)
Now, we will show that the bound in (17) can be improved by using Corollary 3. More precisely, we prove the following result:
Theorem 2. If P(z) = Xlj=0 bjZj e P°(p) where p ^ 1, then for 0 ^ p <8 and 0 ^ t ^ 1 :
n \\((p) + z\\p . , tmpn((p) — 1
\ \ p'(z)\\ »^ t^T i,:;1 , \p(z)\ +
\ \ pn + Z \ \ p \ \ 1 + Z \ \ p
where
pn 1 + ((p)
:18)
¡I x \6o\ + Pn+1\bg\ + tm .
((p) = ——:-—- and m = min\F (z)\. (19)
( ) Pn\bn\+p\bo\+ptm \z\-p \ ( y '
The result is sharp and equality in (18) holds for P(z) = zn + pn.
Remark 2. Since all the zeros of P(z) are in \z\ ^ p, p^ 1, it can be easily seen that ((p) ^ 1. In view of this, Theorem 2 is a refinement of the inequality (17).
The following result is obtained by letting p ^ 8 in the Theorem 2.
Corollary 4. If P(z) = YTj-0 bjzj e P(0(p), where p ^ 1, then for 0 ^ p <8 and 0 ^ t ^ 1
!№)!!» s {l?^ + , (20)
where ( ( ) is given by (19). The result is the best possible as shown by p (z) = zn + pn.
p
Since <fr(p) > 1, inequality (20) improves the result by N. K. Govil [11], which states that if P(z) e P%(p), p> 1, then
IIP'II >-^l|P|| I I P I I 8 > i + pn\ I P 11 8
2. Lemmas. For the proof of our results, we need the following lemmas. The first lemma is a well-known generalization of the Schwarz lemma by Osserman [16].
Lemma 1. Let F(z) be analytic in \z\ < 1 with F(0) = 0, and \F(z)\ < 1 for \ z\ < 1; then
Lemma 2. Let a, b be complex numbers independent of a, where a is real. Then for each p > 0 :
2n 2-k
V i lll_l , w\„ia\P
\a + beia\P da = J \\a\ + \b\eia[ da.
o o
Using periodicity, it is easy to verify the lemma, so we omit the details. The following Lemma is by Aziz and Rather [6]:
Lemma 3. If A, B, C are non-negative real numbers and B + C ^ A, then for every real number a
\(A - C)eia + (B + C)\ ^ \Aeza + B\.
The next lemma is by Gulzar and Rather [12]:
n
Lemma 4. If P(z) = bjZJ e and m = min \P(z)\, then
j-o \z\"1
\bn\ > \bo\ + m.
The next lemma is a consequence of the result by Arestov [ [3], Theorem 4]. Yet, here we deduce it from inequality (7) due to Arestov [2].
n
Lemma 5. If P(z) = bjZJ e P8, then for every p > 0, r < 1 and
j-0
real ft:
J\P(reie) + eifirnP(eie/r)\P d0 ^ \rneifi + 1\P J\P(eie)\P d9. oo
Proof. For 0 < r ^ 1 and \z\ > 1, \z + r\> \rz + 1|. This gives,
\z + r\n >\rz + 1\n, \z\ > 1,
which implies that the polynomial ei3 (z + r)n + (rz + 1)n has no zeros in \z\ > 1. Hence, all the zeros of ei3(z + r)n + (rz + 1)n lie in \z\ ^ 1 for
0 < r ^ 1. Setting Q(z) = ei3(z + r)n + (rz + 1)n and noting that by (1),
1 I Q||o = \rnei3 + 1\, we obtain by invoking inequality (7),
| | P(rz) + ei3rnP(z/r)IIP ^ \rnei3 + 1\|}P(z)}}p, p^ 0. That proves Lemma 5. □
Definition 1. [17, pp. 36]. Let f and g be analytic in \z\ < 1. We say that the function f is subordinate to g, if there exists a function w, analytic in \ z\ < 1 with w(0) = 0 and \w(z)\ < 1 for \z\ < 1, such that
f(z)=g(w(z)) (\z\ < 1).
Lemma 6. [17, pp. 36]. Let f and g be analytic for \ z\ ^ 1 and such that f is subordinate to g. In addition, if g is univalent in the same disc, then for each p > 0 we have:
j\f(eid)\pd9 ^ J \g(eid)\>d9. oo
3. Proofs of the theorems.
Proof of Theorem 1. By the assumption, all the zeros of polynomial
n
P(z) " X! bvzv lie in \z\ ^ 1; therefore, the conjugate polynomial
u-0
__n _
Q(z) = znP(1/z) = bjZn~j has all its zeros in ^ 1
j-0
and m = min \ P(z)\ = min \ Q(z)\, which implies
\z\ = l \z\ = l
\z\nm ^ \Q(z)\ for \z\ = 1. By the Maximum Modulus principle, we have
\z\nm <\Q(z)\ for \z\ < 1. (21)
So, for any a e C with \a| ^ 1, the polynomial G(z) = Q(z) + amzn does not vanish in \z\ < 1. Indeed, if G(z) = Q(z) + amzn has a zero in \z\ < 1 at z = z0, then
G(zo) = Q(zo) + amz" = 0, \zo\ < 1
This implies
\Q( Z0)\=m\a\\Z0\n < m\Z0\n for \ Z0\ < 1,
contradicting (21). Hence, we conclude that @a e C with \a| ^ 1, the polynomial G(z) = Q(z) + amzn = (b0 + am)zn + ^"¡=1 bjZn~j has all its
zeros in \z\ ^ 1. Let H(z) = znG(1/~z) = P(z) + am, then the function
z H(z) . „ . . ,T . . b0 + am
r (z) = satisfies the assumption of Lemma 1 with r '(0) =
G( )
and, therefore,
\F(z)\ ^ \ 4
W +
bn + am
n
1
bn + am
n
| |
This gives
iu-/M/\bn\\z\ + \b0 + am\ , ,
\H(z)\ ^ ' , .. —_ ,, \G(z)\ for \z\ < 1.
| n| + | 0 + am | |
n
(22)
Setting z = re% where 0 ^ 9 ^ 2n and r < 1 in (22), we get
e* ^ rlbn + \b, + am\ eid
\ bn \ +r \ b0 + am\
(23)
The function f(x) = ^ \ ^ \ + X is non-decreasing for x ^ 0. Using the fact
that for any a e C
\ bn \ + rx
\b0 + am\ ^ \b0\ + \a\m,
we get from inequality (23) that for every a e C with \a\ ^ 1, r < 1 and \ * \" 1:
\ H(re«)\ < M +j'°\\+ \G(re«)\.
\ bn \ + r \ b0 \ + r \ a\m
Equivalently,
Pr\P(reid) + am\ ^ \rnP(ete/r) + amrn\. (24)
where pr = \ ^—\. 0\ + \. \.—. Choosing the argument of a in the left-r\bn\ + \ bo \ + \a\m
hand side of (24) such that
\P (r eid) + am\ = \P (r eld )\ + \a\m,
we get
Pr{\P(reid)\ + \a\m} ^ \rnP(eid/r)\ + \a\rnm.
This gives
Pr\P(rel9)\ + \a\m(nr - rn) ^ \rnP(e/r)\,
equivalently,
pA\P(reie)\ + \a\m^r—-} ^ \rnP(el9/r)\- \a\mPr—-. (25) ( 1 + pr ) 1 + pr
Since, by Remark 1, pr > 1, therefore, we have
n n
\P(re10)\ + \a\mpr—- ^ \rnP(et&/r)\- \a\mpr—-. (26) 1 + pr 1 + pr
Taking A = \rnP(e%d/r)\, B = \P(re%d)\, and C = \a\m—-in Lemma 3
1 + pr
and noting by (26) that B + C ^ A - C ^ A, we get for every real ft:
(\rnP(eie/r)\- + (\P(rel°)\ + )
1 + pr ) V 1 + pr
^ \\rnP(e/r)\e113 + \P(rei&
This yields, for each > 0,
+ e13N(0)\pd9 ^ f \\rnP(e*/r)\e13 + \P(re10)\\Pd9, (27)
o
where
II _ iy*'° II _ iV>'</
M(0) = \P(reie)\ + \a\m——-, N(0) = \rnP(eid/r)\- \a\m-—-.
1 + -r 1 + -r
Integrating both sides of (27) with respect to 3 from 0 to 2ir and using Lemma 2, we get
2k 2K
\ M(0) + eil3N(0)\pd0d3 ^
00
2K 2K
r.np(J0 lr\\J3 , \p(rJ8\I \pr
^ | J \ \ rnP(eM/r)\eil3 + \P(re )\\Pd0dp = 0
2k
| f \ \ rnP(eM/r)\ei3 + \P(reiB)\\Pdp^d0 =
00
2k 2K
\ P (r e *) + rn e i33P (eie/r)\Pd0dp. 00
Combining this with Lemma 5, we have
2K 2K 2K 2K
\M(0) + ei33N(0)\Pd0dp ^ J\rnei33 + 1\Pdp J \P(eid\vd0. (28) 0 0 0 0
Now, for every real 3 and r0 ^ r1 ^ 1, we have
\ r0 + ei3 \ ^ \ r1 + ei3 \ , which implies, for each > 0,
2K 2K
r0 + ,3\>da >J\n + e«rda.
00
If \ M(0)\ * 0, we take ro = \ N(0)\/\M(0)\ and n = -r; then, by (25), t0 ^ r1 ^ 1, and we get, by using Lemma 2:
ift
ift
J | M(9) + el"rnN(9)\J dp = \M(9)\ 0 0
1 + e ^
N( )
ift
= \ m (6) r
e ^ +
N( )
ift
M( )
dp = \m (d) f
M( )
dp = n(d) p
M( )
dp ^
ift
^ \ M(9) \PJ | e%p + Pr \1Jdp. 0
For \M(9)\ = 0, this inequality is trivially true. Using this inequality in (28), we obtain for each p > 0, r < 1, and real P:
ift
e 13 + pr Y dp
1ft r*
! \p (
j 0
ift
0
\ P (r eie) \ + \ a\m
^ | \rnel3 + 1\P dp j \P(e19)\J d0 = 0
pr — r'~
1 + pr
de <
ift
\p
ift
ift
= J \rn + e113\P dp J \P(e19)\Pdd. 00
This gives
\ P (re19) \ + \ a\m
Pr — r
1 + pr
^ -n-1
p \\ Pr + z\\
P( )
(29)
which proves the desired result for p > 0. To prove the result for p = 0, we simply let p ^ 0+ in (29). □
Proof of Theorem 2. By the assumption, all the zeros of P(z) lie in \z\ ^ p, where p ^ 1. Therefore, all the zeros of T(z) = P(pz) are in \z\ ^ 1 and, consequently, the zeros of polynomial R(z) = znT(1/z) are outside \z\ < 1. If z1, z2, ..., zn are the zeros of R(z), then \Zj\ ^ 1, j = 1, 2,... ,n, and
zR' (z)
R(z) j" Z — Zj
2
p
rn + z
p
p
p
This gives, for the points eie, 0 ^ 9 < 2n with R(eie) ^ 0:
Re e-RM^ = £ Re
Ad
R{e * )
£
1 n
3 = 1
eid - z^ H 2 2 "
3 = 1
This implies
ezdR' Peld )
nR{ eid )
ezdR' Peld )
nR{ eid )
R(e10) ^ 0.
Equivalently,
| R' pe10) | ^ | nR{e10)- e10R' {ety)|
(30)
for the points e%d, 0 ^ 9 < 2n, which are not zeros of R(z). This inequality is also true, even if e%e is a zero of R(z). It follows that
| R'Pz) | ^ | nR{z) - zR'Pz) | for |z | = 1.
(31)
Since all the zeros of TPz) are in \z\ ^ 1, by the Gauss-Lucas theorem the zeros of T'Pz) also lie in \z\ ^ 1. This implies that the polynomial
zn-iT' ^ =nRPz)-zR' Pz)
(32)
does not vanish in \z\ < 1. Therefore, in view of (30), we conclude that the function
fPz) =
R P )
n RP ) - R P )
is analytic for \z\ ^ 1 and \fPz)\ ^ 1 for \z\ = 1. Moreover, fP0) = 0. Therefore, it follows that the function 1 + fPz) is subordinate to the univalent function 1 + z for \z\ ^ 1. Hence, by Lemma 6, we obtain
1-K
|1 + fPeie)|pd9 ^J |1 + ewfd9, p > 0. 0 0
Now,
1 + fPz) =
n RP )
n RP ) - R P )
1
This gives, for \z\ = 1, with the help of (22), for each p > 0
np\R(etd)\p = \1 + f(eie)\p\nR(ez&) — et&R'(ez&)\p = = \1 + f(e * )\p\e <n+iqeT{^\p =
= \1 + f(etd )\P\T' (eld )\p. (34)
inequality (33) in conjunction with (34) gives, for each p > 0:
1-K ift p
np [\R(eie)\pdd ^ J\1 + e*\pdd ^max \T(35) 0 0
Equivalently, for each p > 0:
n\\R(z)\\p ^ \\1 + z\\p max \T(z)\. (36)
\z\ = l
As the polynomial R(z) does not vanish in \z\ < 1, we can apply Corollary 3 to R(z) with R = p and obtain
\\\ R(p z)\ + t m* pl~ 11 ^ + \\ R(z)\\p, (37)
1 + 9(p) W(p) + 4j}
where
\ 6o\+ pn+l \ bn\+ tm* *
Q(p) = —-;—:-r—- and m = mm\R(z)\.
pn\bn\+p\bo\+tpm* \*\=i \ ( )\
Again, since R(z) = znT(1/z) = znP(p/z), we see that for 0 ^ 9 < 2n \R(peid)\ = pn\P(eid)\ and m* = min\R(z)\ = min \T(z)\ = min\P(z)\.
\z\"1 \z\"1 \z\=p
Combining this with (36) and (37), we get:
nilpn\P(z)\ + tmp]m 1 \\p ^ \\pn + ZW*max \T'(Z)\. (38)
1 + 9(p) W(p) + A\p M=1
Applying inequality (2) to the polynomial T'(z) = pP'(pz) of degree at most n — 1, where p ^ 1, we have
max \T'(z)\ = pmax\P'(pz)\ = pmax\P'(z)\ ^ pn max\P'(z)\. (39)
\z\"1 \z\"1 \z\=p \z\"1
By using inequality (39) in (38), we finally obtain
-IN < —11 + Z
pn y 1 + 111 < _
n
trni pnM - 1 \
^p z)| + - V 1 + M )
\ \ np) + z \ \ p
□
Acknowledgment. The authors are extremely grateful to the anonymous referee for valuable suggestions regarding the paper, which helped us to improve the quality of the manuscript.
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Received April 25, 2024. In revised form, September 17, 2024. Accepted September 19, 2024. Published online October 22, 2024.
Department of Mathematics, University of Kashmir, Srinagar-190006, India N. A. Rather
E-mail: [email protected] Naseer Wani
E-mail: [email protected] Aijaz Bhat
E-mail: [email protected]