DOI: 10.15415/mjis.2014.31001
Some Inequalities for Polynomials Vanishing Inside a Circle ABDULLAH MIR* AND BILAL AHMAD DAR† Department of Mathematics, University of Kashmir, Srinagar-190006, India Email: mabdullah_mir@yahoo.co.in; †Email: darbilal67@gmail.com
*
Received: March 22, 2013| Revised: February 13, 2014| Accepted: February 13, 2014 Published online: September 20, 2014 The Author(s) 2014. This article is published with open access at www.chitkara.edu.in/publications
Abstract: If P ( z ) = Σnj =0 a j z j is a polynomial of degree n having all its zeros in | z |≤ k, k ≥1 , then it is known that n Max|z|=1 | P ( z ) | Max|z|=1 | P ′(z ) |≥ 1+ kn In this paper we obtain a generalization as well as an improvement of the above inequality. Besides this some other results are also obtained. Keywords and Phrases: Inequalities in the complex domain, Polynomials, Extremal problems, Zeros. Mathematics Subject Classification (2010): 30A10, 30C10, 30C15. 1. Introduction and Statement of Results If P(z) is a polynomial of degree n, then concerning the estimate of | P ′( z ) | , it was shown by Turan [8] that if P(z) has all its zeros in |z| ≤ 1, then
n Max|z|=1 | P ′(z ) |≥ Max|z|=1 | P ( z ) | 2
(1)
In (1) equality holds if all the zeros of P(z) lie in |z| = 1. More generally if the polynomial P(z) has all its zeros in |z| ≤ k ≤ 1, it was proved by Malik [7] that the inequality (1) can be replaced by
Max|z|=1 | P ′(z ) |≥
n Max|z|=1 | P ( z ) | 1+ k
(2)
The case when P(z) has all its zeros in | z |≤ k, k ≥1 , was setteled by Govil [5], who proved
Mathematical Journal of Interdisciplinary Sciences Vol. 3, No. 1, September 2014 pp. 1–14
Mir, A Dar, BA
Theorem A. If P ( z ) = Σnj =0 a j z j is a polynomial of degree n havihg all its zeros in |z| ≤ k,k ≥ 1, then Max|z|=1 | P ′(z ) |≥
n Max|z|=1 | P ( z ) | 1+ kn
(3)
The result is best possible and equality holds for the polynomial P(z) = zn + kn. Aziz [1] improved upon the bound in (3) by taking into account the location of all the zeros of the polynomial P(z) instead of concerning only the zero of largest modulus. More precisely, he proved the following result. Theorem B. If all the zeros of the polynomial P ( z ) = Π nj =1 ( z − z j ) of degree n lie in |z| ≤ k, k ≥ 1, then Max|z|=1 | P ′(z ) |≥
2 n k n ∑ 1 + k j =1 k + | z j
Max|z|=1 | P ( z ) | |
(4)
The result is best possible and equality holds in (4) for P(z) = zn + kn. Govil [3] also obtained the following improvement of Theorem B. Theorem C. Let P ( z ) = ∑ nj =0 a j z j = an ∏ j =1 ( z − z j ), an ≠ 0 , be a polynomial n
of degree n ≥ 2,| z j |≤ k j ,1 ≤ j ≤ n , and let k = max{k1 , k2 ,…, kn } ≥ 1 . Then Max|z|=1 | p ′( z ) |≥
2 n k max|z|=1 | P( z ) | ∑ 1 + k n j =1 k + k j
+
2 | an−1 | n 1 k n − 1 k n−2 − 1 − ∑ n − 2 1 + k n j =1 k + k j n
(5)
1 + | a1 | 1 − 2 k if n > 2 and Max|z|=1 | p ′( z ) |≥
2 n k max|z|=1 | P( z ) | ∑ 1 + k n j =1 k + k j
+
n (k − 1)n 1 a | | 1 ∑ n 1+ k j =1 k + k j
1 +1 − | a1 | if n = 2. k In (5) and (6) equality holds for P(z) = zn + kn.
2
(6)
In this paper we prove the following result which gives an improvement of Theorem C, and thus as well of inequality (4). Theorem D. Let P ( z ) = ∑ nj =0 a j z j = an ∏ n ( z − z j ), an ≠ 0 be a polynomial of j =1 degree n ≥ 2 and P (0) ≠ 0,| z j |≤ k j , 1 ≤ j ≤ n , and let k = max{k1 , k2 ,…, kn } ≥ 1. Then 2 n k Max|z|=1 | P ′( z ) |≥ Max|z|=1 | P ( z ) | ∑ 1 + k n j =1 k + k j k n − 1 n k + n min|z|=k | P ( z ) | n ∑ k (1 + k ) j =1 k + k j +
k k n − 1 k n−2 − 1 k | P ′(0) | +k n−1 | Q ′(0) | n ∑ k + k n − n − 2 k n (1 + k n ) j =1 j
1 +1 − 2 | P ′(0) | if n > 2. k
(7)
and Max|z|=1 | P ′( z ) |≥
2 n k Max|z|=1 | P ( z ) | ∑ 1 + k n j =1 k + k j
+
k n − 1 n k min|z|=k | P ( z ) | n ∑ k (1 + k ) j =1 k + k j
+
k (k − 1)n k | P ′(0) | +k n−1 | Q ′(0) | n ∑ k + k n k n (1 + k n ) j =1 j
n
1 +1 − | P ′(0) | if n = 2. k
(8)
1 where Q( z ) = z n P . z The result is best possible and equality holds in (7) and (8) for P(z) n = z + kn. 1 k ≥ for 1 ≤ j ≤ n, then above theorem gives in particular. Since k + kj 2 Corollary 1. If P ( z ) = an Π nj =1 ( z − z j ), an ≠ 0 , is a polynomial of degree n ≥ 2 having all its zeros in |z| ≤ k, k ≥ 1, P(0) ≠ 0, then
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Some Inequalities for Polynomials Vanishing Inside a Circle
Mir, A Dar, BA
n(k n − 1) n Max P z min|z|=k | P( z ) | | ( ) | + | z | = 1 1+ kn 2 k n (1 + k n ) n(k | P ′(0) | +k n−1 | Q ′(0) |) k n − 1 k n−2 − 1 + − n n − 2 2 k n (1 + k n )
Max|z|=1 | P ′( z ) |≥
1 +1 − 2 | P ′(0) | if n > 2, k
(9)
and n(k n − 1) n Max P z min|z|=k | P ( z ) | | ( ) | + | z | = 1 1+ kn 2 k n (1 + k n ) (k − 1)n (k | P ′(0) | +k n−1 | Q ′(0) |) (10) + 2 k n (1 + k n ) 1 +1 − | P ′(0) | if n = 2, k
Max|z|=1 | P ′( z ) |≥
1 where Q( z ) = z n P . z Equality holds in (9) and (10) for P(z) = zn + kn. k n − 1 k n−2 − 1 > 0 , It is easy to verify that if k > 1 and n > 2, then − n n − 2 hence for polynomials of degree > 1, (9) and (10) together provide a refinement of Theorem A. 1 Now, on applying above theorem to Q( z ) = z n P , we obtain the z following result which gives a generalization as well as an improvement upon some well known inequalities. Corollary 2. If P ( z ) = an Π nj =1 ( z − z j ), an ≠ 0 , is a polynomial of degree n ≥ 2 and |zj| ≥ kj and let k = min{k1, k2, ...,kn} ≤ 1, then if | P ′( z ) | and | Q ′( z ) | attain the maximum at the same point on |z| = 1, we have
Max| z |=1 | P ′( z ) |≤
n k −k j Max | P ( z ) | n − k n ∑ | z |=1 1 + k n j =1 k j + k
1
1 − k n n k j min |z |= k | P ( z ) | 1 + k n ∑ j =1 k + kj
−
4
−
( k | P ′(0) | +k n−1 | Q ′(0) |)
n k j 1 − k n k 2 (1 − k n−2 ) − ∑ n − 2 (11) j =1 k + kj n
1+ kn
−(1 − k 2 ) | Q ′(0) | if n > 2, and Max|z|=1 | P ′( z ) |≤
n k −k 1 j n n k − Max|z|=1 | P ( z ) | ∑ n 1 + k j =1 k j + k
1 − k n n k j min|z|=k | P ( z ) | − n ∑ 1 + k j =1 k + kj −
(12)
(k | P ′(0) | +k n−1 | Q ′(0) |) n k j (1 − k )n ∑ j =1 k + kj n 1+ kn
−(1 − k ) | Q ′(0) | if n = 2, 1 n where Q( z ) = z P . z The result is best possible for k j = k,1 ≤ j ≤ n and equality holds in (11) and (12) for P(z) = zn + kn. Proof of Corollary 2. Since the zeros of P(z) satisfy | z j |≥ k j ,1 ≤ j ≤ n , such 1 that k = min{k1 , k2 ,…, kn } . It follows that the zeros of Q( z ) = z n P satisfy z 1 1 1 1 1 1 ≤ ,1 ≤ j ≤ n , such that = max , ,…, or equivalently k = k1 k2 k kn zj kj min{k1, k2,...,kn}. Applying inequality (7) to the polynomial Q(z) and for the case n > 2, we have 1 n 2 k Max | Q( z ) | Max|z|=1 | Q ′( z ) |≥ | z |=1 n ∑ 1 j =1 1 + 1 1 + k k j k 1 1 − 1 n k n ∑ k min 1 | Q( z ) | + | z |= 1 1 k 1 + 1 1 j =1 + n n k k j k k
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Some Inequalities for Polynomials Vanishing Inside a Circle
Mir, A Dar, BA
1 1 1 | Q ′(0) | + n n−1 k k | P ′( 0 ) | ∑ k + j =1 1 1 1 + 1 1 + n n k k j k k n−2 1 n − 1 1 − 1 k k 1 | Q ′(0) | + 1 − × − n n − 2 1 2 k
=
k 2 k n n Max|z|=1 | P ( z ) | ∑ 1 + k n j =1 k + k j
+
k n (k | P ′(0) | +k n−1 | Q ′(0) | n k j ∑ j =1 k + k j (1 + k n )
1 − k n 1 − k n−2 + (1 − k 2 ) | Q ′(0) | . × − n n−2 − nk n k ( 2 )
(13)
Let Max|z|=1 | P ′( z ) =| P ′(eiα ) | , where 0 ≤ α ≤ 2π. Since | P ′( z ) | and | Q ′( z ) | attains their maximum at the same point on |z| = 1, it follows that Max|z|=1 | Q ′( z ) =| Q ′(eiα ) | . We have by Lemma 1(stated in section 2) that | P ′(eiα ) | + | Q ′(eiα ) |≤ nMax|z|=1 | P ( z ) |
(14)
Using (13) in (14), we get nMax|z|=1 | P ( z ) |≥| P ′(eiα ) | +
2 k n n k j ∑ Max|z|=1 | P ( z ) | 1 + k n j =1 k + k j
1 − k n n k j min|z|=k | P ( z ) | + 1 + K n ∑ j =1 k + k j +
(k | P ′(0) | +k n−1 | Q ′(0) |) n k j 1 − k n k 2 (1 − k n−2 ) − ∑ n n − 2 (1 + k n ) j =1 k + k j
+(1 − k 2 ) | Q ′(0) | implying
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Some Inequalities for Polynomials Vanishing Inside a Circle
2 k n n k j | P ′(eiα ) |≤ n − Max|z|=1 | P ( z ) | ∑ 1 + k n j =1 k + k j 1 − k n n k j − min|z|=k | P ( z ) | n ∑ 1 + K j =1 k + k j
−
(k | P ′(0) | +k n−1 | Q ′(0) |) n k j 1 − k n k 2 (1 − k n−2 ) − ∑ n n − 2 (1 + k n ) j =1 k + k j
−(1 − k 2 ) | Q ′(0) | which proves the desired inequality for n > 2. The result for n = 2 follows on the same lines as that for n > 2 but instead of using (7), we use (8). As it is easy to see, our Corollary 2 provides a generalization as well as an improvement of the following result due to A. Aziz and N.Ahmad [2]. Theorem D. Let P ( z ) = Π nj =1 ( z − z j ) be a polynomial of degree n which does 1 not vanish in |z| < k, where k ≤ 1, and let Q( z ) = z n P . If | P ′( z ) | and z | Q ′( z ) | become maximum at the same point on |z| = 1, then n | z | −k 1 j n Max|z|=1 | P ′( z ) |≤ n k (15) − Max|z|=1 | P ( z ) | . ∑ n 1 + k j =1 | z j | +k The result is best possible and equality holds for the polynomial P(z) = zn + kn. 2. Lemmas For the proof of the theorem, we need the following lemmas. Lemma 1. If P(z) is a polynomial of degree n, then on |z| = 1
| P ′( z ) | + | Q ′( z ) |≤ nMax|z|=1 | P ( z ) |
(16)
1 where Q( z ) = z n P . z This is a special case of a result due to Govil and Rahman [6, Lemma10]. Lemma 2. If P(z) is a polynomial of degree n, then for R > 1,
Max|z|= R | P ( z ) |≤ R n Max|z|=1 | P ( z ) | −( R n − R n−2 ) | P(0) | for n > 1, (17)
and
Max|z|= R | P ( z ) |≤ RMax|z|=1 | P ( z ) | −( R − 1) | P (0) | for n = 1.
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(18)
Mir, A Dar, BA
The above result is due to Frappier, Rahman and Ruscheweyh [3, Theorem2]. Lemma 3. If P(z) is a polynomial of degree n ≥ 2 and P(0) ≠ 0 , then for |z| = 1 and R ≥ 1, | P ( Rz ) − P ( z ) | + | Q( Rz ) − Q( z ) |
≤ ( R n − 1) Max|z|=1 | P ( z ) | R n − 1 R n−2 − 1 for n > 2, −(| P ′(0) | + | Q ′(0) |) − n − 2 n
(19)
and | P ( Rz ) − P ( z ) | + | Q( Rz ) − Q( z ) | ≤ ( R n − 1) Max|z|=1 | P ( z ) |
−(| P ′(0) | + | q ′(0) |)
(20)
n
( R − 1) for n = 2, n
1 n where Q( z ) = z P . z Proof of Lemma 3. If P(z) is a polynomial of degree n and p(0) ≠ 0 , then 1 Q( z ) = z n P is also a polynomial of degree n. Applying inequality (17) of z Lemma 2 to the polynomials P ′( z ) and Q ′( z ) , which are of degree (n — 1), we obtain for all t ≥1and 0 ≤ θ < 2π ,
| P ′(teiθ ) |≤ t n−1 | P ′(eiθ ) | −(t n−1 − t n−3 ) | P ′(0) | for n > 2
(21)
| Q ′(teiθ ) |≤ t n−1 | Q ′(eiθ ) | −(t n−1 − t n−3 ) | Q ′(0) | for n > 2.
(22)
and
Adding inequalities (21) and (22), we get for n > 2,
| P ′(teiθ ) | + | Q ′(teiθ ) |≤ t n−1 (| P ′(eiθ )+ | Q ′(eiθ )) −(t n−1 − t n−3 )(| P ′(0) | + | Q ′(0) |).
Using Lemma 1 in above inequality, we get
| P ′(teiθ ) | + | Q ′(teiθ ) |≤ nt n−1 Max|z|=1 | P ( z ) | −(t n−1 − t n−3 )(| P ′(0)) | + | Q ′(0) |)
for 0 ≤ θ < 2π and t ≥1.
8
(23)
Some Inequalities for Polynomials Vanishing Inside a Circle
Also for each θ,0 ≤ θ < 2π and R > 1, we have R
P (Re iθ ) − P (eiθ ) = ∫ eiθ P ′(teiθ )dt
1
and R
Q(Re iθ ) − Q(eiθ ) = ∫ eiθQ ′(teiθ )dt
1
Hence for n > 2
R
| P (Re iθ ) − P (eiθ ) | + | Q(Reiθ ) − Q(eiθ ) |≤ ∫ (| P ′(teiθ ) | + | Q ′(teiθ ) |)dt. 1
On combining inequality (23) with the above inequality, we get for n > 2 | P (Re iθ ) − P (eiθ ) | + | Q(Reiθ ) − Q(eiθ ) |
R
≤ ∫ [ nt n−1 Max|z|=1 | P ( z ) | −(t n−1 − t n−3 )(| P ′(0) | + | Q ′(0) |] dt 1
R n − 1 R n−2 − 1 − = ( R n − 1) Max|z|=1 | P ( z ) | −(| P ′(0) | + | Q ′(0) |) n n − 2 for 0 ≤ θ < 2π and R > 1, which is equivalent to the desired result. The proof of (20) follows on the same lines as the proof of (19), but instead of using (17), it uses (18). Lemma 4. If P(z) is a polynomial of degree n ≥ 2 having all its zeros in | z |≤ k, k ≥1 , and P(0) ≠ 0 , then 2k n Max|z|=1 | P ( z ) | 1+ kn (k | P ′(0) | +k n−1 | q ′(0) |) + (1 + k n ) k n − 1 k n−2 − 1 for n > 2, × − n − 2 n
Max|z|=k | P ( z ) |≥
(24)
and
2k n Max|z|=1 | P ( z ) | 1+ kn n n−1 (k | P ′(0) | +k | q ′(0) |) (k − 1) + for n = 2, (1 + k n ) n
Max|z|=k | P ( z ) |≥
9
(25)
Mir, A Dar, BA
1 where Q( z ) = z n P . z
Proof of Lemma 4. Since P(z) has all its zeros in | z |≤ k, k ≥1 , we write n
(
p( z ) = an ∏ z − rj e
j =1
iθ j
),
where an≠ 0 and rj ≤ k, j = 1, 2,…, n . iθ Then, clearly for points e ,0 ≤ θ < 2π , other than zeros of P(z), we have
(
iθ j
n k 2 eiθ − rj e P (k 2 eiθ ) =∏ iθ P (eiθ ) j =1 eiθ − rj e j
(
)
)
k 4 + rj2 − 2 k 2 rj cos(θ − θ j ) = ∏ 2 j =1 1 + rj − 2rj cos(θ − θ j ) n
n
≥ ∏k = kn. j =1
This implies, | P (k 2 eiθ ) |≥ k n | P (eiθ ) |
for points eiθ ,0 ≤ θ < 2π , other than the zeros of P(z). Since this inequality is trivial for points eiθ , which are the zeros of P(z), it follows that
| P (k 2 z ) |≥ k n | P ( z ) | for | z |= 1.
(26)
1 k Let G(z) = P(kz) and H ( z ) = z n G = z n P . Applying inequality (19) to z z the polynomial G(z), we get for | z |= 1 and k ≥1 , | G (kz ) | + | H (kz ) |≤ (k n + 1) Max|z|=1 | G ( z ) |
k n − 1 k n−2 − 1 for n > 2. − −(| G ′(0) | + | H ′(0) |) n n − 2
Equivalently, for |z| = 1 2 n n | P (k z ) | +k | P ( z ) |≤ (1 + k ) Max|z|=k | P ( z ) |
k n − 1 k n−2 − 1 for n > 2. −(k | P ′(0) | +k n−1 | Q ′(0) |) − n − 2 n
10
This gives with the help of (26) for |z| = 1
2 k n | P ( z ) |≤ (1 + k n ) Max|z|=k | P ( z ) | −(k | P ′(0) | +k
n−1
k n − 1 k n−2 − 1 , − | Q ′(0) |) n n − 2
Some Inequalities for Polynomials Vanishing Inside a Circle
From which inequality (24) follows for n > 2. The proof of the lemma for n = 2 follows on the same lines as that for n > 2, but using inequality (20) instead of inequality (19). Next we prove the following result which gives an improvement of the above lemma by involving the term min|z|=k | P ( z ) | as follows. Lemma 5. If P(z) is a polynomial of degree n ≥ 2 having all its zeros in |z| < k, k > 1, and P(0) ≠ 0 , then Max|z|=k | P ( z ) |≥
k n − 1 2k n Max P z | ( ) | + | z |=1 k n + 1 Min|z|=k | P ( z ) | 1+ kn
(k | P ′(0) | +k n−1 | Q ′(0) |) (1 + k n ) k n − 1 k n−2 − 1 for n > 2, × − n n − 2 +
and
Max|z|=k | P ( z ) |≥ +
(27)
k n − 1 2k n Min|z|=k | P ( z ) | (28) Max|z|=1 | P ( z ) | + n n 1+ k k + 1 (k | P ′(0) | +k n−1 | Q ′(0) |) (k − 1)n for n = 2, n (1 + k n )
1 n where Q( z ) = z P . z Proof of Lemma 5. Since P(z) has all of its zeros in | z |≤ k, k ≥1 , then by Rouche’s theorem, the polynomial F(z) = P(z) + λm with | λ |≤1 , where m = Min|z|=k | P ( z ) | , also has all its zeros in | z |≤ k, k ≥1 . So, on applying Lemma 4 for n > 2 to the polynomial F(z), we have
2k n Max|z|=1 | F ( z ) | 1+ kn (k | F ′(0) | +k n−1 | T ′(0) |) k n − 1 k n−2 − 1 , + − n − 2 n (1 + k n )
Max|z|=k | F ( z ) |≥
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Mir, A Dar, BA
1 1 n n n Where T ( z ) = z F = z P + λm = Q( z ) + λ mz z z This implies for n > 2 2k n Max|z|=1 | P ( z ) + λm | 1+ kn (k | P ′(0) | +k n−1 | Q ′(0) |) + (1 + k n ) k n − 1 k n−2 − 1 , × − n n − 2
Max|z|=k | P ( z ) | + | λ | m ≥
(29)
If we choose z0 so that Max|z|=1 | P ( z ) |=| P ( z0 ) | and also choose argument of λ suitably, such that
Max|z|=1 | P ( z ) + λm |=| P ( z0 ) | + | λ | m,
the inequality (29) becomes 2k n { P ( z0 ) + λ m } 1+ kn (k P ′ (0) + k n−1 Q ′ (0) ) k n −1 k n−2 −1 + , − n − 2 n (1 + k n )
Max z =k P ( z) + λ m ≥
Which on simplification gives,
2k n (k n − 1) m Max P | ( z ) | + | λ | | z | = 1 1+ kn (k n + 1) (k | P ′(0) | +k n−1 | Q ′(0) |) k n − 1 k n−2 − 1 . + − n − 2 (1 + k n ) n
Max|z|=k | P ( z ) |≥
Finally letting λ→1, we get the desired result for n > 2. The proof of inequality (28) follows on the same lines as the proof of (27), but instead of inequality(24) of Lemma 4, we use the inequality (25) of the same lemma. 3. Proof of the Theorem The polynomial G ( z ) = P (kz ) = an ∏ j =1 (kz − z j ) has all its zeros in |z| ≤ 1 and we have n
12
n G ′( z ) =∑ G( z) j =1
Some Inequalities for Polynomials Vanishing Inside a Circle
1 z j z − k
so that for 0 ≤ θ < 2π n G ′(eiθ ) eiθ G ′(eiθ ) Re Re ≥ = ∑ G (eiθ ) G (eiθ ) j =1
This implies n
Max|z|=1 | G ′( z ) |≥ ∑
j =1
n k eiθ ≥∑ z + | zj | k j =1 j iθ e − k
k Max|z|=1 | G ( z ) |, k+ | z j |
equivalently n
kMax|z|=1 | P ′(kz ) |≥ ∑
j =1
k Max|z|=1 | P (kz ) | . k+ | z j |
(30)
Since P'(z) is a polynomial of degree n — 1, it follows by (17) that n
k (k n−1 Max|z|=1 | P ′( z ) | −(k n−1 − k n−3 ) | P ′(0) |) ≥ ∑ j =1
k Max|z|=k | P ( z ) | . k+ | z j |
Now using inequality (27) of Lemma 5 in above inequality, we get
k n Max | P ′( z ) | −(k n − k n − 2 | P ′(0) | | z |= 1 n k 2 k n | P( z) | ≥ ∑ Max | z |= 1 1 + k n k + | z | j =1 j n k − 1 min + | P( z) | | z |= k n 1 + k +
(k | P ′(0) | +k n − 1 | Q ′(0) |) (1 + k n )
n k − 1 k n − 2 − 1 × − n n − 2 implying
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Mir, A Dar, BA
k 2 Max|z|=1 | P ( z ) | 1 + k n j =1 k + | z j | (k n − 1) min|z|=k | P ( z ) | + n k (1 + k n ) n−1 n n − 2 (k | P ′(0) | +k | Q ′(0) |) k − 1 k − 1 + − n − 2 k n (1 + k n ) n 1 +1 − 2 | P ′(0) | k n
Max|z|=1 | P ′( z ) |≥ ∑
from which the theorem follows for n > 2. The proof of the theorem for n = 2 follows on the same lines as that of n > 2, but instead of using inequalities (17) and(27) of Lemma 2 and Lemma 5 respectively, we use inequalities (18) and (28) of the respective lemmas. References [1] Aziz, A. Inequalities for the derivative of a polynomial, Proc. Amer. Math. Soc., 89 (1983), 259-266. http://dx.doi.org/10.1090/S0002-9939-1983-0712634-5 [2] Aziz, A. and Rather, N. A. Inequalities for the derivative of a polynomial, Proc. Indian Acad. Sci.(Math. Sci.), 107, 2 (1997), 189-196. [3] Frappier, C. Rahman Q. I. and Ruscheweyh, St. New inequalities for polynomials, Trans. Amer. Math. Soc., 288 (1985), 69-99. http://dx.doi.org/10.1090/S0002-9947-1985-0773048-1 [4] Govil, N. K. Inequalities for the derivative of a polynomial, J. Approx. Theory, 63 (1990), 65-71. http://dx.doi.org/10.1016/0021-9045(90)90114-6 [5] Govil, N. K. On the derivative of a polynomial, Proc. Amer. Math. Soc., 41 (1973), 543-546. http://dx.doi.org/10.1090/S0002-9939-1973-0325932-8 [6] Govil N. K. and Rahman, Q. I. Functions of exponential type not vanishing in a half-plane and related polynomials, Trans. Amer. Math. Soc., 137 (1969), 501517. http://dx.doi.org/10.1090/S0002-9947-1969-0236385-6 [7] Malik, M. A. On the derivative of a polynomial, J. London Math. Soc., 2 (1969), 57-60. http://dx.doi.org/10.1112/jlms/s2-1.1.57 [8] Turan, P. Uber die ableitung von polynomen, Composition Math., 7 (1939). 89-95.
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