# Integral mean estimates for polynomials whose zeros are within a circle

## Abstract

Let P(z) be a polynomial of degree n having all its zeros in |z| ≤ K ≤ 1, then for each δ > 0, p > 1, q > 1 with $1 p + 1 q =1$, Aziz and Ahmad (Glas Mat Ser III 31:229-237, 1996) proved that

$n ∫ 0 2 π | P ( e i θ ) | δ d θ 1 δ ≤ ∫ 0 2 π | 1 + K e i θ | q δ d θ 1 q δ ∫ 0 2 π | P ′ ( e i θ ) | p δ d θ 1 p δ .$

In this paper, we extend the above inequality to the class of polynomials $P ( z ) := a n z n + ∑ j = μ n a n - j z n - j$, 1 ≤ μn, having all its zeros in |z|K ≤ 1, and obtain a generalization as well as refinement of the above result.

Mathematics Subject Classification (2000)

30A10, 30C10, 30C15

## 1 Introduction and statement of results

Let P(z) be a polynomial of degree n and P′(z) be its derivative. If P(z) has all its zeros in |z| ≤ 1, then it was shown by Turan  that

$M a x | z | = 1 | P ′ ( z ) | ≥ n 2 M a x | z | = 1 | P ( z ) | .$
(1)

Inequality (1) is best possible with equality for P(z) = αzn + β, where |α| = |β|. As an extension of (1), Malik  proved that if P(z) has all its zeros in |z| ≤ K, where K ≤ 1, then

$M a x | z | = 1 | P ′ ( z ) | ≥ n 1 + K M a x | z | = 1 | P ( z ) | .$
(2)

Malik  also obtained a generalization of (1) in the sense that the right-hand side of (1) is replaced by a factor involving the integral mean of |P(z)| on |z| = 1. In fact, he proved the following:

Theorem A. If P(z) has all its zeros in |z| ≤ 1, then for each δ > 0

$n ∫ 0 2 π | P ( e i θ ) | δ d θ 1 δ ≤ ∫ 0 2 π | 1 + e i θ | δ d θ 1 δ M a x | z | = 1 | P ′ ( z ) | .$
(3)

The result is sharp, and equality in (3) holds for P(z) = (z + 1) n . If we let δ → ∞ in (3), we get (1).

As a generalization of Theorem A, Aziz and Shah  proved the following:

Theorem B. If$P ( z ) := a n z n + ∑ j = μ n a n - j z n - j$, 1 ≤ μn is a polynomial of degree n having all its zeros in the disk |z| ≤ K, K ≤ 1, then for each δ > 0,

$n ∫ 0 2 π | P ( e i θ ) | δ d θ 1 δ ≤ ∫ 0 2 π | 1 + K μ e i θ | δ d θ 1 δ M a x | z | = 1 | P ′ ( z ) | .$
(4)

Aziz and Ahmad  generalized (3) in the sense that Max|z|=1|P′(z)| on |z| = 1 on the right-hand side of (3) is replaced by a factor involving the integral mean of |P′(z)| on |z| = 1 and proved the following:

Theorem C. If P(z) is a polynomial of degree n having all its zeros in |z| ≤ K ≤ 1, then for δ > 0, p > 1, q > 1 with$1 p + 1 q =1$,

$n ∫ 0 2 π | P ( e i θ ) | δ d θ 1 δ ≤ ∫ 0 2 π | 1 + K e i θ | q δ d θ 1 q δ ∫ 0 2 π | P ′ ( e i θ ) | p δ d θ 1 p δ .$
(5)

If we let p → ∞ (so that q → 1) in (5), we get (3).

In this paper, we consider a class of polynomials $P ( z ) := a n z n + ∑ j = μ n a n - j z n - j$, 1 ≤ μn, having all the zeros in |z| ≤ K ≤ 1, and thereby obtain a more general result by proving the following:

Theorem 1. If$P ( z ) := a n z n + ∑ j = μ n a n - j z n - j$, 1 ≤ μn is a polynomial of degree n having all its zeros in the disk |z| ≤ K, K ≤ 1, then for each δ > 0, q > 1, p > 1 with$1 p + 1 q =1$and for every complex number λ with |λ| < 1

$n ∫ 0 2 π | P ( e i θ ) + λ m | δ d θ 1 δ ≤ ∫ 0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | e i θ | q δ d θ 1 q δ ∫ 0 2 π | P ′ ( e i θ ) | p δ d θ 1 p δ ,$
(6)

where m = Min|z|=K|P (z)|.

If we take λ = 0 in Theorem 1, we get the following:

Corollary 1. If$P ( z ) := a n z n + ∑ j = μ n a n - j z n - j$, 1 ≤ μn is a polynomial of degree n having all its zeros in the disk |z| ≤ K, K ≤ 1, then for each δ > 0, q > 1, p > 1 with$1 p + 1 q =1$, ,

$n ∫ 0 2 π | P ( e i θ ) | δ d θ 1 δ ≤ ∫ 0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | e i θ | q δ d θ 1 q δ ∫ 0 2 π | P ′ ( e i θ ) | p δ d θ 1 p δ ,$
(7)

For μ = 1 in Theorem 1, we have the following:

Corollary 2. If$P ( z ) := ∑ j = 0 n a j z j$is a polynomial of degree n having all its zeros in the disk |z| ≤ K, K ≤ 1, then for each δ > 0, q > 1, p > 1 with$1 p + 1 q =1$,

$n ∫ 0 2 π | P ( e i θ ) + λ m | δ d θ 1 δ ≤ ∫ 0 2 π | 1 + n | a n | K 2 + | a n - 1 | n | a n | + | a n - 1 | e i θ | q δ d θ 1 q δ ∫ 0 2 π | P ′ ( e i θ ) | p δ d θ 1 p δ ,$
(8)

where m = Min|z|=K|P (z)|.

Remark 1: Since all the zeros of P(z) lie in |z| ≤ K, therefore, $1 n | a n - 1 a n |≤KK≤1$, it can be easily verified that

$n | a n | K 2 + | a n - 1 | n | a n | + | a n - 1 | ≤ K .$

It shows that for λ = 0, Corollary 2 provides a refinement of the result of Aziz and Ahmad .

The next result immediately follows from Theorem 1, if we let p → ∞ (so that q → 1)

Corollary 3. If$P ( z ) := a n z n + ∑ j = μ n a n - j z n - j$, 1 ≤ μn is a polynomial of degree n having all its zeros in the disk |z| ≤ K, K ≤ 1, then for each δ > 0 and for every complex number λ with |λ| < 1

$n ∫ 0 2 π | P ( e i θ ) + λ m | δ d θ 1 δ ≤ ∫ 0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | e i θ | δ d θ 1 δ M a x | z | = 1 | P ′ ( z ) | .$
(9)

Also if we let δ in the Corollary 3 and note that

$lim δ → ∞ { 1 2 π ∫ 0 2 π | P ( e i θ ) | δ d θ } 1 δ = M a x | z | = 1 | P ( z ) | ,$

we get from (9)

$n M a x | z | = 1 | P ( z ) + λ m | ≤ n | a n | ( K 2 μ + K μ - 1 ) + μ | a n - μ | ( 1 + K μ - 1 ) n | a n | K μ - 1 + μ | a n - μ | M a x | z | = 1 | P ′ ( z ) | f o r | z | = 1 .$
(10)

If z0 be such that Max|z|=1|P (z)| = |P (z0)|, then from (10), we have

$n | P ( z 0 ) + λ m | ≤ n | a n | ( K 2 μ + K μ - 1 ) + μ | a n - μ | ( 1 + K μ - 1 ) n | a n | K μ - 1 + μ | a n - μ | M a x | z | = 1 | P ′ ( z ) | f o r | z | = 1 .$

Choosing an argument of λ such that

$| P ( z 0 ) + λ m | = | P ( z 0 ) | + | λ | m ,$

we get

$n ( | P ( z 0 ) | + | λ | m ) ≤ n | a n | ( K 2 μ + K μ - 1 ) + μ | a n - μ | ( 1 + K μ - 1 ) n | a n | K μ - 1 + μ | a n - μ | M a x | z | = 1 | P ′ ( z ) | .$
(11)

From inequality (11), we conclude the following:

Corollary 4. If$P ( z ) := a n z n + ∑ j = μ n a n - j z n - j$, 1 ≤ μn is a polynomial of degree n having all its zeros in the disk |z| ≤ K, K ≤ 1, then for 0 ≤ t ≤ 1, we have

$M a x | z | = 1 | P ′ ( z ) | ≥ n n | a n | K μ - 1 + μ | a n - μ | n | a n | ( K 2 μ + K μ - 1 ) + μ | a n - μ | ( 1 + K μ - 1 ) { M a z | z | = 1 | P ( z ) | + t M i n | z | = K | P ( z ) | } .$

Further, if we take K = t = μ = 1 in the Corollary 4, we get a result of Aziz and Dawood .

## 2. Lemmas

For the proof of this theorem, we need the following lemmas.

The first lemma is due to Qazi .

Lemma 1. If $P ( z ) := a 0 + ∑ j = μ n a j z j ,1≤μ≤n$ is a polynomial of degree n having no zeros in the disk|z| < K, K ≥ 1, then

$n | a 0 | K μ + 1 + μ | a μ | K 2 μ n | a 0 | + μ | a μ | K μ + 1 | P ′ ( z ) | ≤ | Q ′ ( z ) | f o r | z | = 1 ,$

where $Q ( z ) = z n P ( 1 z ̄ ) ¯ and μ n | a μ a 0 | K μ ≤1$.

Lemma 2. If $P ( z ) := a n z n + ∑ j = μ n a n - j z n - j$ is a polynomial of degree n having all its zeros in the disk|z| ≤ K ≤ 1, then

$| Q ′ ( z ) |≤ n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | | P ′ ( z ) |for|z|=1,1≤μ≤n,$

where $Q ( z ) = z n P ( 1 z ̄ ) ¯$.

### Proof of Lemma 2

Since all the zeros of P(z) lie in |z| ≤ K ≤ 1, therefore all the zeros of $Q ( z ) = z n P ( 1 z ̄ ) ¯$ lie in $|z|≥ 1 K ≥1$. Hence, applying Lemma 1 to the polynomial $Q ( z ) := ā n + ∑ j = μ n ā n - j z j$, we get

$n | a n | 1 K μ + 1 + μ | a n - μ | 1 K 2 μ n | a n | + μ | a n - μ | 1 K μ + 1 | Q ′ ( z ) |≤| P ′ ( z ) |.$

Or, equivalently

$| Q ′ ( z ) |≤ n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | | P ′ ( z ) |,|z|=1.$

This proves Lemma 2.

Remark 1: Lemma 3 of Govil and Mc Tume  is a special case of this lemma when μ = 1.

### Proof of Theorem 1

Let $Q ( z ) = z n P ( 1 z ̄ ) ¯$z, we have $P ( z ) = z n Q ( 1 z ̄ ) ¯$. This gives

$P ′ ( z ) =n z n - 1 Q ( 1 z ̄ ) ¯ - z n - 2 Q ′ ( 1 z ̄ ) ¯ .$
(12)

Equivalently,

$z P ′ ( z ) =n z n Q ( 1 z ̄ ) ¯ - z n - 1 Q ′ ( 1 z ̄ ) ¯ .$
(13)

This implies

$| P ′ ( z ) |=|nQ ( z ) -z Q ′ ( z ) |for|z|=1.$
(14)

Let m = Min|z|=K|P (z)|, so that m ≤ |P (z)| for |z| = K. Therefore, for every complex number λ with |λ| < 1, we have || < |P(z)| on |z| = K. Since P(z) has all its zeros in |z| ≤ K ≤ 1, by Rouche's theorem, it follows that all the zeros of the polynomial G(z) = P(z) + λm lie in |z| ≤ K ≤ 1.

If $H ( z ) = z n G ( 1 z ̄ ) ¯ =Q ( z ) +m λ ̄ z n$, then by applying Lemma 2 to the polynomial G(z) = P(z) + λm, we have for |z| = 1

$| H ′ ( z ) |≤ n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | | G ′ ( z ) |,1≤μ≤n.$

This gives

$| Q ′ ( z ) +nm λ ̄ z n - 1 |≤ n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | | P ′ ( z ) |,1≤μ≤n.$
(15)

Using (14) in (15), we get

$| Q ′ ( z ) +nm λ ̄ z n - 1 |≤ n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | |nQ ( z ) -z Q ′ ( z ) |for|z|=1,1≤μ≤n.$
(16)

Since P(z) has all its zeros in |z| ≤ K ≤ 1, by Gauss{Lucas theorem so does P(z). It follows that nQ(z) - zQ(z), which is simply (see (12))

$z n - 1 P ′ ( 1 z ̄ ) ¯ ,$

has all its zeros in |z| ≥ $1 K$ ≥ 1. Hence,

$W ( z ) = n | a n | K μ - 1 + μ | a n - μ | n | a n | K 2 μ + μ | a n - μ | K μ - 1 . z ( Q ′ ( z ) + n m λ ̄ z n - 1 ) ( n Q ( z ) - z Q ′ ( z ) )$
(17)

is analytic for |z| < 1, |W(z)| ≤ 1 for |z| = 1 and W(0)=0. Thus, the function

$1+ n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | .W ( z )$

is subordinate to the function

$1+ n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | z$

for |z| < 1. Hence, by a property of subordination (for reference see [, p. 36, Theorem 1.6.17] or [, p. 454] or ), we have for each δ > 0 and 0 ≤ θ < 2π,

$∫ 0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | W ( e i θ ) | δ d θ ≤ ∫ 0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | e i θ | δ d θ .$
(18)

Also from (17), we have

$1+ n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | W ( z ) = n ( Q ( z ) + m λ ̄ z n ) n Q ( z ) - z Q ′ ( z ) .$

Therefore,

$n|Q ( z ) +m λ ̄ z n |=|1+ n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | W ( z ) ||nQ ( z ) -z Q ′ ( z ) |,$
(19)

which implies

$n|H ( z ) |=|1+ n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | W ( z ) ||nQ ( z ) -z Q ′ ( z ) |.$
(20)

Using (14) and the fact that |H(z)| = |G(z)| = |P(z) + λm| for |z| = 1, we get from (20)

$n|P ( z ) +λm|=|1+ n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | W ( z ) || P ′ ( z ) |for|z|=1.$

Hence, for each δ > 0 and 0 ≤ θ < 2π, we have

$n δ ∫ 0 2 π | P ( e i θ ) + λ m | δ d θ = ∫ 0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - v | W ( e i θ ) | δ | P ′ ( e i θ ) | δ d θ .$
(21)

This gives with the help of Hölder's inequality for p > 1, q > 1, with $1 p + 1 q =1$and for every δ > 0,

$n δ ∫ 0 2 π | P ( e i θ ) + λ m | δ d θ ≤ ∫ 0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | W ( e i θ ) | q δ d θ 1 q ∫ 0 2 π | P ′ ( e i θ ) | p δ d θ 1 p$
(22)

Combining (18) and (22), we get for δ > 0 and 0 ≤ θ < 2π,

$n δ ∫ 0 2 π | P ( e i θ ) + λ m | δ d θ ≤ ∫ 0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | e i θ | q δ d θ 1 q ∫ 0 2 π | P ′ ( e i θ ) | p δ d θ 1 p$
(23)

This is equivalent to

$n ∫ 0 2 π | P ( e i θ ) + λ m | δ d θ 1 δ ≤ ∫ 0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | e i θ | q δ d θ 1 q δ ∫ 0 2 π | P ′ ( e i θ ) | p δ d θ 1 p δ$
(24)

which proves the desired result.

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## Acknowledgements

The authors are grateful to the referee for useful comments.

## Author information

Authors

### Corresponding author

Correspondence to Gulshan Singh.

### Competing interests

The authors declare that they have no competing interests.

### Authors' contributions

GS studied the related literature under the supervision of WMS and jointly developed the idea and drafted the manuscript. GS made the text _le and communicated the manuscript. GS also revised it as per the directions of the referee under the guidance of WMS. Both the authors read and approved the final manuscript.

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Singh, G., Shah, W. Integral mean estimates for polynomials whose zeros are within a circle. J Inequal Appl 2011, 35 (2011). https://doi.org/10.1186/1029-242X-2011-35 