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 $\frac{1}{p}+\frac{1}{q}=1$, Aziz and Ahmad (Glas Mat Ser III 31:229-237, 1996) proved that

$$n{\left\{\underset{0}{\overset{2\pi }{\int }}|P\left(e^{i\theta }\right){|}^{\delta }d\theta \right\}}^{\frac{1}{\delta }}\le {\left\{\underset{0}{\overset{2\pi }{\int }}|1+K{e}^{i\theta }{|}^{q\delta }d\theta \right\}}^{\frac{1}{q\delta }}{\left\{\underset{0}{\overset{2\pi }{\int }}|P^{\prime }\left(e^{i\theta }\right){|}^{p\delta }d\theta \right\}}^{\frac{1}{p\delta }}.$$

In this paper, we extend the above inequality to the class of polynomials $P\left(z\right):=a_n{z}^n+{\sum }_{j=\mu }^n{a}_{n-j}{z}^{n-j}$, 1 ≤ μ ≤ n, 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 [1] that

$$Ma{x}_{\left|z\right|=1}\left|P^{\prime }\left(z\right)\right|\ge \frac{n}{2}Ma{x}_{\left|z\right|=1}\left|P\left(z\right)\right|.$$

(1)

Inequality (1) is best possible with equality for P(z) = αzⁿ + β, where |α| = |β|. As an extension of (1), Malik [2] proved that if P(z) has all its zeros in |z| ≤ K, where K ≤ 1, then

$$Ma{x}_{\left|z\right|=1}\left|P^{\prime }\left(z\right)\right|\ge \frac{n}{1+K}Ma{x}_{\left|z\right|=1}\left|P\left(z\right)\right|.$$

(2)

Malik [3] 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

$$n{\left\{\underset{0}{\overset{2\pi }{\int }}|P\left(e^{i\theta }\right){|}^{\delta }d\theta \right\}}^{\frac{1}{\delta }}\le {\left\{\underset{0}{\overset{2\pi }{\int }}|1+e^{i\theta }{|}^{\delta }d\theta \right\}}^{\frac{1}{\delta }}Ma{x}_{\left|z\right|=1}\left|P^{\prime }\left(z\right)\right|.$$

(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 [4] proved the following:

Theorem B. If$P\left(z\right):=a_n{z}^n+{\sum }_{j=\mu }^n{a}_{n-j}{z}^{n-j}$, 1 ≤ μ ≤ n is a polynomial of degree n having all its zeros in the disk |z| ≤ K, K ≤ 1, then for each δ > 0,

$$n{\left\{\underset{0}{\overset{2\pi }{\int }}|P\left(e^{i\theta }\right){|}^{\delta }d\theta \right\}}^{\frac{1}{\delta }}\le {\left\{\underset{0}{\overset{2\pi }{\int }}|1+K^{\mu }{e}^{i\theta }{|}^{\delta }d\theta \right\}}^{\frac{1}{\delta }}Ma{x}_{\left|z\right|=1}\left|P^{\prime }\left(z\right)\right|.$$

(4)

Aziz and Ahmad [5] 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$\frac{1}{p}+\frac{1}{q}=1$,

$$n{\left\{\underset{0}{\overset{2\pi }{\int }}|P\left(e^{i\theta }\right){|}^{\delta }d\theta \right\}}^{\frac{1}{\delta }}\le {\left\{\underset{0}{\overset{2\pi }{\int }}|1+K{e}^{i\theta }{|}^{q\delta }d\theta \right\}}^{\frac{1}{q\delta }}{\left\{\underset{0}{\overset{2\pi }{\int }}|P^{\prime }\left(e^{i\theta }\right){|}^{p\delta }d\theta \right\}}^{\frac{1}{p\delta }}.$$

(5)

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

In this paper, we consider a class of polynomials $P\left(z\right):=a_n{z}^n+{\sum }_{j=\mu }^n{a}_{n-j}{z}^{n-j}$, 1 ≤ μ ≤ n, having all the zeros in |z| ≤ K ≤ 1, and thereby obtain a more general result by proving the following:

Theorem 1. If$P\left(z\right):=a_n{z}^n+{\sum }_{j=\mu }^n{a}_{n-j}{z}^{n-j}$, 1 ≤ μ ≤ n 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$\frac{1}{p}+\frac{1}{q}=1$and for every complex number λ with |λ| < 1

$$\begin{array}{c}n{\left\{\underset{0}{\overset{2\pi }{\int }}|P\left(e^{i\theta }\right)+\lambda m{|}^{\delta }d\theta \right\}}^{\frac{1}{\delta }}\\ \le {\left\{\underset{0}{\overset{2\pi }{\int }}|1+\left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]e^{i\theta }{|}^{q\delta }d\theta \right\}}^{\frac{1}{q\delta }}{\left\{\underset{0}{\overset{2\pi }{\int }}|P^{\prime }\left(e^{i\theta }\right){|}^{p\delta }d\theta \right\}}^{\frac{1}{p\delta }},\end{array}$$

(6)

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

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

Corollary 1. If$P\left(z\right):=a_n{z}^n+{\sum }_{j=\mu }^n{a}_{n-j}{z}^{n-j}$, 1 ≤ μ ≤ n 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$\frac{1}{p}+\frac{1}{q}=1$, ^,

$$\begin{array}{c}n{\left\{\underset{0}{\overset{2\pi }{\int }}|P\left(e^{i\theta }\right){|}^{\delta }d\theta \right\}}^{\frac{1}{\delta }}\\ \le {\left\{\underset{0}{\overset{2\pi }{\int }}|1+\left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]e^{i\theta }{|}^{q\delta }d\theta \right\}}^{\frac{1}{q\delta }}{\left\{\underset{0}{\overset{2\pi }{\int }}|P^{\prime }\left(e^{i\theta }\right){|}^{p\delta }d\theta \right\}}^{\frac{1}{p\delta }},\end{array}$$

(7)

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

Corollary 2. If$P\left(z\right):={\sum }_{j=0}^n{a}_j{z}^j$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$\frac{1}{p}+\frac{1}{q}=1$,

$$\begin{array}{c}n{\left\{\underset{0}{\overset{2\pi }{\int }}|P\left(e^{i\theta }\right)+\lambda m{|}^{\delta }d\theta \right\}}^{\frac{1}{\delta }}\\ \le {\left\{\underset{0}{\overset{2\pi }{\int }}|1+\left[\frac{n\left|a_n\right|K^2+\left|a_{n-1}\right|}{n\left|a_n\right|+\left|a_{n-1}\right|}\right]e^{i\theta }{|}^{q\delta }d\theta \right\}}^{\frac{1}{q\delta }}{\left\{\underset{0}{\overset{2\pi }{\int }}|P^{\prime }\left(e^{i\theta }\right){|}^{p\delta }d\theta \right\}}^{\frac{1}{p\delta }},\end{array}$$

(8)

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

Remark 1: Since all the zeros of P(z) lie in |z| ≤ K, therefore, $\frac{1}{n}\left|\frac{a_{n-1}}{a_n}\right|\le KK\le 1$, it can be easily verified that

$$\frac{n\left|a_n\right|K^2+\left|a_{n-1}\right|}{n\left|a_n\right|+\left|a_{n-1}\right|}\le K.$$

It shows that for λ = 0, Corollary 2 provides a refinement of the result of Aziz and Ahmad [5].

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

Corollary 3. If$P\left(z\right):=a_n{z}^n+{\sum }_{j=\mu }^n{a}_{n-j}{z}^{n-j}$, 1 ≤ μ ≤ n 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

$$\begin{array}{c}n{\left\{\underset{0}{\overset{2\pi }{\int }}|P\left(e^{i\theta }\right)+\lambda m{|}^{\delta }d\theta \right\}}^{\frac{1}{\delta }}\\ \le {\left\{\underset{0}{\overset{2\pi }{\int }}|1+\left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]e^{i\theta }{|}^{\delta }d\theta \right\}}^{\frac{1}{\delta }}Ma{x}_{\left|z\right|=1}\left|P^{\prime }\left(z\right)\right|.\end{array}$$

(9)

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

$$\underset{\delta \to \infty }{lim}{\left\{\frac{1}{2\pi }{\int }_0^{2\pi }|P\left({\mathsf{\text{e}}}^{i\theta }\right){|}^{\delta }\mathsf{\text{d}}\theta \right\}}^{\frac{1}{\delta }}=Ma{x}_{\left|z\right|=1}\left|P\left(z\right)\right|,$$

we get from (9)

$$nMa{x}_{\left|z\right|=1}|P\left(z\right)+\lambda m|\le \frac{n\left|a_n\right|\left(K^{2\mu }+K^{\mu -1}\right)+\mu \left|a_{n-\mu }\right|\left(1+K^{\mu -1}\right)}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}Ma{x}_{\left|z\right|=1}\left|P^{\prime }\left(z\right)\right|for\left|z\right|=1.$$

(10)

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

$$\begin{array}{c}n|P\left(z_0\right)+\lambda m|\le \\ \frac{n\left|a_n\right|\left(K^{2\mu }+K^{\mu -1}\right)+\mu \left|a_{n-\mu }\right|\left(1+K^{\mu -1}\right)}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}Ma{x}_{\left|z\right|=1}\left|P^{\prime }\left(z\right)\right|for\left|z\right|=1.\end{array}$$

Choosing an argument of λ such that

$$|P\left(z_0\right)+\lambda m|=\left|P\left(z_0\right)\right|+\left|\lambda \right|m,$$

we get

$$n\left(\left|P\left(z_0\right)\right|+\left|\lambda \right|m\right)\le \frac{n\left|a_n\right|\left(K^{2\mu }+K^{\mu -1}\right)+\mu \left|a_{n-\mu }\right|\left(1+K^{\mu -1}\right)}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}Ma{x}_{\left|z\right|=1}\left|P^{\prime }\left(z\right)\right|.$$

(11)

From inequality (11), we conclude the following:

Corollary 4. If$P\left(z\right):=a_n{z}^n+{\sum }_{j=\mu }^n{a}_{n-j}{z}^{n-j}$, 1 ≤ μ ≤ n is a polynomial of degree n having all its zeros in the disk |z| ≤ K, K ≤ 1, then for 0 ≤ t ≤ 1, we have

$$\begin{array}{c}Ma{x}_{\left|z\right|=1}\left|P^{\prime }\left(z\right)\right|\ge \\ n\frac{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}{n\left|a_n\right|\left(K^{2\mu }+K^{\mu -1}\right)+\mu \left|a_{n-\mu }\right|\left(1+K^{\mu -1}\right)}\left\{Ma{z}_{\left|z\right|=1}\left|P\left(z\right)\right|+tMi{n}_{\left|z\right|=K}\left|P\left(z\right)\right|\right\}.\end{array}$$

Further, if we take K = t = μ = 1 in the Corollary 4, we get a result of Aziz and Dawood [6].

2. Lemmas

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

The first lemma is due to Qazi [7].

Lemma 1. If $P\left(z\right):=a_0+{\sum }_{j=\mu }^n{a}_j{z}^j,1\le \mu \le n$ is a polynomial of degree n having no zeros in the disk|z| < K, K ≥ 1, then

$$\left[\frac{n\left|a_0\right|K^{\mu +1}+\mu \left|a_{\mu }\right|K^{2\mu }}{n\left|a_0\right|+\mu \left|a_{\mu }\right|K^{\mu +1}}\right]\left|P^{\prime }\left(z\right)\right|\le \left|Q^{\prime }\left(z\right)\right|for\left|z\right|=1,$$

where $Q\left(z\right)=z^n\overline{P\left(\frac{1}{\bar{z}}\right)}and\frac{\mu }{n}\left|\frac{a_{\mu }}{a_0}\right|K^{\mu }\le 1$.

Lemma 2. If $P\left(z\right):=a_n{z}^n+{\sum }_{j=\mu }^n{a}_{n-j}{z}^{n-j}$ is a polynomial of degree n having all its zeros in the disk|z| ≤ K ≤ 1, then

$$\left|Q^{\prime }\left(z\right)\right|\le \left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]\left|P^{\prime }\left(z\right)\right|for\left|z\right|=1,1\le \mu \le n,$$

where $Q\left(z\right)=z^n\overline{P\left(\frac{1}{\bar{z}}\right)}$.

Proof of Lemma 2

Since all the zeros of P(z) lie in |z| ≤ K ≤ 1, therefore all the zeros of $Q\left(z\right)=z^n\overline{P\left(\frac{1}{\bar{z}}\right)}$ lie in $\left|z\right|\ge \frac{1}{K}\ge 1$. Hence, applying Lemma 1 to the polynomial $Q\left(z\right):={ā}_n+{\sum }_{j=\mu }^n{ā}_{n-j}{z}^j$, we get

$$\left[\frac{n\left|a_n\right|\frac{1}{K^{\mu +1}}+\mu \left|a_{n-\mu }\right|\frac{1}{K^{2\mu }}}{n\left|a_n\right|+\mu \left|a_{n-\mu }\right|\frac{1}{K^{\mu +1}}}\right]\left|Q^{\prime }\left(z\right)\right|\le \left|P^{\prime }\left(z\right)\right|.$$

Or, equivalently

$$\left|Q^{\prime }\left(z\right)\right|\le \left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]\left|P^{\prime }\left(z\right)\right|,\left|z\right|=1.$$

This proves Lemma 2.

Remark 1: Lemma 3 of Govil and Mc Tume [8] is a special case of this lemma when μ = 1.

Proof of Theorem 1

Let $Q\left(z\right)=z^n\overline{P\left(\frac{1}{\bar{z}}\right)}$z, we have $P\left(z\right)=z^n\overline{Q\left(\frac{1}{\bar{z}}\right)}$. This gives

$$P^{\prime }\left(z\right)=n{z}^{n-1}\overline{Q\left(\frac{1}{\bar{z}}\right)}-z^{n-2}\overline{Q^{\prime }\left(\frac{1}{\bar{z}}\right)}.$$

(12)

Equivalently,

$$z{P}^{\prime }\left(z\right)=n{z}^n\overline{Q\left(\frac{1}{\bar{z}}\right)}-z^{n-1}\overline{Q^{\prime }\left(\frac{1}{\bar{z}}\right)}.$$

(13)

This implies

$$\left|P^{\prime }\left(z\right)\right|=|nQ\left(z\right)-z{Q}^{\prime }\left(z\right)|for\left|z\right|=1.$$

(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 |mλ| < |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 $H\left(z\right)=z^n\overline{G\left(\frac{1}{\bar{z}}\right)}=Q\left(z\right)+m\bar{\lambda }{z}^n$, then by applying Lemma 2 to the polynomial G(z) = P(z) + λm, we have for |z| = 1

$$\left|H^{\prime }\left(z\right)\right|\le \left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]\left|G^{\prime }\left(z\right)\right|,1\le \mu \le n.$$

This gives

$$|Q^{\prime }\left(z\right)+nm\bar{\lambda }{z}^{n-1}|\le \left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]\left|P^{\prime }\left(z\right)\right|,1\le \mu \le n.$$

(15)

Using (14) in (15), we get

$$|Q^{\prime }\left(z\right)+nm\bar{\lambda }{z}^{n-1}|\le \left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]|nQ\left(z\right)-z{Q}^{\prime }\left(z\right)|for\left|z\right|=1,1\le \mu \le n.$$

(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))

$$z^{n-1}\overline{P^{\prime }\left(\frac{1}{\bar{z}}\right)},$$

has all its zeros in |z| ≥ $\frac{1}{K}$ ≥ 1. Hence,

$$W\left(z\right)=\left[\frac{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}\right].\frac{z\left(Q^{\prime }\left(z\right)+nm\bar{\lambda }{z}^{n-1}\right)}{\left(nQ\left(z\right)-z{Q}^{\prime }\left(z\right)\right)}$$

(17)

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

$$1+\left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right].W\left(z\right)$$

is subordinate to the function

$$1+\left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]z$$

for |z| < 1. Hence, by a property of subordination (for reference see [[9], p. 36, Theorem 1.6.17] or [[10], p. 454] or [11]), we have for each δ > 0 and 0 ≤ θ < 2π,

$$\begin{array}{c}\underset{0}{\overset{2\pi }{\int }}|1+\left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]W\left({\mathsf{\text{e}}}^{i\theta }\right){|}^{\delta }\mathsf{\text{d}}\theta \\ \le \underset{0}{\overset{2\pi }{\int }}|1+\left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]{\mathsf{\text{e}}}^{i\theta }{|}^{\delta }\mathsf{\text{d}}\theta .\end{array}$$

(18)

Also from (17), we have

$$1+\left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]W\left(z\right)=\frac{n\left(Q\left(z\right)+m\bar{\lambda }{z}^n\right)}{nQ\left(z\right)-z{Q}^{\prime }\left(z\right)}.$$

Therefore,

$$n|Q\left(z\right)+m\bar{\lambda }{z}^n|=|1+\left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]W\left(z\right)||nQ\left(z\right)-z{Q}^{\prime }\left(z\right)|,$$

(19)

which implies

$$n\left|H\left(z\right)\right|=|1+\left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]W\left(z\right)||nQ\left(z\right)-z{Q}^{\prime }\left(z\right)|.$$

(20)

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

$$n|P\left(z\right)+\lambda m|=|1+\left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]W\left(z\right)|\left|P^{\prime }\left(z\right)\right|for\left|z\right|=1.$$

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

$$\begin{array}{c}{n}^{\delta }\underset{0}{\overset{2\pi }{\int }}|P\left({\mathsf{\text{e}}}^{i\theta }\right)+\lambda m{|}^{\delta }\mathsf{\text{d}}\theta \\ =\underset{0}{\overset{2\pi }{\int }}|1+\left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-v}\right|}\right]W\left({\mathsf{\text{e}}}^{i\theta }\right){|}^{\delta }|P^{\prime }\left({\mathsf{\text{e}}}^{i\theta }\right){|}^{\delta }\mathsf{\text{d}}\theta .\end{array}$$

(21)

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

$$\begin{array}{c}{n}^{\delta }\underset{0}{\overset{2\pi }{\int }}|P\left({\mathsf{\text{e}}}^{i\theta }\right)+\lambda m{|}^{\delta }\mathsf{\text{d}}\theta \\ \le {\left\{\underset{0}{\overset{2\pi }{\int }}|1+\left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]W\left({\mathsf{\text{e}}}^{i\theta }\right){|}^{q\delta }\mathsf{\text{d}}\theta \right\}}^{\frac{1}{q}}{\left\{\underset{0}{\overset{2\pi }{\int }}|P^{\prime }\left({\mathsf{\text{e}}}^{i\theta }\right){|}^{p\delta }\mathsf{\text{d}}\theta \right\}}^{\frac{1}{p}}\end{array}$$

(22)

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

$$\begin{array}{c}{n}^{\delta }\underset{0}{\overset{2\pi }{\int }}|P\left({\mathsf{\text{e}}}^{i\theta }\right)+\lambda m{|}^{\delta }\mathsf{\text{d}}\theta \\ \le {\left\{\underset{0}{\overset{2\pi }{\int }}|1+\left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]{\mathsf{\text{e}}}^{i\theta }{|}^{q\delta }\mathsf{\text{d}}\theta \right\}}^{\frac{1}{q}}{\left\{\underset{0}{\overset{2\pi }{\int }}|P^{\prime }\left({\mathsf{\text{e}}}^{i\theta }\right){|}^{p\delta }\mathsf{\text{d}}\theta \right\}}^{\frac{1}{p}}\end{array}$$

(23)

This is equivalent to

$$\begin{array}{c}n{\left\{\underset{0}{\overset{2\pi }{\int }}|P\left({\mathsf{\text{e}}}^{i\theta }\right)+\lambda m{|}^{\delta }\mathsf{\text{d}}\theta \right\}}^{\frac{1}{\delta }}\\ \le {\left\{\underset{0}{\overset{2\pi }{\int }}|1+\left[\frac{n\left|a_n\right|K^{2\mu }+\mu \left|a_{n-\mu }\right|K^{\mu -1}}{n\left|a_n\right|K^{\mu -1}+\mu \left|a_{n-\mu }\right|}\right]{\mathsf{\text{e}}}^{i\theta }{|}^{q\delta }\mathsf{\text{d}}\theta \right\}}^{\frac{1}{q\delta }}{\left\{\underset{0}{\overset{2\pi }{\int }}|P^{\prime }\left({\mathsf{\text{e}}}^{i\theta }\right){|}^{p\delta }\mathsf{\text{d}}\theta \right\}}^{\frac{1}{p\delta }}\end{array}$$

(24)

which proves the desired result.