On Brannan’s Conjecture

Abstract

We prove the Brannan conjecture for particular values of the parameter. The basic tool of the study is an integral representation published in a recent work (Barnard et al., in Proc Am Math Soc 143(5):2117–2128, 2015).

1 Introduction

We consider the following Mac-Laurin development

$$\begin{aligned} \frac{(1+xz)^\alpha }{(1-z)^\beta }=\sum _{n=0}^\infty {A_n(\alpha ,\beta ,x)}z^n, \end{aligned}$$

(1)

where $\alpha>0, \ \beta >0, \ x=\mathrm{e}^{i\theta }, \ \theta \in [-\pi ,\pi ],$ and $z\in {U.}$ The radius of convergence of the series (1) is equal to 1. The author conjectured in [5] that if $\alpha>0, \ \beta >0$ and $|x|=1,$ then

$$\begin{aligned} |A_{2n-1}(\alpha ,\beta ,x)|\le {A_{2n-1}(\alpha ,\beta ,1)}, \end{aligned}$$

where n is a natural number. Partial results regarding this question were already proved in [1, 2, 5, 8].

The case $\beta =1,$ $\alpha \in (0,1)$ is still open. Regarding this case, partial results were obtained in [3, 4, 6, 7]. We will prove some partial results regarding the case $\beta =1,$ and $\alpha \in (0,1).$ We will use an integral representation proved in [3], and the fact that the conjecture was proved for $n\le 26$ in [6].

2 Preliminaries

We introduce the notation:

$$\begin{aligned} A_n(\alpha ,1,x)=A_n(\alpha ,x). \end{aligned}$$

It is easily seen that

$$\begin{aligned}&A_{2n-1}(\alpha ,x)=\sum _{k=0}^{2n-1}\frac{(-\alpha )_k(-x)^k}{k!}=1+\frac{\alpha }{1!}x-\frac{\alpha (1-\alpha )}{2!}x^2+\frac{\alpha (1-\alpha )(2-\alpha )}{3!}x^3\nonumber \\&\quad -\frac{\alpha (1-\alpha )(2-\alpha )(3-\alpha )}{4!}x^4+\cdots + \frac{\alpha (1-\alpha )(2-\alpha )\ldots (2n-2-\alpha )}{(2n-1)!}x^{2n-1}.\nonumber \\ \end{aligned}$$

(2)

We denote

$$\begin{aligned} B(t,\theta )= & {} \frac{\cos \theta +t+t^{2n-1}\cos 2n\theta +t^{2n}\cos (2n-1)\theta }{1+t^2+2t\cos \theta }, \\ C(t,\theta )= & {} \frac{\sin \theta +t^{2n-1}\sin 2n\theta +t^{2n}\sin (2n-1)\theta }{1+t^2+2t\cos \theta }. \end{aligned}$$

Lemma 1

For $\alpha \in (0,1),$ the following equality holds:

$$\begin{aligned}&-\Gamma (\alpha )\Gamma (-\alpha )A_{2n-1}(\alpha ,\mathrm{e}^{i\theta })=\int _0^1F(t)\Bigg [\frac{1}{\alpha }+\sum _{k=1}^{2n-1}(-t)^{k-1}\mathrm{e}^{ki\theta }\Bigg ]\mathrm{d}t \\&\quad =\int _0^1F(t)\Bigg [\frac{1}{\alpha }+B(t,\theta )+iC(t,\theta )\Bigg ]\mathrm{d}t, \end{aligned}$$

where $F'(t)=-t^{-1-\alpha }(1-t)^{\alpha -1},$ and $F(1)=0.$

Proof

We use two known equalities to prove the assertion of the lemma. The first one is the following:

$$\begin{aligned} A_n(\alpha ,x)=1+\frac{1}{\Gamma (\alpha )\Gamma (-\alpha )}\int _0^1\Bigg (\sum _{k=1}^n\frac{(-tx)^k}{k}\Bigg )t^{-\alpha -1}(1-t)^{\alpha -1}\mathrm{d}t, \end{aligned}$$

(3)

which has been deduced in [3], while the second one is the well-known equality $B(p,q)=\int _0^1t^{p-1}(1-t)^{q-1}\mathrm{d}t=\frac{\Gamma (p)\Gamma (q)}{\Gamma (p+q)}.$ Replacing $p=\alpha$ and $q=1-\alpha ,$ we get

$$\begin{aligned} \int _0^1t^{\alpha -1}(1-t)^{-\alpha }\mathrm{d}t=\frac{\Gamma (\alpha )\Gamma (1-\alpha )}{\Gamma (1)}=-\alpha \Gamma (\alpha )\Gamma (-\alpha )=\int _0^1t^{-\alpha }(1-t)^{\alpha -1}\mathrm{d}t. \end{aligned}$$

This equality and (3) imply that

$$\begin{aligned} \Gamma (\alpha )\Gamma (-\alpha )A_n(\alpha ,x)=\frac{1}{\alpha }\int _0^1tF'(t)\mathrm{d}t-\int _0^1\Bigg (\sum _{k=1}^n\frac{(-tx)^k}{k}\Bigg ){F'(t)}\mathrm{d}t, \end{aligned}$$

or equivalently

$$\begin{aligned} -\Gamma (\alpha )\Gamma (-\alpha )A_n(\alpha ,x)=\int _0^1{F'(t)}\Bigg (\frac{-t}{\alpha }+\sum _{k=1}^n\frac{(-tx)^k}{k}\Bigg )\mathrm{d}t. \end{aligned}$$

Now integrating by parts and using that $\lim _{t\rightarrow 0}F(t)(\frac{t}{\alpha }+\sum _{k=1}^n\frac{(-tx)^k}{k})=0$ and $\lim _{t\rightarrow 1}F(t)(\frac{t}{\alpha }+\sum _{k=1}^n\frac{(-tx)^k}{k})=0,$ we infer that

$$\begin{aligned} -\Gamma (\alpha )\Gamma (-\alpha )A_n(\alpha ,x)=\int _0^1{F(t)}\Bigg (\frac{1}{\alpha }+\sum _{k=1}^n{(-t)^{k-1}}x^k\Bigg )\mathrm{d}t. \end{aligned}$$

We get the desired equality replacing n by $2n-1$ and $x=\mathrm{e}^{i\theta }$.

$$\begin{aligned} -\Gamma (\alpha )\Gamma (-\alpha )A_{2n-1}(\alpha ,\mathrm{e}^{i\theta })&=\int _0^1{F(t)}\Bigg (\frac{1}{\alpha }+\sum _{k=1}^{2n-1}{(-t)^{k-1}}x^k\Bigg )\mathrm{d}t\\&=\int _0^1{F(t)}\Bigg (\frac{1}{\alpha }+\frac{x+x^{2n}t^{2n-1}}{1+xt}\Bigg )\mathrm{d}t\\&=\int _0^1{F(t)}\Bigg (\frac{1}{\alpha }+\frac{\mathrm{e}^{i\theta }+t+\mathrm{e}^{2ni\theta }t^{2n-1}+\mathrm{e}^{(2n-1)i\theta }t^{2n}}{1+t^2+2t\cos \theta }\Bigg )\mathrm{d}t. \end{aligned}$$

$\square$

Remark 1

We note that the condition $\alpha \in (0,1)$ implies the existence of the integrals in the previous lemma and its proof.

We denote

$$\begin{aligned} \Phi (\theta ):= & {} -\Gamma (\alpha )\Gamma (-\alpha )A_{2n-1}(\alpha ,\mathrm{e}^{i\theta })\nonumber \\= & {} \int _0^1F(t)\Bigg [\frac{1}{\alpha }+\frac{\cos \theta +t+t^{2n-1}\cos 2n\theta +t^{2n}\cos (2n-1)\theta }{1+t^2+2t\cos \theta }\nonumber \\&\quad +i\frac{\sin \theta +t^{2n-1}\sin 2n\theta +t^{2n}\sin (2n-1)\theta }{1+t^2+2t\cos \theta }\Bigg ]\mathrm{d}t. \end{aligned}$$

(4)

Since $|\Phi (\theta )|^2\in {\mathbb {R}},$ it follows that

$$\begin{aligned} |\Phi (\theta )|^2&=\Phi (\theta )\overline{\Phi (\theta )} \nonumber \\&=\Bigg [\int _0^1F(t)\Bigg (\frac{1}{\alpha } +\frac{\cos \theta +t+t^{2n-1}\cos 2n\theta +t^{2n}\cos (2n-1)\theta }{1+t^2+2t\cos \theta } \nonumber \\&\quad +i\frac{\sin \theta +t^{2n-1}\sin 2n\theta +t^{2n}\sin (2n-1)\theta }{1+t^2+2t\cos \theta }\Bigg )\mathrm{d}t\Bigg ] \nonumber \\&\Bigg [\int _0^1F(v)\Bigg (\frac{1}{\alpha }+\frac{\cos \theta +v+v^{2n-1}\cos 2n\theta +v^{2n}\cos (2n-1)\theta }{1+v^2+2v\cos \theta } \nonumber \\&\quad -i\frac{\sin \theta +v^{2n-1}\sin 2n\theta +v^{2n}\sin (2n-1)\theta }{1+v^2+2v\cos \theta }\Bigg )dv\Bigg ] \nonumber \\&=\int _0^1\int _0^1F(t)F(v)\Bigg [\Bigg (\frac{1}{\alpha }+\frac{\cos \theta +t+t^{2n-1}\cos 2n\theta +t^{2n}\cos (2n-1)\theta }{1+t^2+2t\cos \theta }\Bigg ) \nonumber \\&\Bigg (\frac{1}{\alpha }+\frac{\cos \theta +v+v^{2n-1}\cos 2n\theta +v^{2n}\cos (2n-1)\theta }{1+v^2+2v\cos \theta }\Bigg ) \nonumber \\&\quad +\Bigg (\frac{\sin \theta +t^{2n-1}\sin 2n\theta +t^{2n}\sin (2n-1)\theta }{1+t^2+2t\cos \theta }\Bigg ) \nonumber \\&\Bigg (\frac{\sin \theta +v^{2n-1}\sin 2n\theta +v^{2n}\sin (2n-1)\theta }{1+v^2+2v\cos \theta }\Bigg )\Bigg ]\mathrm{d}t\mathrm{d}v. \end{aligned}$$

(5)

Lemma 2

(a) Let $f,g:[0,1]\rightarrow {\mathbb {R}}$ be two continuous functions. If f is a decreasing function, and if there is a point ${t^*}\in (0,1)$ such that $g(t)\ge 0, \ t\in (0,t^*)$ and $g(t)\le 0, \ t\in (t^*,1),$ then

$$\begin{aligned} \int _0^1f(t)g(t)\mathrm{d}t\ge {f(t^*)}\int _0^1g(t)\mathrm{d}t. \end{aligned}$$

(b) The Chebyshev inequality for integrals. If f and g are monotonic functions with different monotony, then

$$\begin{aligned} \int _0^1f(t)g(t)\mathrm{d}t\le \int _0^1f(t)\mathrm{d}t\int _0^1g(t)\mathrm{d}t, \end{aligned}$$

and in case of the same monotony we have

$$\begin{aligned} \int _0^1f(t)g(t)\mathrm{d}t\ge \int _0^1f(t)\mathrm{d}t\int _0^1g(t)\mathrm{d}t. \end{aligned}$$

Proof

We have

$$\begin{aligned} \int _0^1f(t)g(t)\mathrm{d}t= & {} \int _0^{t^*}f(t)g(t)\mathrm{d}t+\int _{t^*}^1f(t)g(t)\mathrm{d}t\ge \int _0^{t^*}f(t^*)g(t)\mathrm{d}t\nonumber \\&+\int _{t^*}^1f(t^*)g(t)\mathrm{d}t=f(t^*)\int _{0}^1g(t)\mathrm{d}t. \end{aligned}$$

$\square$

Lemma 3

If $\theta \in [0,\frac{\pi }{2}],$ and $n\ge 27,$ then

$$\begin{aligned}&\mathrm{(a)}\ \ \ \ \ \int _0^1F(t)\Bigg (B(t,0)-B(t,\theta )\Bigg )\mathrm{d}t \nonumber \\&\quad \ge \int _0^1F(t)\frac{\frac{1}{2}(1-\cos {\theta })+t^{2n-1}(1-\cos 2n\theta )+t^{2n}(1-\cos (2n-1)\theta )}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t,\nonumber \\\end{aligned}$$

(6)

$$\begin{aligned}&\mathrm{(b)} \ \ \ \ \ \int _0^1F(t)\Bigg (1+B(t,\theta )\Bigg )\mathrm{d}t \nonumber \\&\quad \ge \int _0^1F(t)\frac{(1+t)(1+\cos {\theta })+t^{2n-1}(1+\cos 2n\theta )+t^{2n}(1+\cos (2n-1)\theta )}{1+t^2+2t\cos \theta }\mathrm{d}t.\nonumber \\ \end{aligned}$$

(7)

Proof

According to Lemma 2 (b), we have

$$\begin{aligned} \int _0^1\frac{t^{2n}}{1+t}\mathrm{d}t\le \int _0^1{t^{2n}}\mathrm{d}t\int _0^1\frac{1}{1+t}\mathrm{d}t=\frac{\ln 2}{2n+1}. \end{aligned}$$

(8)

We use assertion (a) of Lemma 2 putting $f(t)=\frac{F(t)}{1+t^2+2t\cos \theta }$ and $g(t)=\frac{\frac{1}{2}-t-2t^{2n}}{1+t}.$ If $\theta \in [0,\frac{\pi }{2}],$ then the mapping $f:[0,1]\rightarrow [0,\infty )$ is strictly decreasing and we get

$$\begin{aligned}&\int _0^1F(t)\frac{\frac{1}{2}-t-2t^{2n}}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t\ge \frac{F(t_n)}{1+t_n^2+2t_n\cos \theta }\nonumber \\&\quad \int _0^1\Bigg (\frac{\frac{1}{2}-t}{1+t}-2\frac{t^{2n}}{1+t}\Bigg )\mathrm{d}t\ge \frac{F(t_n)}{1+t_n^2+2t_n\cos \theta }\Bigg (\frac{3}{2}\ln 2-1-\frac{2\ln 2}{2n+1}\Bigg )>0,\nonumber \\ \end{aligned}$$

(9)

where $t_n$ denotes the unique root of the equation $\frac{1}{2}-t-2t^{2n}=0,$ in the interval $\ t\in (0,1).$ The following equality holds:

$$\begin{aligned}&\int _0^1F(t)\Bigg (B(t,0)-B(t,\theta )\Bigg )\mathrm{d}t =\int _0^1F(t)\frac{\frac{1}{2}(1-\cos {\theta })}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t \nonumber \\&\quad +\int _0^1F(t)\frac{(t^{2n}+t^{2n-1})(1-\cos 2n\theta )}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t \nonumber \\&\quad +\int _0^1F(t)\frac{(t^{2n}+ t^{2n+1})(1-\cos (2n-1)\theta )}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t \nonumber \\&\quad +\int _0^1F(t)\frac{(\frac{1}{2}-t-2t^{2n})(1-\cos \theta )}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t. \end{aligned}$$

(10)

The equality (10) and the inequality (9) imply that

$$\begin{aligned}&\int _0^1F(t)\Bigg (B(t,0)-B(t,\theta )\Bigg )\mathrm{d}t \ge \int _0^1F(t)\frac{\frac{1}{2}(1-\cos {\theta })}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t \nonumber \\&\quad +\int _0^1F(t)\frac{t^{2n-1}(1-\cos 2n\theta )}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t+\int _0^1F(t)\frac{t^{2n}(1-\cos (2n-1)\theta )}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t. \nonumber \\ \end{aligned}$$

(11)

We use again assertion (a) of Lemma 2 putting $f(t)=\frac{F(t)}{1+t^2+2t\cos \theta }$ and $g(t)={t^2-2t^{2n}-t^{2n-1}}.$ If $\theta \in [0,\frac{\pi }{2}],$ then f is strictly decreasing and we get

$$\begin{aligned}&\int _0^1F(t)\frac{t^{2}-t^{2n}-t^{2n-1}}{1+t^2+2t\cos \theta }\mathrm{d}t\ge \frac{{F}(t_n)}{1+t_n^2+2t_n\cos \theta }\int _0^1({t^{2}-t^{2n}-t^{2n-1}})\mathrm{d}t\nonumber \\&\quad =\frac{{F}(t_n)}{1+t_n^2+2t_n\cos \theta }\Bigg (\frac{1}{3}-\frac{1}{2n+1}-\frac{1}{2n}\Bigg )>0. \end{aligned}$$

(12)

To finish the proof of the second inequality, we take notice of the fact that in case $\theta \in [0,\frac{\pi }{2}]$ each member of the following sum is positive:

$$\begin{aligned}&\int _0^1F(t)\Big (1+B(t,\theta )\Big )\mathrm{d}t= \int _0^1F(t)\frac{(1+t)(1+\cos {\theta })}{1+t^2+2t\cos \theta }\mathrm{d}t \\&\quad +\int _0^1F(t)\frac{t^{2n-1}(1+\cos 2n\theta )}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t+\int _0^1F(t)\frac{t^{2n}(1+\cos (2n-1)\theta )}{1+t^2+2t\cos \theta }\mathrm{d}t \\&\quad +\int _0^1F(t)\frac{t^{2}-t^{2n}-t^{2n-1}}{1+t^2+2t\cos \theta }\mathrm{d}t+ \int _0^1F(t)\frac{t\cos \theta }{1+t^2+2t\cos \theta }\mathrm{d}t. \end{aligned}$$

Thus, we get

$$\begin{aligned}&\int _0^1F(t)\Big (1+B(t,\theta )\Big )\mathrm{d}t\ge \int _0^1F(t)\frac{(1+t)(1+\cos {\theta })}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t \\&\quad +\int _0^1F(t)\frac{t^{2n-1}(1+\cos 2n\theta )}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t+\int _0^1F(t)\frac{t^{2n}(1+\cos (2n-1)\theta )}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t, \end{aligned}$$

and the proof is done. $\square$

Lemma 4

If $\theta \in [\frac{\pi }{2},\frac{2\pi }{3}],$ then

$$\begin{aligned} \frac{5}{2}+\frac{t+\cos \theta }{1+t^2+2t\cos \theta }\ge \frac{50}{23}\frac{(1+t)(1+\cos \theta )}{1+t^2+2t\cos \theta }, \ \ (\forall ) \ t\in [0,1]. \end{aligned}$$

(13)

Proof

The inequality (13) is equivalent to

$$\begin{aligned} f(t)=\frac{5}{2}t^2+t\Bigg (\frac{65}{23}\cos \theta -\frac{27}{23}\Bigg )+\frac{15}{46}-\frac{27}{23}\cos \theta \ge 0, \ \ (\forall ) \ t\in [0,1]. \end{aligned}$$

The function f has a minimum point at $t^*=-\frac{65}{115}\cos \theta +\frac{27}{115}\in (0,1),$ for every $\theta \in [\frac{\pi }{2},\frac{2\pi }{3}].$ Thus, we get

$$\begin{aligned} f(t)\ge {f(t^*)}=\frac{15}{46}-\frac{27}{23}\cos \theta -\frac{1}{10}\Bigg (\frac{27-65\cos \theta }{23}\Bigg )^2, \ (\forall ) \ t\in [0,1] \ \text {and} \ \theta \in \Bigg [\frac{\pi }{2},\frac{2\pi }{3}\Bigg ]. \end{aligned}$$

Consequently, to prove (13) we have to show that

$$\begin{aligned} \frac{15}{46}-\frac{27}{23}\cos \theta -\frac{1}{10}\Bigg (\frac{27-65\cos \theta }{23}\Bigg )^2\ge 0, \ (\forall ) \ \ \theta \in \Bigg [\frac{\pi }{2},\frac{2\pi }{3}\Bigg ]. \end{aligned}$$

Using the notation $x=\frac{27-65\cos \theta }{23},$ we get

$$\begin{aligned}&\frac{15}{46}-\frac{27}{23}\cos \theta -\frac{1}{10}\Bigg (\frac{27-65\cos \theta }{23}\Bigg )^2=-\frac{483}{2990}+\frac{27}{65}x-\frac{1}{10}x^2. \nonumber \\&\quad \theta \in \Bigg [\frac{\pi }{2},\frac{2\pi }{3}\Bigg ]\Rightarrow \ x\in \Bigg [\frac{27}{23},\frac{119}{46}\Bigg ]. \end{aligned}$$

(14)

The equality 14 implies that

$$\begin{aligned}&\frac{5}{12}-\frac{13}{12}\cos \theta -\frac{5}{2}\Bigg (\frac{13-35\cos \theta }{60}\Bigg )^2\ge \min _{ x\in [\frac{27}{23},\frac{119}{46}\Bigg ]}\Bigg \{-\frac{483}{2990}+\frac{27}{65}x-\frac{1}{10}x^2\Bigg \} \\&\quad \ge \min _{ x\in \Bigg [\frac{3}{2},\frac{13}{5}\Bigg ]}\Bigg \{-\frac{1}{5}+\frac{5}{13}x-\frac{1}{10}x^2\Bigg \}=\min \Bigg \{g\Bigg (\frac{3}{2}\Bigg ),g\Bigg (\frac{13}{5}\Bigg )\Bigg \} \\&\quad =g\Bigg (\frac{13}{5}\Bigg )=\frac{31}{250}, \ \ \theta \in \Bigg [\frac{\pi }{2},\frac{2\pi }{3}\Bigg ], \end{aligned}$$

where $g(x)=-\frac{1}{5}+\frac{5}{13}x-\frac{1}{10}x^2,$ and consequently (13) holds. $\square$

Lemma 5

If $\theta \in [\frac{\pi }{2},\frac{2\pi }{3}],$ and $n\ge 27,$ then

$$\begin{aligned}&\int _0^1F(t)(B(t,0)-B(t,\theta ))\mathrm{d}t \nonumber \\&\quad \ge \int _0^1F(t)\frac{\frac{27}{50}(1-\cos {\theta })+t^{2n-1}(1-\cos 2n\theta )+t^{2n}(1-\cos (2n-1)\theta )}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t\nonumber \\ \end{aligned}$$

(15)

$$\begin{aligned}&\int _0^1F(t)(2+B(t,0)+B(t,\theta ))\mathrm{d}t\ge \int _0^1F(t) \nonumber \\&\quad \frac{\frac{50}{27}(1+t)(1+\cos {\theta })+2t^{2n-1}(1+\cos 2n\theta )+2t^{2n}(1+\cos (2n-1)\theta )}{1+t^2+2t\cos \theta }\mathrm{d}t.\nonumber \\ \end{aligned}$$

(16)

Proof

We use assertion (a) of Lemma 2, putting $f(t)=\frac{F(t)}{(1+t)(1+t^2+2t\cos \theta )}$ and $g(t)={\frac{27}{50}-t-2t^{2n}}.$ If $\theta \in [\frac{\pi }{2},\frac{2\pi }{3}],$ the mapping $f:[0,1]\rightarrow [0,\infty )$ is strictly decreasing and we get

$$\begin{aligned} \begin{aligned}&\int _0^1F(t)\frac{\frac{27}{50}-t-2t^{2n}}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t \\&\quad \ge \frac{F(t_n)}{(1+t_n)(1+t_n^2+2t_n\cos \theta )}\int _0^1\Bigg ({\frac{27}{50}-t} -2{t^{2n}}\Bigg )\mathrm{d}t\\&\quad>\frac{F(t_n)}{(1+t_n)(1+t_n^2+2t_n\cos \theta )}\Bigg (\frac{27}{50}-\frac{1}{2}-\frac{2}{2n+1}\Bigg )>0 \end{aligned}, \end{aligned}$$

(17)

where $t_n$ denotes the unique root of the equation $\frac{27}{50}-t-2t^{2n}=0,$ in the interval $\ (0,1).$

$$\begin{aligned} \int _0^1F(t)(B(t,0)-B(t,\theta ))\mathrm{d}t= & {} \int _0^1F(t)\frac{\frac{23}{50}(1-\cos {\theta })}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t\nonumber \\&+\int _0^1F(t)\frac{(t^{2n}+t^{2n-1})(1-\cos 2n\theta )}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t\nonumber \\&+\int _0^1F(t)\frac{(t^{2n}+t^{2n+1})(1-\cos (2n-1)\theta )}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t \nonumber \\&+\int _0^1F(t)\frac{\Bigg (\frac{27}{50}-t-2t^{2n}\Bigg )(1-\cos \theta )}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t.\nonumber \\ \end{aligned}$$

(18)

The equality (18) and the inequality (17) imply (15).

To prove the second inequality, we remark that

$$\begin{aligned} \begin{aligned}&\int _0^1F(t)(2+B(t,0)+B(t,\theta ))\mathrm{d}t= \int _0^1F(t)\Bigg (\frac{5}{2}+\frac{1+t^{2n-1}}{1+t}-\frac{1}{2}+B(t,\theta )\Bigg )\mathrm{d}t\\&\quad = \int _0^1F(t)\Bigg (\frac{5}{2}+\frac{1+2t^{2n-1}-t}{2(1+t)}\\&\qquad +\frac{t+\cos \theta +t^{2n}\cos (2n-1)\theta +t^{2n-1}\cos (2n\theta )}{1+t^2+2t\cos \theta }\Bigg )\mathrm{d}t \\&\quad = \int _0^1F(t)\Bigg (\frac{5}{2}+\frac{t+\cos \theta }{1+t^2+2t\cos \theta } \\&\qquad +\frac{1+2t^{2n-1}-t}{2(1+t)}+\frac{t^{2n}\cos (2n-1)\theta +t^{2n-1}\cos (2n\theta )}{1+t^2+2t\cos \theta }\Bigg )\mathrm{d}t \\&\quad = \int _0^1F(t)\Bigg (\frac{5}{2}+\frac{t+\cos \theta }{1+t^2+2t\cos \theta } \\&\qquad +\frac{2t^{2n}(1+\cos (2n-1)\theta )+2t^{2n-1}(1+\cos (2n\theta ))}{1+t^2+2t\cos \theta }\Bigg )\mathrm{d}t\\&\qquad + \int _0^1F(t)\frac{2t^{2n-1}}{1+t}\mathrm{d}t\\&\qquad + \int _0^1F(t)\Bigg (\frac{1-t}{2(1+t)}-\frac{t^{2n}(2+\cos (2n-1)\theta )+t^{2n-1}(2+\cos (2n\theta ))}{1+t^2+2t\cos \theta }\Bigg )\mathrm{d}t, \end{aligned} \end{aligned}$$

(19)

and

$$\begin{aligned}&\int _0^1F(t)\Bigg (\frac{1-t}{2(1+t)}-\frac{t^{2n}(2+\cos (2n-1)\theta )+t^{2n-1}(2+\cos (2n\theta ))}{1+t^2+2t\cos \theta }\Bigg )\mathrm{d}t \nonumber \\&\quad \ge \int _0^1F(t)\Bigg (\frac{1-t}{2(1+t)}-\frac{2(t^{2n-1}+t^{2n})+t^{2n-1}\Bigg |\cos (2n-1)\theta +t\cos (2n\theta )\Bigg |}{1+t^2-t}\Bigg )\mathrm{d}t.\nonumber \\ \end{aligned}$$

(20)

If $\cos (2n-1)\theta$ and $\cos 2n\theta$ have different signs, then we have $|\cos (2n-1)\theta +t\cos (2n\theta )|\le 1.$ If $\cos (2n-1)\theta$ and $\cos 2n\theta$ have the same sign, then it follows that $|\cos (2n-1)\theta +t\cos (2n\theta )|\le |\cos (2n-1)\theta +\cos (2n\theta )|=2|\cos (2n-1/2)\theta |\cos \theta /2\le 2\cos \theta /2\le \sqrt{2}$ for every $\theta \in [\frac{\pi }{2},\frac{2\pi }{3}].$ Thus, we infer

$$\begin{aligned}&\int _0^1F(t)\Bigg (\frac{1-t}{2(1+t)}-\frac{2(t^{2n-1}+t^{2n})+t^{2n-1}\Bigg |\cos (2n-1)\theta +t\cos (2n\theta )\Bigg |}{1+t^2-t}\Bigg )\mathrm{d}t\nonumber \\&\quad \ge \int _0^1F(t)\Bigg (\frac{1-t}{2(1+t)}-\frac{(4+\sqrt{2})t^{2n-1}}{1+t^2-t}\Bigg )\mathrm{d}t\nonumber \\&\quad = \int _0^1 \frac{F(t)}{2(1+t^3)} (1-2t+2t^2-t^3-2(4+\sqrt{2})t^{2n-1}(1+t))\mathrm{d}t\nonumber \\&\quad \ge \int _0^1 \frac{F(t)}{2(1+t^3)} (1-2t+2t^2-t^3-4(4+\sqrt{2})t^{2n-1})\mathrm{d}t. \end{aligned}$$

(21)

Let $g(t)=1-2t+2t^2-t^3-4(4+\sqrt{2})t^{2n-1}$ and $f(t)=\frac{F(t)}{2(1+t^3)}.$ The functions f and g are strictly decreasing on [0, 1] and according to the assertion (a) of Lemma 2 there is a point $t^*\in (0,1)$ ( the unique root of the equation $g(t)=0$) such that the following inequality holds:

$$\begin{aligned}&\int _0^1 \frac{F(t)}{2(1+t^3)} (1-2t+2t^2-t^3-4(4+\sqrt{2})t^{2n-1})\mathrm{d}t\nonumber \\&\quad \ge \frac{F(t^*)}{2(1+(t^*)^3)} \int _0^1(1-2t+2t^2-t^3-4(4+\sqrt{2})t^{2n-1})\mathrm{d}t \nonumber \\&\quad = \frac{F(t^*)}{2(1+(t^*)^3)} \Bigg [\frac{5}{12}-\frac{2(4+\sqrt{2})}{n}\Bigg ]>0, \quad \theta \in \Bigg [\frac{\pi }{2},\frac{2\pi }{3}\Bigg ], \ n\ge 27. \end{aligned}$$

(22)

Now, inequalities (20)–(22) imply

$$\begin{aligned} \begin{aligned}&\int _0^1F(t)\Bigg (\frac{1-t}{2(1+t)}-\frac{t^{2n}(2+\cos (2n-1)\theta )+t^{2n-1}(2+\cos (2n\theta ))}{1+t^2+2t\cos \theta }\Bigg )\mathrm{d}t \\&\quad \ge \frac{F(t^*)}{2(1+(t^*)^3)} \Bigg [\frac{5}{12}-\frac{2(4+\sqrt{2})}{n}\Bigg ]>0, \quad \theta \in \Bigg [\frac{\pi }{2},\frac{2\pi }{3}\Bigg ], \ n\ge 27. \end{aligned} \end{aligned}$$

(23)

Finally inequality (23), Lemma 4 and equality (19) imply (16), and the proof is done. $\square$

Lemma 6

The following inequality holds:

$$\begin{aligned} 6\bigg (1-\frac{1}{3}\bigg )\bigg (1-\frac{1}{3\cdot 2}\bigg )\bigg (1-\frac{1}{3\cdot 3}\bigg )\ldots \bigg (1-\frac{1}{3\cdot 53}\bigg )<\frac{4}{3}. \end{aligned}$$

Proof

We introduce the notation

$$\begin{aligned} N= & {} \bigg (1-\frac{1}{3}\bigg )\bigg (1-\frac{1}{3\cdot 2}\bigg )\bigg (1-\frac{1}{3\cdot 3}\bigg )\ldots \bigg (1-\frac{1}{3\cdot 53}\bigg )\nonumber \\= & {} \frac{1\cdot 3-1}{1\cdot 3}\cdot \frac{2\cdot 3-1}{2\cdot 3}\cdot \frac{3\cdot 3-1}{3\cdot 3}\cdot \frac{4\cdot 3-1}{4\cdot 3}\cdot \cdots \cdot \frac{53\cdot 3-1}{53\cdot 3}. \end{aligned}$$

(24)

We will use the following implication $\alpha>0, a>b>1\Rightarrow \frac{b}{a}<\frac{b+\alpha }{a+\alpha }.$ Using this implication, we get

$$\begin{aligned} N<\frac{1\cdot 3-1}{1\cdot 3}\cdot \frac{2\cdot 3}{2\cdot 3+1}\cdot \frac{3\cdot 3}{3\cdot 3+1}\cdot \frac{4\cdot 3}{4\cdot 3+1}\cdot \cdots \cdot \frac{53\cdot 3}{53\cdot 3+1}\ \end{aligned}$$

(25)

and

$$\begin{aligned} N<\frac{1\cdot 3-1}{1\cdot 3}\cdot \frac{2\cdot 3+1}{2\cdot 3+2}\cdot \frac{3\cdot 3+1}{3\cdot 3+2}\cdot \frac{4\cdot 3+1}{4\cdot 3+2}\cdot \cdots \cdot \frac{53\cdot 3+1}{53\cdot 3+2}. \end{aligned}$$

(26)

Multiplying (24), (25), and (26), we get

$$\begin{aligned} N^3<\bigg (\frac{2}{3}\bigg )^3\frac{5}{53\cdot 3+2}. \end{aligned}$$

Thus to finish the proof, we have to check the inequality

$$\begin{aligned} \bigg (\frac{2}{3}\bigg )^3\frac{5}{53\cdot 3+2}<\bigg (\frac{2}{9}\bigg )^3 \end{aligned}$$

which is immediate. $\square$

3 Main Result

Theorem 1

If n is a natural number, $n\ge 27$ and $\alpha \in (0,1)$ then the following inequality holds:

$$\begin{aligned} |A_{2n-1}(\alpha ,\mathrm{e}^{i\theta })|\le |A_{2n-1}(\alpha ,1)|, \ \ \text {for all} \ \theta \in \left[ -\frac{\pi }{2},\frac{\pi }{2}\right] . \end{aligned}$$

(27)

Proof

According to (4) and (5), the inequality (27) is equivalent to

$$\begin{aligned} |\Phi (0)|^2\ge |\Phi (\theta )|^2, \ \theta \in \left[ -\frac{\pi }{2},\frac{\pi }{2}\right] \Leftrightarrow |\Phi (0)|^2-|\Phi (\theta )|^2\ge 0, \ \theta \in \left[ -\frac{\pi }{2},\frac{\pi }{2}\right] .\nonumber \\ \end{aligned}$$

(28)

We denote

$$\begin{aligned} B(t,\theta )= & {} \frac{\cos \theta +t+t^{2n-1}\cos 2n\theta +t^{2n}\cos (2n-1)\theta }{1+t^2+2t\cos \theta }, \\ C(t,\theta )= & {} \frac{\sin \theta +t^{2n-1}\sin 2n\theta +t^{2n}\sin (2n-1)\theta }{1+t^2+2t\cos \theta }. \end{aligned}$$

The equality (5) implies that

$$\begin{aligned} \Phi ^2(0)-|\Phi (\theta )|^2= & {} \int _0^1\int _0^1F(t)F(v)\Bigg [\Bigg (\frac{1}{\alpha }+B(t,0)\Bigg )\Bigg (\frac{1}{\alpha }+B(v,0)\Bigg ) \nonumber \\&\quad -\Bigg (\frac{1}{\alpha }+B(t,\theta )\Bigg )\Bigg (\frac{1}{\alpha }+B(v,\theta )\Bigg )-C(t,\theta )C(v,\theta )\Bigg ]\mathrm{d}t\mathrm{d}v\nonumber \\&=\int _0^1\int _0^1F(t)F(v)\Bigg [\frac{1}{\alpha }\Big (B(t,0)-B(t,\theta )\Big )+\frac{1}{\alpha }\Big (B(v,0)-B(v,\theta )\Big )\nonumber \\&\quad +B(t,0)B(v,0)-B(t,\theta )B(v,\theta )-C(t,\theta )C(v,\theta )\Bigg ]\mathrm{d}t\mathrm{d}v.\nonumber \\ \end{aligned}$$

(29)

Since $\Phi ^2(0)-|\Phi (\theta )|^2$ is an even function with respect to $\theta ,$ we have to prove (28) only for $\theta \in [0,\frac{\pi }{2}].$

Lemma 3, (a) implies that in case $\theta \in [0,\frac{\pi }{2}],$ the following inequality holds: $\int _0^1F(t)(B(t,0)-{B}(t,\theta ))\mathrm{d}t\ge 0.$ Thus, the condition $\alpha \in (0,1),$ implies that

$$\begin{aligned}&\Phi ^2(0)-|\Phi (\theta )|^2\ge \Lambda (\theta )=\int _0^1\int _0^1F(t)F(v)[(B(t,0)-B(t,\theta ))\\&\quad +(B(v,0)-B(v,\theta ))+B(t,0)B(v,0)-B(t,\theta )B(v,\theta )-C(t,\theta )C(v,\theta )]\mathrm{d}t\mathrm{d}v. \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \Lambda (\theta )&=\int _0^1\int _0^1F(t)F(v) [ (B(t,0)-B(t,\theta ) ) (1+B(v,\theta ) )\\&\quad + (B(v,0)-B(v,\theta ) ) (1+B(t,\theta ) ) ]\mathrm{d}t\mathrm{d}v\\&\quad -\int _0^1\int _0^1F(t)F(v)C(t,\theta )C(v,\theta )\mathrm{d}t\mathrm{d}v \\&\quad +\int _0^1\int _0^1F(t)F(v) (B(t,0)-B(t,\theta ) ) (B(v,0)-B(v,\theta ) )\mathrm{d}t\mathrm{d}v \\&=\int _0^1\int _0^1F(t)F(v) [ (B(t,0)-B(t,\theta ) ) (1+B(v,\theta ) ) \\&\quad + (B(v,0)-B(v,\theta ) ) (1+B(t,\theta ) ) ]\mathrm{d}t\mathrm{d}v\\&\quad -\int _0^1\int _0^1F(t)F(v)C(t,\theta )C(v,\theta )\mathrm{d}t\mathrm{d}v\\&\quad +\int _0^1F(t) (B(t,0)-B(t,\theta ) )\mathrm{d}t\int _0^1F(v) (B(v,0)-B(v,\theta ) )\mathrm{d}v\\&\ge \int _0^1\int _0^1F(t)F(v) [ (B(t,0)-B(t,\theta ) ) (1+B(v,\theta ) )\\&\quad + (B(v,\theta )-B(v,\theta ) ) (1+B(t,\theta ) ) ]\mathrm{d}t\mathrm{d}v\\&\quad -\int _0^1\int _0^1F(t)F(v)C(t,\theta )C(v,\theta )\mathrm{d}t\mathrm{d}v. \end{aligned}$$

This inequality implies

$$\begin{aligned} \Phi ^2(0)-|\Phi (\theta )|^2\ge & {} \int _0^1F(t) (B(t,0)-B(t,\theta ) )\mathrm{d}t\int _0^1F(v) (1+B(v,\theta ) )\mathrm{d}v \nonumber \\&+ \int _0^1F(t) (1+B(t,\theta ) )\mathrm{d}t\int _0^1F(v) (B(v,0)-B(v,\theta ) )\mathrm{d}v \nonumber \\&-\int _0^1\int _0^1F(t)F(v)C(t,\theta )C(v,\theta )\mathrm{d}t\mathrm{d}v. \end{aligned}$$

(30)

The inequality between the arithmetic and geometric means leads to

$$\begin{aligned} \begin{aligned}&\int _0^1F(t)\Big (B(t,0)-B(t,\theta )\Big )\mathrm{d}t\int _0^1F(v)\Big (1+B(v,\theta )\Big )\mathrm{d}v\\&\qquad + \int _0^1F(t)\Big (1+B(t,\theta )\Big )\mathrm{d}t\int _0^1F(v)\Big (B(v,0)-B(v,\theta )\Big )\mathrm{d}v \\&\qquad -\int _0^1\int _0^1F(t)F(v)C(t,\theta )C(v,\theta )\mathrm{d}t\mathrm{d}v \\&\quad \ge 2\Bigg [\int _0^1F(t)(B(t,0)-B(t,\theta ))\mathrm{d}t\int _0^1F(v)(1+B(v,\theta ))\mathrm{d}v\\&\int _0^1F(t)(1+B(t,\theta ))\mathrm{d}t\int _0^1F(v)\Big (B(v,0)-B(v,\theta )\Big )\mathrm{d}v\Bigg ]^{\frac{1}{2}} \\&\qquad -\int _0^1\int _0^1F(t)F(v)C(t,\theta )C(v,\theta )\mathrm{d}t\mathrm{d}v\\&\quad =2\int _0^1F(t)\Big (1+B(t,\theta )\Big )\mathrm{d}t\int _0^1F(t)\Big (B(t,0)-B(t,\theta )\Big )\mathrm{d}t \\&\qquad -\Bigg (\int _0^1F(t)C(t,\theta )\mathrm{d}t\Bigg )^2. \end{aligned} \end{aligned}$$

(31)

We apply Lemma 3 and get

$$\begin{aligned}&2\int _0^1F(t)\Big (1+B(t,\theta )\Big )\mathrm{d}t\int _0^1F(t)\Big (B(t,0)-B(t,\theta )\Big )\mathrm{d}t -\Bigg (\int _0^1F(t)C(t,\theta )\mathrm{d}t\Bigg )^2 \nonumber \\&\quad \ge 2\Bigg (\int _0^1F(t)\frac{\frac{1}{2}(1-\cos {\theta })+t^{2n-1}(1-\cos 2n\theta )+t^{2n}(1-\cos (2n-1)\theta )}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t\nonumber \\&\int _0^1F(t)\frac{(1+t)(1+\cos {\theta })+t^{2n-1}(1+\cos 2n\theta )+t^{2n}(1+\cos (2n-1)\theta )}{1+t^2+2t\cos \theta }\mathrm{d}t\nonumber \\&\qquad -\Bigg (\int _0^1F(t)C(t,\theta )\mathrm{d}t\Bigg )^2. \end{aligned}$$

(32)

The Cauchy–Schwarz inequality for integrals implies

$$\begin{aligned}&2\int _0^1F(t)\frac{\frac{1}{2}(1-\cos {\theta })+t^{2n-1}(1-\cos 2n\theta )+t^{2n}(1-\cos (2n-1)\theta )}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t\nonumber \\&\int _0^1F(t)\frac{(1+t)(1+\cos {\theta })+t^{2n-1}(1+\cos 2n\theta )+t^{2n}(1+\cos (2n-1)\theta )}{1+t^2+2t\cos \theta }\mathrm{d}t\nonumber \\&\qquad -\Bigg (\int _0^1F(t)C(t,\theta )\mathrm{d}t\Bigg )^2\nonumber \\&\quad \ge 2\Bigg (\int _0^1F(t)\sqrt{\frac{\frac{1}{2}(1-\cos {\theta })+t^{2n-1}(1-\cos 2n\theta )+t^{2n}(1-\cos (2n-1)\theta )}{(1+t)(1+t^2+2t\cos \theta )}}\nonumber \\&\sqrt{\frac{(1+t)(1+\cos {\theta })+t^{2n-1}(1+\cos 2n\theta )+t^{2n}(1+\cos (2n-1)\theta )}{1+t^2+2t\cos \theta }}\mathrm{d}t\Bigg )^2 \nonumber \\&\qquad -\Bigg (\int _0^1F(t)C(t,\theta )\mathrm{d}t\Bigg )^2. \end{aligned}$$

(33)

Putting $a_1^2=\frac{\frac{1}{2}(1-\cos {\theta })}{(1+t)(1+t^2+2t\cos \theta )},$$b_1^2=\frac{(1+t)(1+\cos {\theta })}{1+t^2+2t\cos \theta }$ and so on in the inequality

$$\begin{aligned} \sqrt{a_1^2+a_2^2+a_3^2}\sqrt{b_1^2+b_2^2+b_3^2}\ge {|a_1b_1|+|a_2b_2|+|a_3b_3|}, \end{aligned}$$

we get

$$\begin{aligned}&2\Bigg (\int _0^1F(t)\sqrt{\frac{\frac{1}{2}(1-\cos {\theta })+t^{2n-1}(1-\cos 2n\theta )+t^{2n}(1-\cos (2n-1)\theta )}{(1+t)(1+t^2+2t\cos \theta )}}\nonumber \\&\sqrt{\frac{(1+t)(1+\cos {\theta })+t^{2n-1}(1+\cos 2n\theta )+t^{2n}(1+\cos (2n-1)\theta )}{1+t^2+2t\cos \theta }}\mathrm{d}t\Bigg )^2\nonumber \\&\quad -\Bigg (\int _0^1F(t)C(t,\theta )\mathrm{d}t\Bigg )^2 \nonumber \\&\ge 2\left( \int _0^1F(t)\frac{\sqrt{\frac{1}{2}(1+t)}|\sin \theta |+t^{2n-1}|\sin 2n\theta |+t^{2n}|\sin (2n-1)\theta |}{(1+t^2+2t\cos \theta )\sqrt{1+t}} \mathrm{d}t\right) ^2 \nonumber \\&\quad -\Bigg (\int _0^1F(t)C(t,\theta )\mathrm{d}t\Bigg )^2. \end{aligned}$$

(34)

On the other hand, we have

$$\begin{aligned}&2\left( \int _0^1F(t)\frac{\sqrt{\frac{1}{2}(1+t)}|\sin \theta |+t^{2n-1}|\sin 2n\theta |+t^{2n}|\sin (2n-1)\theta |}{(1+t^2+2t\cos \theta )\sqrt{1+t}} \mathrm{d}t\right) ^2\nonumber \\&\quad -\Bigg (\int _0^1F(t)C(t,\theta )\mathrm{d}t\Bigg )^2 \nonumber \\&\ge \Bigg (\int _0^1F(t)\frac{\sqrt{(1+t)}|\sin \theta |+\sqrt{2}t^{2n-1}|\sin 2n\theta |+\sqrt{2}t^{2n}|\sin (2n-1)\theta |}{(1+t^2+2t\cos \theta )\sqrt{1+t}} \mathrm{d}t\Bigg )^2 \nonumber \\&\quad -\Bigg (\int _0^1F(t)C(t,\theta )\mathrm{d}t\Bigg )^2 \nonumber \\&\ge \Bigg (\int _0^1F(t)\frac{|\sin \theta |+t^{2n-1}|\sin 2n\theta |+t^{2n}|\sin (2n-1)\theta |}{1+t^2+2t\cos \theta } \mathrm{d}t\Bigg )^2 \nonumber \\&\quad -\Bigg (\int _0^1F(t)\frac{\sin \theta +t^{2n-1}\sin 2n\theta +t^{2n}\sin (2n-1)\theta }{1+t^2+2t\cos \theta }\mathrm{d}t\Bigg )^2\ge 0, \ \ \theta \in \left[ 0,\frac{\pi }{2}\right] .\nonumber \\ \end{aligned}$$

(35)

Finally, the inequalities (30)–(35) imply that

$$\begin{aligned} \Phi ^2(0)-|\Phi (\theta )|^2\ge 0, \ \theta \in \left[ 0,\frac{\pi }{2}\right] , \end{aligned}$$

and consequently inequalities (28) and (27) hold in case $\theta \in [-\frac{\pi }{2},\frac{\pi }{2}].$ $\square$

Theorem 2

If n is a natural number, $n\ge 27$ and $\alpha \in (0,1),$ then the following inequality holds:

$$\begin{aligned} |A_{2n-1}(\alpha ,\mathrm{e}^{i\theta })|\le |A_{2n-1}(\alpha ,1)|, \ \ \text {for all} \ \theta \in \left[ -\frac{2\pi }{3},-\frac{\pi }{2}\right] \cup \left[ \frac{\pi }{2},\frac{2\pi }{3}\right] .\nonumber \\ \end{aligned}$$

(36)

Proof

Equality (29) can be rewritten as follows:

$$\begin{aligned}&2\Big (\Phi ^2(0)-|\Phi (\theta )|^2\Big ) \nonumber \\&\quad =\int _0^1F(t)\Big (B(t,0)-B(t,\theta )\Big )\mathrm{d}t\int _0^1F(v)\Bigg (\frac{2}{\alpha }+B(v,0) +B(v,\theta )\Bigg )\mathrm{d}v\nonumber \\&\qquad +\int _0^1F(v)\Big (B(v,0)-B(v,\theta )\Big )\mathrm{d}v\int _0^1F(t)\Bigg (\frac{2}{\alpha }+B(t,0)+B(t,\theta )\Bigg )\Bigg ]\mathrm{d}t \nonumber \\&\qquad -2\int _0^1\int _0^1F(t)F(v)C(t,\theta )C(v,\theta )\mathrm{d}t\mathrm{d}v. \end{aligned}$$

(37)

Since $2\Big (\Phi ^2(0)-|\Phi (\theta )|^2\Big )$ defines an even function to prove (36), it is enough to show that

$$\begin{aligned} \Phi ^2(0)-|\Phi (\theta )|^2\ge 0, \ \ \theta \in \Bigg [\frac{\pi }{2},\frac{2\pi }{3}\Bigg ]. \end{aligned}$$

The inequality between the arithmetic and geometric means and the condition $\alpha \in (0,1]$ imply

$$\begin{aligned}&2\Big (\Phi ^2(0)-|\Phi (\theta )|^2\Big )\ge 2\Bigg [\int _0^1F(t)\Bigg (B(t,0)-B(t,\theta )\Bigg )\mathrm{d}t\int _0^1F(v)\Bigg (2+B(v,0) \nonumber \\&\qquad +B(v,\theta )\Bigg )\mathrm{d}v\int _0^1F(v)\Big (B(v,0)-B(v,\theta )\Big )\mathrm{d}v\int _0^1F(t)\Big (2+B(t,0)+B(t,\theta )\Big )\mathrm{d}t\Bigg ]^{\frac{1}{2}} \nonumber \\&\qquad -2\int _0^1\int _0^1F(t)F(v)C(t,\theta )C(v,\theta )\mathrm{d}t\mathrm{d}v \nonumber \\&\quad =2\int _0^1F(t)\Big (B(t,0)-B(t,\theta )\Big )\mathrm{d}t\int _0^1F(t)\Bigg (2+B(t,0)+B(t,\theta \Bigg )\mathrm{d}t \nonumber \\&\qquad -2\Bigg (\int _0^1F(t)C(t,\theta )\mathrm{d}t\Bigg )^2 . \end{aligned}$$

(38)

Lemma 5 and inequality (38) lead to

$$\begin{aligned}&\Phi ^2(0)-|\Phi (\theta )|^2 \nonumber \\&\quad \ge \int _0^1F(t)\frac{\frac{12}{25}(1-\cos {\theta })+t^{2n-1}(1-\cos 2n\theta )+t^{2n}(1-\cos (2n-1)\theta )}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t \nonumber \\&\int _0^1F(t) \frac{\frac{25}{12}(1+t)(1+\cos {\theta })+2t^{2n-1}(1+\cos 2n\theta )+2t^{2n}(1+\cos (2n-1)\theta )}{1+t^2+2t\cos \theta }\mathrm{d}t\nonumber \\&\qquad -\Bigg (\int _0^1F(t)C(t,\theta )\mathrm{d}t\Bigg )^2. \end{aligned}$$

(39)

We apply twice the Cauchy–Schwarz inequality just like in the proof of the previous theorem and we get

$$\begin{aligned}&\int _0^1F(t)\frac{\frac{12}{25}(1-\cos {\theta })+t^{2n-1}(1-\cos 2n\theta )+t^{2n}(1-\cos (2n-1)\theta )}{(1+t)(1+t^2+2t\cos \theta )}\mathrm{d}t \nonumber \\&\int _0^1F(t) \frac{\frac{25}{12}(1+t)(1+\cos {\theta })+2t^{2n-1}(1+\cos 2n\theta )+2t^{2n}(1+\cos (2n-1)\theta )}{1+t^2+2t\cos \theta }\mathrm{d}t\nonumber \\&\quad \ge \Bigg (\int _0^1F(t)\frac{\sqrt{1+t}|\sin {\theta }|+\sqrt{2}t^{2n-1}|\sin {2n}\theta |+\sqrt{2}t^{2n}|\sin ({2n-1})\theta |}{(1+t^2+2t\cos \theta )\sqrt{1+t}}\mathrm{d}t\Bigg )^2.\nonumber \\ \end{aligned}$$

(40)

The following inequality obviously holds:

$$\begin{aligned}&\Bigg (\int _0^1F(t)\frac{\sqrt{1+t}|\sin {\theta }|+\sqrt{2}t^{2n-1}|\sin {2n}\theta |+\sqrt{2}t^{2n}|\sin ({2n-1})\theta |}{(1+t^2+2t\cos \theta )\sqrt{1+t}}\mathrm{d}t\Bigg )^2 \\&\quad \ge \Bigg (\int _0^1F(t)\frac{\sin \theta +t^{2n-1}\sin 2n\theta +t^{2n}\sin (2n-1)\theta }{1+t^2+2t\cos \theta }\mathrm{d}t\Bigg )^2, \end{aligned}$$

and is equivalent to

$$\begin{aligned}&\Bigg (\int _0^1F(t)\frac{\sqrt{1+t}|\sin {\theta }|+\sqrt{2}t^{2n-1}|\sin {2n}\theta |+\sqrt{2}t^{2n}|\sin ({2n-1})\theta |}{(1+t^2+2t\cos \theta )\sqrt{1+t}}\mathrm{d}t\Bigg )^2\nonumber \\&\quad \ge \Bigg (\int _0^1F(t)C(t,\theta )\mathrm{d}t\Bigg )^2. \end{aligned}$$

(41)

Finally, (39)–(41) imply $\Phi ^2(0)-|\Phi (\theta )|^2\ge 0, \theta \in [\frac{\pi }{2},\frac{2\pi }{3}],$ and the proof is done. $\square$

Theorem 3

If n is a natural number, $n\ge 27$ and $\alpha \in [\frac{1}{3},1),$ then the following inequality holds:

$$\begin{aligned} {A_{2n-1}(\alpha ,1)}\ge |A_{2n-1}(\alpha ,\mathrm{e}^{i\theta })|, \ \ \text {for all} \ \theta \in \left[ -\pi ,-\frac{2\pi }{3}\right] \cup \left[ \frac{2\pi }{3},\pi \right] .\qquad \end{aligned}$$

(42)

Proof

We will use the Taylor formula with an integral remainder. Let $f:(-1,1)\rightarrow {\mathbb {R}}$ be a 2n times derivabile function. If $x\in (-1,1),$ then

$$\begin{aligned} f(x)= & {} f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\cdots +\frac{f^{(2n-1)}(0)}{(2n-1)!}x^{2n-1} \\&+\frac{x^{2n}}{(2n-1)!}\int _0^1(1-t)^{2n-1}f^{(2n)}(xt)\mathrm{d}t. \end{aligned}$$

Let f be the function defined by $f:(-1,1)\rightarrow {\mathbb {R}}, \ \ f(x)=(1+x)^\alpha , \ \ \alpha \in (0,1)$ and we get

$$\begin{aligned} (1+x)^\alpha= & {} 1+\frac{\alpha }{1!}x+\frac{\alpha (\alpha -1)}{2!}x^2+\cdots +\frac{\alpha (\alpha -1)\ldots (\alpha -2n+2)}{(2n-1)!}x^{2n-1} \\&+x^{2n}\frac{\alpha (\alpha -1)\ldots (\alpha -2n+1)}{(2n-1)!}\int _0^1(1-t)^{2n-1}(1+xt)^{\alpha -2n}\mathrm{d}t. \end{aligned}$$

Since the mapping $g:U\rightarrow {\mathbb {C}}, \ \ g(z)=(1+z)^\alpha$ is well defined (we take the principal branch of the multi valued function), it follows that the equality

$$\begin{aligned} (1+z)^\alpha= & {} 1+\frac{\alpha }{1!}z+\frac{\alpha (\alpha -1)}{2!}z^2+\cdots +\frac{\alpha (\alpha -1)\ldots (\alpha -2n+2)}{(2n-1)!}z^{2n-1} \\&+z^{2n}\frac{\alpha (\alpha -1)\ldots (\alpha -2n+1)}{(2n-1)!}\int _0^1(1-t)^{2n-1}(1+zt)^{\alpha -2n}\mathrm{d}t \end{aligned}$$

holds for every $z\in {U}.$ The mapping $g:U\rightarrow {\mathbb {C}}, \ \ g(z)=(1+z)^\alpha$ is radially continuous and so we infer that

$$\begin{aligned}&(1+\mathrm{e}^{i\theta })^\alpha =1+\frac{\alpha }{1!}\mathrm{e}^{i\theta }+\frac{\alpha (\alpha -1)}{2!}\mathrm{e}^{2i\theta }+\ldots +\frac{\alpha (\alpha -1)\cdots (\alpha -2n+2)}{(2n-1)!}\mathrm{e}^{(2n-1)i\theta } \\&\quad +\mathrm{e}^{2ni\theta }\frac{\alpha (\alpha -1)\ldots (\alpha -2n+1)}{(2n-1)!}\int _0^1(1-t)^{2n-1}(1+\mathrm{e}^{i\theta }t)^{\alpha -2n}\mathrm{d}t, \ \theta \in (-\pi ,\pi ). \end{aligned}$$

Taking the absolute value, this equality implies that

$$\begin{aligned}&|A_{2n-1}(\alpha ,\mathrm{e}^{i\theta })| \\&\quad =\Bigg |1+\frac{\alpha }{1!}\mathrm{e}^{i\theta }+\frac{\alpha (\alpha -1)}{2!}\mathrm{e}^{2i\theta }+\cdots +\frac{\alpha (\alpha -1)\cdots (\alpha -2n+2)}{(2n-1)!}\mathrm{e}^{(2n-1)i\theta }\Bigg | \\&\quad \le \frac{\alpha (1-\alpha )(2-\alpha )\cdots (2n-1-\alpha )}{(2n-1)!}\int _0^1(1-t)^{2n-1}\Bigg |1+\mathrm{e}^{i\theta }t\Bigg |^{\alpha -2n}\mathrm{d}t \\&\qquad +\Bigg |1+\mathrm{e}^{i\theta } \Bigg |^\alpha \le \Bigg |1+\mathrm{e}^{i\theta } \Bigg |^\alpha +\alpha (1-\alpha )\left( 1-\frac{\alpha }{2}\right) \left( 1-\frac{\alpha }{3}\right) \ldots \left( 1- \frac{\alpha }{2n-1}\right) \\&\qquad \int _0^1(1-t)^{2n-1}\Bigg |1+\mathrm{e}^{i\theta }t\Bigg |^{\alpha -2n}\mathrm{d}t, \ \theta \in (-\pi ,\pi ). \end{aligned}$$

On the other hand, we have

$$\begin{aligned}&A_{2n-1}(\alpha ,1) =1+\frac{\alpha }{1!}-\frac{\alpha (1-\alpha )}{2!}+\frac{\alpha (1-\alpha )(2-\alpha )}{3!} \\&\quad -\frac{\alpha (1-\alpha )(2-\alpha )(3-\alpha )}{4!}+\cdots +\frac{\alpha (1-\alpha )\ldots (2n-2-\alpha )}{(2n-1)!}\ge 1+\frac{\alpha (1+\alpha )}{2}. \end{aligned}$$

Thus to prove (42), we have to show that

$$\begin{aligned}&1+\frac{\alpha (1+\alpha )}{2}\ge \Bigg |1+\mathrm{e}^{i\theta } \Bigg |^\alpha +\alpha (1-\alpha )\left( 1-\frac{\alpha }{2}\right) \left( 1-\frac{\alpha }{3}\right) \cdots \left( 1-\frac{\alpha }{2n-1}\right) \nonumber \\&\int _0^1(1-t)^{2n-1}\Bigg |1+\mathrm{e}^{i\theta }t\Bigg |^{\alpha -2n}\mathrm{d}t, \quad \theta \in \left[ \frac{2\pi }{3},\pi \right] . \end{aligned}$$

(43)

We denote $x=-\cos \theta ,$ and the inequality (43) will be equivalent to

$$\begin{aligned}&1+\frac{\alpha (1+\alpha )}{2}\ge \Bigg (2-2x\Bigg )^\frac{\alpha }{2}+\alpha (1-\alpha )\left( 1-\frac{\alpha }{2}\right) \left( 1-\frac{\alpha }{3}\right) \cdots \left( 1- \frac{\alpha }{2n-1}\right) \\&\int _0^1\Bigg (\frac{1-t}{\sqrt{1+t^2-2tx}}\Bigg )^{2n-1}\Bigg (\sqrt{1+t^2-2tx}\Bigg )^{\alpha -1}\mathrm{d}t, \quad x\in \left[ \frac{1}{2},1\right] . \end{aligned}$$

This inequality can be rewritten in the following form:

$$\begin{aligned}&1+\alpha \ge \frac{(2-2x)^\frac{\alpha }{2}-1}{\frac{\alpha }{2}}+2(1-\alpha )\left( 1-\frac{\alpha }{2}\right) \left( 1-\frac{\alpha }{3}\right) \cdots \left( 1- \frac{\alpha }{2n-1}\right) \nonumber \\&\int _0^1\Bigg (\frac{1-t}{\sqrt{1+t^2-2tx}}\Bigg )^{2n-1}(\sqrt{1+t^2-2tx})^{\alpha -1}\mathrm{d}t,\quad x\in \left[ \frac{1}{2},1\right] . \end{aligned}$$

(44)

It is easily seen that the expression

$$\begin{aligned}&2(1-\alpha )\left( 1-\frac{\alpha }{2}\right) \left( 1-\frac{\alpha }{3}\right) \nonumber \\&\quad \ldots \left( 1- \frac{\alpha }{2n-1}\right) \int _0^1\Bigg (\frac{1-t}{\sqrt{1+t^2-2tx}}\Bigg )^{2n-1}\Bigg (\sqrt{1+t^2-2tx}\Bigg )^{\alpha -1}\mathrm{d}t \end{aligned}$$

is decreasing with respect to n and increasing with respect to x;  moreover, the expression $\frac{(2-2x)^\frac{\alpha }{2}-1}{\frac{\alpha }{2}}$ is negative. Consequently, to prove (44) it is enough to show that the following inequality holds:

$$\begin{aligned} 1+\alpha \ge 2(1-\alpha )\left( 1-\frac{\alpha }{2}\right) \left( 1-\frac{\alpha }{3}\right) \ldots \left( 1-\frac{\alpha }{53}\right) \int _0^1\Bigg (1-t\Bigg )^{\alpha -1}\mathrm{d}t.\qquad \end{aligned}$$

(45)

A simple calculation shows that (45) is equivalent to

$$\begin{aligned} 1+\alpha \ge \frac{2}{\alpha }(1-\alpha )\left( 1-\frac{\alpha }{2}\right) \left( 1-\frac{\alpha }{3}\right) \ldots \left( 1-\frac{\alpha }{53}\right) . \end{aligned}$$

(46)

Since the left hand side is increasing and the right hand side is decreasing provided that $\alpha \in (0,1),$ it follows that to prove inequality (46) for $\alpha \in [\frac{1}{3},1]$ it is enough to be checked in case $\alpha =\frac{1}{3},$ that is,

$$\begin{aligned} \frac{4}{3}\ge 6\left( 1-\frac{1}{3}\right) \left( 1-\frac{1}{3\cdot 2}\right) \left( 1-\frac{1}{3\cdot 3}\right) \cdots \left( 1-\frac{1}{3\cdot 53}\right) . \end{aligned}$$

According to Lemma 6, this inequality holds and the proof is done. $\square$

4 Concluding Remarks

The following two corollaries are the proof of the Brannan conjecture in two different particular cases.

Theorems 1,  2 and the result of [6] imply the following corollary.

Corollary 1

If $x\in {\mathbb {C}}$ with $|\arg {x}|\le \frac{2\pi }{3},$ and $|x|=1,$ then the inequality

$$\begin{aligned} |A_{2n-1}(\alpha ,x)|\le {A_{2n-1}}(\alpha ,1) \end{aligned}$$

holds for every $\alpha \in (0,1), \ n\in {\mathbb {N}}^*.$

Theorems 1,  2 and  3 and the result of [6] imply the following corollary. This corollary is the solution of the Brannan conjecture in case $\alpha \in [\frac{1}{3},1).$

Corollary 2

The inequality

$$\begin{aligned} |A_{2n-1}(\alpha ,x)|\le {A_{2n-1}}(\alpha ,1) \end{aligned}$$

holds for every $\alpha \in [\frac{1}{3},1),$ $n\in \mathbb {N^*}$ and $x\in {\mathbb {C}},$ with $|x|=1.$

The Brannan’s conjecture holds if $\alpha \in [\frac{1}{3},1)$ and remains an open question in case $\alpha \in (0,\frac{1}{3}).$