Bohr-type inequalities of analytic functions

In this paper, we investigate the Bohr-type radii for several different forms of Bohr-type inequalities of analytic functions in the unit disk, we also investigate the Bohr-type radius of the alternating series associated with the Taylor series of analytic functions. We will prove that most of the results are sharp.


Introduction and preliminaries
Bohr's inequality states that if is analytic in the unit disk D = {z ∈ C||z| < 1} and |f (z)| < 1 for all z ∈ D, then ∞ k=0 |a k ||z| k ≤ 1 (1.2) for all |z| ≤ 1 3 . This inequality was discovered by Bohr in 1914 [6]. Bohr actually obtained the inequality for |z| ≤ 1 6 , but subsequently later, Wiener, Riesz and Schur, independently established the inequality for |z| ≤ 1 3 and the constant 1/3 cannot be improved [12,16,17]. Other proofs were also given in [13,14]. The problem was considered by Bohr when he was working on the absolute convergence problem for Dirichlet series of the form a n n -s , but now it has become a very interesting problem. Bohr's idea naturally extends to functions of several complex variables [1,2,5,11] and a variety of results on Bohr's theorem in higher dimensions appeared recently.
The majorant series M f (z) = ∞ k=0 |a k ||z| k belongs to a very important class of series of non-negative terms. In analogy to the Bohr radius, there is also the notion of the Rogosinski radius [10,15], which is described as follows: If f (z) = ∞ k=0 a k z k is an analytic function in D such that |f (z)| < 1 in D, then, for every N ≥ 1, we have |s N (z)| < 1 in the disk |z| We remark that, for N = 1, this quantity is related to the classical Bohr sum in which f (0) is replaced by f (z). More recently, Kayumov and Ponnusamy [9] obtained the following result on the Bohr-Rogosinski radius for analytic functions.
Theorem A ( [9]) Suppose that f (z) = ∞ k=0 a k z k is analytic in the unit disk D and |f (z)| < 1 in D. Then where R N is the positive root of the equation 2(1 + r)r N -(1r) 2 = 0. The radius R N is the best possible. Moreover, where R N is the positive root of the equation (1 + r)r N -(1r) 2 = 0. The radius R N is the best possible.
Remark 1.1 By a simple calculation in Theorem A, we observe that R 1 = √ 5 -2 is unequal to 1 3 when |f (0)| is replaced by |f (z)| in Bohr's inequality. Therefore, it is interesting to note what will happen to the Bohr radius if we use higher order derivatives of f (z) to replace some Taylor coefficients of analytic functions in Bohr's inequality.
In this paper, we mainly study the Bohr-type radii for several forms of Bohr-type inequalities of analytic functions when the Taylor coefficients of classical Bohr inequality are partly replaced and when the Taylor coefficients of the classical Bohr inequality are completely replaced by the higher order derivatives of f (z), respectively. We obtain the Bohr-type radii under certain conditions. Moreover, we also discuss the Bohr-type radius of the alternating series associated with the Taylor series of analytic functions.
In order to establish our main results, we need the following lemmas, which will play the key role in proving the main results of this paper.

Main results
We first provide a result involves computing Bohr-type radius for the analytic functions f (z) for which |a 0 | and |a 1 | are replaced by |f (z)| and |f (z)|, respectively.
The radius r = √ 17-3 4 is the best possible.
Proof By assumption, f (z) = ∞ k=0 a k z k is analytic in D and |f (z)| < 1 in D. Since f (0) = a 0 , by the Schwarz-Pick lemma, we obtain, for z ∈ D, Thus it follows from the above inequality and Lemma 1.2 that, for z = re iθ ∈ D, for k = 1, 2, . . . . Using these inequalities, we have where the second inequality holds for any r ∈ [0, 4 . The first part of the theorem follows.
Next, we discuss the Bohr-type radius when the coefficients of the series of missing series are completely replaced by the higher order derivatives.
The radius R N is the best possible.
Proof By simple calculations we can know that By assumption, f (z) = ∞ k=0 a k z k is analytic in D and |f (z)| < 1 in D. Since f (0) = a 0 , it follows from the Schwarz-Pick lemma and Lemma 1.4 that, for z = re iθ ∈ D, Using these inequalities, we have Now we split all this into two cases to prove that ν N (r) ≤ 0 for r ≤ R N . Case 1. r ≤ R N,1 , where R N,1 is the minimum positive root of the equation ϕ N (r) = (1 -2r)(1r) N-1r N-1 = 0. Since r N-1 -(1 -2r)(1r) N-1 ≤ 0 and |a 0 | < 1, we have The first part of the theorem follows.
Finally, allowing a → 1 in (2.6) shows that Eq. (2.5) is larger than 1 if r > R N . This proves the sharpness.

Corollary 2.3 Suppose that f (z) = ∞ k=0 a k z k is analytic in D and |f (z)| < 1 in D. Then
where R N is the positive root of the equation (1 + r)(1 -2r)(1r) N-1r N = 0. The radius R N is the best possible.
Proof By simple calculations we can know that In analogy to the calculation of Theorem 2.2, we have So (2.7) is smaller than or equal to 1 provided ω N (r) ≤ 1, where where R N is as in the statement of the theorem.
To show the sharpness of the number R N , we let a ∈ [0, 1) and consider the function For this function, we find that

8) is larger than 1 if and only if
1a 2 -1 + 2ar + r 2 -2ar 3 (1ar) N-2 + a N-1 r N > 0. (2.9) In analogy to the processing methods of Theorem 2.2. After elementary calculation, we find that allowing a → 1 in (2.9), it follows that Eq. (2.8) is larger than 1 if r > R N . This proves the sharpness and we complete the proof of Corollary 2.3.
Applying a method similar to Theorem 2.2, we may obtain the following corollary.
The radius r = is the best possible.
In analogy to Theorem C, we now consider the Bohr-type radius when conditions of |f (z)| < 1 are replaced by Re f (z) ≤ 1 and f (0) = a 0 is positive. where R n is the positive root of the equation ϕ n (r) = 0, ϕ n (r) = r n+1 + 3r n + r -1. The radius R n is the best possible.
Finally, allowing a → 1 in (2.14) shows that Eq. (2.13) is larger than 1 if r > R n . This proves the sharpness.
Setting n = 1 in Theorem 2.5, we have the following corollary.
where the radius √ 5 -2 is the best possible.
Remark 2.7 By simple calculation, we can know the Bohr-type radius in Theorem 2.5 with the condition of Re f (z) ≤ 1 and f (0) = a 0 > 0 is the same as the condition of |f (z)| < 1.
Finally, we consider a new Bohr-type radius of the alternating series associated the Taylor series of analytic functions where |a 0 | is replaced by |f (z)|. We have k=0 a k z k is analytic in the unit disk D and |f (z)| < 1 in D. Then The radius r = √ 2 -1 is the best possible.
To find a lower bound for R f (z), we consider the following chain of relations: where the last inequality is obtained by a simple calculation.