Bohr-type inequalities for bounded analytic functions of Schwarz functions

: In this paper, some new versions of Bohr-type inequalities for bounded analytic functions of Schwarz functions are established. Most of these inequalities are sharp. Some previous inequalities are generalized.


Introduction
Let B denote the class of analytic functions f (z) = ∞ k=0 a k z k defined in the unit disk D := {z ∈ C : We call the Bohr sum of f (z).The well known Bohr radius problem is to find r 0 , such that holds for |z| ≤ r 0 .The constant r 0 = 1/3 is sharp, which is called the Bohr radius.The radius was originally obtained in 1914 by Bohr [13] with 1/6.Later, Wiener, Riesz and Schur established the inequality (1.1) for r = |z| ≤ 1/3 and showed that the constant 1/3 cannot be improved [22,25,26].
There are lots of works about the classical Bohr inequality and its generalized forms.Ali et al., [8] and Kayumov and Ponnusamy [15] considered the problem of Bohr radius for the classes of even and odd analytic functions and for alternating series, respectively.In [19], the authors generalized and improved several Bohr inequalities.In [21], several Bohr-type inequalities were obtained when the Taylor coefficients of classical Bohr inequality are partly replaced by higher order derivatives of f .It is worth pointing out that Bohr's radius problem deal with analytic functions from unit disk D into D initially, but later it was generalized to mappings from D to punctured disk [4] or other domains [2].For more discussion on the Bohr radius for analytic functions [3,7,9,27].
The Bohr-Rogosinski sum R f N (z) of f ∈ B is defined by Observe that if N = 1 and f (z) is replaced by f (0), then the Bohr-Rogosinski sum is the Bohr sum.The corresponding Bohr-Rogosinski radius problem is to find R N , such that Recently, Kayumov and Ponnusamy [16] have given the Bohr-Rogosinski radius of f .In [21], the author also solved some problem of the Bohr-Rogosinski radius.Let S N (z) = N−1 k=0 a k z k denotes the partial sums of f .The corresponding Rogosinski radius is [18,23,24].It is obvious that Hence, the Rogosinski radius is related to the Bohr-Rogosinski radius. Let m) (0) 0} be the classes of Schwarz functions, where m ∈ N = {1, 2, • • • }.Our aim of this article is to generalize or improve many versions of Bohr-type inequalities for bounded analytic functions of Schwarz functions.
The paper is organized as follows.In Section 2, we state some lemmas.In Section 3, we present many theorems which improve several versions of Bohr-Rogosinski inequalities and Bohr's type inequalities for bounded analytic functions.There are some corollaries and an open problem in Section 4.

Some Lemmas
In order to establish our main results, we need the following some lemmas which will play the key role in proving the main results of this paper.
and equality holds for distinct z 1 , z 2 ∈ D if and only if φ is a Möbius transformation.In particular, and equality holds for some z ∈ D if and only if φ is a Möbius transformation.
The proof is simple, we omit it.
Lemma 2.4.There is a unique root ξ m of the equation and a unique root α m,n of the equation for r ∈ (0, 1) and m, n ∈ N, respectively.Furthermore, α m,n ≤ ξ m for m ≥ n.
Then we have Observe that k(0)k(1) < 0. Thus the monotonicity of k(r) implies that there is an α m,n that is the unique root of (2.2).

Main results
Theorem 3.1.Suppose that f where R m,n,N is the unique root in (0, 1) of the equation and the radius R m,n,N cannot be improved.
By using inequality (3.3) and Lemma 2.2, we have Obviously, it is enough to show that 2r nN (1 Then it is easy to verify that g(0)g(1) < 0 and g(r) is a continuous and increasing function of r ∈ [0, 1].Thus R m,n,N is the unique root of g(r) and g(r) ≤ 0 holds for r ≤ R m,n,N .
Next we show the radius R m,n,N is sharp.For a ∈ [0, 1), let Taking z = r, substituting (3.4) into the left side of inequality (3.1), then we have Now we just need to show that if r > R m,n,N , then there exists an a, such that the right side of (3.5) is greater than 1.That is Therefore, if r > R m,n,N , then there exists an a, such that inequality (3.6) holds.
where α m,n is the unique root in (0, 1) of the equation The radius α m,n cannot be improved. where and ξ m is the unique root in (0, 1) of the equation It is obvious that Φ m,n (1, r) is monotonically increasing function of r ∈ [0, 1).By the hypothesis or Lemma 2.4, we have Furthermore, observe that Φ m,n (a, r) is a monotonically increasing function of a ∈ [0, 1] for each fixed r ∈ [0, 1).Thus Therefore, by inequalities (3.8) and (3.9), we obtain inequality (3.7).
where s ∈ N, β m,n,s is the unique root in (0, 1) of the equation The radius β m,n,s cannot be improved.Proof.Using inequality (3.3) and Lemma 2.2, we have It is sufficient for us to prove that the right side of (3.14) is less than or equals to 1 for r ≤ β m,n,s .Actually, we just need to prove φ(r) ≤ 0 for r ≤ β m,n,s , where Let h(r) = r m + 3r sn + r m+sn − 1.Then it is easy to verify that h(0)h(1) < 0, h(r) is a continuous and increasing function of r ∈ [0, 1].Thus β m,n,s is unique root of h(r) and h(r) ≤ 0 holds for r ≤ β m,n,s .Thus φ(r) ≤ 0 for r ≤ β m,n,s .
To show the radius β m,n,s is sharp, we consider the functions ω m (z), ω n (z) and f (z) is the same as (3.4).Taking z = r, the left side of inequality (3.12) reduces to Next, we need to show that if r > β m,n,s , then there exists an a, such that the right side of (3.15) is bigger than 1.That is r m+ns a s+1 + 2r ns a s + r ns a s−1 + r m − 1 > 0. (3.16) Let A 3 (a, r) = r m+ns a s+1 + 2r ns a s + r ns a s−1 + r m − 1.
Observe that A 3 (a, r) is a continuous and increasing function for a ∈ [0, 1).It follows that A 3 (a, r) ≤ A 3 (1, r) = r m+ns + 3r ns + r m − 1 = h(r) holds for r ∈ (0, 1).Furthermore, the monotonicity of h(r) leads to that if r > β m,n,s , then A 3 (1, r) > 0. Hence, by the continuity of A 3 (a, r), if r > β m,n,s , we have Therefore, if r > β m,n,s , then there exists an a, such that inequality (3.16) holds.
Proof.On the one hand, by the assumption, we have r 2n 1 − r 2n .Now we need to show that above inequality is smaller than or equal to 1.It is sufficient for us to prove ψ(r) ≤ 0, where Furthermore, it is easy to verify that r m + 3r 2n + r m+2n − 1 is increasing on r ∈ [0, 1) and have a unique zero γ m,n .Therefore, we have ψ(r) ≤ 0 for r ≤ γ m,n .
On the other hand, we have It is obvious that the last item of above is greater than or equals −1 for all r ≤ γ m,n .We complete the proof.
where R m,1,N is the unique root in (0, 1) of the equation and the radius R m,1,N cannot be improved.
where R 1,n,N is the unique root in (0, 1) of the equation and the radius R 1,n,N cannot be improved.
where α m,1 is the unique root in (0, 1) of the equation The radius α m,1 cannot be improved.
where α m,m is the unique root in (0, 1) of the equation The radius α m,m cannot be improved.
where β m,m,s is the unique root in (0, 1) of the equation The radius β m,m,s cannot be improved.

Conclusions
We obtain some new versions of Bohr-type inequalities for bounded analytic functions of Schwarz functions by replacing the variable z by Schwarz functions in function's power series expansions.we conclude that most of the corresponding Bohr radii are exact.These inequalities generalize the classical Bohr inequality and some earlier results on the Bohr inequality.

Theorem 3 . 2 .
Suppose that f (z) = ∞ k=0 a k z k ∈ B, a := |a 0 | and ω m ∈ B m , ω n ∈ B n with m, n ∈ N and m ≥ n.Then we have

Corollary 4 . 1 .
Suppose that f (z) = ∞ k=0 a k z k ∈ B, a := |a 0 | and ω ∈ B m for m ∈ N. Then we have

Corollary 4 . 2 .
Suppose that f (z) = ∞ k=0 a k z k ∈ B, a := |a 0 | and ω ∈ B n for n ∈ N. Then we have

Corollary 4 . 5 .
Suppose that f (z) = ∞ k=0 a k z k ∈ B, a := |a 0 | and ω ∈ B m for m ∈ N. Then we have