# Some Bohr-Type Inequalities for Bounded Analytic Functions

Authors: Hong Wang
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, some new versions of Bohr-type inequalities with one parameter or involving convex combination for bounded analytic functions of Schwarz function are established. Some previous inequalities are generalized. All the results are sharp.


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## Body

## 1. Introduction and Preliminaries

Letℍ denote the class of analytic functions fz=∑k=0∞akzk defined in the unit disk D≔z∈ℂ:z<1 such that fz<1 in D. The Bohr’s theorem states that if fz=Σk=0∞akzk∈ℍ, then(1)∑k=0∞akzk≤1,forz≤13,the value 1/3 is the best possible, and the inequality (1) is called classical Bohr inequality. In 1914, Bohr originally established the inequality (1) for z<1/6 [1]. But subsequently later, Wiener, Riesz, and Schur independently proved the inequality (1) for z<1/3, and the value is proved to be sharp [2–4].Bohr phenomenon and Bohr radius problem are very interesting. Many generalized forms are studied and a lot of Bohr-type inequalities are obtained. Kayumov considered the Bohr-type inequalities for odd analytic functions and some other classes of analytic function in [5]. Bhowmik and Das [6] studied Bohr’s phenomenon for subordinating families of certain univalent functions. Huang et al. [7] refined the Bohr inequality by allowing Schwarz function in place of the initial variable of the functions. Hu et al. [8, 9] established some Bohr inequalities with one parameter or involving convex combination. Abu-Muhanna [10] investigated the Bohr phenomenon for the class of analytic functions from the unit disk into the punctured unit disk.Some authors considered the Bohr phenomenon for functions defined on other domains, such as concave-wedge domain [11, 12], convex domain [13], and the exterior of a compact domain [14].The Bohr-type inequalities were also considered in other branches of mathematics. The analogous Bohr’s radius was also studied for K-quasiconformal harmonic mappings by Liu and Ponnusamy [15]. Djakov and Ramanujan [16], Aizenberg [17], and Bayart et al. [18] generalized the Bohr inequality to the case of higher dimensions. Abu-Muhanna and Gunatillake [19] considered Bohr phenomenon in weighted Hardy-Hilbert spaces. Blasco [20] obtained the Bohr’s radius of a Banach space. In [21], Paulsen explored Bohr’s radius problem in Banach algebras. For more discussion on the Bohr-type inequalities, we can refer to [22–25].We recall some results as follows.Theorem 1 (see [26]).
Supposefz=∑k=0∞akzk∈ℍ, N∈ℕ=1,2,3,…, then(2)fz+∑k=N∞akzk≤1,forz≤α,where α is the positive root of the equation(3)21+rrN−1−r2=0.The radiusα is the best possible. Moreover,(4)fz2+∑k=N∞akrk≤1,forz≤β,where β is the positive root of the equation(5)1+rrN−1−r2=0.The radiusβ is the best possible.Theorem 2 (see [27]).
Supposefz=∑k=0∞akzk∈ℍ, and s∈ℕ, then(6)fz+∑k=1∞askzsk≤1,forz≤γ,where γ is the positive root of the equation(7)rs+1+3rs+r−1.
The radiusγ is the best possible.Theorem 3 (see [27]).
Supposefz=∑k=0∞akzk∈ℍ, then(8)fz+f′z‖z+∑k=2∞ak‖zk≤1,forz≤17−34,where 17−3/4 is the best possible.Theorem 4 (see [28]).
Supposefz=∑k=0∞akzk∈ℍ, a≔a0, then(9)∑k=0∞ak‖zk+11+a+z1−|z|∑k=1∞ak2z2k≤1,forz≤12+a,and the numbers 1/2+a and 1/1+a cannot be improved. Moreover,(10)a2+∑k=1∞ak‖zk+11+a+z1−z∑k=1∞ak2z2k≤1,forz≤12,and the numbers 1/2+a and 1/1+a cannot be improved.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.Lemma 1 (see [27]) (Schwarz-Pick lemma).
Supposefz=∑k=0∞akzk∈ℍ, then(11)fz1−fz21−fz1¯fz2≤z1−z21−z1¯z2,forz1,z2∈D,and the equality holds for distinct z1,z2∈D if and only if fz is a Möbius transformation.In particular,(12)f′z≤1−fz21−z2,forz∈D,and the equality holds for some z∈D, if and only if fz is a Möbius transformation.Lemma 2 (see [29]).
Supposefz=∑k=0∞akzk∈ℍ, then ak≤1−a02, for k∈ℕ.Lemma 3.
For0≤x≤x0≤1, it holds that(13)fx≔x+A1−x2≤fx0,whenever0≤A≤12.The proof is simple. We omit it.Lemma 4 (see [28]).
Supposefz=∑k=0∞akzk∈ℍ, and a≔a0, then(14)∑k=0∞ak‖zk+11+a+z1−z∑k=1∞ak2z2k≤a+1−a2z1−z,forz∈0,1.LetSm≔ω∈ℍ:ω0=⋯=ωm−10=0,andωm0≠0, m∈ℕ=1,2,3,…, be the classes of Schwarz function [30, 31]. The aim of this paper is to establish some new versions of Bohr-type inequalities with one parameter or involving convex combination for bounded analytic functions of Schwarz function. Theorems 1–4 are generalized. All of our results are sharp.


## 2. Main Results

Theorem 5.
Supposefz=∑k=0∞akzk∈ℍ, a≔a0 and ωm∈Sm, ψn∈Sn for m,n∈ℕ. Then, for λ∈0,+∞ and N,s∈ℕ, we have(15)fωmz+λ∑k=N∞ask‖ψnzsk≤1,for z=r≤Rλ,N,m,n,s, where Rλ,N,m,n,s is the unique root in 0,1 of the equation(16)2λrNns+m+2λrNns+rm+rns−rm+ns−1=0,and the radius Rλ,N,m,n,s is the best possible.Proof.
By the Schwarz lemma and the Schwarz-Pick lemma, respectively, we obtain(17)ωmz≤zm,ψnz≤zn,fωmz≤a+rm1+arm,forz∈D.
Then, by Lemma2, we obtain(18)fωmz+λ∑k=N∞ask‖ψnzsk≤a+rm1+arm+λ1−a2rNns1−rns≔AN,m,n,sa,r,λ.
We just need to show thatAN,m,n,sa,r,λ≤1 holds for r≤Rλ,N,m,n,s. That is to prove Aa,r,λ≤0 for r≤Rλ,N,m,n,s, where(19)Aa,r,λ=a+rm1−rns+λ1−a2rNns1+arm−1+arm1−rns=a+rm−1−arm1−rns+λ1−a2rNns1+arm=1−arm−11−rns+λ1+arNns1+arm≤1−arm−11−rns+2λrNns1+rm.
Obviously, it is enough to show thatrm−11−rns+2λrNns1+rm≤0 holds for r≤Rλ,N,m,n,s. Let(20)gr=rm−11−rns+2λrNns1+rm=2λrNns+m+2λrNns+rm+rns−rm+ns−1.
Then, we have(21)g′r=2λNns+mrNns+m−1+2λNnsrNns−1+mrm−1+nsrns−1−m+nsrm+ns−1=2λNns+mrNns+m−1+2λNnsrNns−1+mrm−11−rns+nsrns−11−rm>0.
We claim thatgr is an increasing function of r∈0,1 for fixing λ∈0,+∞, N∈ℕ. Meanwhile, we observe that g0=−1<0 and g1=4λ>0. Then, there is a unique root r∗∈0,1 such that gr∗=0. Hence, gr≤0 holds for r≤r∗≔Rλ,N,m,n,s.
Now, we show that the radiusRλ,N,m,n,s is the best possible. Let a∈0,1,(22)ωmz=zm,ψnz=zn,fz=a+z1+az=a+1−a2∑k=1∞−ak−1zk,z∈D.
Takingz=r∈0,1, then the left side of inequality (15) reduces to(23)frm+λ∑k=N∞askrnsk=a+rm1+arm+λ1−a2aNs−1rNns1−asrns.
Now, we just need to show that ifr>Rλ,N,m,n,s, then there exists an a∈0,1 such that the right side of (23) is greater than 1. That is to prove la,r,λ>0 for r>Rλ,N,m,n,s, where(24)la,r,λ=a+rm1−asrns+λ1−a2aNs−1rNns1+arm−1+arm1−asrns=1−asrnsa+rm−1−arm+λ1−a2aNs−1rNns1+arm=1−arm−11−asrns+λ1+aaNs−1rNns1+arm.
Let(25)ja,r,λ=rm−11−asrns+λ1+aaNs−1rNns1+arm.
Observe that(26)lima⟶1−ja,r,λ=2λrNns+m+2λrNns+rns+rm+−rns+m−1=gr.
Thus, the monotonicity ofgr implies that if r>Rλ,N,m,n,s, then gr>0. Therefore, by the continuity of ja,r,λ for variable a, we have that if r>Rλ,N,m,n,s, there exists an a∈0,1 such that ja,r,λ>0, so does la,r,λ.
In Theorem5, setting s = 1, then we have Corollary 1.Corollary 1.
Supposefz=∑k=0∞akzk∈ℍ, a≔a0 and ωm∈Sm, ψn∈Sn for m,n∈ℕ. Then, for λ∈0,+∞ and N∈ℕ, it holds that(27)fωmz+λ∑k=N∞ak‖ψnzk≤1,for z=r≤Rλ,,N,m,n, where Rλ,N,m,n is the unique root in 0,1 of the equation(28)2λrNn+m+2λrNn+rm+rn−rm+n−1=0,and the radius Rλ,N,m,n is the best possible.Remark 1.
(1)
Ifωmz=ψnz=z, N = 1, and λ=1 in Corollary 1, then it reduces to the first part of Theorem 1.(2)
Ifωmz=ψnz=z, N = 1, and λ=1 in Theorem 5, then it reduces to Theorem 2.Theorem 6.
Supposefz=∑k=0∞akzk∈ℍ, a≔a0 and ωm∈Sm, ψn∈Sn for m,n∈ℕ. Then, for λ∈0,+∞ and N∈ℕ, we have(29)fωmz2+λ∑k=N∞ask‖ψnzsk≤1,for z=r≤Rλ,N,m,n,s∗, where Rλ,N,m,n,s∗ is the unique root in 0,1 of the equation(30)λrNns+m+λrNns+rns+rm−rns+m−1=0,and the radius Rλ,N,m,n,s∗ is the best possible.Proof.
Inequality (17) and Lemma 2 lead to that(31)fωmz2+λ∑k=N∞ask‖ψnzsk≤a+rm1+arm2+λ1−a2rNns1−rns≔BN,m,n,sa,r,λ.
We just need to show thatBN,m,n,sa,r,λ≤1 holds for r≤Rλ,N,m,n,s∗. That is to prove Ba,r,λ≤0 for r≤Rλ,N,m,n,s∗, where(32)Ba,r,λ=a+rm21−rns+λ1−a2rNns1+rm2−1+arm21−rns=1−rnsa+rm+1+arma+rm−1−arm+λ1−a2rNns1+arm2=1−a21−rns1+rmrm−1+λrNns1+arm2≤1−a21+rm1−rnsrm−1+λrNns1+rm.
It is sufficient for us to prove1−rnsrm−1+λrNns1+rm≤0 holds for r≤Rλ,N,m,n,s∗. Let(33)gr=1−rnsrm−1+λrNns1+rm=λrNns+m+λrNns+rm+rns−rns+m−1,then we have(34)g′r=λNns+mrNns+m−1+λNnsrNns−1+mrm−1+nsrns−1−m+nsrm+ns−1=λNns+mrNns+m−1+λNnsrNns−1+nsrns−11−rm+mrm−11−rns>0.
Obviously,gr is an increasing function of r∈0,1 for fix λ∈0,+∞, N,s∈ℕ, and we also have g0=−1<0, g1=2λ>0. Then, there is a unique root r∗∈0,1 such that gr∗=0. Hence, gr≤0 holds for r≤r∗≔Rλ,N,m,n,s∗.
Now, we show that the radiusRλ,N,m,n,s∗ is the best possible. We still consider the functions ωmz, ψnz, fz as in (22). Taking z=r∈0,1, then the left side of inequality (29) reduces to(35)frm2+λ∑k=N∞askrnsk=a+rm1+arm2+λ1−a2aNs−1rNns1−arns.
Now, we just need to show that ifr>Rλ,N,m,n,s∗, then there exists an a∈0,1 such that the right side of (35) is greater than 1. That is to prove la,r,λ>0, where(36)la,r,λ=a+rm21−arns+λ1−a2aNs−1rNns1+arm2−1+arm21−arns=1−arnsa+rm+1+arma+rm−1−arm+λ1−a2aNs−1rNns1+arm2=1−a21+rmrm−11−arns+λaNs−1rNns1+arm.
Let(37)ja,r,λ=rm−11−arns+λaNs−1rNns1+arm=λrNn+maNs+λrNnsaNs−1+rm+arns−arns+m−1.
Observe that(38)lima⟶1−ja,r,λ=λrNn+m+λrNns+rm+rns−rns+m−1=gr.
Thus, the monotonicity ofgr implies that if r>Rλ,N,m,n,s∗, then gr>0. Therefore, by the continuity of ja,r,λ for variable a, we have that if r>Rλ,N,m,n,s∗, there exists an a∈0,1 such that inequality ja,r,λ>0, so does la,r,λ.
In Theorem6, setting s=1, then we have Corollary 2.Corollary 2.
Supposefz=∑k=0∞akzk∈ℍ, a≔a0 and ωm∈Sm, ψn∈Sn for m,n∈ℕ. Then, for λ∈0,+∞ and N∈ℕ, we have(39)fωmz2+λ∑k=N∞ak‖ψnzk≤1,for z=r≤Rλ,N,m,n∗, where Rλ,N,m,n is the unique root in 0,1 of the equation(40)λrNn+m+λrNn+rns+rm−rn+m−1=0,and the radius Rλ,N,m,n∗ is the best possible.Remark 2.
Ifωmz=ψnz=z and λ=1 in Corollary 2, then it reduces to the second part of Theorem 1.Theorem 7.
Supposefz=∑k=0∞akzk∈ℍ, a≔a0, and ωm∈Sm for m∈ℕ. Then, for λ∈0,+∞ and N∈ℕ, we have(41)fωmz+f′ωmz‖ωmz+λ∑k=N∞ak‖ωmzk≤1,for z=r≤Rλ,N,m, where Rλ,N,m is the unique root in 0,2−1m of the equation(42)2λrNm1+rm2+1−rmr2m+2rm−1=0,and the radius Rλ,N,m is the best possible.Proof.
By the Schwarz lemma and the Schwarz-Pick lemma, respectively, we have(43)ωmz≤rm,fωmz≤rm+a1+arm,forz∈D.
Then, by Lemmas1–3, we obtain(44)fωmz+f′ωmzωmz+λ∑k=N∞akωmzk≤fωmz+1−fωmz21−ωmz2ωmz+λ∑k=N∞akωmzk≤a+rm1+arm+1−a+rm1+arm2rm1−r2m+λ1−a2rNm1−rm=a+rm1+arm+1−a2rm1+arm2+λ1−a2rNm1−rm≔CN,ma,r,λ,for r≤2−1m, where 2−1m is the unique root of the equation r2m+2rm−1=0 in 0,1. We just need to show that AN,ma,r,λ≤1 holds for r≤Rλ,N,m. That is to prove Ca,r,λ≤0 for r≤Rλ,N,m, where(45)Ca,r,λ=a+rm1+arm+1−a2rm−1+arm21−rm+λ1−a2rNm1+arm2=1−rm1+arma+rm−1−arm+1−a2rm1−rm+λ1−a2rNm1+arm2=1−a1−rm1+armrm−1+1+arm1−rm+λ1+arNm1+arm2.
Let(46)gr=1−rm1+armrm−1+1+arm1−rm+λ1+arNm1+arm2,then(47)gr=λrNm+2ma3+λrNmr2m+2rma2+1−rmr2m+λrNm2rm+1a+1−rm2rm−1+λrNm≤λrNm+2m+λrNmr2m+2rm+1−rmr2m+λrNm2rm+1+1−rm2rm−1+λrNm=2λrNmrm+12+1−rmr2m+2rm−1≔hr.
It is easy to verify that both2λrNmrm+12 and 1−rmr2m+2rm−1 are continuous and increasing functions for r∈0,2−1m. Thus, hr is a monotonically increasing function for r∈0,2−1m. Furthermore, noting h0=−1<0 and h2−1m=4λ2−1N>0, thus there exists a unique real root r∗∈0,2−1m such that hr∗=0. Hence, it holds that hr≤0, for r≤r∗≔Rλ,N,m.
Now, we show that the radiusRλ,N,m is the best possible. Let a∈0,1,(48)ωmz=zm,fz=a+z1+az=a+1−a2∑k=1∞−ak−1zk,z∈D.
Takingz=r∈0,1, then the left side of inequality (41) reduces to(49)frm+f′rm‖rm+λ∑k=N∞akrmk=a+rm1+arm+1−a2rm1+arm2+λ1−a2aN−1rNm1−arm.
Now, we just need to show that ifr>Rλ,N,m, then there exists an a∈0,1 such that the right side of (49) is greater than 1. That is to prove la,r,λ>0 for r>Rλ,N,m, where(50)la,r,λ=1−arm1+arma+rm−1−arm+1−a2rm1−arm+λaN−1rNm1+arm2=1−arm−11−arm1+arm+1+arm1−arm+λ1+aaN−1rNm1+arm2.
Let(51)ja,r,λ=rm−11−arm1+arm+1+arm1−arm+λ1+aaN−1rNm1+arm2.
Observe that(52)lima⟶1−ja,r,λ=2λrNm1+rm2+1−rmr2m+2rm−1=hr.
Thus, the monotonicity ofhr implies that if r∈Rλ,N,m,2−1m, then hr>0. Therefore, by the continuity of ja,r,λ for variable a, we have that if r∈Rλ,N,m,2−1m, there exists an a∈0,1 such that inequality ja,r,λ>0, so does la,r,λ.Remark 3.
Ifωmz=z, N=2, and λ=1 in Theorem 7, then it reduces to Theorem 3.Theorem 8.
Supposefz=∑k=0∞akzk∈ℍ, a≔a0, and ωm∈Sm, for m∈ℕ. Then, for t∈0,1, we have(53)tfωmz+1−t∑k=0∞ak‖ωmzk+11+a+ωmz1−ωmz∑k=1∞ak2ωmz2k≤1,for z=r≤Rt,m, where(54)Rt,m=11+a1−t+1m,fort∈0,1+a2−11+a2∪1+a2−11+a2,1,12m,fort=1+a2−11+a2,and the radius Rt,m is the best possible.Proof.
Inequalities (43) and Lemmas 1, 2 and 4 lead to that(55)tfωmz+1−t∑k=0∞ak‖ωmzk+11+a+ωmz1−ωmz∑k=1∞ak2ωmz2k≤ta+rm1+arm+1−ta+1−t1−a2rm1−rm≔Dma,r,t.
We just need to show thatDma,r,t≤1 holds for r≤Rt,m. That is to prove Da,r,t≤0 for r≤Rt,m, where(56)Da,r,t=ta+rm1−rm+1−t1+arma1−rm+1−a2rm−1+arm1−rm=1−a−1+a2t+1+a2−1r2m+2rm−1.
Let(57)gr=−1+a2t+1+a2−1r2m+2rm−1.
Then, we divide it into two cases to discuss.Case 1. Ift∈0,1+a2−1/1+a2∪1+a2−1/1+a2,1, then we have gr≤0 for r≤1/1+a1−t+1m, where 1/1+a1−t+1m is the unique root in 0,1 of equation −1+a2t+1+a2−1r2m+2rm−1=0. Thus, Ba,r,t≤0 for r≤1/1+a1−t+1m.Case 2. Ift=1+a2−1/1+a2, then we have gr≤0 for r≤1/2m. Thus, Ba,r,t≤0 for r≤1/2m.
Now, we show that the radiusRt,m is the best possible. Let a∈0,1; we still consider the functions ωmz, fz as in (48). Taking z=r∈0,1, then the left side of inequality (53) reduces to(58)tfrm+1−t∑k=0∞akrmk+11+a+rm1−rm∑k=1∞ak2r2mk=ta+rm1+arm+1−ta+∑k=1∞1−a2ak−1rmk+11+a+rm1−rm∑k=1∞1−a22a2k−2r2mk=ta+rm1+arm+1−ta+1−a2rm1−rm≔Dm∗a,r,t.
Now, we just need to show that ifr>Rt,m, then there exists an a∈0,1 such that Dm∗a,r,t is greater than 1. That is to prove la,r,t>0 for r>Rt,m, where(59)la,r,t=ta+rm1−rm+1−ta1+arm1−rm+rm1+arm−1+arm1−rm=1−a−1+a2t+1+a2−1r2m+2rm−1.
Let(60)hr=−1+a2t+1+a2−1r2m+2rm−1.
Next, we divide it into two cases to discuss.Case 1. Ift∈0,1+a2−1/1+a2∪1+a2−1/1+a2,1, then we have hr≥0 for r≥1/1+a1−t+1m. Thus, la,r,t≥0 for r≥1/1+a1−t+1m.Case 2. Ift=1+a2−1/1+a2, then we have hr≥0 for r≥1/2m. Thus, la,r,t≥0 for r≥1/2m.Remark 4.
Ifωmz=z and t=0 in Theorem 8, then it reduces to Theorem 4.


## 3. Conclusion

We establish some new versions of Bohr-type inequalities with one parameter or involving convex combination for bounded analytic functions of Schwarz function, and we conclude that all of the corresponding Bohr radii in the paper are exact. These inequalities generalize some earlier results on the Bohr radius problem.



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