Bohr--Rogosinski inequalities for bounded analytic functions
Seraj A. Alkhaleefah, Ilgiz R. Kayumov, Saminathan Ponnusamy
aa r X i v : . [ m a t h . C V ] A p r Bohr–Rogosinski inequalities for bounded analytic functions
Seraj A. Alkhaleefah, * Ilgiz R. Kayumov, ** and Saminathan Ponnusamy *** N.I. Lobachevskii Institute of Mathematics and Mechanics, Kazan (Volga Region) Federal University,Kremlevskaya ul. 18, Kazan, Tatarstan, 420008 Russia Department of Mathematics, Indian Institute of Technology Madras, Chennai-600 036, India
Abstract —In this paper we first consider another version of the Rogosinski inequality foranalytic functions f ( z ) = P ∞ n =0 a n z n in the unit disk | z | < , in which we replace the coeffi-cients a n ( n = 0 , , . . . , N ) of the power series by the derivatives f ( n ) ( z ) /n ! ( n = 0 , , . . . , N ) .Secondly, we obtain improved versions of the classical Bohr inequality and Bohr’s inequalityfor the harmonic mappings of the form f = h + g , where the analytic part h is bounded by and that | g ′ ( z ) | ≤ k | h ′ ( z ) | in | z | < and for some k ∈ [0 , . Primary: 30A10, 30B10; 30C62, 30H05,31A05, 41A58; Secondary: 30C75, 40A30
Keywords and phrases:
Bounded analytic function, Bohr inequality, Bohr radius, Rogosinskiinequality, Rogosinski radius, harmonic mappings.
1. INTRODUCTIONLet D = { z ∈ C : | z | < } denote the open unit disk, and A denote the space of analyticfunctions in D with the topology of uniform convergence on compact sets. Define B = { f ∈ A : | f ( z ) | < in D } . Then the Bohr radius is the largest number r > such that if f ∈ B has thepower series expansion f ( z ) = P ∞ n =0 a n z n , then P ∞ n =0 | a n | | z | n ≤ for | z | ≤ r which is called theclassical Bohr inequality for the family B . Rogosinski radius is the largest number r > suchthat, under the previous assumptions, | S N ( z ) | < for | z | < r, where S N ( z ) = P Nn =0 a n z n ( N ≥ denote the partial sums of f . This inequality is called the classical Rogosinski inequality for thefamily B .If B and R denote the Bohr radius and the Rogosinski radius, respectively, then because | S N ( z ) | ≤ P Nn =0 | a n | | z | n ≤ P ∞ n =0 | a n | | z | n , it is clear that B ≤ R . In fact the following two classicalresults are well-known. Theorem A.
Suppose that f ∈ B . Then we have B = 1 / , and (see Rogosinski [1] and also [2, 3]) R = 1 / .There is a long history about the consequences of Bohr’s inequality, in particular. Indeed, Bohr[4] discovered that B ≥ / and the fact that B = 1 / was obtained independently by M. Riesz,I. Schur and N. Weiner. Extensions and modifications of Bohr’s result can be found from [5–7] and the recent articles [8–15]). We refer to [16–19] for the extension of the Bohr inequality toseveral complex variables. More recently, Kayumov and Ponnusamy [20] introduced and investigatedBohr–Rogosinski’s radii for the family B , and they discussed Bohr–Rogosinski’s radius for the classof subordinations. In [21], Aizenberg, et al. generalized the Rogosinski radius for holomorphicmappings of the open unit polydisk into an arbitrary convex domain. In [22], Kayumov et al. * E-mail: [email protected] ** E-mail: [email protected] ***
E-mail: [email protected]
SERAJ A. ALKHALEEFAH ET AL. investigated Bohr’s radius for complex-valued harmonic mappings that are locally univalent in D .Several improved versions of Bohr’s inequality were given by Kayumov and Ponnusamy in [20] andthese were subsequently followed by Evdoridis et al. [23] to obtain improved versions of Bohr’sinequality for the class of harmonic mappings. In [24], Kayumov and Ponnusamy discussed Bohr’sradius for the class of analytic functions g , when g is subordinate to a member of the class of oddunivalent functions. For more information about Bohr’s inequality and further related works, werefer the reader to the recent survey article [25] and the references therein.In this paper we shall introduce and investigate another version of the Rogosinski inequality foranalytic functions defined on the unit disk D by substituting the derivatives of the analytic functioninstead of the coefficients of its power series. We shall also introduce and study several new versionsof the classical Bohr’s inequality.2. AN IMPROVED VERSION OF THE CLASSICAL ROGOSINSKI INEQUALITYWhat could happened to the partial sums of the analytic function in the unit disk if we replacedthe coefficients a , a , . . . , a N − by the functions f ( z ) , f ′ ( z ) , . . . , f ( N − ( z ) ? In this section we givean answer in the following form. Theorem 1.
Suppose that f ∈ B and f ( z ) = P ∞ n =0 a n z n . Then (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =0 f ( k ) ( z ) k ! z k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ n X k =0 (cid:18) − k (cid:19) for all | z | ≤ r ≤ . Proof.
To prove this theorem we will use a modification of Landau’s method (see [26] and [2, p. 26]).We consider the function g : D → D defined by g ( ζ ) = f ( α ( ζ + 1)) , where | α | ≤ / , and use thesubstitution ξ = D ( ζ ) = α ( ζ + 1) . In view of the Cauchy integral formula, integration along a circle γ around the origin lying in its neighborhood, we have f ( k ) ( α ) k ! = 12 πi Z D ( γ ) f ( ξ )( ξ − α ) k +1 dξ = 12 πiα k Z γ g ( ζ ) ζ k +1 dζ and thus, we can write n X k =0 f ( k ) ( α ) k ! α k = 12 πi Z γ g ( ζ ) n X k =0 ζ k +1 ! dζ = 12 πi Z γ g ( ζ ) ζ n +1 n X k =0 ζ k ! dζ. (1)Set ζ + ζ + ζ + · · · = (1 − ζ ) − = K ( ζ ) = ( K n ( ζ )) + O ( ζ n +1 ) , where we write K ( ζ ) = (1 − ζ ) − / = ∞ X k =0 (cid:18) − k (cid:19) ( − ζ ) k and K n ( ζ ) = n X k =0 (cid:18) − k (cid:19) ( − ζ ) k . In view of the above observations, (1) reduces to n X k =0 f ( k ) ( α ) k ! α k = 12 πi Z γ g ( ζ ) ζ n +1 ( K n ( ζ )) dζ (2)and therefore, with ζ = | ζ | e iφ , and | g ( ζ ) | ≤ for all | α | ≤ / and | ζ | ≤ , we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =0 f ( k ) ( α ) k ! α k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ π Z π | ζ | n +1 | K n ( ζ ) | | ζ | dφ = 1 | ζ | n n X k =0 (cid:18) − k (cid:19) | ζ | k . Allowing | ζ | → , we get (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =0 f ( k ) ( α ) k ! α k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ n X k =0 (cid:18) − k (cid:19) for all | α | ≤ which completes the proof of the theorem. OHR–ROGOSINSKI INEQUALITIES 3
3. IMPROVED VERSIONS OF THE CLASSICAL BOHR’S INEQUALITYFor f ∈ B and f ( z ) = P ∞ n =0 a n z n , the following inequalities due to Schwarz-Pick will be usedfrequently: for | z | = r < , | f ( z ) | ≤ r + a ar and | f ′ ( z ) | ≤ − | f ( z ) | − | z | , (3)where | a | = a ∈ [0 , . Also, it is well-known that the Taylor coefficients of f ∈ B satisfy theinequalities: | a k | ≤ − a for each k ≥ . (4)More generally, we have ([27]) the sharp estimate | f ( k ) ( z ) | k ! ≤ − | f ( z ) | (1 − | z | ) k (1 + | z | ) for each k ≥ and z ∈ D , (5)which in particular gives second inequality in (3), and (4) by setting z = 0 in (5). In the followingwe also assume that m ∈ N , and the idea of replacing a k by f ( k ) ( z ) k ! is used in [15]. But our concernhere is slightly different from theirs. Theorem 2.
Suppose that f ∈ B and f ( z ) = P ∞ n =0 a n z n . Then A f ( z ) := | f ( z m ) | + | z m | | f ′ ( z m ) | + ∞ X k =2 | a k | r k ≤ for all r ≤ R m, , (6) where R m, is the maximal positive root of the equation ϕ m ( r ) = 0 with ϕ m ( r ) = (1 − r )( r m + 2 r m −
1) + 2 r (1 + r m ) (7) and the constant R m, cannot be improved. m R m, Table 1. R m, is the maximal positive root of the equation (1 − r )( r m + 2 r m −
1) + 2 r (1 + r m ) = 0 Proof.
Let f ∈ B and | a | = a ∈ [0 , . It is a simple exercise to see that for ≤ x ≤ x ( ≤ and ≤ α ≤ / , we have b ( x ) := x + α (1 − x ) ≤ b ( x ) . This simple fact will be used in the later
SERAJ A. ALKHALEEFAH ET AL. theorems also. Using this inequality and (4), we easily obtain from (3) and (6) that A f ( z ) ≤ | f ( z m ) | + r m − r m (1 − | f ( z m ) | ) + (1 − a ) r − r ≤ r m + a ar m + r m − r m " − (cid:18) r m + a ar m (cid:19) + (1 − a ) r − r = r m + a ar m + (1 − a ) r m (1 + ar m ) + (1 − a ) r − r = 1 − (1 − a )(1 − r m )1 + ar m + (1 − a ) r m (1 + ar m ) + (1 − a ) r − r = 1 + (1 − a ) ϕ m ( a, r )(1 + ar m ) (1 − r ) , where ϕ m ( a, r ) = − (1 − r m )(1 + ar m )(1 − r ) + (1 + a ) r m (1 − r ) + r (1 + a )(1 + ar m ) = (1 − r )( ar m + 2 r m −
1) + r (1 + a )(1 + ar m ) . The second inequality above is justified because of the fact that r m − r m ≤ for r ≤ m p √ − . Also, R m, ≤ m p √ − , where R m, is as in the statement. Now, since ϕ m ( a, r ) is an increasing functionof a in [0 , , it follows that ϕ m ( a, r ) ≤ ϕ m (1 , r ) = (1 − r )( r m + 2 r m −
1) + 2 r (1 + r m ) = ϕ m ( r ) , where ϕ m ( r ) is given by (7). Clearly, A f ( z ) ≤ if ϕ m ( r ) ≤ , which holds for r ≤ R m, .To show the sharpness of the radius R m, , we let a ∈ [0 , and consider the function f ( z ) = a + z az = a + (1 − a ) ∞ X k =1 ( − a ) k − z k , z ∈ D (8)so that f ( k ) ( z ) k ! = (1 − a ) ( − a ) k − (1 + az ) k +1 for k ≥ and z ∈ D .For this function, we observe that for z = r and a ∈ [0 , , | f ( z m ) | + | z m | | f ′ ( z m ) | + ∞ X k =2 | a k | r k = a + 2 r m + ar m (1 + ar m ) + (1 − a ) ar − ar = 1 + (1 − a ) P m ( a, r )(1 + ar m ) (1 − ar ) , (9)where P m ( a, r ) = (1 − ar )( ar m + 2 r m −
1) + ar (1 + a )(1 + ar m ) and the last expression (9) islarger than if and only if P m ( a, r ) > . By a simple calculation, we find that ∂P m ( a, r ) ∂a = r m + 2 a ( r − r m +1 ) + r (1 + r − r m ) + 3 a r m +2 + 4 ar m +2 + 2 ar + 4 a r m +2 + 6 a r m +2 which is clearly non-negative for each r ∈ [0 , and thus, for each r ∈ [0 , , P m ( a, r ) is an increasingfunction of a . This fact gives < r m + r (1 + r − r m ) = P m (0 , r ) ≤ P m ( a, r ) ≤ P m (1 , r ) = ϕ m ( r ) , where ϕ m ( r ) is given by (7). Therefore, the right hand side of (9) is smaller than or equal to forall a ∈ [0 , , only in the case r ≤ R m, . Finally, it also suggests that the right hand side of (9) islarger than if r > R m, . This completes the proof. OHR–ROGOSINSKI INEQUALITIES 5
Remark 1.
In Tables 1 and 2, we listed the values of R m, and R m, for certain values of m . If weallow m → ∞ in Theorem 2 (resp. Theorem 3 below), we note that R m, → / (resp. R m, → / below) and thus if f ∈ B , then we have the inequality | f (0) | + ∞ X k =2 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f ( k ) (0) k ! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r k ≤ for all r ≤ / ,and the number / is sharp. Theorem 3. If f ∈ B , then B f ( z ) := | f ( z m ) | + ∞ X k =2 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f ( k ) ( z m ) k ! (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r k ≤ for all r ≤ R m, , (10) where R m, is the minimum positive root of the equation ψ m ( r ) = 0 with ψ m ( r ) = 2 r − (1 − r m )(1 − r m − r ) (11) and the constant R m, cannot be improved. m R m, Table 2. R m, is the maximal positive root of the equation r − (1 − r m )(1 − r m − r ) = 0 Proof.
As before we let f ∈ B and a = | a | . By assumption, (3) and (5) (with z m in place of z ), wehave B f ( z ) ≤ | f ( z m ) | + 1 − | f ( z m ) | r m ∞ X k =2 (cid:18) r − r m (cid:19) k = | f ( z m ) | + r (1 − r m )(1 − r m − r ) (1 − | f ( z m ) | ) ≤ r m + a ar m + r (1 − r m )(1 − r m − r ) " − (cid:18) r m + a ar m (cid:19) = 1 − (1 − a )(1 − r m )1 + ar m + r (1 − a )(1 + ar m ) (1 − r m − r ) = 1 + (1 − a ) ψ m ( a, r )(1 + ar m ) (1 − r m − r ) , where ψ m ( a, r ) = − (1 − r m )(1 + ar m )(1 − r m − r ) + r (1 + a ) . The second inequality above is a consequence of our earlier observation used in Theorem 2 butthis time with α = r / [(1 − r m )(1 − r m − r )] . It is a simple exercise to see that ψ m ( a, r ) , for each m ≥ , is an increasing function of a in [0 , , and thus, it follows that ψ m ( a, r ) ≤ ψ m (1 , r ) = ψ m ( r ) , where ψ m ( r ) is given by (11). Clearly, ψ m ( a, r ) ≤ if ψ m ( r ) ≤ , which holds for r ≤ R m, , where SERAJ A. ALKHALEEFAH ET AL. R m, is the minimum positive root of the equation ψ m ( r ) = 0 . Thus, B f ( z ) ≤ for r ≤ R m, andthe inequality (10) follows.To show the sharpness of the radius R m, , we let a ∈ (0 , and consider the function g ( z ) = f ( − z ) , where f is given by (8). Then for g , we easily have g ( k ) ( z ) k ! = − (1 − | a | ) a k − (1 − az ) k +1 , z ∈ D . Now, we choose a as close to as we please and set z = r < m √ a . By a simple calculation, thecorresponding B g ( z ) takes the form B g ( z ) = a − r m − ar m + ar (1 − a )(1 − ar m ) (1 − ar m − ar )= 1 − (1 − a )(1 + r m )1 − ar m + ar (1 − a )(1 − ar m ) (1 − ar m − ar ) = 1 + (1 − a ) P m ( a, r )(1 − ar m ) (1 − ar m − ar ) , (12)where P m ( a, r ) = a (1 + a ) r − (1 + r m )(1 − ar m )(1 − ar m − ar ) . Clearly, B g ( z ) < if and only if P m ( a, r ) < . By elementary calculations, we find that ∂P m ( a, r ) ∂a = 2 a [ r − r m (1 + r m )( r m + r )] + r + (1 + r m )(2 r m + r )= 2 a ( r − r m − r m +1 − r m − r m +1 ) + (2 r m + r m +1 + 2 r m + r + r ) which is easily seen to be greater than or equal to for any r ∈ [0 , and m ≥ . Consequently, P m ( a, r ) ≤ P m (1 , r ) = ψ m ( r ) = 2 r − (1 − r m )(1 − r m − r ) . Therefore, the expression on the right of (12) is smaller than or equal to for all a ∈ (0 , , only inthe case when r ≤ R m, . Finally, it also suggests that a → in the right hand side of (12) showsthat the expression (12) is larger than if and only if r > R m, . This completes the proof. Theorem 4.
Suppose that f ∈ B and f ( z ) = P ∞ n =0 a n z n . Then C f ( z ) := | f ( z m ) | + | z | | f ′ ( z m ) | + ∞ X k =2 | a k | r k ≤ for all r ≤ R m, , (13) where R m, is the maximal positive root of the equation Φ m ( r ) = 0 with Φ m ( r ) = 3 r − r m (cid:2) r ( r m + 2) + r m (1 − r ) (cid:3) (14) and the constant R m, cannot be improved. Proof.
As in the proofs of Theorems 2 and 3, it follows from (3), (4) and (5) that C f ( z ) ≤ | f ( z m ) | + r − r m (1 − | f ( z m ) | ) + (1 − a ) r − r ≤ r m + a ar m + r − r m " − (cid:18) r m + a ar m (cid:19) + (1 − a ) r − r = 1 − (1 − a )(1 − r m )1 + ar m + (1 − a ) r (1 + ar m ) + (1 − a ) r − r = a + 2 r m + ar m (1 + ar m ) + (1 − a ) r − r = 1 + (1 − a )Φ m ( a, r )(1 + ar m ) (1 − r ) OHR–ROGOSINSKI INEQUALITIES 7 m R m, Table 3. R m, is the maximal positive root of the equation r − r m [2 r ( r m + 2) + r m (1 − r )] = 0 where Φ m ( a, r ) = − (1 − r m )(1 + ar m )(1 − r ) + r (1 + a )(1 − r ) + r (1 + a )(1 + ar m ) = r (1 + a ) + ar m +2 (1 + a )(2 + ar m ) − (1 − r m )(1 + ar m )(1 − r ) ≤ Φ m (1 , r ) = Φ m ( r ) because Φ m ( a, r ) is seen to be an increasing function of a in [0 , , and Φ m ( r ) is given by (14). Notethat the second inequality above holds since max r − r m < and so for any r < R m , where R m isthe maximal positive root of the equation r − (1 − r m ) = 0 , and R m, < R m for m ∈ N , where R m, is the maximal positive root of the equation Φ m ( r ) = 0 . Since Φ m ( r ) ≤ for r ≤ R m, , weobtain C f ( z ) ≤ for r ≤ R m, and the desired inequality (13) follows.It remains to show the sharpness of the radius R m, . To do this we let a ∈ [0 , and considerthe function f is given by (8). For this function, we observe that for z = r , C f ( z ) = ( r m + a )(1 + ar m ) + r (1 − a )(1 + ar m ) + (1 − a ) ar − ar = 1 + (1 − a ) Q m ( a, r )(1 + ar m ) (1 − r ) , (15)where Q m ( a, r ) = r (1 + a ) + a r m +2 (1 + a )(2 + ar m ) − (1 − r m )(1 + ar m )(1 − ar ) . We see that C f ( z ) > for a ∈ [0 , if and only if Q m ( a, r ) > . By a simple calculation, we findthat Q m ( a, r ) is an increasing function of a in [0 , and therefore, we have Q m ( a, r ) ≤ Q m (1 , r ) = 2 r + 2 r m +2 (2 + r m ) − (1 − r m )(1 + r m )(1 − r ) = Φ m ( r ) , where Φ m ( r ) is given by (14). Therefore, the expression (15) is smaller than or equal to for all a ∈ [0 , , only when r ≤ R m, . Finally, it also suggests that a → in (15) shows that the expression(15) is larger than if r > R m, . This completes the proof. Remark 2.
In Table 3, we listed the values of R m, for certain values of m . If we allow m → ∞ in Theorem 4, we see that R m, → and hence we have the classical Bohr inequality for f ∈ B : | f (0) | + ∞ X k =1 | a k | | z | k ≤ for all r ≤ / , and / is sharp. SERAJ A. ALKHALEEFAH ET AL.
4. TWO IMPROVED VERSIONS OF BOHR’S INEQUALITY FOR HARMONICMAPPINGS
Theorem 5.
Suppose that f ( z ) = h ( z ) + g ( z ) = P ∞ n =0 a n z n + P ∞ n =1 b n z n is a harmonic mappingof D such that | g ′ ( z ) | ≤ k | h ′ ( z ) | for some k ∈ [0 , and h ∈ B . Then we have D f ( z ) := | h ( z m ) | + ∞ X n =1 | a n | r n + ∞ X n =1 | b n | r n ≤ for all r ≤ R km, , (16) where R km, is the maximal positive root of the equation λ m ( r ) = 0 with λ m ( r ) = 2 r (1 + k )(1 + r m ) − (1 − r )(1 − r m ) (17) and the constant R km, cannot be improved. m R m, Table 4. R m, is the maximal positive root of the equation r (1 + r m ) − (1 − r )(1 − r m ) = 0 Proof.
Recall that, as h ∈ B and h ( z ) = P ∞ n =0 a n z n , h ( z ) ≺ a + z a z = a + (1 − | a | ) ∞ X n =1 ( − n − ( a ) n − z n , z ∈ D , which gives [9, 10] ∞ X n =1 | a n | r n ≤ (1 − | a | ) ∞ X n =1 | a | n − r n = (1 − a ) r − ar for all r ≤ , where a = | a | ∈ [0 , . Moreover, by assumption, we obtain that g ′ ( z ) ≺ q kh ′ ( z ) which quicklygives from [9] that ∞ X n =1 n | b n | r n − ≤ ∞ X n =1 kn | a n | r n − for all r ≤ and integrating this with respect to r gives ∞ X n =1 | b n | r n ≤ k ∞ X n =1 | a n | r n for all r ≤ . Here ≺ q denotes the quasi-subordination. Using these and the first inequality in (3) for h ( z ) onecan obtain that for | z | = r ≤ / , D f ( z ) ≤ r m + a ar m + (1 + k ) r − a − ar = 1 + (1 − a ) λ m ( a, r )(1 + ar m )(1 − ar ) , OHR–ROGOSINSKI INEQUALITIES 9 where λ m ( a, r ) = r (1 + k )(1 + a )(1 + ar m ) − (1 − r m )(1 − ar ) , which is indeed an increasing function of a ∈ [0 , so that λ m ( a, r ) ≤ λ m (1 , r ) = λ m ( r ) , where λ m ( r ) is given by (17). We see that D f ( z ) ≤ if λ m ( r ) ≤ , which holds for r ≤ R km, , where R km, is themaximal positive root of the equation λ m ( r ) = 0 . This proves the inequality (16).Finally, to show the sharpness of the radius R km, , we consider the function f ( z ) = h ( z ) + λkh ( z ) , h ( z ) = z + a az , (18)where λ ∈ (0 , . For this function, we get that (for z = r and a ∈ [0 , ) D f ( z ) = r m + a ar m + ( λk + 1) r − a − ra and the last expression shows the sharpness of R km, with λ → . This completes the proof of thetheorem. Remark 3.
In Table 4, we listed the values of R km, for k = 1 and for certain values of m . When m → ∞ , we have from Theorem 5 that R km, → k +6 . Thus, under the hypotheses of Theorem 5, wehave | h (0) | + ∞ X n =1 | a n | r n + ∞ X n =1 | b n | r n ≤ for all r ≤ k + 6 which for k = 0 gives the classical Bohr’s inequality and for k = 1 , this inequality contains the Bohrinequality for sense-preserving harmonic mapping f ( z ) = h ( z ) + g ( z ) of the disk D with the Bohrradius / (see [22]). Theorem 6.
Assume the hypotheses of Theorem 5. Then we have E f ( z ) := | h ( z m ) | + | z m | | h ′ ( z m ) | + ∞ X n =2 | a n | r n + ∞ X n =1 | b n | r n ≤ for all r ≤ R km, , where R km, is the maximal positive root of the equation Λ m ( r ) = 0 with Λ m ( r ) = (1 − r )( r m + 2 r m −
1) + 2 r ( r + k )(1 + r m ) and the constant R km, cannot be improved. m R m, Table 5. R m, is the maximal positive root of the equation (1 − r )( r m + 2 r m −
1) + 2 r ( r + 1)(1 + r m ) = 0 Proof.
As in the proofs of Theorem 5 and earlier theorems, we easily have E f ( z ) ≤ r m + a ar m + r m − r m " − (cid:18) r m + a ar m (cid:19) + (1 − a ) r − r + k (1 − a ) r − r = 1 − (1 − a )(1 − r m )1 + ar m + (1 − a ) r m (1 + ar m ) + (1 − a )( r + k ) r − r = 1 + (1 − a )Λ m ( a, r )(1 + ar m ) (1 − r ) , where Λ m ( a, r ) = − (1 − r m )(1 − r )(1 + ar m ) + r m (1 + a )(1 − r ) + (1 + a ) r ( r + k )(1 + ar m ) = (1 − r )( ar m + 2 r m −
1) + r (1 + a )( r + k )(1 + ar m ) ≤ Λ m (1 , r ) = Λ m ( r ) . The first inequality above is justified with the same reasoning as in the proofs of earlier theorems.Now, we see that E f ( z ) ≤ whenever Λ m ( r ) ≤ , which holds for r ≤ R km, , where R km, is themaximal positive root of the equation Λ m ( r ) = 0 .To show the sharpness of the radius R km, , consider the function f defined by (18) with λ ∈ (0 , .For this function, the corresponding expression for E f ( z ) with z = r turned out to be E f ( z ) = a + 2 r m + ar m (1 + ar m ) + ar (1 − a )( r + λ )1 − ar . (19)The last expression is larger than if and only if P km ( a, r ) > , where P km ( a, r ) = (1 − ar )( ar m + 2 r m −
1) + ar (1 + a )( r + λ )(1 + ar m ) . (20)By a simple calculation, we find that P km ( a, r ) is an increasing function of a ∈ [0 , , and for each r ∈ [0 , , so that P km ( a, r ) ≤ P km (1 , r ) = (1 − r )( r m + 2 r m −
1) + 2 r ( r + λ )(1 + r m ) . Therefore, the expression (19) is smaller than or equal to for all a ∈ [0 , , only in the case when r ≤ R km, ( λ = k ). Finally, it also suggests that a → in (20) shows that the expression (19) islarger than if r > R km, . This completes the proof. Remark 4.
In Table 5, we listed the values of R km, for k = 1 and for certain values of m . If weallow m → ∞ in Theorem 6, we obtain that R km, → R k := 14 (cid:16)p (2 k + 1) + 8 − (2 k + 1) (cid:17) , where R k is the positive root of the equation x ( x + k ) + x − and the conclusion of Theorem 6takes the following form: | h (0) | + ∞ X n =2 | a n | r n + ∞ X n =1 | b n | r n ≤ for all r ≤ (cid:16)p (2 k + 1) + 8 − (2 k + 1) (cid:17) . Acknowledgements
The work of S. Alkhaleefah and I. Kayumov is supported by the Russian Science Foundationunder grant 18-11-00115. The work of the third author is supported by Mathematical ResearchImpact Centric Support of Department of Science & Technology, India (MTR/2017/000367).
OHR–ROGOSINSKI INEQUALITIES 11
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