The sharp estimates of all initial taylor coefficients in the Krzyz's problem
aa r X i v : . [ m a t h . C V ] A p r UDC 517.54D.L. STUPINTver state university
THE SHARP ESTIMATES OF ALL INITIALTAYLOR COEFFICIENTS IN THEKRZYZ’S PROBLEM For each t > , up to the number n = N ( t ) , the exact estimations ofall initial Taylor coefficients in the class B t were found, where B t is a set ofholomorphic in unit disk functions f, < | f | < , f (0) = e − t . Let ∆ := { z ∈ C : | z | < } , where C — a field of all complex numbers.Class B consists of holomorphic in ∆ functions w = f ( z ) , not vanishingand such that | f ( z ) | , z ∈ ∆ . The Krzyz conjecture [1] consists in that for every f ∈ B and all natural n the Taylor coefficients { f } n of function f satisfy to inequality |{ f } n | /e. The extremals can be only rotations of the function F ∗ ( z n , in the planes z and w, where F ∗ ( z, t ) := e − t H ( z ) , H ( z ) := 1 + z − z , t > . (1) The existence of extremals of this problem is obvious, since after additionof function f ( z ) ≡ to the class B it becomes a family of functions, compactin the topology of the locall uniform convergence. However, extremal prob-lems on the class B are very complicated and, at present, the hypothesisremains unconfirmed.Since the class B is invariant under rotations in the planes of the variables z and w, than we can restrict ourselves to studying only of functions, forwhich < { f } < . Further, we can fix the parameter t ∈ (0 , + ∞ ) and set { f } = e − t ; these subclasses we denote by B t . From geometrical considerations it is clear that every function of class B t can be represented in the form f ( z ) = F ∗ ( ω ( z ) , t ) , ω ∈ Ω , (2) where class Ω consists of holomorphic in unit disk ∆ functions ω ( z ) such,that ω (0) = 0 and | ω ( z ) | < , when z ∈ ∆ . Note that for every t > the formula (2) establishes a one-to-one corres-pondence between the classes B t and Ω (see [2]). c (cid:13) D.L. Stupin, 2010. f t ∈ (0 , than the Krzyz conjecture can be refined: if f ∈ B t than,probably, the accurate estimates |{ f } n | |{ F ∗ } ( t ) | = 2 t/e t are correct;the equality is delivered only by functions F ∗ ( e iϕ z n , t ) , n ∈ N , ϕ ∈ R . Let us dwell on representations of the form (2). And let the functions g ( z ) and G ( z ) be holomorphic in ∆ . Function g ( z ) is called subordinate inthe disk ∆ for the function G ( z ) , if it can be represented in ∆ in the form g ( z ) = G ( ω ( z )) , where ω ∈ Ω . Function G ( z ) will be called majorant for the g ( z ) in the domain ∆ . Let g ( z ) = ∞ P n =0 { g } n z n , G ( z ) = ∞ P j =0 { G } j z j , ω ( z ) = ∞ P k =1 { ω } k z k . Then g ( z ) = G ( ω ( z )) = ∞ X j =0 { G } j ω ( z ) j = { G } + ∞ X n =1 n X j =1 { G } j { ω j } n ! z n . Whence it appears { g } = { G } and { g } n ( ω ) = n X j =1 { G } j { ω j } n , n ∈ N , ω ∈ Ω . (3) The class of all functions f ( z ) , regular and univalent in ∆ , with thenormalization f (0) = 0 , f ′ (0) = 1 , mapping the disk ∆ on convex domain,is denoted by S . The set of all functions h ( z ) , with a positive in ∆ real part and with thenormalization h (0) = 1 , mapping the unit disk ∆ into right half-plane, iscalled the Caratheodory class and is denoted by C. Between C and S there is the bijection [2]: h ( z ) = 1 + z f ′′ ( z ) f ′ ( z ) , h ∈ C, f ∈ S . (4) The following simple but important statement is hold [2]:
Lemma 1
If the function s ( z ) = ∞ P n =1 { s } n z n , regular in ∆ , is subordinatedto the function S ( z ) ∈ S , then the sharp estimates |{ s } n | |{ S } | = 1 , n ∈ N , are valid. Equality is achieved only on the functions S ( e iϕ z n ) , ϕ ∈ R , n ∈ N . Carath´eodory and T¨oplitz [3, 4] have completely solved the problem aboutthe possibility of extention of a polynomial up to a function of the class C. Later Shur gave a constructive proof of this theorem [5]. riterion 1 (Carath´eodory, T¨oplitz) Let { h } , . . . , { h } n , n ∈ N , — arefixed complex numbers. Polynom R ( z, n ) := 1 + n X k =1 { h } k z k can be extended up to the function h ( z ) = R ( z, n ) + O ( z n +1 ) ∈ C if and onlyif determinants M k := det { a ij } ki,j =0 , k n,a ii = 2 , a ij = { h } j − i , j > i, a ij = a ji , j < i, either all positive or positive to a certain number, from which are equal tozero. In the latter case, the extension is unique. For further progress we need to study the Taylor coefficients of ourmajorant function F ∗ ( z, t ) of class B t . Everywhere below, we will not beinterested in the zero coefficient of this function, since it is not included inthe formula (3). The first coefficient { F ∗ } ( t ) of function F ∗ ( z, t ) is equal to − t/e t . We normalize the function F ∗ ( z, t ) so, that the first coefficient in itsTaylor expansion becomes equal to . Let us introduce the notation F ( z, t ) := F ∗ ( z, t ) { F ∗ } ( t ) . (5) The question arises: whether convex univalent functions f ∈ S existthere, with some initial Taylor coefficients that match with all first coefficientsof the function F ( z, t ) , except for the coefficient { F } ( t )? It is not hard to check. By substituting f ( z ) = F ( z, t ) to the formula (4),we obtain h ( z ) = 1 + 2 z (cid:18) − z − − z ) t (cid:19) . From which we elementary derive remarkably simple formula { h } j = 2(1 − jt ) , j ∈ N . (6) We use the Carath´eodory-T¨oeplitz extension criterion of polynomials upto a function of class C. Let us compute the principal minors M j − , for all j ∈ N . Here the index j − means, that the dimension of the corresponding tothe minor M j − matrix is equal to j. According to lemma 2, the formulation nd the proof of which can be found at the end of this work (see section 6),we have: M j − = 2 j − t j − (2 − ( j − t ) , j ∈ N . (7) Further, it is obvious that the minors M , . . . , M n − are not negative ifand only if t / ( n − , for n > , and t > , for n = 1 , or n /t + 1 . Note also, that if t = 2 / ( n − , n ∈ N \ { } , the extension is unique.Thus, we have Theorem 1
For every t > and n /t + 1 , n ∈ N , the segment ofTaylor expansion of the function F ( z, t ) , first introduced in the formula (5) ,— polynom P ( z, t, n ) := z + n P k =2 { F } k ( t ) z k — can be extended to the function f ( z ) = P ( z, t, n ) + O ( z n +1 ) ∈ S . For t = 2 / ( n − , n ∈ N \ { } , theextension is unique. From lemma 1, theorem 1, formula (3) and normalization (5) followsa central for this work
Theorem 2
For every t > , arbitrary N /t + 1 , N ∈ N , and each f ∗ ∈ B t , sharp estimates |{ f ∗ } n | |{ F ∗ } ( t ) | = 2 te t , n ∈ { , . . . , N } , (8) are correct. Extremals in the estimates (8) are only the functions F ∗ ( e iϕ z n , t ) , ϕ ∈ R , where the function F ∗ defined by (1) . Proof.
We fix ω ∈ Ω , t > and N /t + 1 , N ∈ N . Let us take a natural number n, not exceeding the number N. Usingformula (3), we write n -th coefficient of function f ( z ) := F ( ω ( z ) , t ) , where F is defined in formula (5), in the form { f } n = n X j =1 { F } j { ω j } n . Now we apply theorem 1 to n -th segment of Taylor expansion of function F ( z, t ) , which we have denoted by P ( z, t, n ) . Let S ( z ) — be an extention of olynom P ( z, t, n ) to function of class S . Then, using the formula (3), n -thcoefficient of function s ( z ) := S ( ω ( z ) , t ) can be written as { s } n = n X j =1 { S } j { ω j } n . From which, by lemma 1, we find that |{ s } n | . But { S } j := { F } j ( t ) , where j ∈ { , . . . , n } , therefore { f } n = { s } n , on basisof which we conclude, that |{ f } n | . Remembering about the normalization (5), we obtain the estimates (8). Ac-curacy of the estimates (8) and the form of extremal functions follow fromlemma 1.The theorem has been completely proved.From theorem 2 it implies, that the smaller the number t > we fix, themore Taylor coefficients we can estimate on the class B t . In this case, ourestimates are sharp in the sense, that the equality, in the inequality (8), isattained on functions F ∗ ( e iϕ z n , t ) . For example, if t > we can estimate only one coefficient, for t = 2 —two coefficients, for t = 1 — three coefficients, and at t = 1 / — five. Andso on. Similar results were obtained in [6]. We eliminate the blank in the arguments given above. To do this wemust only prove the validity of the formula (7).
Lemma 2
If the coefficients { h } j , j ∈ N , are defined by formula (6) , thenfor all integers n > M n = 2 n t n (2 − nt ) . Proof.
The minor M n / n +1 is equal to the determinant (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − t − t . . . − ( n − t − nt − t − t . . . − ( n − t − ( n − t − t − t . . . − ( n − t − ( n − t ... ... ... . . . ... ... − ( n − t − ( n − t − ( n − t . . . − t − t − nt − ( n − t − ( n − t . . . − t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . ubtracting from each row, except the first one, the previous one we obtain (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − t − t . . . − ( n − t − nt − t t t . . . t t − t − t t . . . t t ... ... ... . . . ... ... − t − t − t . . . t t − t − t − t . . . − t t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . For each column, except for the last one, we add the last column (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − nt − ( n + 1) t − ( n + 2) t . . . − (2 n − t − nt t t . . . t t t . . . t t ... ... ... . . . ... ... . . . t t . . . t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == 2 n − t n (2 − nt ) . We present an example of an extention. Let t = 1 / . By theorem 1,the desired extension is unique. Setting in the formula (4) f ( z ) = z + 12 z + 16 z − z − z + . . . ∈ S or using the formula (6), we obtain that h ( z ) = 1 + z − z − z + . . . ∈ C. We know that the function ω ( z ) = 1 − h ( z )1 + h ( z ) = − z + 14 z + 38 z + 916 z + . . . belongs to the class Ω . It is also known (see [2, 5]) that ω ( z ) = λ α n − + α n − z + . . . + α z n − α + α z + . . . + α n − z n − . In this case λ = 1 (see [2, 5]). Having found the parameters α , . . . , α n − , we find ω ( z ) = z − z − z − − z + z . hence h ( z ) = 1 + z − z − z z . Now we use the formula (4), by substituting the obtained expression for h ( z ) there. Well f ( z ) = z Z (cid:16) v + √ v +1 v −√ v +1 (cid:17) √ / √ v dv. CITATION1. Krzyz J.G.
Coefficient problem for bounded nonvanishing functions //Ann. Polon. Math. 1968. V. 70. P. 314.2. Golusin G.M.
Geometric theory of functions of a complex variable.
Engl. transl.: American Mathematical Society, Providence, RI, 1969.3. Carath´eodory C. ¨Uber die Variabilit¨atsbereich des Fourierschen Kon-stanten von Positiv Harmonischen Funktion // Rendiconti Circ. Mat.di Palermo. 1911. V. 32. P. 193–217.4. T¨oplitz O. ¨Uber die Fouriersche Entwicklung Positiver Funktionen //Rendiconti. Circ. Mat. di Palermo. 1911. V. 32. P. 191–192.5. Schur I. ¨Uber potenzreihen, die in Innern des Einheitskrises Beschr¨anktSind // Reine Angew. Math. 1917. V. 147. P. 205–232.6. Peretz R.
Applications of subordination theory to the class of boundednonvanishing functions // Compl. Var. 1992. V. 17. P. 213–222.April 2010// Compl. Var. 1992. V. 17. P. 213–222.April 2010