The Landau problem for nonvanishing functions with real coecients

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1 Journal of Computational Applied Mathematics 139 (2002) The Lau problem for nonvanishing functions with real coecients Philippos Koulorizos, Nikolas Samaris Department of Mathematics, University of Patras, University Campus, GR Patras, Greece Received 25 November 2000 Abstract For the class of holomorphic functions f(z) =f 0 + f 1z + f 2z 2 + in the unit disk z 1 with 0 f(z) 1 f n R, n =0; 1; 2;::: we prove that max(f 0 + f f 4) = max f 0 + f f 4 =1:46109 ::: min(f 0 + f f 4)= 0: :::. c 2002 Elsevier Science B.V. All rights reserved. MSC: 30C50 Keywords: Real coecient nonvanishing function; The Lau problem Let H(D) be the class of holomorphic functions in the unit disk D = {z: z 1}. We consider the following classes of functions: B = {f H(D): f(z)=f 0 + f 1 z + f 2 z 2 + ; f(z) 1; z D}; B 0 = {f B: f(z) 0}; W = {f B: f(0)=0}; P = {p H(D): p(z)=1+p 1 z + p 2 z 2 + ; Rp(z) 0; z D}; P R = {p P: p n R; n=1; 2;:::}: Corresponding author. address: samaris@math.upatras.gr (N. Samaris) /02/$ - see front matter c 2002 Elsevier Science B.V. All rights reserved. PII: S (01)

2 130 P. Koulorizos, N. Samaris / Journal of Computational Applied Mathematics 139 (2002) With no loss of generality we may assume that for f B 0 we have the form f =e tp with t 0 p P. In [1] Lau proved that sup n =1+ f B n ( ) 1 3 ::: (2k 1) 2 :=G n ; n N; 2 4 ::: 2k k=1 with this result being sharp while G n (1=) log n, n. In [2] Lewowski Szynal consider the Lau problem for the class B 0 they prove that max f B 0 f 0 + f 1 6 2e 1=2 1:21 ::: max f B 0 f 0 + f 1 + f 2 1:33 :::; while in both cases the extremal functions are of the form f =e tp, with t 0 p P R. They also conjecture that for any f B 0 there exists a constant L 1 n N such that max f B0 f 0 + f 1 + f f n 6 L. In [3] Ganczar et al., consider the Lau problem for the class B 0 (R) which consists of the functions f B 0 with real Taylor coecients they prove that S3 =1:40315 ::: where S n = max f 0 + f f n ; (n =2; 3;:::): f B 0(R) In this paper, we calculate S 4, S 4 s 4 where S n = max (f 0 + f f n ); f B 0(R) s n = min (f 0 + f f n ); (n =2; 3;:::): f B 0(R) In the present paper, we face a problem involving the estimation of quantities which depend on the Taylor coecients of functions belonging to the class P R. Our rst idea is to use the Caratheodory Toeplitz conditions as they are the strongest relations between the Taylor coecients of the class P R. A second idea is to express these relations in such a way that each Taylor coecient can be converted separately to a polynomial of several variables. Combining these two ideas, we transform the initial problem into nding the max (or min) of a polynomial of several variables, dened in a closed interval [0; 1] k, k 6 4. A serious problem in this paper is the size of the polynomials which are involved in the elementary calculations. Using the computer algebra system Mathematica 4:0, we obtained all necessary results. We will need the following Lemmas. Lemma 1. Let K n (P R ) be the set of x=(x 1 ;x 2 ;:::;x n ) R n for which there exists a q(z)=1+q 1 z+ q 2 z 2 + P R having q 1 = x 1 ;q 2 = x 2 ;:::;q n = x n. Let also A n be the set of x =(x 1 ;x 2 ;:::;x n ) R n

3 P. Koulorizos, N. Samaris / Journal of Computational Applied Mathematics 139 (2002) such that D k (x 1 ;x 2 ;:::;x k ) 0; k=1; 2;:::;n where: 2 x 1 x 2 ::: x k x 1 2 x 1 ::: x k 1 D k (x 1 ;x 2 ;:::;x k )= x 2 x 1 2 ::: x k 2 : x k x k 1 x k 2 ::: 2 If A n is the closure of A n then A n = K n (P R ). The above Lemma is a part of the Caratheodory Toeplitz Theorem (see [4,5]). Lemma 2. If x =(x 1 ;x 2 ;:::;x n ) R n (n 6 4) the following propositions are equivalent. (i) x K n (P R ) (ii) there exists a (t 1 ;t 2 ;:::;t n ) [0; 1] n such that: x 1 =p 1 (t 1 );x 2 =p 2 (t 1 ;t 2 );:::; x n =p n (t 1 ;t 2 ;:::;t n ) where p 1 (t 1 )= 2+4t 1 ; p 2 (t 1 ;t 2 )=2+16t 1 ( 1+t 1 + t 2 t 1 t 2 ); p 3 (t 1 ;t 2 ;t 3 )= 2+4t 1 ( 3+4t 1 ) 2 32t 1 t 2 (1 5t 1 +4t 2 1) 64t 2 1t 2 2(1 + t 1 ) + 64( 1+t 1 )t 1 (1 t 2 )t 2 t 3 ; p 4 (t 1 ;t 2 ;t 3 ;t 4 )=2(1+32t 1 ( 1+5t 1 ) + 128t 3 1( 2+t 1 )+32t 1 t 2 (1 9t 1 +20t t 3 1) + 128t 2 1t 2 2(1 4t 1 +3t 2 1) + 128t 3 1t 3 2(1 t 1 ) + 128t 1 t 2 t 3 (1 3t 1 +2t 2 1 t 2 ) + 128t 2 1t 2 2t 3 (5 4t 1 2t 2 + t 3 +2t 1 t 2 t 2 t 3 ) + 128t 1 t 2 2t 2 3( 1+t 2 ) + 128t 1 t 2 t 3 t 4 ( 1+t 1 )( 1+t 2 )( 1+t 3 )): Proof. The quantity D k (x 1 ;x 2 ;:::;x k ) can be written as polynomial of second degree in x k form: of the D k 2 (x 1 ;x 2 ;:::;x k 2 )x 2 k + ; (D k = 1 for k 6 0): If k k (x 1 ;x 2 ;:::;x k 1 ) k k (x 1;x 2 ;:::;x k 1 ) are the roots of the above polynomial it is easy to see that the relation x A n is equivalent to min( k ; k ) x k max( k ; k )or x k = k + t k ( k k ); t k (0; 1); k =1; 2;:::;n: (1) For k = 1 we get 1 = 2 1 = 2. Therefore x 1 = p 1 (t 1 ); t 1 (0; 1): (2)

4 132 P. Koulorizos, N. Samaris / Journal of Computational Applied Mathematics 139 (2002) For k = 2 by the equation D 2 (x 1 ;x 2 ) = 0 we obtain 2 = 2+x1 2 2 = 2. Thus combining (1) with (2) we get x 2 = p 2 (t 1 ;t 2 ); (t 1 ;t 2 ) (0; 1) 2 : (3) For k = 3 through equation D 3 (x 1 ;x 2 ;x 3 ) = 0 we obtain 3 = 4 2x 1 (x 1 x 2 ) 2 3 = 4+2x 1 (x 1 + x 2 ) 2 : 2+x 1 2+x 1 Consequently, combining (1) with (2) (3) we obtain after the calculations x 3 = p 3 (t 1 ;t 2 ;t 3 ); (t 1 ;t 2 ;t 3 ) (0; 1) 3 : (4) In the same manner we can see that x 4 = p 4 (t 1 ;t 2 ;t 3 ;t 4 ). Summarizing, we have that the transform (t 1 ;t 2 ;:::;t n ) (p 1 (t 1 );p 2 (t 1 ;t 2 );:::;p n (t 1 ;t 2 ;:::;t n )) is one-to-one from (0; 1) n onto A n. After the above observation the rest of the proof is straightforward. Lemma 3. It holds that S 4 = max{l 4 (t; t 1 ;:::;t 4 ): (t; t 1 ;:::;t 4 ) (0; + ) [0; 1] 4 }; S 4 = max{ L 4 (t; t 1 ;:::;t 4 ) : (t; t 1 ;:::;t 4 ) (0; + ) [0; 1] 4 } s 4 = min{l 4 (t; t 1 ;:::;t 4 ): (t; t 1 ;:::;t 4 ) (0; + ) [0; 1] 4 } where L 4 (t; t 1 ;t 2 ;t 3 ;t 4 )=e t [1 (p 1 (t 1 )+p 2 (t 1 ;t 2 )+p 3 (t 1 ;t 2 ;t 3 )+p 4 (t 1 ;t 2 ;t 3 ;t 4 ))t +( 1 2 (p2 1(t 1 )+p 2 2(t 1 ;t 2 )) + p 1 (t 1 )(p 2 (t 1 ;t 2 )+p 3 (t 1 ;t 2 ;t 3 )))t ( 1 3 p3 1(t 1 )+p 2 1(t 1 )p 2 (t 1 ;t 2 ))t p4 1(t 1 )t 4 ]: Proof. Without loss of generality we suppose that f B 0 (R) if can be written in the form f =e tp with t 0 p P R. Therefore from Lemma 2 it follows that f(z)=e t[1+p1(t1)z+p2(t1;t2)z2 + +p n(t 1;t 2;:::;t n)z n + ] : (5) By Taylor expansion of the previous form of f after elementary calculations we obtain f 0 + f f 3 = L 3 f 0 + f f 4 = L 4. For the next Theorem we will need the following polynomials: 0 (t)= t t t t t t t t t t t 11 ; (6)

5 P. Koulorizos, N. Samaris / Journal of Computational Applied Mathematics 139 (2002) (t) = t t t t t t t t t 9 ; (7) 1(t)=4t( t t t t t t t t 8 ); (8) 2 (t; t 1 )= 11 4t 2 (1 2t 1 ) 2 16t 1 ( 7+8t 1 )+4t(5 30t 1 +32t1); 2 (9) 2(t; t 1 ) = 64( 2+t)( 1+t 1 )t 1 ; (10) 3 (t; t 1 ;t 2 )=1+2t +4t 1 (2 t 2t 2 ); (11) 3(t; t 1 ;t 2 ) = 8(1 t 2 ); (12) 0(t)=3( 17+4t) 2 ( t t t t t 5 ); 1(t)= t t t t 4 ; (14) 1(t)=4t( t 2000t t 3 ); (15) 2(t; t 1 )= 3+4t(1 t) 8tt 1 (7 2t 8t 1 +2tt 1 ); (16) (13) 2(t; t 1 )=64t( 1+t 1 )t 1 ; (17) 3(t; t 1 ;t 2 )=3 2t +4t 1 ( 2+t +2t 2 ) (18) 3(t; t 1 ;t 2 )=8t 2 : Considering the Lau problem for B 0 (R) we nd the following partial results: (19) Theorem 1. For S 4 ;S4 s 4 it holds that: (i) S 4 =1:46109 ::: is the value of the function L 4 (t; t 1 ;t 2 ;t 3 ; 1) for t = t 0 0: :::; while t 0 is the root of the equation 0 (t)=0; t 0 t 1 = 1(t) 1(t) ; t 2 = 2(t; t 1 ) 2(t; t 1 ) (ii) S4 = S 4 t 3 = 3(t; t 1 ;t 2 ) 3(t; t 1 ;t 2 ) ; (iii) s 4 = 0: ::: is the value of the function L 4 (t; t 1 ;t 2 ;t 3 ; 0) for t = t0 1:11167 :::; while is the root of the equation 0 (t)=0; t 1 = 1 (t) 1 (t); t 2 = 2 (t; t 1 ) 2 (t; t 1) t 3 = 3 (t; t 1;t 2 ) 3 (t; t 1;t 2 ) :

6 134 P. Koulorizos, N. Samaris / Journal of Computational Applied Mathematics 139 (2002) Proof. At rst, we remark that: 1. If we set in any function L k (k =1;:::;4), t i =0 or t i =1 (i =1; 2; 3; 4), the value of the function does not depend on the variables t j when j i. 2. The polynomial L 4 is of rst degree in t 4 with corresponding coecient 256e t t( 1 + t 1 )t 1 ( 1+t 2 )t 2 ( 1+t 3 )t 3 which is non-negative. Therefore in order to nd S 4 we set t 4 = 1 for s 4 we set t 4 =0. 3. In order to calculate value S 4 we consider all the restrictions of the function L 4 for t 4 =1, t i =1 t i =0, i =1; 2; 3. We then nd all critical points in the interior of the denition domains of each restriction their corresponding values. The larger of these values coincides with S 4. In a similar way we are working for s Let =( 0 ; 1 ;:::; n ) C n, b =(b 0 ;b 1 ;:::;b k ) C k, P (x)= x + :::; n x n P b (x)= b 0 + b 1 x + + b k x k. The equations P (x) =0 P b (x) = 0 yield two relations of the form A(; b)x + B(; b) =0 C(; b) = 0 where A(; b), B(; b) C(; b) are polynomials in each element of, b, with the following procedure. The rst polynomial P (x) is divided with the second P b (x), proceeding with the division of P b (x) with the remainder of the initial polynomial division. The nal elimination of x is achieved by recursions of the procedure described, which will be used several times through this paper. For t 4 =1: 4 =@t 3 = 0 gives h 1 (t; t 1 ;t 2 )( 1 2t 8t 1 +4tt 1 +8t 1 t 2 +8t 3 8t 2 t 3 )=0; (20) where h 1 (t; t 1 ;t 2 )= e t 64t( 1+t 1 )t 1 ( 1+t 2 )t 2. Since h 1 (t; t 1 ;t 2 ) 0, then t 3 3(t; t 1 ;t 2 ) 3 (t; t 1 ;t 2 )=0: (21) Eliminating t 3 between 4 =@t 2 = 0 we get h 2 (t; t 1 )(11 20t +4t 2 112t tt 1 16t 2 t t tt t 2 t t 1 t 2 64tt 1 t 2 128t1t tt1t 2 2 ) = 0 (22) where h 2 (t; t 1 )=e t 4t( 1+t 1 )t 1. Since h 2 (t; t 1 ) 0, then t 2 2(t; t 1 ) 2 (t; t 1 )=0: (23) Eliminating t 2 4 =@t 1 4 =@t = 0 (23) we have, respectively, h 3 (t; t 1 ) ( 2+t) =0; (24) 2 where h 3 (t; t 1 )=( t 5331t t 3 888t 4 +16t 6 16t 1 ( t 1431t t 3 74t 4 52t 5 +8t 6 )+96t 2 1( t +20t t 3 44t 4 +4t 5 ) 256t 2 t 3 1 ( t +93t 2 27t 3 +2t 4 ) 256t 3 t 4 1(64+50t 14t 2 + t 3 )) h 4 (t; t 1 ) ( 2+t) =0; (25)

7 P. Koulorizos, N. Samaris / Journal of Computational Applied Mathematics 139 (2002) where h 4 (t; t 1 )=t( t + 174t 2 +4t 3 8t 4 12tt 1 ( 32+37t +16t 2 )+t 2 t 2 1( 2 15t +2t 2 ) +8t 3 t 3 1( 64+8t)): Applying the procedure described in Remark 4 to (24) (25) with respect to t 1 we get h 5 (t; t 1 ) ( 8+t) 2 ( 2+t) =0; 2 where (26) h 5 (t; t 1 )= t 17268t t t t 5 364t 6 +16t 7 8tt 1 ( t t t t 4 186t 5 +8t 6 ) +16t 2 t 2 1( t 3560t t 3 95t 4 +4t 5 )) ( 8+t) 2 ( 2+t)[t 1 1(t) 1 (t)]=0: (27) Therefore for t 2 t 8 by (26) (27) we obtain 3( t 3560t t 3 95t 4 +4t 5 ) 2 0 (t)=0: (28) Solving Eq. (28) with Mathematica 4:0 we succeed in nding all its roots which are: 1 = 2 = 1:54984 :::; 3 = 4 =2:63304 :::; 5 = 6 =8:31271 :::; 7 = 8 =5:6272 ::: i 1:36096 :::; 9 = 10 = 5:6272 :::+ i 1:36096 :::; 11 =0: :::; 12 =1:20285 :::; 13 =1:67847 :::; 14 =1:81244 :::; 15 =2:3582 :::; 16 =2:98161 :::; 17 =4:12162 :::; 18 =4:46949 ::: i 0: :::; 19 =4:46949 ::: + i 0: :::; 20 =6:33369 ::: i 1:39775 :::; 21 =6:33369 :::+ i 1:39775 :::. For t =2: the equation h 5 (2;t 1 ) = 0 has the roots r 1 =(4 3)=8 r 2 =(4+ 3)=8. Solving the system of equations h 5 (2;t 1 4 =@t 1 = 0 (21) with respect to t 1, t 2 t 3 we obtain as solutions the points ( (t 1 ;t 2 ;t 3 )= ; ) 3 ; 4( 3+ 3)( ) (0: :::;0: :::;0: :::) ( (t 1 ;t 2 ;t 3 )= 4+ 3 ; ; (3 + 3)( ) ) (0: :::;0: :::;0: :::):

8 136 P. Koulorizos, N. Samaris / Journal of Computational Applied Mathematics 139 (2002) For t =8: in the same manner we obtain the point ( (t 1 ;t 2 ;t 3 )= ; 32 3( )( ) ; ( 42 + ) 615)( ) 4( ) (0: :::;0: :::;0: :::): Using the roots of Eq. (28) the solutions of the systems formed in the cases t = 2 t =8, after the calculations we nd that all the critical points of function L 4 (t; t 1 ;t 2 ;t 3 ; 1) for (t; t 1 ;t 2 ;t 3 ) (0; ) (0; 1) 3, belong to the set A 1 = {(1:54984 :::;0: :::;0: :::;0: :::); (0: :::;0: :::; 0: :::;0: :::); (1:20285 :::;0: :::;0: :::;0: :::); (1:67847 :::;0: :::;0: :::;0: :::); (1:81244 :::;0: :::; 0: :::;0: :::); (2:3582 :::;0: :::;0: :::;0: :::); (2:98161 :::;0: :::;0: :::;0: :::); (4:12162 :::;0: :::; 0: :::;0: :::); (8:31271 :::;0: :::;0: :::;0: :::); (2: :::;0: :::;0: :::;0: :::); (2: :::;0: :::; 0: :::;0: :::); (8: :::;0: :::;0: :::;0: :::)}; max{l 4 (t; t 1 ;t 2 ;t 3 ; 1):(t; t 1 ;t 2 ;t 3 ) A 1 } =1: ::: min{l 4 (t; t 1 ;t 2 ;t 3 ; 1):(t; t 1 ;t 2 ;t 3 ) A 1 } = 0: :::. For t 4 =0: proceeding exactly as in previous case, we derive the set A 2 = {(1:11167 :::;0: :::;0: :::;0: :::); (2:36682 :::;0: :::; 0: :::;0: :::); (2:76576 :::;0:80512 :::;0: :::;0: :::)}; max{l 4 (t; t 1 ;t 2 ;t 3 ; 0): (t; t 1 ;t 2 ;t 3 ) A 2 } = 0:3034 ::: min{l 4 (t; t 1 ;t 2 ;t 3 ; 0):(t; t 1 ;t 2 ;t 3 ) A 2 } = 0: :::. real values for t 1, t 2 t 3. For t 3 =1: we consider the =0: Applying to (29) the procedure of eliminating t 2 t 1, we examine separately as we did previously, some cases of the form q i (t)=0, i =1; 2; 3;:::. Thus we obtain three polynomial equations of the form t 2 2 (t; t 1 ) 2 (t; t 1 )=0; (30) t 1 1 (t) 1 (t) = 0 (31)

9 P. Koulorizos, N. Samaris / Journal of Computational Applied Mathematics 139 (2002) (t)=0: (32) We omit giving explicitly the polynomials q i (t), 1 (t), 2 (t; t 1), 1 (t), 2 (t; t 1), 0 (t) because of their size. Especially the last polynomial 0 (t) can be converted to a product of factors of the form 9( 1+2t) 2 1(t) 2 2(t) 3 (t) 2 4(t); (33) where 1 (t) is of sixth degree, 2 (t) of twelfth degree, 3 (t) of fteenth degree 4 (t) of thirtieth degree with respect to t, respectively. Continuing we nd all one hundred twelve roots of Eq. (32). Using the roots of Eq. (32) the points that satisfy the relations q i (t)=0i =1; 2; 3;:::, after the calculations we nd that all the critical points of function L 4 (t; t 1 ;t 2 ; 1; 1) for (t; t 1 ;t 2 ) (0; ) (0; 1) 2, belong to the set A 3 = {(0: :::;0: :::;0: :::); (0: :::;0: :::;0: :::); (0: :::;0: :::;0: :::); (1: :::;0: :::;0: :::); (1: :::;0: :::;0: :::); (1: :::;0: :::;0: :::); (3: :::;0: :::;0: :::); (0: :::;0: :::;0: :::); (0: :::;0: :::;0: :::); (1: :::;0: :::;0: :::); (1: :::;0: :::;0: :::); (2: :::;0: :::;0: :::); (2: :::;0: :::;0: :::); (3:94974 :::;0: :::;0: :::); (1:07394 :::;0: :::;0: :::); (1:07394 :::;0: :::;0: :::); (6:28933 :::;0: :::;0: :::); (0: :::;0: :::;0: :::); (0: :::;0: :::;0: :::); (2:55783 :::;0: :::;0: :::); (10:243 :::;0: :::;0: :::); (47:9003 :::;0: :::;0: :::); (47:9003 :::;0: :::;0: :::); (0:3984 :::;0: :::;0: :::); (0: :::;0: :::;0: :::); (1:14692 :::;0: :::;0: :::); (1:49999 :::;0: :::;0: :::); (2:40766 :::;0: :::;0: :::); (2:69004 :::;0: :::;0: :::); (0: :::;0: :::;0: :::); (0: :::;0: :::;0: :::)}; max{l 4 (t; t 1 ;t 2 ; 1; 1):(t; t 1 ;t 2 ) A 3 } =1: ::: min{l 4 (t; t 1 ;t 2 ; 1; 1): (t; t 1 ;t 2 ) A 3 } = :25246 :::.

10 138 P. Koulorizos, N. Samaris / Journal of Computational Applied Mathematics 139 (2002) For t 3 =0: proceeding exactly as in previous case we derive A 4 = {(0: :::;0: :::;0: :::); (2: :::;0: :::;0: :::); (4: :::;0: :::;0: :::); (0: :::;0: :::;0: :::); (0: :::;0: :::;0: :::); (0: :::;0: :::;0: :::); (1: :::;0: :::;0: :::); (4: :::;0: :::;0: :::); (1: :::;0: :::;0: :::); (1: :::;0: :::;0: :::); (3: :::;0: :::;0: :::); (3:71729 :::;0: :::;0: :::); (0: :::;0: :::;0: :::); (0: :::;0: :::;0:38198 :::); (8:10891 :::;0: :::;0: :::); (1:67665 :::;0: :::;0: :::); (11:7856 :::;0: :::;0: :::); (49:2188 :::;0: :::;0: :::); (49:2188 :::;0: :::;0: :::); (1:52606 :::;0: :::;0: :::); (1:58586 :::;0: :::;0: :::); (4:27136 :::;0: :::;0: :::)}; max{l 4 (t; t 1 ;t 2 ; 0; 1): (t; t 1 ;t 2 ) A 4 } = 0: ::: min{l 4 (t; t 1 ;t 2 ; 0; 1): (t; t 1 ;t 2 ) A 4 } = 0: :::. For t 2 =1: by the same procedure, through simpler calculations than before we have A 5 = {(0: :::;0: :::); (1:3839 :::;0: :::); (1:66779 :::;0: :::); (2:89925 :::;0: :::); (3:13721 :::;0: :::); (3:81812 :::;0: :::); (3: :::;0: :::)}; max{l 4 (t; t 1 ; 1; 1; 1): (t; t 1 ) A 5 } = 0: ::: min{l 4 (t; t 1 ; 1; 1; 1): (t; t 1 ) A 5 } = 0: :::. For t 2 =0: in the same manner we get A 6 = {(0: :::;0: :::); (0: :::;0:40951 :::); (0: :::;0: :::); (0: :::;0:016 :::); (0: :::;0: :::); (0: :::;0: :::); (1:23589 :::;0: :::); (2:04847 :::;0: :::); (3:65511 :::;0: :::); (5:16722 :::;0: :::); (0: :::;0: :::); (0: :::;0:49065 :::); (1:14202 :::;0: :::); (3:61717 :::;0: :::); (8:07298 :::;0:54914 :::); (0: :::;0: :::); (0: :::;0: :::); (3:30541 :::;0:18893 :::); (7:75877 :::;0: :::)};

11 P. Koulorizos, N. Samaris / Journal of Computational Applied Mathematics 139 (2002) max{l 4 (t; t 1 ; 0; 1; 1): (t; t 1 ) A 6 } = 1:34838 ::: min{l 4 (t; t 1 ; 0; 1; 1): (t; t 1 ) A 6 } = 0: :::. For t 1 =1: the 4 =@t = 0 yields 1 3 e t (27 96t +84t 2 24t 3 +2t 4 )=0; (34) therefore A 7 = {0: :::;1:44592 :::;3:29095 :::;6:8489 :::}, max{l 4 (t; 1; 1; 1; 1): t A 7 } =0: ::: min{l 4 (t; 1; 1; 1; 1): t A 7 } = 0: :::. For t 1 =0: similarly we get 1 3 e t (3 24t +36t 2 16t 3 +2t 4 )=0; (35) with A 8 = {0: :::;0: :::;2:26831 :::;4:69754 :::}, max{l 4 (t; 0; 1; 1; 1): t A 8 } = 1: ::: min{l 4 (t; 0; 1; 1; 1): t A 8 } =0: :::. Comparing the occurring values for the maximum the minimum of L 4 of the relative sets A i (i =1;:::;8), we obtain the result needed. Remark. If we had considered initially the class B 4 = {f B 0: f i R, i =1;:::;4} instead of the class B 0 (R), all our evaluations wouldn t have been modied. References [1] E. Lau, Abschatzung der koezientensummme einer potenzreihe, Arch. Math. Phys. 21 (1913) [2] Z. Lewowski, J. Szynal, The Lau problem for bounded nonvanishing functions, J. Comput. Appl. Math. 105 (1999) [3] A. Ganczar, M. Michalska, J. Szynal, On the Lau problem for bounded nonvanishing functions, Ann. Univ. Mariae Curie Sklodowska Sect. A 52 (2) (1998) [4] C. Caratheodory, Uber den variabilitatsbereich der koezienten von potenzreihen die gegebene Werte nicht annehmen, Math. Ann. 64 (1907) [5] O. Toeplitz, Uber die Fourier sche entwickelung positiver funktionen, Rend. Circ. Mat. Palermo 32 (1911)