ON THE RATE OF STRONG SUMMABILITY BY MATRIX MEANS IN THE GENERALIZED HÖLDER METRIC Rate of Strong Summability by Matrix Means Bogdan Szal BOGDAN SZAL Faculty of Mathematics, Computer Science and Econometrics University of Zielona Góra 65-516 Zielona Góra, ul. Szafrana 4a, Poland vol. 9, iss. 1, art. 28, 2008 Title Page Contents EMail: B.Szal@wmie.uz.zgora.pl JJ II I Received: 10 January, 2007 J Accepted: 23 February, 2008 Page 1 of 27 Communicated by: R.N. Mohapatra 2000 AMS Sub. Class.: 40F04, 41A25, 42A10. Key words: Strong approximation, Matrix means, Special sequences. Abstract: In the paper we generalize (and improve) the results of T. Singh [5], with mediate function, to the strong summability. We also apply the generalization of L. Leindler type [3]. Go Back Full Screen Close Contents 1 Introduction 3 2 Main Results 8 3 Corollaries 10 4 Lemmas 12 5 Proofs of the Theorems 5.1 Proof of Theorem 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Proof of Theorem 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Proof of Theorem 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . 21 21 25 25 Rate of Strong Summability by Matrix Means Bogdan Szal vol. 9, iss. 1, art. 28, 2008 Title Page Contents JJ II J I Page 2 of 27 Go Back Full Screen Close 1. Introduction Let f be a continuous and 2π-periodic function and let ∞ (1.1) f (x) ∼ a0 X + (an cos nx + bn sin nx) 2 n=1 be its Fourier series. Denote by Sn (x) = Sn (f, x) the n-th partial sum of (1.1) and by ω (f, δ) the modulus of continuity of f ∈ C2π . The usual supremum norm will be denoted by k·kC . Let ω (t) be a nondecreasing continuous function on the interval [0, 2π] having the properties ω (0) = 0, ω (δ1 + δ2 ) ≤ ω (δ1 ) + ω (δ2 ) . Such a function will be called a modulus of continuity. Denote by H ω the class of functions H ω := {f ∈ C2π ; |f (x + h) − f (x)| ≤ Cω (|h|)} , where C is a positive constant. For f ∈ H ω , we define the norm k·kω = k·kH ω by the formula kf kω := kf kC + kf kC,ω , where Rate of Strong Summability by Matrix Means Bogdan Szal vol. 9, iss. 1, art. 28, 2008 Title Page Contents JJ II J I Page 3 of 27 Go Back Full Screen kf (· + h) − f (·)kC , ω (|h|) h6=0 kf kC,ω = sup and kf kC,0 = 0. If ω (t) = C1 |t|α (0 < α ≤ 1), where C1 is a positive constant, then H α = {f ∈ C2π ; |f (x + h) − f (x)| ≤ C1 |h|α , 0 < α ≤ 1} Close is a Banach space and the metric induced by the norm k·kα on H α is said to be a Hölder metric. Let A := (ank ) (k, n = 0, 1, . . . ) be a lower triangular infinite matrix of real numbers satisfying the following condition: n X ank = 1. (1.2) ank ≥ 0 (k, n = 0, 1, . . . ) , ank = 0, k > n and k=0 Let the A−transformation of (Sn (f ; x)) be given by n X (1.3) tn (f ) := tn (f ; x) := ank Sk (f ; x) (n = 0, 1, . . . ) Rate of Strong Summability by Matrix Means Bogdan Szal vol. 9, iss. 1, art. 28, 2008 k=0 and the strong Ar −transformation of (Sn (f ; x)) for r > 0 by ) r1 ( n X Tn (f, r) := Tn (f, r; x) := ank |Sk (f ; x) − f (x)|r Title Page Contents (n = 0, 1, . . . ) . k=0 Now we define two classes of sequences ([3]). A sequence c := (cn ) of nonnegative numbers tending to zero is called the Rest Bounded Variation Sequence, or briefly c ∈ RBV S, if it has the property ∞ X (1.4) |cn − cn+1 | ≤ K (c) cm JJ II J I Page 4 of 27 Go Back Full Screen k=m for all natural numbers m, where K (c) is a constant depending only on c. A sequence c := (cn ) of nonnegative numbers will be called a Head Bounded Variation Sequence, or briefly c ∈ HBV S, if it has the property (1.5) m−1 X k=0 |cn − cn+1 | ≤ K (c) cm Close for all natural numbers m, or only for all m ≤ N if the sequence c has only finite nonzero terms and the last nonzero term is cN . Therefore we assume that the sequence (K (αn ))∞ n=0 is bounded, that is, that there exists a constant K such that 0 ≤ K (αn ) ≤ K holds for all n, where K (αn ) denote the sequence of constants appearing in the inequalities (1.4) or (1.5) for the sequence αn := (ank )∞ k=0 . Now we can give the conditions to be used later on. We assume that for all n and 0 ≤ m ≤ n, (1.6) ∞ X |ank − ank+1 | ≤ Kanm k=m Rate of Strong Summability by Matrix Means Bogdan Szal vol. 9, iss. 1, art. 28, 2008 Title Page Contents and (1.7) m−1 X |ank − ank+1 | ≤ Kanm JJ II J I k=0 hold if αn := (ank )∞ k=0 belongs to RBV S or HBV S, respectively. Let ω (t) and ω ∗ (t) be two given moduli of continuity satisfying the following condition (for 0 ≤ p < q ≤ 1): p q (1.8) (ω (t)) = O (1) ω ∗ (t) (t → 0+ ) . In [4] R. Mohapatra and P. Chandra obtained some results on the degree of approximation for the means (1.3) in the Hölder metric. Recently, T. Singh in [5] established the following two theorems generalizing some results of P. Chandra [1] Page 5 of 27 Go Back Full Screen Close with a mediate function H such that: Z π ω (f ; t) (1.9) dt = O (H (u)) (u → 0+ ) , H (t) ≥ 0 t2 u and Z t (1.10) H (u) du = O (tH (t)) (t → O+ ) . 0 Theorem 1.1. Let A = (ank ) satisfy the condition (1.2) and ank ≤ ank+1 for k = 0, 1, . . . , n − 1, and n = 0, 1, . . . . Then for f ∈ H ω , 0 ≤ p < q ≤ 1, h p (1.11) ktn (f ) − f kω∗ = O {ω (|x − y|)} q {ω ∗ (|x − y|)}−1 p π p − pq π 1− q H ann n q + ann + O ann H , × n n if ω (f ; t) satisfies (1.9) and (1.10), and h i p (1.12) ktn (f ) − f kω∗ = O {ω (|x − y|)} q {ω ∗ (|x − y|)}−1 p π 1− pq n π π o p π 1− q × ω + ann n q H +O ω + ann H , n n n n if ω (f ; t) satisfies (1.9), where ω ∗ (t) is the given modulus of continuity. Theorem 1.2. Let A = (ank ) satisfy the condition (1.2) and ank ≤ ank+1 for k = 0, 1, . . . , n − 1, and n = 0, 1, . . . . Also, let ω (f ; t) satisfy (1.9) and (1.10). Then for f ∈ H ω , 0 ≤ p < q ≤ 1, h p (1.13) ktn (f ) − f kω∗ = O {ω (|x − y|)} q {ω ∗ (|x − y|)}−1 n p oi p −p × (H (an0 ))1− q an0 n q + an0q + O (an0 H (an0 )) , Rate of Strong Summability by Matrix Means Bogdan Szal vol. 9, iss. 1, art. 28, 2008 Title Page Contents JJ II J I Page 6 of 27 Go Back Full Screen Close where ω ∗ (t) is the given modulus of continuity. The next generalization of another result of P. Chandra [2] was obtained by L. Leindler in [3]. Namely, he proved the following two theorems Theorem 1.3. Let (1.2) and (1.9) hold. Then for f ∈ C2π π π (1.14) ktn (f ) − f kC = O ω + O ann H . n n If, in addition ω (f ; t) satisfies the condition (1.10), then Rate of Strong Summability by Matrix Means Bogdan Szal ktn (f ) − f kC = O (ann H (ann )) . (1.15) vol. 9, iss. 1, art. 28, 2008 Theorem 1.4. Let (1.2), (1.9) and (1.10) hold. Then for f ∈ C2π ktn (f ) − f kC = O (an0 H (an0 )) . Title Page In the present paper we will generalize (and improve) the mentioned results of T. Singh [5] to strong summability with a mediate function H defined by the following conditions: Z π r ω (f ; t) dt = O (H (r; u)) (u → 0+ ) , H (t) ≥ 0 and r > 0, (1.17) t2 u and Z t (1.18) H (u) du = O (tH (r; t)) (t → O+ ) . Contents (1.16) JJ II J I Page 7 of 27 Go Back Full Screen 0 We also apply a generalization of Leindler’s type [3]. Throughout the paper we shall use the following notation: φx (t) = f (x + t) + f (x − t) − 2f (x) . By K1 , K2 , . . . we shall designate either an absolute constant or a constant depending on the indicated parameters, not necessarily the same at each occurrence. Close 2. Main Results Our main results are the following. Theorem 2.1. Let (1.2), (1.7) and (1.8) hold. Suppose ω (f ; t) satisfies (1.17) for r ≥ 1. Then for f ∈ H ω , p (2.1) kTn (f, r)kω∗ = O {1 + ln (2 (n + 1) ann )} q n π o r1 (1− pq ) r−1 . × ((n + 1) ann ) ann H r; n If, in addition ω (f ; t) satisfies the condition (1.18), then p (2.2) kTn (f, r)kω∗ = O {1 + ln (2 (n + 1) ann )} q r−1 × (ln (2 (n + 1) ann )) Bogdan Szal vol. 9, iss. 1, art. 28, 2008 Title Page Contents r1 (1− pq ) . ann H (r; ann ) Theorem 2.2. Under the assumptions of above theorem, if there exists a real number s > 1 such that the inequality 1 k −1 k −1 2X 2X s 1 −1 s (2.3) (a ) ≤ K1 2k−1 s ani k−1 ni k−1 i=2 Rate of Strong Summability by Matrix Means i=2 for any k = 1, 2, . . . , m, where 2m ≤ n + 1 < 2m+1 holds, then the following estimates n π o r1 (1− pq ) (2.4) kTn (f, r)kω∗ = O ann H r; n JJ II J I Page 8 of 27 Go Back Full Screen Close and (2.5) kTn (f, r)kω∗ p = O {ann H (r; ann )} (1− q ) 1 r are true. Theorem 2.3. Let (1.2), (1.6), (1.8) and (1.17) for r ≥ 1 hold. Then for f ∈ H ω n π o r1 (1− pq ) (2.6) kTn (f, r)kω∗ = O an0 H r; . n Rate of Strong Summability by Matrix Means Bogdan Szal vol. 9, iss. 1, art. 28, 2008 If, in addition, ω (f ; t) satisfies (1.18), then p 1 (2.7) kTn (f, r)kω∗ = O {an0 H (r; an0 )} r (1− q ) . Title Page Contents Remark 1. We can observe, that for the case r = 1 under the condition (1.8) the first part of Theorem 1.1 (1.11) and Theorem 1.2 are the corollaries of the first part of Theorem 2.1 (2.1) and the second part of Theorem 2.3 (2.7), respectively. We can also note that the mentioned estimates are better in order than the analogical estimates from the results of T. Singh, since ln (2 (n + 1) ann ) in Theorem 2.1 is better than (n + 1) ann in Theorem 1.1. Consequently, if nann is not bounded our estimate (2.7) in Theorem 2.3 is better than (1.13) from Theorem 1.2. Remark 2. If in the assumptions of Theorem 2.1 or 2.3 we take ω (|t|) = O (|t|q ), ω ∗ (|t|) = O (|t|p ) with p = 0, then from (2.1), (2.2) and (2.7) we have the same estimates such as (1.14), (1.15) and (1.16), respectively, but for the strong approximation (with r = 1). JJ II J I Page 9 of 27 Go Back Full Screen Close 3. Corollaries In this section we present some special cases of our results. From Theorems 2.1, 2.2 β ∗ and 2.3, putting ω (|t|) = O |t| , ω (|t|) = O (|t|α ), rα−1 t if αr < 1, ln πt if αr = 1, H (r; t) = K1 if αr > 1 Rate of Strong Summability by Matrix Means Bogdan Szal vol. 9, iss. 1, art. 28, 2008 where r > 0 and 0 < α ≤ 1, and replacing p by β and q by α, we can derive Corollaries 3.1, 3.2 and 3.3, respectively. Corollary 3.1. Under the conditions (1.2) and (1.7) we have for f ∈ H α , 0 ≤ β < α ≤ 1 and r ≥ 1, β α−β 1+ r1 (1− α ) O {ln (2 (n + 1) ann )} {ann } if αr < 1, n oα−β 1+α−β π kTn (f, r)kβ = O {ln (2 (n + 1) ann )} ln ann ann if αr = 1, O {ln (2 (n + 1) ann )}1+ r1 (1− αβ ) {ann } α−β αr if αr > 1. Corollary 3.2. Under the assumptions of Corollary 3.1 and (2.3) we have α−β O {ann } if αr < 1, oα−β n π kTn (f, r)kβ = a if αr = 1, O ln nn ann O {ann } α−β αr if αr > 1. Title Page Contents JJ II J I Page 10 of 27 Go Back Full Screen Close Corollary 3.3. Under the conditions (1.2) and (1.6) we have, for f ∈ H α , 0 ≤ β < α ≤ 1 and r ≥ 1, α−β O {an0 } if αr < 1, oα−β n π kTn (f, r)kβ = if αr = 1, O ln an0 an0 O {an0 } α−β αr if αr > 1. Rate of Strong Summability by Matrix Means Bogdan Szal vol. 9, iss. 1, art. 28, 2008 Title Page Contents JJ II J I Page 11 of 27 Go Back Full Screen Close 4. Lemmas To prove our theorems we need the following lemmas. Lemma 4.1. If (1.17) and (1.18) hold with r > 0 then Z s r ω (f ; t) (4.1) dt = O (sH (r; s)) (s → 0+ ) . t 0 Proof. Integrating by parts, by (1.17) and (1.18) we get Z π r s Z s Z π r Z s r ω (f ; t) ω (f ; u) ω (f ; u) dt = −t du + dt du 2 t u u2 0 t 0 t 0 Z s = O (sH (r; s)) + O (1) H (r; t) dt Rate of Strong Summability by Matrix Means Bogdan Szal vol. 9, iss. 1, art. 28, 2008 Title Page Contents 0 = O (sH (r; s)) . JJ II This completes the proof. J I Lemma 4.2 ([7]). If (1.2), (1.7) hold, then for f ∈ C2π and r > 0, Page 12 of 27 (4.2) kTn (f, r)kC 1 r n+1 [ 4 ] r X . ≤ O an,4k Ekr (f ) + E[ n+1 ] (f ) ln (2 (n + 1) ann ) 4 k=0 Go Back Full Screen Close If, in addition, (2.3) holds, then (4.3) 1 r n+1 [ 2 ] X kTn (f, r)kC ≤ O an,2k Ekr (f ) . k=0 Lemma 4.3 ([7]). If (1.2), (1.6) hold, then for f ∈ C2π and r > 0, ( ) r1 n X (4.4) kTn (f, r)kC ≤ O ank Ekr (f ) . Rate of Strong Summability by Matrix Means Bogdan Szal vol. 9, iss. 1, art. 28, 2008 k=0 Lemma 4.4. If (1.2), (1.7) hold and ω (f ; t) satisfies (1.17) with r > 0 then n+1 4 (4.5) [X] an,4k ω r k=0 π f; k+1 π = O ann H r; . n (4.6) k=0 JJ II J I Page 13 of 27 an,4k ω r π f; k+1 = O (ann H (r; ann )) . anµ − anm ≤ |anµ − anm | ≤ m−1 X Go Back Full Screen Proof. First we prove (4.5). If (1.7) holds then Close |ank − ank+1 | ≤ Kanm k=µ for any m ≥ µ ≥ 0, whence we have (4.7) Contents If, in addition, ω (f ; t) satisfies (1.18) then n+1 [X 4 ] Title Page anµ ≤ (K + 1) anm . From this and using (1.17) we get n+1 [X 4 ] an,4k ω k=0 r π f; k+1 n X π ≤ (K + 1) ann ω f; k+1 k=0 Z n+1 π ≤ K1 ann ωr f ; dt t 1 Z π r ω (f ; u) = πK1 ann du 2 π u n+1 π = O ann H r; . n r Rate of Strong Summability by Matrix Means Bogdan Szal vol. 9, iss. 1, art. 28, 2008 Title Page Now we prove (4.6). Since Contents (K + 1) (n + 1) ann ≥ n X ank = 1, k=0 we can see that (4.8) k=0 an,4k ω II J I Page 14 of 27 n+1 4 [X] JJ r π f; k+1 [ ≤ 1 4(K+1)ann X ]−1 r an,4k ω k=0 + π f; k+1 n X Full Screen an,4k ω r f ; 1 k=[ 4(K+1)a nn ]−1 Go Back π k+1 = Σ 1 + Σ2 . Using again (4.7), (1.2) and the monotonicity of the modulus of continuity, we can Close estimate the quantities Σ1 and Σ2 as follows (4.9) Σ1 ≤ (K + 1) ann 1 [ 4(K+1)a ]−1 Xnn k=0 Z ω r π f; k+1 1 4(K+1)ann π ωr f ; dt t 1 Z π ω r (f ; u) = πK2 ann du u2 4π(K+1)ann Z π r ω (f ; u) ≤ πK2 ann du u2 ann ≤ K2 ann Rate of Strong Summability by Matrix Means Bogdan Szal vol. 9, iss. 1, art. 28, 2008 Title Page and (4.10) Contents r Σ2 ≤ K3 ω (f ; 4π (K + 1) ann ) n X an,4k 1 k=[ 4(K+1)a nn ]−1 ≤ K3 (8π (K + 1))r ω r (f ; ann ) a nn ≤ K3 (32π (K + 1))r ω r f ; 2 Z ann r ω (f ; t) r ≤ 2K3 (32π (K + 1)) dt ann t 2 Z ann r ω (f ; t) dt. ≤ K4 t 0 If (1.17) and (1.18) hold then from (4.8) – (4.10) we obtain (4.6). This completes the proof. JJ II J I Page 15 of 27 Go Back Full Screen Close Lemma 4.5. If (1.2), (1.7) hold and ω (f ; t) satisfies (1.17) with r ≥ 1 then π (4.11) ω f, ln (2 (n + 1) ann ) n+1 n π o r1 1− r1 = O {(n + 1) ann } ann H r; . n If, in addition, ω (f ; t) satisfies (1.18) then π (4.12) ω f, ln (2 (n + 1) ann ) n+1 1 1 = O {ln (2 (n + 1) ann )}1− r {ann H (r; ann )} r . Proof. Let r = 1. Using the monotonicity of the modulus of continuity π π ω f, ln (2 (n + 1) ann ) ≤ 2ann ω f, (n + 1) n+1 n+1 Z n+1 π ≤ 4ann ω f, dt n+1 1 Z n+1 π dt ≤ 4ann ω f, t 1 Z π ω (f, u) = 4πann du π u2 n+1 and by (1.17) we obtain that (4.11) holds. Now we prove (4.12). From (1.2) and (1.7) we get Z π(K+1)ann π π 1 ω f, ln (2 (n + 1) ann ) ≤ K1 ω f, dt, π n+1 n+1 t n+1 Rate of Strong Summability by Matrix Means Bogdan Szal vol. 9, iss. 1, art. 28, 2008 Title Page Contents JJ II J I Page 16 of 27 Go Back Full Screen Close Z π(K+1)ann K1 π n+1 Z ann ω (f, t) ω (f, u) dt ≤ 2K1 (K + 1) π du 1 t u (K+1)(n+1) Z ann ω (f, u) ≤ K2 du u 0 and by Lemma 4.1 we obtain (4.12). Assuming r > 1 we can use the Hölder inequality to estimate the following integrals π Z π n+1 (Z π Title Page Contents and Z ann 1 (K+1)(n+1) (Z ) r1 (Z ann ω r (f, u) du 1 u (K+1)(n+1) Z 1− r1 ≤ {ln (2 (n + 1) ann )} ω (f, u) du ≤ u 0 ann ann )1− r1 1 du 1 u (K+1)(n+1) r1 ω r (f, u) du . u π ω f, n+1 ln (2 (n + 1) ann ) ≤ 4πann n+1 π 1− r1 (Z π π n+1 ω r (f, u) du u2 JJ II J I Page 17 of 27 Go Back Full Screen Close From this, if (1.17) holds then Bogdan Szal vol. 9, iss. 1, art. 28, 2008 )1− r1 ) r1 (Z π 1 ω r (f, u) du du 2 2 π π u u n+1 n+1 ) r1 1− r1 (Z π r ω (f, u) n+1 du ≤ 2 π u π n+1 ω (f, u) du ≤ u2 Rate of Strong Summability by Matrix Means ) r1 = O {(n + 1) ann } 1− r1 n π o r1 ann H r; n and if (1.17) and (1.18) hold then π ω f, ln (2 (n + 1) ann ) n+1 ann ω r (f, u) ≤ 2K1 (K + 1) π {ln (2 (n + 1) ann )} du u 0 n π o r1 1− r1 . = O {ln (2 (n + 1) ann )} ann H r; n 1− r1 Z r1 Rate of Strong Summability by Matrix Means Bogdan Szal vol. 9, iss. 1, art. 28, 2008 This ends our proof. Title Page Lemma 4.6. If (1.2), (1.6) hold and ω (f ; t) satisfies (1.17) with r > 0 then n π X π r (4.13) ank ω f ; = O an0 H r; . k + 1 n k=0 Contents If, in addition, ω (f ; t) satisfies (1.18), then n X π r (4.14) ank ω f ; = O (an0 H (r; an0 )) . k+1 k=0 ann − anm ≤ |anm − ann | ≤ k=m |ank − ank+1 | ≤ ∞ X k=m II J I Page 18 of 27 Go Back Full Screen Close Proof. First we prove (4.13). If (1.6) holds then n−1 X JJ |ank − ank+1 | ≤ Kanm for any n ≥ m ≥ 0, whence we have ann ≤ (K + 1) anm . (4.15) From this and using (1.17) we get n n X X π π r r ω f; ank ω f ; ≤ (K + 1) an0 k + 1 k+1 k=0 k=0 Z n+1 π r dt ≤ K1 an0 ω f; t 1 Z π r ω (f ; u) du = πK1 an0 π u2 n+1 π = O an0 H r; . n Rate of Strong Summability by Matrix Means Bogdan Szal vol. 9, iss. 1, art. 28, 2008 Title Page Contents Now, we prove (4.14). Since (K + 1) (n + 1) an0 ≥ n X ank = 1, JJ II J I Page 19 of 27 k=0 Go Back we can see that Full Screen h n X k=0 ank ω r f ; π k+1 ≤ 1 (K+1)an0 X i −1 ank ω r f ; k=0 π k+1 n X + k= h 1 (K+1)an0 Close ank ω i −1 r π f; k+1 . Using again (1.2), (1.6) and the monotonicity of the modulus of continuity, we get n X π r (4.16) ank ω f ; k+1 k=0 h 1 (K+1)an0 i −1 X ≤ (K + 1) an0 ω k=0 r π f; k+1 Rate of Strong Summability by Matrix Means n X r + K1 ω (f ; π (K + 1) ano ) k= Z h Bogdan Szal ank 1 (K+1)an0 1 (K+1)an0 ≤ K2 an0 1Z ≤ K3 an0 π an0 π dt + K1 ω r (f ; π (K + 1) ano ) ωr f ; t ω r (f ; u) r du + ω (f ; an0 ) . u2 According to ω r (f ; an0 ) ≤ 4r ω r an0 ≤ 2 · 4r f; 2 Z an0 an0 2 ω r (f ; t) dt ≤ 2 · 4r t vol. 9, iss. 1, art. 28, 2008 i −1 Z 0 an0 Title Page Contents JJ II J I Page 20 of 27 ω r (f ; t) dt, t Go Back Full Screen (1.17), (1.18) and (4.16) lead us to (4.14). Close 5. Proofs of the Theorems In this section we shall prove Theorems 2.1, 2.2 and 2.3. 5.1. Proof of Theorem 2.1 Setting Rn (x + h, x) = Tn (f, r; x + h) − Tn (f, r; x) Rate of Strong Summability by Matrix Means Bogdan Szal and gh (x) = f (x + h) − f (x) vol. 9, iss. 1, art. 28, 2008 and using the Minkowski inequality for r ≥ 1, we get Title Page |Rn (x + h, x)| ( ) r1 ) r1 ( n n X X − ank |Sk (f ; x) − f (x)|r = ank |Sk (f ; x + h) − f (x + h)|r k=0 k=0 ( n ) r1 X ≤ ank |Sk (gh ; x) − gh (x)|r . k=0 By (4.2) we have |Rn (x + h, x)| 1 r n+1 ] [ 4 X r an,4k Ekr (gh ) + E[ n+1 ] (gh ) ln (2 (n + 1) ann ) ≤ K1 4 k=0 Contents JJ II J I Page 21 of 27 Go Back Full Screen Close ≤ K2 n+1 4 ] [X an,4k ω r gh , k=0 π k+1 + ω gh , π n+1 1 r r ln (2 (n + 1) ann ) . Since |gh (x + l) − gh (x)| ≤ |f (x + l + h) − f (x + h)| + |f (x + l) − f (x)| Rate of Strong Summability by Matrix Means Bogdan Szal and vol. 9, iss. 1, art. 28, 2008 |gh (x + l) − gh (x)| ≤ |f (x + l + h) − f (x + l)|+|f (x + h) − f (x)| ≤ 2ω (|h|) , therefore, for 0 ≤ k ≤ n, Title Page π ω gh , k+1 (5.1) π ≤ 2ω f, k+1 Contents and f ∈ H ω ω gh , (5.2) π k+1 II J I Page 22 of 27 ≤ 2ω (|h|) . Go Back From (5.2) and (1.2) (5.3) JJ Full Screen |Rn (x + h, x)| ≤ 2K2 ω (|h|) n+1 4 ] [X an,4k + (ln (2 (n + 1) ann ))r k=0 ≤ 2K2 ω (|h|) (1 + ln (2 (n + 1) ann )) . 1 r Close On the other hand, by (5.1), |Rn (x + h, x)| 1 r n+1 ] [ 4 r X π π ≤ 2K2 an,4k ω r f, + ω f, ln (2 (n + 1) ann ) . k+1 n+1 k=0 (5.4) Using (5.3) and (5.4) we get (5.5) Rate of Strong Summability by Matrix Means Bogdan Szal vol. 9, iss. 1, art. 28, 2008 kTn (f, r; · + h) − Tn (f, r; ·)kC ω (|h|) h6=0 sup p p (kRn (· + h, ·)kC ) q = sup (kRn (· + h, ·)kC )1− q ω (|h|) h6=0 p q ≤ K3 (1 + ln (2 (n + 1) ann )) n+1 4 ] [X π r × an,4k ω f, k+1 k=0 + ω f, π n+1 Title Page Contents JJ II J I Page 23 of 27 r ) r1 (1− pq ) ln (2 (n + 1) ann ) . Go Back Full Screen Close Similarly, by (4.2) we have (5.6) kTn (f, r)kC ≤ K4 n+1 4 ] [X k=0 an,4k ω r π f, k+1 + ω f, ≤ K4 n+1 4 ] [X an,4k ω π n+1 r k=0 r ) r1 ln (2 (n + 1) ann ) π f, k+1 r ) r1 pq π + ω f, ln (2 (n + 1) ann ) n+1 × n+1 4 ] [X an,4k ω r k=0 π f, k+1 Bogdan Szal vol. 9, iss. 1, art. 28, 2008 Title Page Contents r ) r1 (1− pq ) π ln (2 (n + 1) ann ) + ω f, n+1 p q ≤ K5 (1 + ln (2 (n + 1) ann )) n+1 4 ] [X π r × an,4k ω f, k+1 k=0 + ω f, Rate of Strong Summability by Matrix Means π n+1 r ) r1 (1− pq ) ln (2 (n + 1) ann ) . Collecting our partial results (5.5), (5.6) and using Lemma 4.4 and Lemma 4.5 we obtain that (2.1) and (2.2) hold. This completes our proof. JJ II J I Page 24 of 27 Go Back Full Screen Close 5.2. Proof of Theorem 2.2 Using (4.3) and the same method as in the proof of Lemma 4.4 we can show that (5.7) n+1 [X 2 ] an,2k ω k=0 r π f, k+1 π = O ann H r; n Rate of Strong Summability by Matrix Means holds, if ω (t) satisfies (1.17) and (1.18), and Bogdan Szal (5.8) n+1 [X 2 ] an,2k ω r f, k=0 π k+1 vol. 9, iss. 1, art. 28, 2008 = O (ann H (r; ann )) Title Page if ω (t) satisfies (1.17). The proof of Theorem 2.2 is analogously to the proof of Theorem 2.1. The only difference being that instead of (4.2), (4.5) and (4.6) we use (4.3), (5.7) and (5.8) respectively. This completes the proof. 5.3. Proof of Theorem 2.3 Contents JJ II J I Page 25 of 27 Using the same notations as in the proof of Theorem 2.1, from (4.4) and (5.2) we get ( n ) r1 X (5.9) |Rn (x + h, x)| ≤ K1 ank Ekr (gh ) k=0 ≤ K2 ( n X ank ω r k=0 ≤ 2K2 ω (|h|) . π gh , k+1 ) r1 Go Back Full Screen Close On the other hand, by (4.4) and (5.1), we have (5.10) |Rn (x + h, x)| ≤ K2 ( n X ank ω r k=0 ≤ 2K2 ( n X ank ω r k=0 π gh , k+1 ) r1 π f, k+1 ) r1 . Rate of Strong Summability by Matrix Means Bogdan Szal Similarly, we can show that (5.11) kTn (f, r)kC ≤ K3 vol. 9, iss. 1, art. 28, 2008 ( n X k=0 ank ω r f, π k+1 ) r1 . Title Page Contents Finally, using the same method as in the proof of Theorem 2.1 and Lemma 4.6, (2.6) and (2.7) follow from (5.9) – (5.11). JJ II J I Page 26 of 27 Go Back Full Screen Close References [1] P. CHANDRA, On the degree of approximation of a class of functions by means of Fourier series, Acta Math. Hungar., 52 (1988), 199–205. [2] P. CHANDRA, A note on the degree of approximation of continuous function, Acta Math. Hungar., 62 (1993), 21–23. [3] L. LEINDLER, On the degree of approximation of continuous functions, Acta Math. Hungar., 104(1-2), (2004), 105–113. Rate of Strong Summability by Matrix Means Bogdan Szal vol. 9, iss. 1, art. 28, 2008 [4] R.N. MOHAPATRA AND P. CHANDRA, Degree of approximation of functions in the Hölder metric, Acta Math. Hungar., 41(1-2) (1983), 67–76. Title Page [5] T. SINGH, Degree of approximation to functions in a normed spaces, Publ. Math. Debrecen, 40(3-4) (1992), 261–271. [6] XIE-HUA SUN, Degree of approximation of functions in the generalized Hölder metric, Indian J. Pure Appl. Math., 27(4) (1996), 407–417. [7] B. SZAL, On the strong approximation of functions by matrix means in the generalized Hölder metric, Rend. Circ. Mat. Palermo (2), 56(2) (2007), 287– 304. Contents JJ II J I Page 27 of 27 Go Back Full Screen Close