CUBO A Mathematical Journal Vol.15, No¯ 02, (01–19). June 2013 Uniform Convergence With Rates of General Singular Operators George A. Anastassiou and Razvan A. Mezei The University of Memphis Department of Mathematical Sciences, Memphis, TN 38152, U.S.A. ganastss@memphis.edu, rmezei@memphis.edu ABSTRACT In this article we study the approximation properties of general singular integral oper- ators over the real line. We establish their convergence to the unit operator with rates. The estimates are mostly sharp and they are pointwise or uniform. The established inequalities involve the higher order modulus of smoothness. We apply this theory to the trigonometric singular operators. RESUMEN En este art́ıculo estudiamos propiedades de aproximación de operadores integrales singulares generales sobre la recta real. Establecemos su convergencia al operador unidad con las tasas correspondientes. Las estimaciones son mayormente ajustadas y son tanto puntuales como uniformes. Las desigualdades encontradas involucran el módulo de suavidad de alto orden. Aplicamos esta teoŕıa a los operadores singulares trigonométricos. Keywords and Phrases: Best constant, general singular integral, trigonometric singular integral, modulus of smoothness, sharp inequality. 2010 AMS Mathematics Subject Classification: 26A15, 41A17, 41A35. 2 George A. Anastassiou and Razvan A. Mezei CUBO 15, 2 (2013) 1 Introduction The rate of convergence of singular integrals has been studied earlier in [5], [12], [13], [15], [3], [6], [7], [8], [9] and these articles motivate this work. Here we consider some very general singular integral operators over R and we study the degree of approximation to the unit operator with rates over smooth functions. We establish related inequalities involving the higher modulus of smoothness with respect to ‖ · ‖∞. The estimates are pointwise and uniform. Most of the times these are optimal in the sense that the inequalities are attained by basic functions. We give particular applications of these operators to the trigonometric singular integral operators over R. The discussed operators are not in general positive. Other motivation comes from [1], [2]. 2 Main Results In the next we study the following smooth general singular integral operators Θr,ξ(f, x) defined as follows. Let ξ > 0 and let µξ be Borel probability measures on R. For r ∈ N and n ∈ Z+ we put αj =    (−1)r−j ( r j ) j−n, j = 1, . . . , r, 1− r∑ j=1 (−1)r−j ( r j ) j−n, j = 0, (1) that is r∑ j=0 αj = 1. Let f : R → R be Borel measurable, we define for x ∈ R, the integral Θr,ξ(f; x) := ∫∞ −∞   r∑ j=0 αjf(x + jt)  dµξ(t). (2) We suppose that Θr,ξ(f; x) ∈ R for all x ∈ R. We will use also that Θr,ξ(f; x) = r∑ j=0 αj (∫∞ −∞ f(x + jt)dµξ(t) ) . (3) We notice that Θr,ξ(c, x) = c, c constant, and Θr,ξ(f; x) − f(x) = r∑ j=0 αj (∫∞ −∞ f(x + jt) − f(x) ) dµξ(t). (4) Let f ∈ Cn (R) , n ∈ Z+ with the rth modulus of smoothness finite, i.e. ωr(f (n), h) := sup |t|≤h ‖∆rtf (n)(x)‖∞,x < ∞, h > 0, (5) CUBO 15, 2 (2013) Uniform Convergence With Rates of General Singular Operators 3 where ∆rtf (n)(x) := r∑ j=0 (−1)r−j ( r j ) f(n)(x + jt), (6) see [10], p. 44. We need to introduce δk := r∑ j=1 αjj k, k = 1, . . . , n ∈ N, (7) and the even function Gn(t) := ∫ |t| 0 (|t| − w)n−1 (n − 1)! ωr(f (n), w)dw, n ∈ N (8) with G0(t) := ωr(f, |t|), t ∈ R. (9) Denote by ⌊·⌋ the integral part. We present our first result Theorem 1. The integrals ck,ξ := ∫∞ −∞ tkdµξ(t), k = 1, . . . , n, are assumed to be finite. Then ∣ ∣ ∣ ∣ ∣ Θr,ξ(f; x) − f(x) − n∑ k=1 f(k)(x) k! δkck,ξ ∣ ∣ ∣ ∣ ∣ ≤ ∫∞ −∞ Gn(t)dµξ(t). (10) Proof. By Taylor’s formula we obtain f(x + jt) = n−1∑ k=0 f(k)(x) k! (jt)k + ∫jt 0 (jt − z)n−1 (n − 1)! f(n)(x + z)dz = n−1∑ k=0 f(k)(x) k! (jt)k + jn ∫t 0 (t − w)n−1 (n − 1)! f(n)(x + jw)dw. (11) Multiplying both sides of (11) by αj and summing up we get r∑ j=0 αj(f(x + jt) − f(x)) = n∑ k=1 f(k)(x) k! δkt k + Rn(0, t), (12) where Rn(0, t) := ∫t 0 (t − w)n−1 (n − 1)! τ(w)dw, (13) with τ(w) := r∑ j=0 αjj nf(n)(x + jw) − δnf (n)(x). 4 George A. Anastassiou and Razvan A. Mezei CUBO 15, 2 (2013) Notice also that − r∑ j=1 (−1)r−j ( r j ) = (−1)r ( r 0 ) . (14) According to [3], p. 306, [1], we get τ(w) = ∆rwf (n)(x). (15) Therefore |τ(w)| ≤ ωr(f (n), |w|), (16) all w ∈ R independently of x. We do have after integration, see also (4), that Θr,ξ(f; x) − f(x) = ∫∞ −∞   r∑ j=0 αj(f(x + jt) − f(x))  dµξ(t) = ∫∞ −∞ ( n∑ k=1 f(k)(x) k! δkt k + Rn(0, t) ) dµξ(t) = n∑ k=1 f(k)(x) k! δk (∫∞ −∞ tkdµξ(t) ) + R∗n, (17) where R∗n := ∫∞ −∞ Rn(0, t)dµξ(t). (18) Here by (8) and (13) we get |Rn(0, t)| ≤ ∫ |t| 0 (|t| − w)n−1 (n − 1)! |τ (sign(t)w) |dw ≤ Gn(t), (19) see [5]. Hence by (18) we find |R∗n| ≤ ∫∞ −∞ Gn(t)dµξ(t). (20) We also have Θr,ξ(f; x) − f(x) − n∑ k=1 f(k)(x) k! δkck,ξ = R ∗ n. (21) Inequality (10) is now clear via (21) and (20). � Corollary 2. Assume ωr(f, ξ) < ∞, ξ > 0. Then it holds for n = 0 that |Θr,ξ(f; x) − f(x)| ≤ ∫∞ −∞ ωr (f, |t|) dµξ (t) . (22) CUBO 15, 2 (2013) Uniform Convergence With Rates of General Singular Operators 5 Proof. We observe that Θr,ξ(f; x) − f(x) = ∫∞ −∞   r∑ j=1 αj(f(x + jt) − f(x))  dµξ (t) = ∫∞ −∞   r∑ j=1 (−1)r−j ( r j ) (f(x + jt) − f(x))  dµξ (t) = ∫∞ −∞   r∑ j=1 (−1)r−j ( r j ) f(x + jt) −   r∑ j=1 (−1)r−j ( r j )  f(x)  dµξ (t) (14) = ∫∞ −∞   r∑ j=1 (−1)r−j ( r j ) f(x + jt) + (−1)r ( r 0 ) f(x) ) dµξ (t) = ∫∞ −∞   r∑ j=0 (−1)r−j ( r j ) f(x + jt)  dµξ (t) (6) = ∫∞ −∞ ((∆rtf)(x)) dµξ (t) . I.e. we have proved Θr,ξ(f; x) − f(x) = ∫∞ −∞ (∆rtf(x)) dµξ (t) . (23) Hence by (23) we derive |Θr,ξ(f; x) − f(x)| ≤ ∫∞ −∞ |∆rtf(x)|dµξ (t) ≤ ∫∞ −∞ ωr(f, |t|)dµξ (t) . That is proving (22). � Inequality (10) is sharp. Theorem 3. Inequality ( 10) at x = 0 is attained by f(x) = xr+n, r, n ∈ N with r + n even. Proof. As in [3], p. 307, [1], [16], p. 54 and (5), (6) we obtain ωr(f (n), t) = (r + n)(r + n − 1) · . . . · (r + 1)r!tr, 6 George A. Anastassiou and Razvan A. Mezei CUBO 15, 2 (2013) t > 0. And Gn(t) = r!t r+n, t ∈ R. Also we have f(k)(0) = 0, k = 0, 1, . . . , n. Thus the right hand side of (10) equals r! ∫∞ −∞ tr+ndµξ (t) . (24) The left hand side of (10) equals |Θr,ξ(f; 0)| = ∣ ∣ ∣ ∣ ∣ ∣ ∫∞ −∞   r∑ j=0 αjf(jt)  dµξ (t) ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∫∞ −∞   r∑ j=1 αj(jt) r+n  dµξ (t) ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∫∞ −∞   r∑ j=1 (−1)r−j ( r j ) j−n(jt)r+n  dµξ (t) ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣   r∑ j=0 (−1)r−j ( r j ) jr   (∫∞ −∞ tr+ndµξ (t) ) ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ (∆r1x r)(0) ∫∞ −∞ tr+ndµξ (t) ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ r! ∫∞ −∞ tr+ndµξ (t) ∣ ∣ ∣ ∣ = r! ∫∞ −∞ tr+ndµξ (t) . I.e. we have established |Θr,ξ(f; 0)| = r! ∫∞ −∞ tr+ndµξ (t) . (25) Thus by (24) and (25) we have established the claim of the theorem. � Corollary 4. Inequality ( 22) is sharp, that is attained at x = 0 by f(x) = xr, r even. Proof. Notice that ∆rtx r = r!tr and ωr(f (n), t) = r!tr, t > 0. Thus R.H.S.(22) = r! ∫∞ −∞ trdµξ (t) . CUBO 15, 2 (2013) Uniform Convergence With Rates of General Singular Operators 7 Also f(0) = 0. Therefore L.H.S.(22) = |Θr,ξ(f; 0)| = ∣ ∣ ∣ ∣ ∣ ∣ ∫∞ −∞   r∑ j=1 αjj rtr  dµξ (t) ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∫∞ −∞   r∑ j=0 (−1)r−j ( r j ) jr  trdµξ (t) ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ (∆r1x r)(0) ∫∞ −∞ trdµξ (t) ∣ ∣ ∣ ∣ = r! ∫∞ −∞ trdµξ (t) , proving the claim. � Remark 5. On inequalities (10) and (22). We have the uniform estimates ∥ ∥ ∥ ∥ ∥ Θr,ξ(f; x) − f(x) − n∑ k=1 f(k)(x) k! δkck,ξ ∥ ∥ ∥ ∥ ∥ ∞,x ≤ ∫∞ −∞ Gn(t)dµξ(t), n ∈ N, (26) and ‖Θr,ξ(f; x) − f(x)‖∞,x ≤ ∫∞ −∞ ωr(f, |t|)dµξ(t), n = 0. (27) Remark 6. The following regards the convergence of operators Θr,ξ. From (8) we have Gn(t) ≤ |t|n n! ωr(f (n), |t|). (28) Therefore by (26) we get ∥ ∥ ∥ ∥ ∥ Θr,ξ(f; x) − f(x) − n∑ k=1 f(k)(x) k! δkck,ξ ∥ ∥ ∥ ∥ ∥ ∞,x ≤ 1 n! ∫∞ −∞ |t|nωr(f (n), |t|)dµξ(t). (29) Next using ωr(f, λt) ≤ (λ + 1) r ωr(f, t), λ, t > 0, we get 1 n! ∫∞ −∞ |t|nωr(f (n), |t|)dµξ(t) = 1 n! ∫∞ −∞ |t|nωr ( f(n), ξ · |t| ξ ) dµξ(t) ≤ ωr(f (n), ξ) n! ∫∞ −∞ |t|n ( 1 + |t| ξ )r dµξ(t). So we have proved that K1 : = ∥ ∥ ∥ ∥ ∥ Θr,ξ(f; x) − f(x) − n∑ k=1 f(k)(x) k! δkck,ξ ∥ ∥ ∥ ∥ ∥ ∞,x ≤ ωr(f (n), ξ) n! ∫∞ −∞ |t|n ( 1 + |t| ξ )r dµξ(t). (30) 8 George A. Anastassiou and Razvan A. Mezei CUBO 15, 2 (2013) Similarly we get ∫∞ −∞ ωr(f, |t|)dµξ(t) = ∫∞ −∞ ωr ( f, ξ · |t| ξ ) dµξ(t) ≤ ωr(f, ξ) ∫∞ −∞ ( 1 + |t| ξ )r dµξ(t). So that we have K2 : = ‖Θr,ξ(f; x) − f(x)‖∞ ≤ ωr(f, ξ) ∫∞ −∞ ( 1 + |t| ξ )r dµξ(t). (31) In case that ∫∞ −∞ |t|n ( 1 + |t| ξ )r dµξ(t) ≤ λ1, ∀ξ > 0, we get that K1 → 0, as ξ → 0. In case that ∫∞ −∞ ( 1 + |t| ξ )r dµξ(t) ≤ λ2, ∀ξ > 0, we get that K2 → 0, that is Θr,ξ → I (unit operator) uniformly, as ξ → 0. Note 7. The operators Θr,ξ are not in general positive and they are of convolution type. Let r = 2, n = 3. Then α0 = 23 8 , α1 = −2, α2 = 1 8 . Consider f(t) = t2 ≥ 0 and x = 0. Then Θ2,ξ(t 2; 0) = −1.5 ∫∞ −∞ t2dµξ(t) < 0, where we assumed that ∫∞ −∞ t2dµξ(t) < ∞. Remark 8. From [5] we get that Gn(t) ≤ ωr(f (n), ξ) ξr { n−1∑ k=0 (−1)k k!(n − k − 1)!(k + r + 1) · [ (ξ + |t|) n+r − ξr+k+1 (ξ + |t|) n−k−1 ]} , (32) for ξ > 0, ∀t ∈ R. So by (26) we obtain ∥ ∥ ∥ ∥ ∥ Θr,ξ(f; x) − f(x) − n∑ k=1 f(k)(x) k! δkck,ξ ∥ ∥ ∥ ∥ ∥ ∞,x ≤ ωr(f (n), ξ) ξr { n−1∑ k=0 (−1)k k!(n − k − 1)!(k + r + 1) · [∫∞ −∞ (ξ + |t|) n+r dµξ(t) − ξ r+k+1 ∫∞ −∞ (ξ + |t|) n−k−1 dµξ(t) ]} . (33) CUBO 15, 2 (2013) Uniform Convergence With Rates of General Singular Operators 9 So from Remarks 5, 6 we derive Theorem 9. Let f ∈ Cn(R), n ∈ Z+. Set ck,ξ := ∫∞ −∞ tkdµξ(t), k = 1, . . . , n. Assume also ωr(f (n), h) < ∞, ∀h > 0. It is also supposed that ∫∞ −∞ |t|n ( 1 + |t| ξ )r dµξ(t) < ∞. Then ∥ ∥ ∥ ∥ ∥ Θr,ξ(f; x) − f(x) − n∑ k=1 f(k)(x) k! δkck,ξ ∥ ∥ ∥ ∥ ∥ ∞,x ≤ ωr(f (n), ξ) n! ∫∞ −∞ |t|n ( 1 + |t| ξ )r dµξ(t). (34) When n = 0 the sum in L.H.S.(34) collapses. 3 Applications to General Trigonometric Singular Opera- tors We make Remark 10. We need the following preliminary result. Let p and m be integers with 1 ≤ p ≤ m. We define the integral I(m; p) := ∫∞ −∞ (sin x)2m x2p dx = 2 ∫∞ 0 (sin x)2m x2p dx. (35) This is an (absolutely) convergent integral. According to [11], page 210, item 1033, we obtain I(m; p) = π (−1)p(2m)! 4m−p(2p − 1)! m∑ k=1 (−1)k k2p−1 (m − k)!(m + k)! . (36) In particular, for p = m the above formula becomes ∫∞ 0 (sin x)2m x2m dx = π(−1)mm m∑ k=1 (−1)k k2m−1 (m − k)!(m + k)! . (37) In this section we apply the general theory of this article to the trigonometric smooth general singular integral operators Tr,ξ(f, x) defined as follows. Let ξ > 0. 10 George A. Anastassiou and Razvan A. Mezei CUBO 15, 2 (2013) Let f : R → R be Borel measurable, we define for x ∈ R and β ∈ N, the integral Tr,ξ(f; x) := 1 W ∫∞ −∞   r∑ j=0 αjf(x + jt)   ( sin (t/ξ) t )2β dt, (38) where W = ∫∞ −∞ ( sin (t/ξ) t )2β dt = 2ξ1−2β ∫∞ 0 ( sin t t )2β dt (37) = 2ξ1−2βπ(−1)ββ β∑ k=1 (−1)k k2β−1 (β − k)!(β + k)! . (39) We suppose that Tr,ξ(f; x) ∈ R for all x ∈ R. We present our first result of this section Theorem 11. Let 1 ≤ n ≤ 2β − 2 and k = 1, . . . , n. The integrals ck,ξ : = 1 W ∫∞ −∞ tk ( sin (t/ξ) t )2β dt =    0, for k odd ξ k (−1) k 2 (2β−1)! 2k(2β−k−1)! ∑β j=1 (−1) j j 2β−k−1 (β−j)!(β+j)! ∑β j=1 (−1)j j2β−1 (β−j)!(β+j)! , for k even , are finite. Moreover, it holds ∣ ∣ ∣ ∣ ∣ ∣ Tr,ξ(f; x) − f(x) − ⌊n/2⌋∑ k=1 f(2k)(x) (2k)! δ2kc2k,ξ ∣ ∣ ∣ ∣ ∣ ∣ ≤ 1 W ∫∞ −∞ Gn(t) ( sin (t/ξ) t )2β dt. (40) When n = 1 the sum in the L.H.S.(40) colapses. Proof. We used Theorem 1 and relations (36) and (39). � Corollary 12. Assume ωr(f, ξ) < ∞, ξ > 0. Then it holds for n = 0 and β > 1 that |Tr,ξ(f; x) − f(x)| ≤ 2 W ∫∞ 0 ωr (f, t) ( sin (t/ξ) t )2β dt. (41) Proof. We are applying Corollary 2 here. � Note 13. The operators Tr,ξ are not in general positive and they are of convolution type. CUBO 15, 2 (2013) Uniform Convergence With Rates of General Singular Operators 11 Let r = 2, n = 3, and β ≥ 2. Then α0 = 23 8 , α1 = −2, α2 = 1 8 . Consider f(t) = t2 ≥ 0 and x = 0. Then T2,ξ(t 2; 0) = −1.5 1 W ∫∞ −∞ t2 ( sin (t/ξ) t )2β dt < 0, since 1 W ∫∞ −∞ t2 ( sin (t/ξ) t )2β dt = ξ2(−1)(2β − 1) (β − 1) 2 ∑β j=1(−1) j j 2β−3 (β−j)!(β+j)! ∑β j=1(−1) j j 2β−1 (β−j)!(β+j)! < ∞, by Theorem 11. � Theorem 14. Let f ∈ Cn(R), n ∈ Z+, and β ≥ 1+ ⌊ n+r+1 2 ⌋ . Assume also ωr(f (n), h) < ∞, ∀h > 0. Then ∥ ∥ ∥ ∥ ∥ ∥ Tr,ξ(f; x) − f(x) − ⌊n/2⌋∑ k=1 f(2k)(x) (2k) ! δ2kc2k,ξ ∥ ∥ ∥ ∥ ∥ ∥ ∞,x ≤ ξn n! ωr(f (n), ξ) π(−1)ββ [∑β k=1(−1) k k 2β−1 (β−k)!(β+k)! ] · [∫∞ 0 tn (1 + t) r ( sin t t )2β dt ] . (42) When n = 0, 1 the sum in L.H.S.(42) collapses. Proof. For i = 0, ..., r we have that ∫∞ 0 tn+i ( sin t t )2β dt = π(−1)β− n+i 2 (2β)! 2n+i+1 [2β − n − i − 1] ! β∑ k=1 (−1)k k2β−n−i−1 (β − k)!(β + k)! < ∞ in the case n + i even (because β ≥ 1 + ⌊ n+r+1 2 ⌋ ≥ 1 + n+i 2 ). 12 George A. Anastassiou and Razvan A. Mezei CUBO 15, 2 (2013) Furthermore, when n + i is odd we get ∫∞ 0 tn+i ( sin t t )2β dt = ∫1 0 tn+i ( sin t t )2β dt + ∫∞ 1 tn+i ( sin t t )2β dt ≤ ∫1 0 tn+i−1 ( sin t t )2β dt + ∫∞ 1 tn+i+1 ( sin t t )2β dt ≤ ∫∞ 0 tn+i−1 ( sin t t )2β dt + ∫∞ 0 tn+i+1 ( sin t t )2β dt = π(−1)β− n+i−1 2 (2β)! 2n+i [2β − n − i] ! β∑ k=1 (−1)k k2β−n−i (β − k)!(β + k)! + π(−1)β− n+i+1 2 (2β)! 2n+i+2 [2β − n − i − 2] ! β∑ k=1 (−1)k k2β−n−i−2 (β − k)!(β + k)! < ∞, by β ≥ 1 + ⌊ n+r+1 2 ⌋ ≥ 1 + n+i+1 2 . Therefore it holds ∫∞ 0 r∑ i=0 ( r i ) tn+i ( sin t t )2β dt = r∑ i=0 ( r i )∫∞ 0 tn+i ( sin t t )2β dt < ∞. Hence 1 W ∫∞ −∞ |t|n ( 1 + |t| ξ )r ( sin (t/ξ) t )2β dt = 2 W ∫∞ 0 tn ( 1 + t ξ )r ( sin (t/ξ) t )2β dt = 2ξn−2β+1 W ∫∞ 0 tn (1 + t) r ( sin t t )2β dt (39) = ξn πβ [ (−1)β ∑β k=1(−1) k k 2β−1 (β−k)!(β+k)! ] ∫∞ 0 tn (1 + t) r ( sin t t )2β dt = ξn πβ [ (−1)β ∑β k=1(−1) k k 2β−1 (β−k)!(β+k)! ] ∫∞ 0 r∑ i=0 ( r i ) tn+i ( sin t t )2β dt < ∞, which implies that 1 W ∫∞ −∞ |t|n ( 1 + |t| ξ )r ( sin (t/ξ) t )2β dt < ∞. Applying Theorem 9 here we obtain the inequality (42). � CUBO 15, 2 (2013) Uniform Convergence With Rates of General Singular Operators 13 4 Applications to Particular Trigonometric Singular Oper- ators In this section we work on the approximation results given in the previous section, for some particular values of n and β. Case β = 1. We have the following value corresponding to formula (39) W = πξ−1. None of the previous results hold in this case. Case β = 2. We have the following value corresponding to formula (39) W = 2π 3 ξ−3. Theorem 15. It holds, for n = 1 |Tr,ξ(f; x) − f(x)| ≤ 3 2π ξ3 ∫∞ −∞ G1(t) ( sin (t/ξ) t )4 dt, (43) and, for n = 2 ∣ ∣ ∣ ∣ Tr,ξ(f; x) − f(x) − f′′(x) 2 δ2c2,ξ ∣ ∣ ∣ ∣ ≤ 3 2π ξ3 ∫∞ −∞ G2(t) ( sin (t/ξ) t )4 dt. (44) Proof. By Theorem 11, with β = 2, n = 1, 2. � Corollary 16. Assume ωr(f, ξ) < ∞, ξ > 0. Then it holds for n = 0 that |Tr,ξ(f; x) − f(x)| ≤ 3 π ξ3 ∫∞ 0 ωr (f, t) ( sin (t/ξ) t )4 dt. (45) Proof. By Corollary 12. � Theorem 17. Let f ∈ C1(R). Assume ω1(f ′, h) < ∞, ∀h > 0. Then ‖T1,ξ(f) − f‖∞ ≤ 3ξ π [ ln 2 + π 4 ] ω1(f ′, ξ). (46) 14 George A. Anastassiou and Razvan A. Mezei CUBO 15, 2 (2013) Proof. We are applying Theorem 14 here for n = r = 1, etc. � Case β = 3. We have the following value corresponding to formula (39) W = 11π 20 ξ−5. We have Theorem 18. It holds |Tr,ξ(f; x) − f(x)| ≤ 20 11π ξ5 ∫∞ −∞ G1(t) ( sin (t/ξ) t )6 dt. (47) ∣ ∣ ∣ ∣ Tr,ξ(f; x) − f(x) − f′′(x) 2 δ2c2,ξ ∣ ∣ ∣ ∣ ≤ 20 11π ξ5 ∫∞ −∞ G2(t) ( sin (t/ξ) t )6 dt. (48) ∣ ∣ ∣ ∣ Tr,ξ(f; x) − f(x) − f′′(x) 2 δ2c2,ξ ∣ ∣ ∣ ∣ ≤ 20 11π ξ5 ∫∞ −∞ G3(t) ( sin (t/ξ) t )6 dt. (49) ∣ ∣ ∣ ∣ Tr,ξ(f; x) − f(x) − f′′(x) 2 δ2c2,ξ − f(4)(x) 24 δ4c4,ξ ∣ ∣ ∣ ∣ ≤ 20 11π ξ5 ∫∞ −∞ G4(t) ( sin (t/ξ) t )6 dt. (50) Proof. By Theorem 11, for β = 3, and n = 1, 2, 3, 4. � Corollary 19. Assume ωr(f, ξ) < ∞, ξ > 0. Then it holds for n = 0 that |Tr,ξ(f; x) − f(x)| ≤ 40 11π ξ5 ∫∞ 0 ωr (f, t) ( sin (t/ξ) t )6 dt. (51) Proof. We are applying Corollary 12 here. � Theorem 20. Let f ∈ Cn(R), n ∈ Z+, and n+r = 1, 2, 3, 4. Assume also ωr(f (n), h) < ∞, ∀h > 0. Then ∥ ∥ ∥ ∥ ∥ ∥ Tr,ξ(f; x) − f(x) − ⌊n/2⌋∑ k=1 f(2k)(x) (2k) ! δ2kc2k,ξ ∥ ∥ ∥ ∥ ∥ ∥ ∞,x (52) ≤ 40 11π [∫∞ 0 tn (1 + t) r ( sin t t )6 dt ] ωr(f (n), ξ)ξn n! . When n = 0, 1 the sum in L.H.S.(52) collapses. CUBO 15, 2 (2013) Uniform Convergence With Rates of General Singular Operators 15 Proof. We are applying Theorem 14 here. � Case β = 4. We have the following value corresponding to formula (39) W = 151π 315 ξ−7. Theorem 21. It holds |Tr,ξ(f; x) − f(x)| ≤ 315 151π ξ7 ∫∞ −∞ G1(t) ( sin (t/ξ) t )8 dt. (53) ∣ ∣ ∣ ∣ Tr,ξ(f; x) − f(x) − f′′(x) 2 δ2c2,ξ ∣ ∣ ∣ ∣ ≤ 315 151π ξ7 ∫∞ −∞ G2(t) ( sin (t/ξ) t )8 dt. (54) ∣ ∣ ∣ ∣ Tr,ξ(f; x) − f(x) − f′′(x) 2 δ2c2,ξ ∣ ∣ ∣ ∣ ≤ 315 151π ξ7 ∫∞ −∞ G3(t) ( sin (t/ξ) t )8 dt. (55) ∣ ∣ ∣ ∣ Tr,ξ(f; x) − f(x) − f′′(x) 2 δ2c2,ξ − f(4)(x) 24 δ4c4,ξ − f(6)(x) 720 δ6c6,ξ ∣ ∣ ∣ ∣ (56) ≤ 315 151π ξ7 ∫∞ −∞ G6(t) ( sin (t/ξ) t )8 dt. Proof. We used Theorem 11, with β = 4, and n = 1, 2, 3, 6. � Corollary 22. Assume ωr(f, ξ) < ∞, ξ > 0. Then for n = 0 it holds |Tr,ξ(f; x) − f(x)| ≤ 630 151π ξ7 ∫∞ 0 ωr (f, t) ( sin (t/ξ) t )8 dt. (57) Proof. We are applying Corollary 12 here. � Theorem 23. Let f ∈ Cn(R), n ∈ Z+, and n + r = 1, 2, 3, 4, 5, 6. Assume also ωr(f (n), h) < ∞, ∀h > 0. Then ∥ ∥ ∥ ∥ ∥ ∥ Tr,ξ(f; x) − f(x) − ⌊n/2⌋∑ k=1 f(2k)(x) (2k) ! δ2kc2k,ξ ∥ ∥ ∥ ∥ ∥ ∥ ∞,x (58) ≤ 630 151π [∫∞ 0 tn (1 + t) r ( sin t t )8 dt ] ωr(f (n), ξ)ξn n! . 16 George A. Anastassiou and Razvan A. Mezei CUBO 15, 2 (2013) When n = 0, 1 the sum in L.H.S.(58) collapses. Proof. We are applying Theorem 14 here. � Case β = 6. We have the following value corresponding to formula (39) W = ∫∞ −∞ ( sin (t/ξ) t )12 dt = 655177π 1663200 ξ−11. Theorem 24. It holds |Tr,ξ(f; x) − f(x)| ≤ 1663200 655177π ξ11 ∫∞ −∞ G1(t) ( sin (t/ξ) t )12 dt. (59) ∣ ∣ ∣ ∣ Tr,ξ(f; x) − f(x) − f′′(x) 2 δ2c2,ξ ∣ ∣ ∣ ∣ ≤ 1663200 655177π ξ11 ∫∞ −∞ G2(t) ( sin (t/ξ) t )12 dt. (60) ∣ ∣ ∣ ∣ Tr,ξ(f; x) − f(x) − f′′(x) 2 δ2c2,ξ ∣ ∣ ∣ ∣ ≤ 1663200 655177π ξ11 ∫∞ −∞ G3(t) ( sin (t/ξ) t )12 dt. (61) ∣ ∣ ∣ ∣ Tr,ξ(f; x) − f(x) − f′′(x) 2 δ2c2,ξ − f(4)(x) 24 δ4c4,ξ − f(6)(x) 720 δ6c6,ξ ∣ ∣ ∣ ∣ (62) ≤ 1663200 655177π ξ11 ∫∞ −∞ G6(t) ( sin (t/ξ) t )12 dt. ∣ ∣ ∣ ∣ Tr,ξ(f; x) − f(x) − f′′(x) 2 δ2c2,ξ − f(4)(x) 24 δ4c4,ξ − f(6)(x) 720 δ6c6,ξ − f(8)(x) 8! δ8c8,ξ − f(10)(x) 10! δ10c10,ξ ∣ ∣ ∣ ∣ (63) ≤ 1663200 655177π ξ11 ∫∞ −∞ G10(t) ( sin (t/ξ) t )12 dt. (64) Proof. We used Theorem 11, with β = 6, and n = 1, 2, 3, 6, 10. � Corollary 25. Assume ωr(f, ξ) < ∞, ξ > 0. Then for n = 0 it holds |Tr,ξ(f; x) − f(x)| ≤ 3326400 655177π ξ11 ∫∞ 0 ωr (f, t) ( sin (t/ξ) t )12 dt. (65) Proof. We are applying Corollary 12 here. � CUBO 15, 2 (2013) Uniform Convergence With Rates of General Singular Operators 17 Theorem 26. Let f ∈ Cn(R), n ∈ Z+, and n + r ∈ {1, 2, 3, . . . , 10} . Assume also ωr(f (n), h) < ∞, ∀h > 0. Then ∥ ∥ ∥ ∥ ∥ ∥ Tr,ξ(f; x) − f(x) − ⌊n/2⌋∑ k=1 f(2k)(x) (2k) ! δ2kc2k,ξ ∥ ∥ ∥ ∥ ∥ ∥ ∞,x ≤ 3326400 655177π (66) · [∫∞ 0 tn (1 + t) r ( sin t t )12 dt ] ωr(f (n), ξ)ξn n! . When n = 0, 1 the sum in L.H.S.(46) collapses. Proof. We are applying Theorem 14 here. � Case β = 10. We have the following value corresponding to formula (39) W = ∫∞ −∞ ( sin (t/ξ) t )20 dt = 37307713155613π 121645100408832 ξ−19. Theorem 27. It holds |Tr,ξ(f; x) − f(x)| ≤ 121645100408832 37307713155613π ξ19 (67) · ∫∞ −∞ G1(t) ( sin (t/ξ) t )20 dt. ∣ ∣ ∣ ∣ Tr,ξ(f; x) − f(x) − f′′(x) 2 δ2c2,ξ ∣ ∣ ∣ ∣ ≤ 121645100408832 37307713155613π ξ19 (68) · ∫∞ −∞ G2(t) ( sin (t/ξ) t )20 dt. ∣ ∣ ∣ ∣ Tr,ξ(f; x) − f(x) − f′′(x) 2 δ2c2,ξ ∣ ∣ ∣ ∣ ≤ 121645100408832 37307713155613π ξ19 (69) · ∫∞ −∞ G3(t) ( sin (t/ξ) t )20 dt. ∣ ∣ ∣ ∣ Tr,ξ(f; x) − f(x) − f′′(x) 2 δ2c2,ξ − f(4)(x) 24 δ4c4,ξ − f(6)(x) 720 δ6c6,ξ ∣ ∣ ∣ ∣ (70) ≤ 121645100408832 37307713155613π ξ19 ∫∞ −∞ G6(t) ( sin (t/ξ) t )20 dt. 18 George A. Anastassiou and Razvan A. Mezei CUBO 15, 2 (2013) ∣ ∣ ∣ ∣ Tr,ξ(f; x) − f(x) − f′′(x) 2 δ2c2,ξ − f(4)(x) 24 δ4c4,ξ − f(6)(x) 720 δ6c6,ξ − f(8)(x) 8! δ8c8,ξ − f(10)(x) 10! δ10c10,ξ ∣ ∣ ∣ ∣ ≤ 121645100408832 37307713155613π ξ19 ∫∞ −∞ G10(t) ( sin (t/ξ) t )20 dt. (71) Proof. We used Theorem 11, with β = 10, and n = 1, 2, 3, 6, 10. � Corollary 28. Assume ωr(f, ξ) < ∞, ξ > 0. Then for n = 0 it holds |Tr,ξ(f; x) − f(x)| ≤ 243290200817664 37307713155613π ξ19 ∫∞ 0 ωr (f, t) ( sin (t/ξ) t )20 dt. (72) Proof. We are applying Corollary 12 here. � Theorem 29. Let f ∈ Cn(R), n ∈ Z+, and n + r = {1, 2, . . . , 18} . Assume also ωr(f (n), h) < ∞, ∀h > 0. Then ∥ ∥ ∥ ∥ ∥ ∥ Tr,ξ(f; x) − f(x) − ⌊n/2⌋∑ k=1 f(2k)(x) (2k) ! δ2kc2k,ξ ∥ ∥ ∥ ∥ ∥ ∥ ∞,x ≤ 243290200817664 37307713155613π (73) · [∫∞ 0 tn (1 + t) r ( sin t t )20 dt ] ωr(f (n), ξ)ξn n! . When n = 0, 1 the sum in L.H.S.(73) collapses. Proof. We are applying Theorem 14 here. � Acknowledgement. The authors would like to thank Professor V. Papanicolaou of National Technical University of Athens, Greece, for having fruitful discussions during the preparation of this article. Received: February 2011. Accepted: April 2011. References [1] G.A. Anastassiou, Rate of convergence of non-positive linear convolution type operators. A sharp inequality, J. Math. Anal. and Appl., 142 (1989), 441–451. [2] G.A. 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