() CUBO A Mathematical Journal Vol.13, No¯ 03, (17–48). October 2011 Applications and Lipschitz results of Approximation by Smooth Picard and Gauss-Weierstrass Type Singular Integrals Razvan A. Mezei The University of Memphis, Department of Mathematical Sciences, Memphis, TN 38152, U.S.A. email: rmezei@memphis.edu ABSTRACT We continue our studies in higher order uniform convergence with rates and in Lp convergence with rates. Namely, in this article we establish some Lipschitz type results for the smooth Picard type singular integral operators and for the smooth Gauss- Weierstrass type singular integral operators. RESUMEN Continuamos nuestros estudios sobre convergencia uniforme de orden superior con ra- dios y sobre convergencia Lp con radios. Concretamente, en este art́ıculo establecemos algunos resultados de tipo Lipschitz para operadores integrales suves del tipo Picard singulares y para operadores integrales singulares de tipo Gauss-Weierstrass. Keywords: Smooth Picard Type singular integral, Smooth Gauss-Weierstrass Type singular in- tegral, modulus of smoothness, rate of convergence, Lp convergence, Higher Order Uniform Con- vergence with Rates, sharp inequality, Lipschitz functions. Mathematics Subject Classification: 26A15, 26D15, 41A17, 41A35, 41A60, 41A80. 18 Razvan A. Mezei CUBO 13, 3 (2011) 1. Introduction We are motivated by [1], [2], [3] and [4]. We denote by Lp, 1 ≤ p < ∞, the classes of functions f (x) , integrable in −∞ < x < ∞ with the norm ‖f‖ p = [∫ ∞ −∞ |f (u)| p du ] 1 p . (1.1) The Picard singular integral Pξ(f; x) corresponding to the function f (x) , is defined as follows Pξ(f; x) = 1 2ξ ∫ ∞ −∞ f(x + y)e−|y|/ξdy, for all x ∈ R, ξ > 0. (1.2) The Gauss Weierstrass singular integral Wξ(f; x) corresponding to the function f (x) , is de- fined as follows Wξ(f; x) = 1√ πξ ∫ ∞ −∞ f(x + y)e−y 2 /ξ dy, for all x ∈ R, ξ > 0. (1.3) 2. Convergence with Rates of Smooth Picard Singular In- tegral Operators In the next we deal with the following smooth Picard singular integral operators Pr,ξ(f; x) defined as follows. For r ∈ N and n ∈ Z+ we set αj =    (−1)r−j ( r j ) j −n , j = 1, . . . , r, 1 − r∑ j=1 (−1)r−j ( r j ) j −n , j = 0, (2.1) that is r∑ j=0 αj = 1. Let f : R → R be Lebesgue measurable, we define for x ∈ R, ξ > 0 the Lebesgue integral Pr,ξ(f; x) := 1 2ξ ∫ ∞ −∞   r∑ j=0 αjf(x + jt)  e −|t|/ξ dt. (2.2) We assume that Pr,ξ(f; x) ∈ R for all x ∈ R. We mention the useful here formula ∫ ∞ 0 t k e −t/ξ dt = Γ (k + 1) ξk+1, k > −1. (2.3) CUBO 13, 3 (2011) Applications and Lipschitz results . . . 19 We need to introduce δk := r∑ j=1 αjj k , k = 1, . . . , n ∈ N. (2.4) Denote by ⌊·⌋ the integral part. We give a special related result. Proposition 1. Let f be defined as above in this section. It holds that |P2,ξ(f; x) − f(x)| ≤ 1 ξ ∫ ∞ 0 (∫ |t| 0 ω2(f ′ , w)dw ) e −t/ξ dt. (2.5) Proof. In Theorem 1 of [1] we use n = 1, r = 2. � We also present the Lipschitz type result corresponding to the Theorem 1 of [1]. Theorem 2. Let f be defined as above in this section, with n ∈ N. Furthermore we assume the following Lipschitz condition: ωr ( f(n), δ ) ≤ Kδr−1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then it holds that ∣ ∣ ∣ ∣ ∣ ∣ Pr,ξ(f; x) − f(x) − ⌊ n 2 ⌋ ∑ m=1 f (2m)(x)δ2mξ 2m ∣ ∣ ∣ ∣ ∣ ∣ ≤ KΓ (γ + r) ξn+r+γ−1. (2.6) In L.H.S.(2.6) the sum collapses when n = 1. Proof. As in the proof of Theorem 1, of [1], we get again that Pr,ξ(f; x) − f(x) = n∑ k=1 f(k)(x) k! δk 1 2ξ (∫ ∞ −∞ t k e −|t|/ξ dt ) + R∗n, (2.7) where R∗n := 1 2ξ ∫ ∞ −∞ Rn(0, t)e−|t|/ξdt, (2.8) with Rn(0, t) := ∫ t 0 (t − w)n−1 (n − 1)! τ(w)dw, (2.9) and τ(w) := r∑ j=0 αjj n f (n)(x + jw) − δnf (n)(x). Also we get |Rn(0, t)| ≤ ∫ |t| 0 (|t| − w)n−1 (n − 1)! ωr(f (n) , w)dw. (2.10) 20 Razvan A. Mezei CUBO 13, 3 (2011) Using the Lipschitz type condition we obtain |Rn(0, t)| ≤ ∫ |t| 0 (|t| − w)n−1 (n − 1)! Kw r−1+γ dw = K|t|n+r+γ−2 (n − 1)! ∫ |t| 0 ( 1 − w |t| )n−1 ( w |t| )r−1+γ dw = K|t|n+r+γ−1 (n − 1)! ∫ 1 0 (1 − y) n−1 y r−1+γ dy = K|t|n+r+γ−1Γ (γ + r) Γ (n + γ + r) . (2.11) Then, by (2.3), we obtain |R∗n| ≤ 1 2ξ ∫ ∞ −∞ K|t|n+r+γ−1Γ (γ + r) Γ (n + γ + r) e −|t|/ξ dt = K 2ξ Γ (γ + r) Γ (n + γ + r) ∫ ∞ −∞ |t| n+r+γ−1 e −|t|/ξ dt = K ξ Γ (γ + r) Γ (n + γ + r) ∫ ∞ 0 t n+r+γ−1 e −t/ξ dt (2.3) = KΓ (γ + r) ξn+r+γ−1. (2.12) We also notice that Pr,ξ(f; x) − f(x) − n∑ k=1 f(k)(x) k! δk 1 2ξ (∫ ∞ −∞ t k e −|t|/ξ dt ) = Pr,ξ(f; x) − f(x) − ⌊ n 2 ⌋ ∑ m=1 f (2m)(x)δ2mξ 2m = R∗n. (2.13) By (2.12) and (2.13) we complete the proof of the theorem. � Corollary 3. Let f be defined as above in this section. Furthermore we assume the following Lipschitz condition ω2 (f ′, δ) ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then |P2,ξ(f; x) − f(x)| ≤ KΓ (γ + 2) ξ2+γ. (2.14) Proof. In Theorem 2 we use n = 1, r = 2. � For the case n = 0 we have Theorem 4. Let f be defined as above in this section, with n = 0. Furthermore we assume the following Lipschitz condition: ωr (f, δ) ≤ Kδr−1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. It holds that |Pr,ξ(f; x) − f(x)| ≤ KΓ (r + γ) ξr+γ−1. (2.15) CUBO 13, 3 (2011) Applications and Lipschitz results . . . 21 Proof. As in the proof of Corollary 1, of [1], with n = 0, using the Lipschitz type condition, we get that |Pr,ξ(f; x) − f(x)| ≤ 1 ξ ∫ ∞ 0 ωr(f, t)e −t/ξ dt ≤ 1 ξ ∫ ∞ 0 Kt r−1+γ e −t/ξ dt (2.3) = KΓ (r + γ) ξr+γ−1 (2.16) This completes the proof of Theorem 4. � Corollary 5. Let f be defined as above in this section, with n = 0. Furthermore we assume the following Lipschitz condition: ω2 (f, δ) ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then |P2,ξ(f; x) − f(x)| ≤ KΓ (2 + γ) ξγ+1. (2.17) Proof. In Theorem 4 we use r = 2. � In the next we consider f ∈ Cn(R), n ≥ 2 even and the simple smooth singular operator of symmetric convolution type Pξ(f, x0) := 1 2ξ ∫ ∞ −∞ f(x0 + y)e −|y|/ξ dy, for all x0 ∈ R, ξ > 0. (2.18) That is Pξ(f; x0) = 1 2ξ ∫ ∞ 0 ( f(x0 + y) + f(x0 − y) ) e −y/ξ dy, for all x0 ∈ R, ξ > 0. (2.19) We assume that f is such that Pξ(f; x0) ∈ R, ∀x0 ∈ R, ∀ξ > 0 and ω2(f(n), h) < ∞, h > 0. Note that P1,ξ = Pξ and if Pξ(f; x0) ∈ R then Pr,ξ(f; x0) ∈ R. Proposition 6. Assume ω2(f, h) < ∞, h > 0. Furthermore we assume the following Lipschitz condition: ω2 (f, δ) ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then ‖Pξ(f) − f‖∞ ≤ KΓ (2 + γ) 2 ξ γ+1 . (2.20) Proof. Using Proposition 1 of [1] we obtain |Pξ(f; x0) − f(x0)| ≤ 1 2ξ ∫ ∞ 0 ω2(f, y)e −y/ξ dy ≤ 1 2ξ ∫ ∞ 0 Ky 1+γ e −y/ξ dy (2.3) = KΓ (2 + γ) 2 ξ γ+1 , (2.21) 22 Razvan A. Mezei CUBO 13, 3 (2011) proving the claim of the proposition. � Let K2(x0) := Pξ(f; x0) − f(x0) − n/2∑ ρ=1 f (2ρ)(x0)ξ 2ρ . (2.22) We give Theorem 7. Let f ∈ Cn(R), n even, Pξ(f) real valued. Furthermore we assume the following Lipschitz condition: ω2 ( f(n), δ ) ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then |K2(x0)| ≤ KΓ (n + γ + 2) 2n! ξ n+γ+1 . (2.23) Proof. Using Theorem 6 of [1] we obtain |K2(x0)| ≤ 1 2ξn! ∫ ∞ 0 ω2(f (n) , y)yne−y/ξdy ≤ 1 2ξn! ∫ ∞ 0 Ky 1+γ y n e −y/ξ dy (2.3) = KΓ (n + γ + 2) 2n! ξ n+γ+1 , (2.24) proving the claim of the theorem. � In particular we have Corollary 8. Let f ∈ C4(R) such that Pξ(f) is real valued. Furthermore we assume the following Lipschitz condition: ω2 ( f(4), δ ) ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then |K2(x0)| ≤ KΓ (γ + 6) 48 ξ γ+5 . (2.25) Proof. In Theorem 7 we use n = 4. � We also give Corollary 9. Let f ∈ C2(R), such that ω2(f ′′ , |y|) ≤ 2A|y|γ, 0 < γ ≤ 2, A > 0. Then for x0 ∈ R we have ∣ ∣Pξ(f; x0) − f(x0) − f ′′(x0)ξ 2 ∣ ∣ ≤ Γ (α + 1)Aξγ+2. (2.26) CUBO 13, 3 (2011) Applications and Lipschitz results . . . 23 Inequality (2.16 ) is sharp, namely it is attained at x0 = 0 by f∗(y) = A|y|γ+2 (γ + 1)(γ + 2) . Proof. In Theorem 7 of [1] we use n = 2. � We also give Corollary 10. Assume that ω2(f, ξ) < ∞ and n = 0. Then ‖P2,ξ(f) − f‖∞ ≤ 5ω2(f, ξ), (2.27) and as ξ → 0, P2,ξ u → I with rates. Proof. By formula (37) of [1] with r = 2. � Next let K1 := ∥ ∥ ∥ ∥ ∥ ∥ Pr,ξ(f; x) − f(x) − ⌊n/2⌋∑ m=1 [ f (2m)(x)δ2mξ 2m ] ∥ ∥ ∥ ∥ ∥ ∥ ∞ ,x . (2.28) We present Corollary 11. Assuming f ∈ C2(R) and ω2(f′′, ξ) < ∞, ξ > 0 we have K1 = ∥ ∥P2,ξ(f; x) − f(x) − f ′′(x)δ2ξ 2 ∥ ∥ ∞ ,x ≤ 21 4 ξ 2 ω2(f ′′ , ξ). (2.29) That is as ξ → 0 we get P2,ξ → I, pointwise with rates, given that ‖f′′‖∞ < ∞. Proof. In Theorem 11 of [1] we use r = n = 2. � We also present Corollary 12. Assuming f ∈ C2(R) and ω2(f′′, ξ) < ∞, ξ > 0 we have ‖K2(x)‖∞ ,x = ∥ ∥Pξ(f; x0) − f(x0) − f ′′(x0)ξ 2 ∥ ∥ ∞ ,x ≤ 21 8 ξ 2 ω2(f ′′ , ξ). (2.30) That is as ξ → 0 we get Pξ → I, pointwise with rates, given that ‖f′′‖∞ < ∞. Proof. In Theorem 12 of [1] we use n = 2. � 24 Razvan A. Mezei CUBO 13, 3 (2011) 3. Lp Convergence with Rates of Smooth Picard Singular Integral Operators For r ∈ N and n ∈ Z+ we let αj as in (2.1). Let f ∈ Cn(R) and f(n) ∈ Lp(R), 1 ≤ p < ∞, we define for x ∈ R, ξ > 0 the Lebesgue integral Pr,ξ(f; x) as in (2.2). We need the rth Lp-modulus of smoothness ωr(f (n) , h)p := sup |t|≤h ‖∆rtf(n)(x)‖p,x, h > 0, (3.1) where ∆ r tf (n)(x) := r∑ j=0 (−1)r−j ( r j ) f (n)(x + jt), (3.2) Here we have that ωr(f (n), h)p < ∞, h > 0. We need to introduce δk’s as in (2.4). We define ∆(x) := Pr,ξ(f; x) − f(x) − ⌊n/2⌋∑ m=1 f (2m)(x)δ2mξ 2m . (3.3) We have the following results. Corollary 13. Let n ∈ N and the rest as above in this section. Then ‖∆(x)‖2 ≤ √ 2τξn √ (2r + 1)(4n − 2)(n − 1)! ωr(f (n) , ξ)2, (3.4) where 0 < τ := [∫ ∞ 0 (1 + u)2r+1u2n−1e−udu − (2n − 1)! ] < ∞. (3.5) Hence as ξ → 0 we obtain ‖∆(x)‖2 → 0. If additionally f(2m) ∈ L2(R), m = 1, 2, . . . , ⌊ n 2 ⌋ then ‖Pr,ξ(f) − f‖2 → 0, as ξ → 0. Proof. In Theorem 1 of [2], we place p = q = 2. � Corollary 14. Let f be as above in this section. In particular, for n = 1, we have ‖Pr,ξ(f; ·) − f‖2 ≤ √ τξ √ (2r + 1) ωr(f ′ , ξ)2, (3.6) CUBO 13, 3 (2011) Applications and Lipschitz results . . . 25 where 0 < τ := [∫ ∞ 0 (1 + u)2r+1ue−udu − 1 ] < ∞. (3.7) Hence as ξ → 0 we obtain ‖Pr,ξ(f; ·) − f‖2 → 0. Proof. In Theorem 1 of [2], we place p = q = 2, n = 1. � Corollary 15. Let f be as above in this section and n = 2. Then ‖Pr,ξ(f; x) − f(x) − f′′(x)δ2ξ2‖2 ≤ √ 2τξ2 √ 6(2r + 1) ωr(f ′′ , ξ)2, (3.8) where 0 < τ := [∫ ∞ 0 (1 + u)2r+1u3e−udu − 6 ] < ∞. (3.9) Hence as ξ → 0 we obtain ‖∆(x)‖2 → 0. If additionally f′′ ∈ L2(R), then ‖Pr,ξ(f) − f‖2 → 0, as ξ → 0. Proof. In Theorem 1 of [2], we place p = q = n = 2. � Next we present the Lipschitz type result corresponding to Theorem 1 of [2]. Theorem 16. Let p, q > 1 such that 1 p + 1 q = 1, n ∈ N, and the rest as above in this section. Furthermore we assume the following Lipschitz condition: ωr ( f(n), δ ) p ≤ Kδr−1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0.Then ‖∆(x)‖p ≤ (Γ (p (r − 1 + γ + n) + 1)) 1 p 2(r+γ+n)K [ (n − 1)!q 1 q p r− 1 q +γ+n (q(n − 1) + 1) 1 q (p (r − 1 + γ) + 1) 1 p ]ξ (r−1+γ+n) . (3.10) Hence as ξ → 0 we obtain ‖∆(x)‖p → 0. If additionally f(2m) ∈ Lp(R), m = 1, 2, . . . , ⌊ n 2 ⌋ then ‖Pr,ξ(f) − f‖p → 0, as ξ → 0. Proof. As in the proof of Theorem 1, [2], we get again I : = ∫ ∞ −∞ |∆(x)|pdx ≤ c1 (∫ ∞ −∞ ((∫ |t| 0 ωr(f (n) , w)ppdw ) |t| np−1 e −|pt|/2ξ ) dt ) , (3.11) where c1 := 2p−2 ξqp−1((n − 1)!)p(q(n − 1) + 1)p/q . (3.12) 26 Razvan A. Mezei CUBO 13, 3 (2011) Using the Lipschitz condition, we obtain I ≤ c1 (∫ ∞ −∞ (∫ |t| 0 ( Kw r−1+γ )p dw ) |t| np−1 e −p|t|/2ξ dt ) = c1K p (p (r − 1 + γ) + 1) (∫ ∞ −∞ |t| p(r−1+γ+n) e −p|t|/2ξ dt ) = 2c1K p (p (r − 1 + γ) + 1) (∫ ∞ 0 t p(r−1+γ+n) e −pt/2ξ dt ) = 2c1K p (p (r − 1 + γ) + 1) ( 2 p )p(r−1+γ+n)+1 (∫ ∞ 0 z p(r−1+γ+n) e −z/ξ dz ) (2.3) = 2c1K pΓ (p (r − 1 + γ + n) + 1) (p (r − 1 + γ) + 1) ( 2 p )p(r−1+γ+n)+1 ξ p(r−1+γ+n)+1 . (3.13) Thus we obtain I ≤ Γ (p (r − 1 + γ + n) + 1) qp−1((n − 1)!)p(q(n − 1) + 1)p/qpp(r−1+γ+n)+1 2p(r+γ+n)Kp (p (r − 1 + γ) + 1) ξ p(r−1+γ+n) . (3.14) That is finishing the proof of the theorem. � In particular we have Corollary 17. Let f such that the following Lipschitz condition holds: ω7 ( f(4), δ ) 2 ≤ Kδ6+γ, K > 0, 0 < γ ≤ 1, for any δ > 0, and the rest as above in this section. Then ‖∆(x)‖2 ≤ K 6 √ (Γ (2γ + 21)) 7 (2γ + 13) ξ (γ+10) . (3.15) Hence as ξ → 0 we obtain ‖∆(x)‖2 → 0. If additionally f(2m) ∈ L2(R), m = 1, 2, then ‖P7,ξ(f) − f‖2 → 0, as ξ → 0. Proof. In Theorem 16 we place p = q = 2, n = 4, and r = 7. � The counterpart of Theorem 16 follows, case of p = 1. Theorem 18. Let f ∈ Cn(R) and f(n) ∈ L1(R), n ∈ N. Furthermore we assume the following Lipschitz condition: ωr ( f(n), δ ) 1 ≤ Kδr−1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0.Then ‖∆(x)‖1 ≤ K (n − 1)! (r + γ) Γ (r + γ + n) ξr+γ+n−1. (3.16) Hence as ξ → 0 we obtain ‖∆(x)‖1 → 0. CUBO 13, 3 (2011) Applications and Lipschitz results . . . 27 If additionally f(2m) ∈ L1(R), m = 1, 2, . . . , ⌊ n 2 ⌋ then ‖Pr,ξ(f) − f‖1 → 0, as ξ → 0. Proof. As in the proof of Theorem 2 of [2] we get ‖∆(x)‖1 ≤ 1 2ξ(n − 1)! (∫ ∞ −∞ (∫ |t| 0 ωr(f (n) , w)1dw ) |t| n−1 e −|t|/ξ dt ) . (3.17) Consequently we have ‖∆(x)‖1 ≤ 1 2ξ(n − 1)! (∫ ∞ −∞ (∫ |t| 0 Kw r−1+γ dw ) |t| n−1 e −|t|/ξ dt ) (3.18) = K 2ξ(n − 1)! (∫ ∞ −∞ ( |t|r+γ r + γ ) |t| n−1 e −|t|/ξ dt ) = K 2ξ(n − 1)! (r + γ) (∫ ∞ −∞ |t| r+γ+n−1 e −|t|/ξ dt ) = K ξ(n − 1)! (r + γ) (∫ ∞ 0 t r+γ+n−1 e −t/ξ dt ) (2.3) = K (n − 1)! (r + γ) Γ (r + γ + n) ξr+γ+n−1, (3.19) proving (3.16). � Corollary 19. Let f ∈ C2(R) and f′′ ∈ L1(R). Furthermore we assume the following Lipschitz condition: ω2 (f ′′, δ) 1 ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0.Then ‖∆(x)‖1 ≤ K (2 + γ) Γ (4 + γ) ξγ+3. (3.20) Hence as ξ → 0 we obtain ‖∆(x)‖1 → 0. If additionally f′′ ∈ L1(R),then ‖P2,ξ(f) − f‖1 → 0, as ξ → 0. Proof. In Theorem 18 we place n = r = 2. � Next, when n = 0 we get Proposition 20. Let r ∈ N and the rest as above. Then ‖Pr,ξ(f) − f‖2 ≤ θ1/2ωr(f, ξ)2, (3.21) where 0 < θ := ∫ ∞ 0 (1 + x)2re−xdx < ∞. (3.22) Hence as ξ → 0 we obtain Pr,ξ → unit operator I in the L2 norm. 28 Razvan A. Mezei CUBO 13, 3 (2011) Proof. In the proof of Proposition 1 of [2] we use p = q = 2. � We continue with Proposition 21. Let p, q > 1 such that 1 p + 1 q = 1 and the rest as above. Furthermore we assume the following Lipschitz condition: ωr (f, δ)p ≤ Kδr−1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then ‖Pr,ξ(f) − f‖p ≤ p √ Γ (p (r − 1 + γ) + 1) K q1/q 2(r+γ)ξ(r+γ−1) p( r−1+γ+ 1 p ) . (3.23) Hence as ξ → 0 we obtain Pr,ξ → unit operator I in the Lp norm, p > 1. Proof. As in the proof of Proposition 1 of [2] we find ∫ ∞ −∞ |Pr,ξ(f; x) − f(x)| p dx ≤ 1 2p−1ξp ( 4ξ q )p/q (∫ ∞ 0 ωr(f, t) p pe −pt/(2ξ) dt ) ≤ 1 2p−1ξp ( 4ξ q )p/q (∫ ∞ 0 ( Kt r−1+γ )p e −pt/(2ξ) dt ) (2.3) = Kp qp−1 Γ (p (r − 1 + γ) + 1) 2p(r+γ)ξ(r−1+γ)p p(p(r+γ−1)+1) . (3.24) We have established the claim of the proposition. � Corollary 22. Let f such that the following Lipschitz condition holds: ω4 (f, δ)2 ≤ Kδ3+γ, K > 0, 0 < γ ≤ 1, for any δ > 0, and the rest as above in this section. Then ‖P4,ξ(f) − f‖2 ≤ √ Γ (2γ + 7)Kξ(3+γ). (3.25) Hence as ξ → 0 we obtain P4,ξ → unit operator I in the L2 norm. Proof. In Proposition 21 we place p = q = 2 and r = 4. � In general, for the L1 case, n = 0 we have Proposition 23. It holds ‖P2,ξf − f‖1 ≤ 5ω2(f, ξ)1. (3.26) Hence as ξ → 0 we get P2,ξ → I in the L1 norm. Proof. In the proof of Proposition 2 of [2] we use r = 2. � Proposition 24. We assume the following Lipschitz condition: ωr (f, δ)1 ≤ Kδr−1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then ‖Pr,ξf − f‖1 ≤ KΓ (r + γ) ξr−1+γ. (3.27) CUBO 13, 3 (2011) Applications and Lipschitz results . . . 29 Hence as ξ → 0 we get Pr,ξ → I in the L1 norm. Proof. As in the proof of Proposition 2 of [2] we get ∫ ∞ −∞ |Pr,ξ(f; x) − f(x)| dx ≤ 1 ξ ∫ ∞ 0 ωr(f, t)1e −t/ξ dt ≤ K ξ ∫ ∞ 0 t r−1+γ e −t/ξ dt = KΓ (r + γ) ξr−1+γ, (3.28) proving the claim. � Corollary 25. Assume the following Lipschitz condition: ω2 (f, δ)1 ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then ‖P2,ξf − f‖1 ≤ KΓ (2 + γ) ξ1+γ. (3.29) Hence as ξ → 0 we get P2,ξ → I in the L1 norm. Proof. In Proposition 24 we place r = 2. � In the next we consider f ∈ Cn(R) and f(n) ∈ Lp(R), n = 0 or n ≥ 2 even, 1 ≤ p < ∞ and the similar smooth singular operator of symmetric convolution type Pξ(f; x) = 1 2ξ ∫ ∞ −∞ f(x + y)e−|y|/ξdy, for all x ∈ R, ξ > 0. (3.30) Denote K(x) := Pξ(f; x) − f(x) − n/2∑ ρ=1 f (2ρ)(x)ξ2ρ. (3.31) We give Theorem 26. Let n ≥ 2 even and the rest as above. Then ‖K(x)‖2 ≤ ( √ τ̃ 20(2n − 1) ) ξn (n − 1)! ω2(f (n) , ξ)2, (3.32) where 0 < τ̃ = (∫ ∞ 0 (1 + x)5x2n−1e−xdx − (2n − 1)! ) < ∞. (3.33) Hence as ξ → 0 we get ‖K(x)‖2 → 0. If additionally f(2m) ∈ L2(R), m = 1, 2, . . . , n2 then ‖Pξ(f) − f‖2 → 0, as ξ → 0. Proof. In the proof of Theorem 3 of [2] we use p = q = 2. � 30 Razvan A. Mezei CUBO 13, 3 (2011) It follows a Lipschitz type approximation result. Theorem 27. Let p, q > 1 such that 1 p + 1 q = 1, n ≥ 2 even and the rest as above. Furthermore we assume the following Lipschitz condition: ω2 ( f(n), δ ) p ≤ Kδγ+1, K > 0, 0 < γ ≤ 1, for any δ > 0.Then ‖K(x)‖p ≤ ( 2 p )(γ+n+1) K [Γ (p (γ + n + 1) + 1)] 1/p (n − 1)!q1/qp1/p(q(n − 1) + 1)1/q [p (γ + 1) + 1] 1/p ξ γ+n+1 . (3.34) Hence as ξ → 0 we get ‖K(x)‖p → 0. If additionally f(2m) ∈ Lp(R), m = 1, 2, . . . , n2 then ‖Pξ(f) − f‖p → 0, as ξ → 0. Proof. As in the proof of Theorem 3, of [2] we find ∫ ∞ −∞ |K(x)|pdx ≤ c2 (∫ ∞ 0 (∫ y 0 ω2(f (n) , t)ppdt ) y pn−1 e −py/(2ξ) dy ) ≤ Kpc2 (∫ ∞ 0 ( yp(γ+1)+1 p (γ + 1) + 1 ) y pn−1 e −py/(2ξ) dy ) = Kpc2 p (γ + 1) + 1 ( 2 p )p(γ+n+1)+1 (∫ ∞ 0 z p(γ+n+1) e −z/ξ dz ) (2.3) = Kpc2Γ (p (γ + n + 1) + 1) p (γ + 1) + 1 ( 2 p )p(γ+n+1)+1 ξ p(γ+n+1)+1 . (3.35) where here we denoted c2 := 1 2ξqp/q((n − 1)!)p(q(n − 1) + 1)p/q . (3.36) We have established the claim of the theorem. � Corollary 28. Assume the following Lipschitz condition: ω2 (f ′′, δ) 2 ≤ Kδγ+1, K > 0, 0 < γ ≤ 1, for any δ > 0, and the rest as above in this section.Then ‖K(x)‖2 ≤ √ Γ (2γ + 7) 6γ + 9 K 2 ξ γ+3 . (3.37) Hence as ξ → 0 we get ‖K(x)‖2 → 0. If additionally f′′ ∈ L2(R), then ‖Pξ(f) − f‖2 → 0, as ξ → 0. Proof. In Theorem 27 we place p = q = n = 2. � Theorem 29. Let f ∈ C2(R) and f′′ ∈ L1(R). Here K(x) = Pξ(f; x) − f(x) − f′′(x)ξ2. Then ‖K(x)‖1 ≤ 8ω2(f′′, ξ)1ξ2. (3.38) Hence as ξ → 0 we obtain ‖K(x)‖1 → 0. CUBO 13, 3 (2011) Applications and Lipschitz results . . . 31 Also ‖Pξ(f) − f‖1 → 0, as ξ → 0. Proof. In the proof of Theorem 4 of [2] we use n = 2. � The Lipschitz case of p = 1 follows. Theorem 30. Let f ∈ Cn(R) and f(n) ∈ L1(R), n ≥ 2 even. Furthermore we assume the following Lipschitz condition: ω2 ( f(n), δ ) 1 ≤ Kδγ+1, K > 0, 0 < γ ≤ 1, for any δ > 0. Then ‖K(x)‖1 ≤ Γ (γ + n + 2) K 2(n − 1)! (γ + 2) ξ γ+n+1 . (3.39) Hence as ξ → 0 we obtain ‖K(x)‖1 → 0. If additionally f(2m) ∈ L1(R), m = 1, 2, . . . , n2 then ‖Pξ(f) − f‖1 → 0, as ξ → 0. Proof. As in the proof of Theorem 4 of [2] we have ‖K(x)‖1 ≤ 1 2ξ (∫ ∞ 0 (∫ y 0 ω2(f (n) , t)1dt ) yn−1 (n − 1)! e −y/ξ dy ) ≤ 1 2ξ (∫ ∞ 0 (∫ y 0 Kt γ+1 dt ) yn−1 (n − 1)! e −y/ξ dy ) = K 2ξ(n − 1)! (γ + 2) (∫ ∞ 0 y γ+n+1 e −y/ξ dy ) (2.3) = Γ (γ + n + 2) K 2(n − 1)! (γ + 2) ξ γ+n+1 . (3.40) We have proved the claim of the theorem. � Corollary 31. Let f ∈ C6(R) and f(6) ∈ L1(R). Furthermore we assume the following Lipschitz condition: ω2 ( f(6), δ ) 1 ≤ Kδγ+1, K > 0, 0 < γ ≤ 1, for any δ > 0. Then ‖K(x)‖1 ≤ Γ (γ + 8) K 240 (γ + 2) ξ γ+7 . (3.41) Hence as ξ → 0 we obtain ‖K(x)‖1 → 0. If additionally f(2m) ∈ L1(R), m = 1, 2, 3 then ‖Pξ(f) − f‖1 → 0, as ξ → 0. Proof. In Theorem 30 we place n = 6. � The case of n = 0 follows. Proposition 32. Let f as above in this section. Then ‖Pξ(f) − f‖2 ≤ √ 65 2 ω2(f, ξ)2. (3.42) Hence as ξ → 0 we obtain Pξ → I in the L2 norm. 32 Razvan A. Mezei CUBO 13, 3 (2011) Proof. In the proof of Proposition 3 of [2] we use p = q = 2. � The related Lipschitz case for n = 0 comes next. Proposition 33. Let p, q > 1 such that 1 p + 1 q = 1 and the rest as above. Furthermore we assume the following Lipschitz condition: ω2 (f, δ)p ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then ‖Pξ(f) − f‖p ≤ ( 2 p )1+γ [Γ ((1 + γ) p + 1)] 1/p K q1/qp1/p ξ 1+γ . (3.43) Hence as ξ → 0 we obtain Pξ → I in the Lp norm, p > 1. Proof. As in the proof of Proposition 3 of [2] we get ∫ ∞ −∞ |Pξ(f; x) − f(x)| p dx ≤ 1 2ξqp/q (∫ ∞ 0 ω2(f, y) p pe −py/(2ξ) dy ) ≤ Kp 2ξqp/q (∫ ∞ 0 y (1+γ)p e −py/(2ξ) dy ) (2.3) = Kp qp/qp ( 2 p )(1+γ)p Γ ((1 + γ) p + 1) ξ(1+γ)p. (3.44) The proof of the claim is now completed. � A particular example follows Corollary 34. Let f as above in this section. Furthermore we assume the following Lipschitz condition: ω2 (f, δ)2 ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then ‖Pξ(f) − f‖2 ≤ K 2 √ Γ (3 + 2γ)ξ1+γ. (3.45) Hence as ξ → 0 we obtain Pξ → I in the L2 norm. Proof. In Proposition 33 we place p = q = 2. � It follows the Lipschitz type result Proposition 35. Assume the following Lipschitz condition: ω2 (f, δ)1 ≤ Kδγ+1, K > 0, 0 < γ ≤ 1, for any δ > 0. It holds, ‖Pξf − f‖1 ≤ K 2 Γ (γ + 2) ξγ+1. (3.46) Hence as ξ → 0 we get Pξ → I in the L1 norm. Proof. As in the proof of Proposition 4 of [2] we derive ∫ ∞ −∞ |Pξ(f; x) − f(x)|dx ≤ 1 2ξ ∫ ∞ 0 ω2(f, y)1e −y/ξ dy ≤ 1 2ξ ∫ ∞ 0 Ky γ+1 e −y/ξ dy (2.3) = K 2 Γ (γ + 2) ξγ+1, (3.47) CUBO 13, 3 (2011) Applications and Lipschitz results . . . 33 proving the claim. � 4. Convergence with Rates of Smooth Gauss Weierstrass Singular Integral Operators In the next we deal with the following smooth Gauss Weierstrass singular integral operators Wr,ξ(f; x) defined as follows. For r ∈ N and n ∈ Z+ we set αj’s as in (2.1). Let f : R → R be Lebesgue measurable, we define for x ∈ R, ξ > 0 the Lebesgue integral Wr,ξ(f; x) := 1√ πξ ∫ ∞ −∞   r∑ j=0 αjf(x + jt)  e −t 2 /ξ dt. (4.1) We assume that Wr,ξ(f; x) ∈ R for all x ∈ R. We mention the useful here formula ∫ ∞ 0 t k e −t 2 /ξ dt = 1 2 Γ ( k + 1 2 ) ξ k+1 2 , for any k > −1. (4.2) We also need to introduce δk’s as in (2.4). Proposition 36. Let f ∈ C1(R) be defined as above in this section, and assume that W2,ξ(f; x) ∈ R for all x ∈ R. Then |W2,ξ(f; x) − f(x)| ≤ 2√ πξ ∫ ∞ 0 (∫ |t| 0 ω2(f ′ , w)dw ) e − t 2 ξ dt. (4.3) Proof. In Theorem 1 of [3] we use n = 1, r = 2. � We present the Lipschitz type result corresponding to the Theorem 1 of [3]. Theorem 37. Let f ∈ Cn(R), n ∈ Z+ and assume that Wr,ξ(f; x) ∈ R for all x ∈ R. Furthermore we assume the following Lipschitz condition: ωr ( f(n), δ ) ≤ Kδr−1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then it holds that ∣ ∣ ∣ ∣ ∣ ∣ Wr,ξ(f; x) − f(x) − ⌊n/2⌋∑ m=1 f (2m)(x)δ2m 1 m! ( ξ 4 )m ∣ ∣ ∣ ∣ ∣ ∣ ≤ K√ π Γ (γ + r) Γ (n + γ + r) Γ ( n + r + γ 2 ) ξ n+r+γ−1 2 . (4.4) 34 Razvan A. Mezei CUBO 13, 3 (2011) In L.H.S.(4.4) the sum collapses when n = 1. Proof. As in the proof of Theorem 1, of [3], we get again that Wr,ξ(f; x) − f(x) = n∑ k=1 f(k)(x) k! δk 1√ πξ (∫ ∞ −∞ t k e − t 2 ξ dt ) + R∗n, (4.5) where R∗n := 1√ πξ ∫ ∞ −∞ Rn(0, t)e− t2 ξ dt, (4.6) with Rn(0, t) := ∫ t 0 (t − w)n−1 (n − 1)! τ(w)dw, (4.7) and τ(w) := r∑ j=0 αjj n f (n)(x + jw) − δnf (n)(x). Also we get |Rn(0, t)| ≤ ∫ |t| 0 (|t| − w)n−1 (n − 1)! ωr(f (n) , w)dw. (4.8) Using the Lipschitz type condition we obtain again |Rn(0, t)| ≤ K|t|n+r+γ−1Γ (γ + r) Γ (n + γ + r) , (4.9) and, using (4.2), we obtain |R∗n| ≤ 1√ πξ ∫ ∞ −∞ K|t|n+r+γ−1Γ (γ + r) Γ (n + γ + r) e − t 2 ξ dt = K√ πξ Γ (γ + r) Γ (n + γ + r) ∫ ∞ −∞ |t| n+r+γ−1 e − t 2 ξ dt = 2K√ πξ Γ (γ + r) Γ (n + γ + r) ∫ ∞ 0 t n+r+γ−1 e − t 2 ξ dt (4.2) = K√ π Γ (γ + r) Γ (n + γ + r) Γ ( n + r + γ 2 ) ξ n+r+γ−1 2 . (4.10) We notice also that Wr,ξ(f; x) − f(x) − n∑ k=1 f(k)(x) k! δk 1√ πξ (∫ ∞ −∞ t k e − t 2 ξ dt ) = Wr,ξ(f; x) − f(x) − ⌊ n 2 ⌋ ∑ m=1 [ f(2m)(x) (2m)! √ π δ2mΓ ( 2m + 1 2 ) ξ m ] = R∗n. (4.11) CUBO 13, 3 (2011) Applications and Lipschitz results . . . 35 Furthermore we have that 1 (2m)! √ π Γ ( 2m + 1 2 ) = = 1 (2m) · (2m − 1) · ... · 3 · 2 · 1 · 1√ π · 2m − 1 2 · 2m − 3 2 · ... · 3 2 · 1 2 Γ ( 1 2 ) = 1 m! ( 1 4 )m . (4.12) By (4.10), (4.11) and (4.12) we complete the proof of the theorem. � Corollary 38. Let f ∈ C1(R), and assume that W2,ξ(f; x) ∈ R for all x ∈ R. Furthermore we assume the following Lipschitz condition: ω2 (f ′, δ) ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then |W2,ξ(f; x) − f(x)| ≤ K (γ + 2) √ π Γ ( 3 + γ 2 ) ξ 2+γ 2 . (4.13) Proof. In Theorem 37 we use n = 1, r = 2. � For the case n = 0 we have Theorem 39. Let f be defined as above in this section, with n = 0. Furthermore we assume the following Lipschitz condition: ωr (f, δ) ≤ Kδr−1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. It holds that |Wr,ξ(f; x) − f(x)| ≤ K√ π Γ ( r + γ 2 ) ξ r+γ−1 2 . (4.14) Proof. As in the proof of Corollary 1, of [3], with n = 0, using the Lipschitz type condition, we get that |Wr,ξ(f; x) − f(x)| ≤ 2√ πξ ∫ ∞ 0 ωr(f, t)e − t 2 ξ dt ≤ 2√ πξ ∫ ∞ 0 Kt r−1+γ e − t 2 ξ dt (4.2) = K√ π Γ ( r + γ 2 ) ξ r+γ−1 2 . (4.15) This completes the proof of Theorem 39. � Corollary 40. Let f be defined as above in this section, with n = 0. Furthermore we assume the following Lipschitz condition: ω2 (f, δ) ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then |W2,ξ(f; x) − f(x)| ≤ K√ π Γ ( 2 + γ 2 ) ξ γ+1 2 . (4.16) Proof. In Theorem 39 we use r = 2. � 36 Razvan A. Mezei CUBO 13, 3 (2011) In the next we consider f ∈ Cn(R), n ≥ 2 even and the simple smooth singular operator of symmetric convolution type Wξ(f, x0) := 1√ πξ ∫ ∞ −∞ f(x0 + y)e −y 2 /ξ dy, for all x0 ∈ R, ξ > 0. (4.17) That is Wξ(f; x0) = 1√ πξ ∫ ∞ 0 (f(x0 + y) + f(x0 − y)) e −y 2 /ξ dy, for all x0 ∈ R, ξ > 0. (4.18) We assume that f is such that Wξ(f; x0) ∈ R, ∀x0 ∈ R, ∀ξ > 0 and ω2(f(n), h) < ∞, h > 0. Note that W1,ξ = Wξ and if Wξ(f; x0) ∈ R then Wr,ξ(f; x0) ∈ R. Proposition 41. Assume f ∈ Cn(R), ω2(f, h) < ∞, h > 0. Furthermore we assume the following Lipschitz condition: ω2 (f, δ) ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then ‖Wξ(f) − f‖∞ ≤ K 2 √ π Γ ( 2 + γ 2 ) ξ γ+1 2 . (4.19) Proof. Using Proposition 1 of [3] we obtain |Wξ(f; x0) − f(x0)| ≤ 1√ πξ ∫ ∞ 0 ω2(f, y)e −y 2 /ξ dy ≤ 1√ πξ ∫ ∞ 0 Ky 1+γ e −y 2 /ξ dy (4.2) = K 2 √ π Γ ( 2 + γ 2 ) ξ γ+1 2 , (4.20) proving the claim of the proposition. � Define the quantity K2(x0) := Wξ(f; x0) − f(x0) − n/2∑ ρ=1 f (2ρ)(x0) 1 ρ! ( ξ 4 )ρ . (4.21) We give Theorem 42. Let f ∈ Cn(R), n even, Wξ(f) real valued. Furthermore we assume the following Lipschitz condition: ω2 ( f(n), δ ) ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then |K2(x0)| ≤ K n!2 √ π Γ ( n + γ + 2 2 ) ξ n+γ+1 2 . (4.22) CUBO 13, 3 (2011) Applications and Lipschitz results . . . 37 Proof. Using Theorem 6 of [3] we obtain |K2(x0)| ≤ 1 n! √ πξ ∫ ∞ 0 ω2(f (n) , y)yne−y 2 /ξ dy ≤ 1 n! √ πξ ∫ ∞ 0 Ky 1+γ y n e −y 2 /ξ dy (4.2) = K n!2 √ π Γ ( n + γ + 2 2 ) ξ n+γ+1 2 , (4.23) proving the claim of the theorem. � In particular we have Corollary 43. Let f ∈ C4(R) such that Wξ(f) is real valued. Furthermore we assume the following Lipschitz condition: ω2 ( f(4), δ ) ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then |K2(x0)| ≤ K 48 √ π Γ ( γ + 6 2 ) ξ γ+5 2 . (4.24) Proof. In Theorem 42 we use n = 4. � We also give Corollary 44. Let f ∈ C2(R), such that ω2(f ′′ , |y|) ≤ 2A|y|γ, 0 < γ ≤ 2, A > 0. Then for x0 ∈ R we have ∣ ∣ ∣ ∣ Wξ(f; x0) − f(x0) − f′′(x0)ξ 4 ∣ ∣ ∣ ∣ ≤ A (γ + 1)(γ + 2) √ π Γ ( 3 + γ 2 ) ξ 2+γ 2 . (4.25) Inequality (4.25) is sharp, namely it is attained at x0 = 0 by f∗(y) = A|y|γ+2 (γ + 1)(γ + 2) . Proof. In Theorem 7 of [3] we use n = 2. � We also give Corollary 45. Assume that ω2(f, ξ) < ∞ and n = 0. Then ‖W2,ξ(f) − f‖∞ ≤ [ 2√ π + 3 2 ] ω2(f, √ ξ). (4.26) 38 Razvan A. Mezei CUBO 13, 3 (2011) and as ξ → 0, W2,ξ u → I with rates. Proof. By formula (37) of [3] with r = 2. � Define the quantity K1 := ∥ ∥ ∥ ∥ ∥ ∥ Wr,ξ(f; x) − f(x) − ⌊n/2⌋∑ m=1 f (2m)(x)δ2m 1 m! ( ξ 4 )m ∥ ∥ ∥ ∥ ∥ ∥ ∞ ,x . (4.27) We present Corollary 46. Assuming f ∈ C2(R) and ω2(f′′, ξ) < ∞, ξ > 0 we have K1 = ∥ ∥ ∥ ∥ W2,ξ(f; x) − f(x) − f ′′(x)δ2 ξ 4 ∥ ∥ ∥ ∥ ∞ ,x ≤ { 1 3 √ π + 5 16 } ω2(f ′′ , √ ξ)ξ. (4.28) That is as ξ → 0 we get W2,ξ → I, pointwise with rates, given that ‖f′′‖∞ < ∞. Proof. In Theorem 11 of [3] we use r = n = 2. � We also present Corollary 47. Assuming f ∈ C2(R) and ω2(f′′, ξ) < ∞, ξ > 0 we have ∥ ∥K2(x) ∥ ∥ ∞ ,x = ∥ ∥ ∥ ∥ Wξ(f; x0) − f(x0) − f ′′(x0) ξ 4 ∥ ∥ ∥ ∥ ∞ ,x ≤ { 1 6 √ π + 5 32 } ω2(f ′′ , √ ξ)ξ. (4.29) That is as ξ → 0 we get Wξ → I, pointwise with rates, given that ‖f′′‖∞ < ∞. Proof. In Theorem 12 of [3] we use n = 2. � 5. Lp Convergence with Rates of Smooth Gauss Weierstrass Singular Integral Operators For r ∈ N and n ∈ Z+ we let αj as in (2.1). CUBO 13, 3 (2011) Applications and Lipschitz results . . . 39 Let f ∈ Cn(R) and f(n) ∈ Lp(R), 1 ≤ p < ∞, we define for x ∈ R, ξ > 0 the Lebesgue integral Wr,ξ(f; x) as in (4.1). The rth Lp-modulus of smoothness ωr(f (n), h)p was defined in (3.1). Here we have that ωr(f (n), h)p < ∞, h > 0. The δk’s were introduced in (2.4). We define ∆(x) := Wr,ξ(f; x) − f(x) − ⌊n/2⌋∑ m=1 f (2m)(x)δ2m 1 m! ( ξ 4 )m . (5.1) We have the following results. Corollary 48. Let n ∈ N and the rest as above in this section. Then ‖∆(x)‖2 ≤ √ 2τξ n 2 (n − 1)! 4 √ π √ (2r + 1) (2n − 1) ωr(f (n) , √ ξ)2, (5.2) where 0 < τ := [∫ ∞ 0 (1 + u) 2r+1 u 2n−1 e −u 2 du − ∫ ∞ 0 u 2n−1 e −u 2 du ] < ∞. (5.3) Hence as ξ → 0 we obtain ‖∆(x)‖2 → 0. If additionally f(2m) ∈ L2(R), m = 1, 2, . . . , ⌊ n 2 ⌋ then ‖Wr,ξ(f) − f‖2 → 0, as ξ → 0. Proof. In Theorem 1 of [4], we place p = q = 2. � Corollary 49. Let f be as above in this section. In particular, for n = 1, we have ‖Wr,ξ(f; ·) − f‖2 ≤ √ 2τ 4 √ π √ (2r + 1) √ ξωr(f ′ , √ ξ)2, (5.4) where 0 < τ := [∫ ∞ 0 (1 + u) 2r+1 ue −u 2 du − 1 2 ] < ∞. (5.5) Hence as ξ → 0 we obtain ‖Wr,ξ(f; ·) − f‖2 → 0. Proof. In Theorem 1 of [4], we place p = q = 2, n = 1. � Corollary 50. Let f be as above in this section and n = 2. Then ‖Wr,ξ(f; x) − f(x) − f′′(x)δ2 4 ξ‖2 ≤ √ 2τ 4 √ π √ 3 (2r + 1) ξωr(f ′′ , √ ξ)2, (5.6) 40 Razvan A. Mezei CUBO 13, 3 (2011) where 0 < τ := [∫ ∞ 0 (1 + u) 2r+1 u 3 e −u 2 du − 1 2 ] < ∞. (5.7) Hence as ξ → 0 we obtain ‖∆(x)‖2 → 0. If additionally f′′ ∈ L2(R), then ‖Wr,ξ(f) − f‖2 → 0, as ξ → 0. Proof. In Theorem 1 of [4], we place p = q = n = 2. � Next we present the Lipschitz type result corresponding to Theorem 1 of [4]. Theorem 51. Let p, q > 1 such that 1 p + 1 q = 1, n ∈ N, and the rest as above in this section. Furthermore we assume the following Lipschitz condition: ωr ( f(n), δ ) p ≤ Kδr−1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0.Then ‖∆(x)‖p ≤ ( Γ ( p(r−1+γ+n)+1 2 )) 1 p 2 (r+γ+n) 2 Kξ (r−1+γ+n) 2 [ (n − 1)!p r− 1 q +γ+n 2 q 1 2q π 1 2p (q(n − 1) + 1) 1 q (p (r − 1 + γ) + 1) 1 p ]. (5.8) Hence as ξ → 0 we obtain ‖∆(x)‖p → 0. If additionally f(2m) ∈ Lp(R), m = 1, 2, . . . , ⌊ n 2 ⌋ then ‖Wr,ξ(f) − f‖p → 0, as ξ → 0. Proof. As in the proof of Theorem 1, [4], we get again I := ∫ ∞ −∞ |∆(x)|pdx ≤ c1 (∫ ∞ −∞ (∫ |t| 0 ωr(f (n) , w)ppdw ) |t| np−1 e − pt2 2ξ dt ) , (5.9) where c1 := 2 p−1 2 q p−1 2 √ πξ((n − 1)!)p(q(n − 1) + 1)p/q . (5.10) Using the Lipschitz condition, we obtain I ≤ c1 (∫ ∞ −∞ (∫ |t| 0 ( Kw r−1+γ )p dw ) |t| np−1 e − pt2 2ξ dt ) = c1K p (p (r − 1 + γ) + 1) (∫ ∞ −∞ |t| p(r−1+γ+n) e − pt2 2ξ dt ) = 2c1K p (p (r − 1 + γ) + 1) (∫ ∞ 0 t p(r−1+γ+n) e − pt2 2ξ dt ) (4.2) = c1K pΓ ( p(r−1+γ+n)+1 2 ) (p (r − 1 + γ) + 1) ( 2 p ) p(r−1+γ+n)+1 2 ξ p(r−1+γ+n)+1 2 . (5.11) CUBO 13, 3 (2011) Applications and Lipschitz results . . . 41 Thus we obtain I ≤ Kp2 p(r+γ+n) 2 Γ ( p(r−1+γ+n)+1 2 ) ξ p(r−1+γ+n) 2 q p−1 2 √ π((n − 1)!)p(q(n − 1) + 1)p/q (p (r − 1 + γ) + 1) p p(r−1+γ+n)+1 2 . (5.12) That is finishing the proof of the theorem. � In particular we have Corollary 52. Let f such that the following Lipschitz condition holds: ω7 ( f(4), δ ) 2 ≤ Kδ6+γ, K > 0, 0 < γ ≤ 1, for any δ > 0, and the rest as above in this section. Then ‖∆(x)‖2 ≤ K 6 √ √ √ √ Γ ( 2γ+21 2 ) 7 √ π (2γ + 13) ξ (γ+10) 2 . (5.13) Hence as ξ → 0 we obtain ‖∆(x)‖2 → 0. If additionally f(2m) ∈ L2(R), m = 1, 2, then ‖W7,ξ(f) − f‖2 → 0, as ξ → 0. Proof. In Theorem 51 we place p = q = 2, n = 4, and r = 7. � The counterpart of Theorem 51 follows, case of p = 1. Theorem 53. Let f ∈ Cn(R) and f(n) ∈ L1(R), n ∈ N. Furthermore we assume the following Lipschitz condition: ωr ( f(n), δ ) 1 ≤ Kδr−1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0.Then ‖∆(x)‖1 ≤ K (n − 1)! (r + γ) √ π Γ ( r + γ + n 2 ) ξ r+γ+n−1 2 . (5.14) Hence as ξ → 0 we obtain ‖∆(x)‖1 → 0. If additionally f(2m) ∈ L1(R), m = 1, 2, . . . , ⌊ n 2 ⌋ then ‖Wr,ξ(f) − f‖1 → 0, as ξ → 0. Proof. As in the proof of Theorem 2, [4] we get ‖∆(x)‖1 ≤ 1 (n − 1)! √ πξ (∫ ∞ −∞ (∫ |t| 0 ωr(f (n) , w)1dw ) |t| n−1 e −t 2 /ξ dt ) . (5.15) Consequently we have 42 Razvan A. Mezei CUBO 13, 3 (2011) ‖∆(x)‖1 ≤ 1 (n − 1)! √ πξ (∫ ∞ −∞ (∫ |t| 0 Kw r−1+γ dw ) |t| n−1 e −t 2 /ξ dt ) = K (n − 1)! √ πξ (∫ ∞ −∞ ( |t|r+γ r + γ ) |t| n−1 e −t 2 /ξ dt ) = K (n − 1)! (r + γ) √ πξ (∫ ∞ −∞ |t| r+γ+n−1 e −t 2 /ξ dt ) = 2K (n − 1)! (r + γ) √ πξ (∫ ∞ 0 t r+γ+n−1 e −t 2 /ξ dt ) (4.2) = K (n − 1)! (r + γ) √ πξ Γ ( r + γ + n 2 ) ξ r+γ+n 2 . (5.16) We have gotten that ‖∆(x)‖1 ≤ K (n − 1)! (r + γ) √ π Γ ( r + γ + n 2 ) ξ r+γ+n−1 2 . (5.17) Hence the validity of (5.14). � Corollary 54. Let f ∈ C2(R) and f′′ ∈ L1(R). Furthermore we assume the following Lipschitz condition: ω2 (f ′′, δ) 1 ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0.Then ‖∆(x)‖1 ≤ K (2 + γ) √ π Γ ( 4 + γ 2 ) ξ γ+3 2 . (5.18) Hence as ξ → 0 we obtain ‖∆(x)‖1 → 0. Also we get ‖W2,ξ(f) − f‖1 → 0, as ξ → 0. Proof. In Theorem 53 we place n = r = 2. � Next, when n = 0 we get Proposition 55. Let r ∈ N and the rest as above. Then ‖Wr,ξ(f) − f‖2 ≤ 2 3 4 θ 1 2 q 1 4 π 1 4 ωr(f, √ ξ)2, (5.19) where 0 < θ := ∫ ∞ 0 (1 + t) 2r e −t 2 dt < ∞. (5.20) Hence as ξ → 0 we obtain Wr,ξ → unit operator I in the L2 norm, p > 1. Proof. In the proof of Proposition 1 of [4] we use p = q = 2. � We continue with CUBO 13, 3 (2011) Applications and Lipschitz results . . . 43 Proposition 56. Let p, q > 1 such that 1 p + 1 q = 1 and the rest as above. Furthermore we assume the following Lipschitz condition: ωr (f, δ)p ≤ Kδr−1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then ‖Wr,ξ(f) − f‖p ≤ p √ Γ ( p(r − 1 + γ) + 1 2 )( 2 p ) r+γ 2 ( p q ) 1 2q K p √ π ξ (r−1+γ) 2 . (5.21) Hence as ξ → 0 we obtain Wr,ξ → unit operator I in the Lp norm, p > 1. Proof. As in the proof of Proposition 1 of [4] we find ∫ ∞ −∞ |Wr,ξ(f; x) − f(x)| p dx ≤ 2 (πξ) p 2 ( 2πξ q ) p 2q ∫ ∞ 0 ωr(f, t) p pe − pt2 2ξ dt ≤ 2Kp (πξ) p 2 ( 2πξ q ) p 2q ∫ ∞ 0 t p(r−1+γ) e − pt2 2ξ dt (4.2) = Kp π p 2 ( 2π q ) p 2q ( 2 p ) p(r−1+γ)+1 2 Γ ( p(r − 1 + γ) + 1 2 ) ξ p(r−1+γ) 2 . (5.22) We have established the claim of the proposition. � Corollary 57. Let f such that the following Lipschitz condition holds: ω4 (f, δ)2 ≤ Kδ3+γ, K > 0, 0 < γ ≤ 1, for any δ > 0, and the rest as above in this section. Then ‖W4,ξ(f) − f‖2 ≤ √ Γ ( 2γ + 7 2 ) K√ π ξ (3+γ) 2 . (5.23) Hence as ξ → 0 we obtain W4,ξ → unit operator I in the L2 norm. Proof. In Proposition 56 we place p = q = 2 and r = 4. � In the L1 case, n = 0 we have Proposition 58. It holds ‖W2,ξf − f‖1 ≤ ( 2√ π + 3 2 ) ω2(f, √ ξ)1. (5.24) Hence as ξ → 0 we get W2,ξ → I in the L1 norm. Proof. In the proof of Proposition 2 of [4] we use r = 2. � Proposition 59. We assume the following Lipschitz condition: ωr (f, δ)1 ≤ Kδr−1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then ‖Wr,ξf − f‖1 ≤ K√ π Γ ( r + γ 2 ) ξ r−1+γ 2 . (5.25) 44 Razvan A. Mezei CUBO 13, 3 (2011) Hence as ξ → 0 we get Wr,ξ → I in the L1 norm. Proof. As in the proof of Proposition 2 of [4] we get ∫ ∞ −∞ |Wr,ξ(f; x) − f(x)| dx ≤ 1√ πξ ∫ ∞ −∞ ωr(f, |t|)1e −t 2 /ξ dt ≤ 1√ πξ ∫ ∞ −∞ K|t| r−1+γ e −t 2 /ξ dt = 2K√ πξ ∫ ∞ 0 t r−1+γ e −t 2 /ξ dt (4.2) = K√ π Γ ( r + γ 2 ) ξ r−1+γ 2 . (5.26) We have proved the claim of the proposition. � Corollary 60. Assume the following Lipschitz condition: ω2 (f, δ)1 ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then ‖W2,ξf − f‖1 ≤ K√ π Γ ( 2 + γ 2 ) ξ 1+γ 2 . (5.27) Hence as ξ → 0 we get W2,ξ → I in the L1 norm. Proof. In Proposition 59 we place r = 2. � In the next we consider f ∈ Cn(R) and f(n) ∈ Lp(R), n = 0 or n ≥ 2 even, 1 ≤ p < ∞ and the similar smooth singular operator of symmetric convolution type Wξ(f; x) = 1√ πξ ∫ ∞ −∞ f(x + y)e−y 2 /ξ dy, for all x ∈ R, ξ > 0. (5.28) Denote K(x) := Wξ(f; x) − f(x) − n/2∑ ρ=1 f(2ρ)(x) ρ! · ( ξ 4 )ρ . (5.29) We give Theorem 61. Let n ≥ 2 even and the rest as above. Then ‖K(x)‖2 ≤ √ τ̃ 10 √ π (2n − 1) ξ n 2 (n − 1)! ω2(f (n) , √ ξ)2, (5.30) where 0 < τ̃ = ∫ ∞ 0 ( (1 + u) 5 − 1 ) u 2n−1 e −u 2 du < ∞. (5.31) Hence as ξ → 0 we get ‖K(x)‖2 → 0. CUBO 13, 3 (2011) Applications and Lipschitz results . . . 45 If additionally f(2m) ∈ L2(R), m = 1, 2, . . . , n2 then ‖Wξ(f) − f‖2 → 0, as ξ → 0. Proof. In the proof of Theorem 3 of [4] we use p = q = 2. � It follows a Lipschitz type approximation result. Theorem 62. Let p, q > 1 such that 1 p + 1 q = 1, n ≥ 2 even and the rest as above. Furthermore we assume the following Lipschitz condition: ω2 ( f(n), δ ) p ≤ Kδγ+1, K > 0, 0 < γ ≤ 1, for any δ > 0.Then ‖K(x)‖p ≤ K [ Γ ( p(γ+n+1)+1 2 )] 1 p √ 2π 1 2p (n − 1)!p 1 2p q 1 2q [q(n − 1) + 1] 1 q [p (γ + 1) + 1] 1 p ( 2 p ) (γ+n+1) 2 ξ (γ+n+1) 2 . (5.32) Hence as ξ → 0 we get ‖K(x)‖p → 0. If additionally f(2m) ∈ Lp(R), m = 1, 2, . . . , n2 then ‖Wξ(f) − f‖p → 0, as ξ → 0. Proof. As in the proof of Theorem 3, of [4] we find ∫ ∞ −∞ |K(x)|pdx ≤ c2 (∫ ∞ 0 (∫ y 0 ω2(f (n) , t)ppdt ) y pn−1 e − py2 2ξ dy ) ≤ Kpc2 (∫ ∞ 0 ( yp(γ+1)+1 p (γ + 1) + 1 ) y pn−1 e − py2 2ξ dy ) = Kpc2 p (γ + 1) + 1 (∫ ∞ 0 y p(γ+n+1) e − py2 2ξ dy ) (4.2) = Kpc2 p (γ + 1) + 1 ( 2 p ) p(γ+n+1)+1 2 · 1 2 Γ ( p (γ + n + 1) + 1 2 ) ξ p(γ+n+1)+1 2 . (5.33) where here we denoted c2 := 1 2 p 2q q p 2q (q(n − 1) + 1)p/q ((n − 1)!) p √ πξ . (5.34) We have established the claim of the theorem. � Corollary 63. Assume the following Lipschitz condition: ω2 (f ′′, δ) 2 ≤ Kδγ+1, K > 0, 0 < γ ≤ 1, for any δ > 0, and the rest as above in this section.Then ‖K(x)‖2 ≤ √ √ √ √ [ Γ ( 2γ+7 2 )] √ π [6γ + 9] K 2 ξ (γ+3) 2 . (5.35) Hence as ξ → 0 we get ‖K(x)‖2 → 0. If additionally f′′ ∈ L2(R), then ‖Wξ(f) − f‖2 → 0, as ξ → 0. 46 Razvan A. Mezei CUBO 13, 3 (2011) Proof. In Theorem 62 we place p = q = n = 2. � Theorem 64. Let f ∈ C2(R) and f′′ ∈ L1(R). Here K(x) = Wξ(f; x) − f(x) − f ′′ (x) 4 ξ. Then ‖K(x)‖1 ≤ ( 1 2 √ π + 3 8 ) ω2(f ′′ , √ ξ)1ξ. (5.36) Hence as ξ → 0 we obtain ‖K(x)‖1 → 0. Also ‖Wξ(f) − f‖1 → 0, as ξ → 0. Proof. In the proof of Theorem 4 of [4] we use n = 2. � The Lipschitz case of p = 1 follows. Theorem 65. Let f ∈ Cn(R) and f(n) ∈ L1(R), n ≥ 2 even. Furthermore we assume the following Lipschitz condition: ω2 ( f(n), δ ) 1 ≤ Kδγ+1, K > 0, 0 < γ ≤ 1, for any δ > 0. Then ‖K(x)‖1 ≤ Γ ( γ+n+2 2 ) K (n − 1)! (γ + 2) 2 √ π ξ γ+n+1 2 . (5.37) Hence as ξ → 0 we obtain ‖K(x)‖1 → 0. If additionally f(2m) ∈ L1(R), m = 1, 2, . . . , n2 then ‖Wξ(f) − f‖1 → 0, as ξ → 0. Proof. As in the proof of Theorem 4 of [4] we have ‖K(x)‖1 ≤ 1√ πξ ∫ ∞ 0 ((∫ y 0 ω2(f (n) , t)1dt ) yn−1 (n − 1)! e −y 2 /ξ ) dy ≤ 1√ πξ ∫ ∞ 0 ((∫ y 0 Kt γ+1 dt ) yn−1 (n − 1)! e −y 2 /ξ ) dy = K (n − 1)! (γ + 2) √ πξ ∫ ∞ 0 ( y γ+n+1 e −y 2 /ξ ) dy (4.2) = Γ ( γ+n+2 2 ) K (n − 1)! (γ + 2) 2 √ π ξ γ+n+1 2 . (5.38) We have proved the claim of the theorem. � Corollary 66. Let f ∈ C6(R) and f(6) ∈ L1(R). Furthermore we assume the following Lipschitz condition: ω2 ( f(6), δ ) 1 ≤ Kδγ+1, K > 0, 0 < γ ≤ 1, for any δ > 0. Then ‖K(x)‖1 ≤ Γ ( γ+8 2 ) K 240 (γ + 2) √ π ξ γ+7 2 . (5.39) Hence as ξ → 0 we obtain ‖K(x)‖1 → 0. If additionally f(2m) ∈ L1(R), m = 1, 2, 3 then ‖Wξ(f) − f‖1 → 0, as ξ → 0. CUBO 13, 3 (2011) Applications and Lipschitz results . . . 47 Proof. In Theorem 65 we place n = 6. � The case of n = 0 follows. Proposition 67. Let f as above in this section. Then ‖Wξ(f) − f‖2 ≤ √ 2√ π + 19 16 ω2(f, √ ξ)2. (5.40) Hence as ξ → 0 we obtain Wξ → I in the L2 norm. Proof. In the proof of Proposition 3 of [4] we use p = q = 2. � The related Lipschitz case for n = 0 comes next. Proposition 68. Let p, q > 1 such that 1 p + 1 q = 1 and the rest as above. Furthermore we assume the following Lipschitz condition: ω2 (f, δ)p ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then ‖Wξ(f) − f‖p ≤ ( 2 p ) (1+γ) 2 [ Γ ( (1+γ)p+1 2 )] 1 p K π 1 2p p 1 2p q 1 2q √ 2 ξ (1+γ) 2 . (5.41) Hence as ξ → 0 we obtain Wξ → I in the Lp norm, p > 1. Proof. As in the proof of Proposition 3 of [4] we get ∫ ∞ −∞ |Wξ(f; x) − f(x)| p dx ≤ 1 √ πξ (2q) p 2q ∫ ∞ 0 ω2(f, y) p pe −py2 2ξ dy ≤ 1 √ πξ (2q) p 2q ∫ ∞ 0 ( Ky 1+γ )p e −py2 2ξ dy (4.2) = Kp √ π (2q) p 2q ( 2 p ) (1+γ)p+1 2 1 2 Γ ( (1 + γ) p + 1 2 ) ξ (1+γ)p 2 . (5.42) The proof of the claim is now completed. � A particular example follows Corollary 69. Let f as above in this section. Furthermore we assume the following Lipschitz condition: ω2 (f, δ)2 ≤ Kδ1+γ, K > 0, 0 < γ ≤ 1, for any δ > 0. Then ‖Wξ(f) − f‖2 ≤ K 2 √ √ √ √ Γ ( 3+2γ 2 ) √ π ξ (1+γ) 2 . (5.43) Hence as ξ → 0 we obtain Wξ → I in the L2 norm. Proof. In Proposition 68 we place p = q = 2. � 48 Razvan A. Mezei CUBO 13, 3 (2011) We finish with the Lipschitz type result Proposition 70. Assume the following Lipschitz condition: ω2 (f, δ)1 ≤ Kδγ+1, K > 0, 0 < γ ≤ 1, for any δ > 0. It holds, ‖Wξf − f‖1 ≤ K 2 √ π Γ ( γ + 2 2 ) ξ γ+1 2 . (5.44) Hence as ξ → 0 we get Wξ → I in the L1 norm. Proof. As in the proof of Proposition 4 of [4] we derive ∫ ∞ −∞ |Wξ(f; x) − f(x)|dx ≤ 1√ πξ ∫ ∞ 0 ω2(f, y)1e −y 2 /ξ dy ≤ 1√ πξ ∫ ∞ 0 Ky γ+1 e −y 2 /ξ dy (4.2) = K 2 √ π Γ ( γ + 2 2 ) ξ γ+1 2 . (5.45) We have established the claim. � Received: September 2009. Revised: July 2010. References [1] George A. Anastassiou, ”Basic Convergence with Rates of Smooth Picard Singular Opera- tors”, J. Comput. Anal. Appl., 8 (2006), 313–334. [2] George A. Anastassiou, ”Lp convergence with rates of smooth Picard singular operators”, Differential & difference equations and applications, Hindawi Publ. Corp., New York, (2006), 31–45. [3] George A. Anastassiou, Razvan A. Mezei, “Uniform Convergence with Rates of Smooth Gauss-Weierstrass Singular Integral Operators”, Applicable Analysis, 88:7 (2009), 1015 — 1037. [4] George A. Anastassiou, Razvan A. Mezei, “Lp Convergence with Rates of Smooth Gauss- Weierstrass Singular Operators”, Nonlinear Studies, accepted 2009. Introduction Convergence with Rates of Smooth Picard Singular Integral Operators Lp Convergence with Rates of Smooth Picard Singular Integral Operators Convergence with Rates of Smooth Gauss Weierstrass Singular Integral Operators Lp Convergence with Rates of Smooth Gauss Weierstrass Singular Integral Operators