CUBO A Mathematical Journal Vol.15, No¯ 01, (97–112). March 2013 Approximation by Discrete Singular Operators George A. Anastassiou University of Memphis Department of Mathematical Sciences Memphis, TN 38152, U.S.A. ganastss@memphis.edu ABSTRACT Here we study basic approximation properties with rates of our discrete versions of Picard, Gauss-Weierstrass, Poisson-Cauchy singular operators and of two other discrete operators. We prove uniform convergence of these operators to the unit operator. Also all these operators fulfill the global smoothness preservation property. The discussed operators act on the space of uniformly continuous functions over the real line. RESUMEN Aqúı estudiamos las propiedades de aproximación básica con cocientes de nuestras versiones discretas de operadores singulares de Picard, Gauss-Weierstrass, Poisson- Cauchy y de otros dos operadores discretos. Probamos la convergencia uniforme de estos operadores al operador unitario. Además, todos estos operadores satisfacen la propiedad de preservación de suavidad global. Los operadores discutidos actúan sobre el espacio de funciones uniformemente continua sobre la recta real. Keywords and Phrases: Discrete singular operator, modulus of continuity, uniform convergence, global smoothness. 2010 AMS Mathematics Subject Classification: 26A15, 26D15, 41A17, 41A25, 41A44. 98 George A. Anastassiou CUBO 15, 1 (2013) 1 Preliminaries Let f : R → R be a function which is uniformly continuous (f ∈ CU (R)). Following [2], p. 40-41, we define the first modulus of continuity, ω1 (f, t) := sup x,y∈R |x−y|≤t |f (x) − f (y)| , t ≥ 0. (1) The function ω1 is continuous at t = 0 if and only if f is uniformly continuous on R. So that here ω1 (f, t) → ω1 (f, 0) = 0, as t → 0. It also holds ω1 (f, λt) ≤ (λ + 1) ω1 (f, t) , λ ≥ 0. (2) Clearly ω1 (f, t) is finite for each t ≥ 0. In [1] we studied extensively the convergence to the unit operator of various integral singular operators. Here we define the discrete analogs of these operators next, and we study their uniform convergence to the unit operator with rates. Let 0 < ξ ≤ 1, such that ξ → 0+, x ∈ R; 1 ξ ≥ 1. i) We define the discrete Picard operators: ( P∗ξf ) (x) := ∑ ∞ ν=−∞ f (x + ν) e− |ν| ξ ∑ ∞ ν=−∞ e− |ν| ξ . (3) ii) We define the discrete Gauss-Weierstrass operators: ( W∗ξf ) (x) := ∑ ∞ ν=−∞ f (x + ν) e− ν2 ξ ∑ ∞ ν=−∞ e− ν2 ξ . (4) iii) We define the general discrete Poisson-Cauchy operators: let α ∈ N, β > 1 α ; ( M∗ξf ) (x) := ∑ ∞ ν=−∞ f (x + ν) ( ν2α + ξ2α )−β ∑ ∞ ν=−∞ (ν2α + ξ2α) −β . (5) iv) We define the basic discrete convolution operators: let ϕ : R → R, with ‖ϕ‖ ∞ := sup x∈R |ϕ (x)| ≤ K, K > 0, β ∈ N − {1}; ( θ∗ξf ) (x) := f (x) + ∑ ν∈Z−{0} f (x + ν) ( ϕ( νξ ) ν ξ )2β 1 + ∑ ν∈Z−{0} ( ϕ( νξ ) ν ξ )2β . (6) CUBO 15, 1 (2013) Approximation by Discrete Singular Operators 99 v) We define the general discrete convolution operators: let ϕ : R → R+ with ϕ (x) ≤ Ax2β, ∀ x ∈ R, β ∈ N − {1}, A > 0; ( T∗ξf ) (x) := f (x) + ∑ ν∈Z−{0} f (x + ν) ϕ( νξ ) ( νξ ) 4β 1 + ∑ ν∈Z−{0} ϕ( νξ ) ( νξ ) 4β . (7) The above operators, as we will see, are well defined and are linear, positive, and bounded when ‖f‖ ∞ := sup x∈R |f (x)| < ∞. Furthermore P∗ξ (1) = W ∗ ξ (1) = M ∗ ξ (1) = θ ∗ ξ (1) = T ∗ ξ (1) = 1, (8) with ∥ ∥P∗ξ ∥ ∥ = ∥ ∥W∗ξ ∥ ∥ = ∥ ∥M∗ξ ∥ ∥ = ∥ ∥θ∗ξ ∥ ∥ = ∥ ∥T∗ξ ∥ ∥ = 1, (9) on continuous bounded functions. In this article we are motivated by [3]. 2 Main Results All here as in Preliminaries earlier. We start with the basic approximation properties of discrete Picard operators. We present Theorem 2.1. It holds ∥ ∥P∗ξf − f ∥ ∥ ∞ ≤   1 + 2e− 1 ξ ( 2ξ + 2 + 1 ξ ) 1 + 2ξe− 1 ξ  ω1 (f, ξ) . (10) The constant in the right hand side of (10) converges to 1 as ξ → 0+. So that P∗ξ → I (unit operator), uniformly with rates, as ξ → 0 + . Proof. We will use a lot ∞∑ ν=1 1 ν2 = π2 6 (Euler, 1741). We see that −∞∑ ν=−1 e− |ν| ξ = ∞∑ ν=1 e− ν ξ < ∞∑ ν=1 1 ν2 = π2 6 , it converges. 100 George A. Anastassiou CUBO 15, 1 (2013) Thus ∞∑ ν=−∞ e− |ν| ξ = 2 ∞∑ ν=1 e− ν ξ + 1 < π2 3 + 1. (11) Using [4] we obtain ∞∑ ν=1 e− ν ξ − e− 1 ξ ≤ ∫ ∞ 1 e− ν ξ dν ≤ ∞∑ ν=1 e− ν ξ . (12) Hence 2 ∫ ∞ 1 e− ν ξ dν + 1 ≤ 2 ∞∑ ν=1 e− ν ξ + 1 = ∞∑ ν=−∞ e− |ν| ξ . (13) Thus 0 < 1 ∑ ∞ ν=−∞ e− |ν| ξ ≤ 1 2 ∫ ∞ 1 e− ν ξ dν + 1 = 1 2ξe− 1 ξ + 1 → 1, as ξ → 0 + . (14) We need to prove that g (ν) = νe− ν ξ is decreasing for ν ≥ 1. Indeed we have that g′ (ν) = e− ν ξ ( 1 − ν ξ ) ≤ 0, by ξ ≤ 1 ≤ ν. So that, again by [4], we get that 1 + 2 ∞∑ ν=1 ( 1 + ν ξ ) e− ν ξ ≤ 1 + 2 [∫ ∞ 1 ( 1 + ν ξ ) e− ν ξ dν + ( 1 + 1 ξ ) e− 1 ξ ] =: (∗) (15) Using integration by parts we have ∫ ∞ 1 ξ xe−xdx = −e−x (x + 1) |∞1 ξ = e− 1 ξ ( 1 ξ + 1 ) . (16) Hence we get ∫ ∞ 1 ( 1 + ν ξ ) e− ν ξ dν = ∫ ∞ 1 e− ν ξ dν + ∫ ∞ 1 ν ξ e− ν ξ dν = ξe− 1 ξ + ξ ∫ ∞ 1 ξ xe−xdx = e− 1 ξ (2ξ + 1) . (17) Therefore (∗) = 1 + 2 [ e− 1 ξ (2ξ + 1) + ( 1 + 1 ξ ) e− 1 ξ ] = 1 + 2e− 1 ξ ( 2ξ + 2 + 1 ξ ) . (18) Consequently we have found that ∞∑ ν=−∞ ( 1 + |ν| ξ ) e− |ν| ξ = 1 + 2 ∞∑ ν=1 ( 1 + ν ξ ) e− ν ξ ≤ 1 + 2e− 1 ξ ( 2ξ + 2 + 1 ξ ) (finite) → 1, as ξ → 0 + . (19) CUBO 15, 1 (2013) Approximation by Discrete Singular Operators 101 Finally we observe ( P∗ξf ) (x) − f (x) = ∑ ∞ ν=−∞ (f (x + ν) − f (x)) e− |ν| ξ ∑ ∞ ν=−∞ e− |ν| ξ . (20) So that ∣ ∣ ( P∗ξf ) (x) − f (x) ∣ ∣ ≤ ∑ ∞ ν=−∞ |f (x + ν) − f (x)| e− |ν| ξ ∑ ∞ ν=−∞ e− |ν| ξ (21) ≤ ∑ ∞ ν=−∞ ω1 (f, |ν|) e − |ν| ξ ∑ ∞ ν=−∞ e− |ν| ξ = ∑ ∞ ν=−∞ ω1 ( f, ξ |ν| ξ ) e− |ν| ξ ∑ ∞ ν=−∞ e− |ν| ξ (by (2)) ≤ ω1 (f, ξ)   ∑ ∞ ν=−∞ ( 1 + |ν| ξ ) e− |ν| ξ ∑ ∞ ν=−∞ e− |ν| ξ   (22) (by (14), (19)) ≤ ω1 (f, ξ) ( 1 + 2e− 1 ξ ( 2ξ + 2 + 1 ξ )) ( 2ξe− 1 ξ + 1 ) . (23) We notice that ( 1 + 2e− 1 ξ ( 2ξ + 2 + 1 ξ )) ( 2ξe− 1 ξ + 1 ) → 1, as ξ → 0 + . We have proved ∣ ∣ ( P∗ξf ) (x) − f (x) ∣ ∣ ≤   1 + 2e− 1 ξ ( 2ξ + 2 + 1 ξ ) 1 + 2ξe− 1 ξ  ω1 (f, ξ) , (24) ∀ x ∈ R. The proof now is completed. Next we prove preservation of global smoothness of P∗ξ. Theorem 2.2. It holds ω1 ( P∗ξf, δ ) ≤ ω1 (f, δ) , ∀ δ > 0. (25) Inequality (25) is sharp, namely it is attained by f (x) = identity (x) = x. Proof. We see that ( P∗ξf ) (x) − ( P∗ξf ) (y) = ∑ ∞ ν=−∞ (f (x + ν) − f (y + ν)) e− |ν| ξ ∑ ∞ ν=−∞ e− |ν| ξ . (26) 102 George A. Anastassiou CUBO 15, 1 (2013) Hence ∣ ∣ ( P∗ξf ) (x) − ( P∗ξf ) (y) ∣ ∣ ≤ ∑ ∞ ν=−∞ |f (x + ν) − f (y + ν)| e− |ν| ξ ∑ ∞ ν=−∞ e− |ν| ξ ≤ ∑ ∞ ν=−∞ ω1 (f, |x − y|) e − |ν| ξ ∑ ∞ ν=−∞ e− |ν| ξ = ω1 (f, |x − y|) . (27) So that for any x, y ∈ R : |x − y| < δ we get (25). If f = id, then trivially we get ( P∗ξid ) (x) − ( P∗ξid ) (y) = x − y = id (x) − id (y) , (28) thus (25) is attained. Next we study the approximation properties of discrete Gauss-Weierstrass operators. Theorem 2.3. Let f ∈ CU (R), 0 < ξ ≤ 1. Then ∥ ∥W∗ξf − f ∥ ∥ ∞ ≤ C (ξ) ω1 ( f, √ ξ ) , (29) where C (ξ) :=  1 +   e− 1 ξ (√ ξ + 2 + 2√ ξ ) √ πξ ( 1 − erf ( 1√ ξ )) + 1     . (30) We have lim ξ→0+ C (ξ) = 1, and by lim ξ→0+ ω1 ( f, √ ξ ) = 0, we get W∗ξ → I uniformly with rates, as ξ → 0 + . Proof. We notice easily that −∞∑ ν=−1 e− ν2 ξ = ∞∑ ν=1 e− ν2 ξ < ∞∑ ν=1 1 ν2 = π2 6 < ∞. (31) So we can write ∞∑ ν=−∞ e− ν2 ξ = 2 ∞∑ ν=1 e− ν2 ξ + 1 < π2 3 + 1. (32) Since e− ν2 ξ is positive, continuous and decreasing, by [4], we get ∞∑ ν=1 e− ν2 ξ − e− 1 ξ ≤ ∫ ∞ 1 e− ν2 ξ dν ≤ ∞∑ ν=1 e− ν2 ξ . (33) So that 2 ∫ ∞ 1 e− ν2 ξ dν + 1 ≤ 2 ∞∑ ν=1 e− ν2 ξ + 1 = ∞∑ ν=−∞ e− ν2 ξ , (34) CUBO 15, 1 (2013) Approximation by Discrete Singular Operators 103 and 0 < 1 ∑ ∞ ν=−∞ e− ν2 ξ ≤ 1 2 ∫ ∞ 1 e− ν2 ξ dν + 1 . (35) We know that ∫ ∞ 0 e−t 2 dt = √ π 2 , and erf (x) := 2√ π ∫x 0 e−t 2 dt, with erf (∞) = 1. Hence 2 ∫ ∞ 1 e− ν2 ξ dν + 1 = 2 √ ξ ∫ ∞ 1 e − ( ν √ ξ ) 2 d ( ν√ ξ ) + 1 = (36) 2 √ ξ ∫ ∞ 1 √ ξ e−θ 2 dθ + 1 = 2 √ ξ [∫ ∞ 0 e−θ 2 dθ − ∫ 1 √ ξ 0 e−θ 2 dθ ] + 1 = 2 √ ξ [√ π 2 − √ π 2 erf ( 1√ ξ )] + 1 = √ πξ ( 1 − erf ( 1√ ξ )) + 1. (37) Therefore 2 ∫ ∞ 1 e− ν2 ξ dν + 1 = √ πξ ( 1 − erf ( 1√ ξ )) + 1 → 1, as ξ → 0 + . (38) So we got that 0 < 1 ∑ ∞ ν=−∞ e− ν2 ξ ≤ 1 √ πξ ( 1 − erf ( 1√ ξ )) + 1 → 1, as ξ → 0 + . (39) Next we prove that g (ν) = νe− ν2 ξ is decreasing for ν ≥ 1. Indeed we have g′ (ν) = e− ν2 ξ ( 1 − 2ν 2 ξ ) ≤ 0, iff 1 − 2ν 2 ξ ≤ 0, iff ξ ≤ 2ν2, which is true. So that we have (by [4]) ∞∑ ν=1 ( 1 + ν√ ξ ) e− ν2 ξ ≤ ∫ ∞ 1 ( 1 + ν√ ξ ) e− ν2 ξ dν + ( 1 + 1√ ξ ) e− 1 ξ = (40) ∫ ∞ 1 e− ν2 ξ dν + ∫ ∞ 1 ν√ ξ e− ν2 ξ dν + e− 1 ξ + e− 1 ξ √ ξ = √ πξ 2 ( 1 − erf ( 1√ ξ )) + √ ξ 2 ∫ ∞ 1 e− ν2 ξ d ( ν2 ξ ) + e− 1 ξ + e− 1 ξ √ ξ = √ πξ 2 ( 1 − erf ( 1√ ξ )) + √ ξ 2 ∫ ∞ 1 ξ e−xdx + e− 1 ξ + e− 1 ξ √ ξ = √ πξ 2 ( 1 − erf ( 1√ ξ )) + √ ξ 2 e− 1 ξ + e− 1 ξ + e− 1 ξ √ ξ . (41) That is ∞∑ ν=1 ( 1 + ν√ ξ ) e− ν2 ξ ≤ √ πξ 2 ( 1 − erf ( 1√ ξ )) + e− 1 ξ ( √ ξ 2 + 1 + 1√ ξ ) (42) (finite) → 0, as ξ → 0 + . 104 George A. Anastassiou CUBO 15, 1 (2013) Since ∞∑ ν=−∞ ( 1 + |ν|√ ξ ) e− ν2 ξ = 2 ∞∑ ν=1 ( 1 + ν√ ξ ) e− ν2 ξ + 1 < ∞, (43) we find ∞∑ ν=−∞ ( 1 + |ν|√ ξ ) e− ν2 ξ ≤ √ πξ ( 1 − erf ( 1√ ξ )) + e− 1 ξ ( √ ξ + 2 + 2√ ξ ) + 1 (44) (is finite) → 1, as ξ → 0 + . Next we observe that ( W∗ξf ) (x) − f (x) = ∑ ∞ ν=−∞ (f (x + ν) − f (x)) e− ν2 ξ ∑ ∞ ν=−∞ e− ν2 ξ . (45) Thus ∣ ∣ ( W∗ξf ) (x) − f (x) ∣ ∣ ≤ ∑ ∞ ν=−∞ |f (x + ν) − f (x)| e− ν2 ξ ∑ ∞ ν=−∞ e− ν2 ξ ≤ (46) ∑ ∞ ν=−∞ ω1 (f, |ν|) e − ν2 ξ ∑ ∞ ν=−∞ e− ν2 ξ = ∑ ∞ ν=−∞ ω1 ( f, √ ξ |ν|√ ξ ) e− ν2 ξ ∑ ∞ ν=−∞ e− ν2 ξ ≤ (47) ω1 ( f, √ ξ )∑ ∞ ν=−∞ ( 1 + |ν|√ ξ ) e− ν2 ξ ∑ ∞ ν=−∞ e− ν2 ξ (48) (by (39), (44)) ≤ ω1 ( f, √ ξ )   √ πξ ( 1 − erf ( 1√ ξ )) + e− 1 ξ (√ ξ + 2 + 2√ ξ ) + 1 √ πξ ( 1 − erf ( 1√ ξ )) + 1   = (49) ω1 ( f, √ ξ )  1 + e− 1 ξ (√ ξ + 2 + 2√ ξ ) √ πξ ( 1 − erf ( 1√ ξ )) + 1   . So we have proved that ∣ ∣ ( W∗ξf ) (x) − f (x) ∣ ∣ ≤ ω1 ( f, √ ξ )  1 + e− 1 ξ (√ ξ + 2 + 2√ ξ ) √ πξ ( 1 − erf ( 1√ ξ )) + 1   , (50) ∀ x ∈ R, any 0 < ξ ≤ 1. The constant in the last inequality converges to 1, as ξ → 0 + . The proof of the theorem is completed. It follows the global smoothness preservation property of W∗ξ. CUBO 15, 1 (2013) Approximation by Discrete Singular Operators 105 Theorem 2.4. It holds ω1 ( W∗ξf, δ ) ≤ ω1 (f, δ) , ∀ δ > 0. (51) Inequality (51) is sharp, attained by f (x) = id (x) = x. Proof. We see that ( W∗ξf ) (x) − ( W∗ξf ) (y) = ∑ ∞ ν=−∞ (f (x + ν) − f (y + ν)) e− ν2 ξ ∑ ∞ ν=−∞ e− ν2 ξ , ∀ x, y ∈ R. (52) Hence ∣ ∣ ( W∗ξf ) (x) − ( W∗ξf ) (y) ∣ ∣ ≤ ∑ ∞ ν=−∞ |f (x + ν) − f (y + ν)| e− ν2 ξ ∑ ∞ ν=−∞ e− ν2 ξ ≤ ω1 (f, |x − y|)   ∑ ∞ ν=−∞ e− ν2 ξ ∑ ∞ ν=−∞ e− ν2 ξ   = ω1 (f, |x − y|) , ∀ x, y ∈ R, (53) proving (51). Sharpness is obvious. Next we study the approximation properties of general discrete Poisson-Cauchy operators. Theorem 2.5. Let f ∈ CU (R), 0 < ξ ≤ 1. Then ∥ ∥M∗ξf − f ∥ ∥ ≤ D (ξ) ω1 (f, ξ) , (54) where D (ξ) := [ 1 + 4ξ2αβ ( αβ 2αβ − 1 ) + ξ2αβ−1 ( 2αβ − 1 αβ − 1 )] . (55) We have lim ξ→0+ D (ξ) = 1, and by lim ξ→0+ ω1 (f, ξ) = 0, we get M ∗ ξ → I uniformly with rates, as ξ → 0 + . Proof. Here 0 < ξ ≤ 1, α ∈ N, β > 1 α , x ∈ R. By [5], p. 397, formula 595, we have ∫ ∞ 0 1 (t2α + ξ2α) β dt = Γ ( 1 2α ) Γ ( β − 1 2α ) 2Γ (β) αξ2αβ−1 . (56) Clearly ( ν2α + ξ2α )−β is decreasing, continuous and positive for ν ∈ [1, ∞). Hence by [4], we get 0 < ∞∑ ν=1 ( ν2α + ξ2α )−β ≤ ( 1 + ξ2α )−β + ∫ ∞ 1 ( ν2α + ξ2α )−β dν ≤ (57) ( 1 + ξ2α )−β + ∫ ∞ 0 ( ν2α + ξ2α )−β dν = ( 1 + ξ2α )−β + Γ ( 1 2α ) Γ ( β − 1 2α ) 2Γ (β) αξ2αβ−1 < ∞, ∀ ξ ∈ (0, 1]. 106 George A. Anastassiou CUBO 15, 1 (2013) Consequently we find convergence of 0 < S1 := ∞∑ ν=−∞ ( ν2α + ξ2α )−β = ξ−2αβ + 2 ∞∑ ν=1 ( ν2α + ξ2α )−β ≤ ξ−2αβ + 2 ( 1 + ξ2α )−β + Γ ( 1 2α ) Γ ( β − 1 2α ) Γ (β) αξ2αβ−1 < ∞, ∀ ξ ∈ (0, 1]. (58) Similarly we have ∞∑ ν=1 ( ν2α + ξ2α )−β ≥ ∫ ∞ 1 ( ν2α + ξ2α )−β dν, (59) and ∞∑ ν=−∞ ( ν2α + ξ2α )−β ≥ ξ−2αβ + 2 ∫ ∞ 1 ( ν2α + ξ2α )−β dν. (60) That is 0 < 1 ∑ ∞ ν=−∞ (ν2α + ξ2α) −β ≤ 1 ξ−2αβ + 2 ∫ ∞ 1 (ν2α + ξ2α) −β dν < ξ2αβ. (61) That 0 < 1 S1 < ξ2αβ → 0, as ξ → 0 + . (62) Hence lim ξ→0+ 1 S1 = 0. (63) Call g (ν) := ν ( ν2α + ξ2α )−β , ν ∈ [1, ∞). We have that g′ (ν) = ( ν2α + ξ2α )−β [ 1 − ( 2αβν2α ν2α + ξ2α )] ≤ 0, (64) iff 1 − ( 2αβν 2α ν2α+ξ2α ) ≤ 0, iff ν2α + ξ2α ≤ 2αβν2α, iff ξ2α ≤ ν2α (2αβ − 1), which is true because 2αβ − 1 ≥ 1 and ν2α (2αβ − 1) ≥ 1 ≥ ξ2α. That is g is decreasing, positive and continuous on [1, ∞). Hence ( 1 + ν ξ ) ( ν2α + ξ2α )−β is decreasing, positive and continuous on [1, ∞). Thus again by [4] we derive ∞∑ ν=1 ( 1 + ν ξ ) ( ν2α + ξ2α )−β ≤ (65) ( 1 + 1 ξ ) ( 1 + ξ2α )−β + ∫ ∞ 1 ( 1 + ν ξ ) ( ν2α + ξ2α )−β dν. We further notice that ∫ ∞ 1 ( 1 + ν ξ ) ( ν2α + ξ2α )−β dν = ∫ ∞ 1 ( ν2α + ξ2α )−β dν + 1 ξ ∫ ∞ 1 ν ( ν2α + ξ2α )−β dν < CUBO 15, 1 (2013) Approximation by Discrete Singular Operators 107 ∫ ∞ 1 ν−2αβdν + 1 ξ ∫ ∞ 1 ν−2αβ+1dν = ( 1 2αβ − 1 ) + (66) ( 1 2ξ (αβ − 1) ) < ∞, ∀ ξ ∈ (0, 1]. So that ∞∑ ν=1 ( 1 + ν ξ ) ( ν2α + ξ2α )−β < (67) ( 1 + 1 ξ ) ( 1 + ξ2α )−β + ( 1 2αβ − 1 ) + ( 1 2ξ (αβ − 1) ) < ∞, ∀ ξ ∈ (0, 1]. Consequently we obtain 0 < S2 := ∞∑ ν=−∞ ( 1 + |ν| ξ ) ( ν2α + ξ2α )−β = (68) ξ−2αβ + 2 ∞∑ ν=1 ( 1 + ν ξ ) ( ν2α + ξ2α )−β < ξ−2αβ + 2 ( 1 + 1 ξ ) ( 1 + ξ2α )−β + ( 2 2αβ − 1 ) + ( 1 ξ (αβ − 1) ) < 1 ξ2αβ + 2 ( 1 + 1 ξ ) + ( 2 2αβ − 1 ) + ( 1 ξ (αβ − 1) ) =: ϕ (ξ) . So that 0 < S2 < ϕ (ξ) < ∞, ∀ ξ ∈ (0, 1], (69) and 0 < S2 S1 (62) < ξ2αβϕ (ξ) = 1 + 2ξ2αβ ( 1 + 1 2αβ − 1 ) + ξ2αβ−1 ( 2 + 1 αβ − 1 ) = [ 1 + 4ξ2αβ ( αβ 2αβ − 1 ) + ξ2αβ−1 ( 2αβ − 1 αβ − 1 )] → 1, as ξ → 0 + . (70) Hence 0 < lim ξ→0+ S2 S1 < 1. (71) Finally we have that M∗ξ (f, x) − f (x) = ∑ ∞ ν=−∞ (f (x + ν) − f (x)) ( ν2α + ξ2α )−β ∑ ∞ ν=−∞ (ν2α + ξ2α) −β , (72) and ∣ ∣M∗ξ (f, x) − f (x) ∣ ∣ ≤ ∑ ∞ ν=−∞ |f (x + ν) − f (x)| ( ν2α + ξ2α )−β ∑ ∞ ν=−∞ (ν2α + ξ2α) −β ≤ (73) 108 George A. Anastassiou CUBO 15, 1 (2013) ∑ ∞ ν=−∞ ω1 ( f, ξ |ν| ξ ) ( ν2α + ξ2α )−β ∑ ∞ ν=−∞ (ν2α + ξ2α) −β ≤ ω1 (f, ξ)   ∑ ∞ ν=−∞ ( 1 + |ν| ξ ) ( ν2α + ξ2α )−β ∑ ∞ ν=−∞ (ν2α + ξ2α) −β   = ( S2 S1 ) ω1 (f, ξ) (70) ≤ (74) ≤ [ 1 + 4ξ2αβ ( αβ 2αβ − 1 ) + ξ2αβ−1 ( 2αβ − 1 αβ − 1 )] ω1 (f, ξ) . We have derived ∣ ∣M∗ξ (f, x) − f (x) ∣ ∣ ≤ [ 1 + 4ξ2αβ ( αβ 2αβ − 1 ) + ξ2αβ−1 ( 2αβ − 1 αβ − 1 )] ω1 (f, ξ) , (75) ∀ x ∈ R, ∀ ξ ∈ (0, 1], proving the claim. It follows the global smoothness preservation property of M∗ξ. Theorem 2.6. It holds ω1 ( M∗ξf, δ ) ≤ ω1 (f, δ) , ∀ δ > 0. (76) Inequality (76) is sharp, attained by f (x) = id (x) = x. Proof. Similar to the proof of Theorem 2.4. We continue with Theorem 2.7. It holds ∥ ∥θ∗ξf − f ∥ ∥ ∞ ≤ ( 2 3 π2K2β ) ξ2β−1ω1 (f, ξ) → 0, as ξ → 0 + . (77) Proof. Here we use a lot ∞∑ ν=1 1 ν2 = π2 6 (Euler, 1741). (78) We have θ∗ξ (f, x) − f (x) = ξ2β ∑ ν∈Z−{0} (f (x + ν) − f (x)) ( ϕ( νξ ) ν )2β 1 + ξ2β ∑ ν∈Z−{0} ( ϕ( νξ ) ν )2β . (79) Hence ∣ ∣θ∗ξ (f, x) − f (x) ∣ ∣ ≤ ξ2β ∑ ν∈Z−{0} |f (x + ν) − f (x)| ( ϕ( νξ ) ν )2β 1 + ξ2β ∑ ν∈Z−{0} ( ϕ( νξ ) ν )2β CUBO 15, 1 (2013) Approximation by Discrete Singular Operators 109 ≤ ξ2β ∑ ν∈Z−{0} ω1 ( f, ξ |ν| ξ ) ( ϕ( νξ ) ν )2β 1 + ξ2β ∑ ν∈Z−{0} ( ϕ( νξ ) ν )2β (80) ≤ ξ2βω1 (f, ξ) ∑ ν∈Z−{0} ( 1 + |ν| ξ ) ( ϕ( νξ ) ν )2β 1 + ξ2β ∑ ν∈Z−{0} ( ϕ( νξ ) ν )2β (81) ≤ ξ2βω1 (f, ξ) K 2β (∑ ν∈Z−{0} ( 1 + |ν| ξ ) 1 ν2β ) 1 + ξ2β ∑ ν∈Z−{0} ( ϕ( νξ ) ν )2β =: (∗) . We observe that 0 ≤ S2 := ∑ ν∈Z−{0}   ϕ ( ν ξ ) ν   2β ≤ K2β ∑ ν∈Z−{0} 1 ν2β = 2K2β ∞∑ ν=1 1 ν2β < 2K2β ∞∑ ν=1 1 ν2 = K2βπ2 3 . (82) Thus 0 ≤ S2 ≤ K2βπ2 3 . (83) So that 1 ≤ 1 + ξ2βS2 ≤ 1 + ξ2βK2βπ2 3 < ∞, ∀ ξ > 0. (84) That is 0 < 1 1 + ξ2βS2 ≤ 1, (85) with lim ξ→0+ 1 1 + ξ2βS2 = 1. (86) Consequently it holds (∗) ≤ 2ξ2βω1 (f, ξ) K2β ( ∞∑ ν=1 ( 1 + ν ξ ) 1 ν2β ) = 2ξ2βω1 (f, ξ) K 2β ( ∞∑ ν=1 1 ν2β + 1 ξ ∞∑ ν=1 1 ν2β−1 ) ≤ 2ξ2βω1 (f, ξ) K2β ( ∞∑ ν=1 1 ν2 + 1 ξ ∞∑ ν=1 1 ν2 ) = ξ2βπ2 3 ω1 (f, ξ) K 2β ( 1 + 1 ξ ) ≤ 2ξ2β−1π2 3 K2βω1 (f, ξ) . (87) 110 George A. Anastassiou CUBO 15, 1 (2013) We have proved that ∣ ∣θ∗ξ (f, x) − f (x) ∣ ∣ ≤ ( 2 3 π2K2β ) ξ2β−1ω1 (f, ξ) → 0, as ξ → 0 + . (88) The proof is completed. Example 2.8. In Theorem 2.7 we can take ϕ to be sine, cosine with K = 1. Theorem 2.9. It holds ω1 ( θ∗ξf, δ ) ≤ ω1 (f, δ) , ∀ δ > 0. (89) Inequality (89) is attained by f = id. We finish by studying T∗ξ. Theorem 2.10. It holds ∥ ∥T∗ξf − f ∥ ∥ ∞ ≤ ( 2π2A 3 ) ξ2β−1ω1 (f, ξ) → 0, as ξ → 0 + . (90) Theorem 2.11. It holds ω1 ( T∗ξf, δ ) ≤ ω1 (f, δ) , ∀ δ > 0. (91) Inequality (91) is attained by f = id. Proof. of Theorem 2.10. We have T∗ξ (f, x) − f (x) = ∑ ν∈Z−{0} (f (x + ν) − f (x)) ( ϕ( νξ ) ( νξ ) 4β ) 1 + ∑ ν∈Z−{0} ( ϕ( νξ ) ( νξ ) 4β ) . (92) Thus ∣ ∣T∗ξ (f, x) − f (x) ∣ ∣ ≤ ∑ ν∈Z−{0} |f (x + ν) − f (x)| ( ϕ( νξ ) ( νξ ) 4β ) 1 + ∑ ν∈Z−{0} ( ϕ( νξ ) ( νξ ) 4β ) ≤ ∑ ν∈Z−{0} ω1 ( f, ξ |ν| ξ ) ( ϕ( νξ ) ( νξ ) 4β ) 1 + ∑ ν∈Z−{0} ( ϕ( νξ ) ( νξ ) 4β ) (93) ≤ ω1 (f, ξ) ∑ ν∈Z−{0} ( 1 + |ν| ξ ) ( ϕ( νξ ) ( νξ ) 4β ) 1 + ∑ ν∈Z−{0} ( ϕ( νξ ) ( νξ ) 4β ) (94) CUBO 15, 1 (2013) Approximation by Discrete Singular Operators 111 ≤ ω1 (f, ξ) ∑ ν∈Z−{0} ( 1 + |ν| ξ ) A ( νξ ) 2β 1 + ∑ ν∈Z−{0} ( ϕ( νξ ) ( νξ ) 4β ) = Aω1 (f, ξ) ξ 2β ∑ ν∈Z−{0} ( 1 + |ν| ξ ) 1 ν2β 1 + ∑ ν∈Z−{0} ( ϕ( νξ ) ( νξ ) 4β ) (95) = 2Aω1 (f, ξ) ξ 2β ∑ ∞ ν=1 ( 1 + ν ξ ) 1 ν2β 1 + ∑ ν∈Z−{0} ( ϕ( νξ ) ( νξ ) 4β ) = 2Aω1 (f, ξ) ξ 2β [∑ ∞ ν=1 1 ν2β + 1 ξ ∑ ∞ ν=1 1 ν2β−1 ] 1 + ∑ ν∈Z−{0} ( ϕ( νξ ) ( νξ ) 4β ) (96) ≤ 2Aω1 (f, ξ) ξ 2β [ π 2 6 + 1 ξ π 2 6 ] ( 1 + ∑ ν∈Z−{0} ( ϕ( νξ ) ( νξ ) 4β )) ≤ ( 2π 2 3 A ) ω1 (f, ξ) ξ 2β−1 1 + ∑ ν∈Z−{0} ϕ( νξ ) ( νξ ) 4β =: (∗) . (97) We see that 1 + ∑ ν∈Z−{0} ϕ ( ν ξ ) ( ν ξ )4β ≤ 1 + A ∑ ν∈Z−{0} 1 ( ν ξ )2β (98) = 1 + Aξ2β ∑ ν∈Z−{0} 1 ν2β = 1 + 2Aξ2β ∞∑ ν=1 1 ν2β < 1 + 2Aξ2β ( ∞∑ ν=1 1 ν2 ) = 1 + Aξ2βπ2 3 < ∞, ∀ ξ > 0. (99) That is 1 ≤ 1 + ∑ ν∈Z−{0} ϕ ( ν ξ ) ( ν ξ )4β < 1 + Aπ2ξ2β 3 < ∞, ∀ ξ > 0. (100) Also 1 + Aπ 2 ξ 2β 3 → 1, as ξ → 0+, that is lim ξ→0+    1 + ∑ ν∈Z−{0} ϕ ( ν ξ ) ( ν ξ )4β    = 1. (101) Furthermore we have 0 < 1 1 + ∑ ν∈Z−{0} ϕ( νξ ) ( νξ ) 4β ≤ 1. (102) 112 George A. Anastassiou CUBO 15, 1 (2013) Hence (∗) ≤ ( 2π2 3 A ) ω1 (f, ξ) ξ 2β−1. (103) We have proved ∣ ∣T∗ξ (f, x) − f (x) ∣ ∣ ≤ ( 2π2A 3 ) ξ2β−1ω1 (f, ξ) → 0, as ξ → 0+, ∀ x ∈ R, (104) proving the claim. Note 2.12. All estimates of this article are also true for f ∈ Cb (R), continuous and bounded functions on R. However the convergences fail if f is not uniformly continuous. Received: October 2012. Revised: March 2013. 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