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Estimates with Optimal Constants Using Weighted K -Functionals on Simplex

Estimates with Optimal Constants Using Weighted K -Functionals on Simplex We present general estimates of the degree of approximation by positive s linear operators with the weighted K-functionals K1,,p , 1 s, p . Mathematics Subject Classification 2010: 41A36, 41A25. Key words: positive linear operators, Peetre's K-functional, degree of approximation. 1. Introduction Let S = x = (x1 , . . . , xd ) Rd |x1 , . . . , xd 0, x1 + . . . + xd 1 be the simplex in Rd , d N. Starting from the weight function used in [4], we consider the function (x) = [(x1 + . . . + xd ) (1 - x1 ) . . . (1 - xd )] , (0, 1). We denote by C (S) = 1 WC (S) = f C(S \ vi , i = 0, d )|() lim f (x)(x) R, i = 0, d xvi and f C(S) | f C (S), i = 1, d , xi where vi , i = 0, d are simplex vertices. We consider the K-functional s 1 K1,p, (f, t) = K s f, t; C(S), WC ,p (S) , t > 0, 1 s, p defined for the Banach space (C(S), · ) and the semi-Banach subspace 1 WC ,p (S), |·|W 1 C ,p , |f |W 1 C ,p = f f f ,..., x1 xd C ,p (S) s, s by K1,p, (f, t) = inf gW1 In Section 2 are given general estimates using the weighted K-functional s K1,p, , 1 s, p , and iroved the optimality of the constants (in the sense from [3]) appearing in these estimates. In Section 3 are established general estimates with optimal constants for the smooth functions in the case d = 1. These results are generalizations of those presented in [5] and [6]. We use the notation e0 for the function e0 : Rd R, e0 (x) = 1 and e1 for the function e1 : Rd Rd , e1 (x) = x. s 2. General estimates with K1,p, , 1 p, s ( f - g , t g p ) 1 s, p . Theorem 1. Let L : C(S) - C(S) a positive linear operator and f C(S). Then for every x S \ vi |i = 0, d and t > 0 we have |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| (1) + 2L (e0 , x) + 1 L ( e1 - xe0 q , x) · 1- t (x) 1 where q is such that p + 1 = 1. q Conversely, if there exists A, B, C 0 such that |L(f, x) - f (x)| A · |f (x)| |L (e0 , x) - 1| (2) + B · L (e0 , x) + C L ( e1 - xe0 q , x) t (x) holds for all positive linear operators, any f C(S), any x S\ vi |i = 0, d 1 and any t > 0, then A 1, C 1- and B 2. 1 Proof. Let g WC (S). For x, y S we denote by u(s) = (1 - s)x + sy, s [0, 1]. In [4], Lemma 2, iroved that the function s 1-s 1 is (u(s)) strictly decreasing on (0, 1). Then 1 1 ds (u(s)) (u(0)) (1 - s)- ds = 1 (1 - ) (x) and |f (y) - f (x)| 2 f - g + |g(y) - g(x)| = 2 f - g + |g(u(1))-g(u(0))| =2 f - g + (g (u(s))) ds 1 d i=1 d =2 f -g + 2 f -g + 2 f -g + g (u(s)) (yi - xi ) ds xi g |yi - xi | xi i=1 1 ds (u(s)) y-x q 1 · · t g p 1- t (x) y-x q 1 2+ · · max { f - g , t g p } . 1- t (x) Since g is arbitrary it follows that |f (y) - f (x)| Then |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| + |L (f - f (x)e0 , x)| |f (x)| · |L (e0 , x) - 1| + L 2e0 + 2+ 1 y-x q · 1- t (x) · K1,p, (f, t). |f (x)| · |L (e0 , x) - 1| + L (|f - f (x)e0 | , x) e1 - xe0 1 · 1- t (x) q , x · K1,p, (f, t) = |f (x)| · |L (e0 , x) - 1| + 2L (e0 , x) + 1 L ( e1 - xe0 q , x) · 1- t (x) which is (1). Now we prove the converse part. If we choose L(h, x) = 0 and f = e0 and replace in (2) we obtain A 1. We choose L(h, x) = h(1, 0, . . . , 0) and f (x) = (1 - x1 )1- . Since f f 1 WC (S) this implie t f p = t x1 = t (1 - ) . For x = (x1 , 0, . . . , 0), x1 (0, 1), we have L (f, x) = 0, L ( e1 - xe0 q , x) = 1 - x1 , f (x) = (1 - x1 )1- and (x) = x (1 - x1 ) . Replace in (2) and 1 we obtain (1 - x1 )1- Bt(1 - ) + C(1 - ) (1 - x1 )1- , ()t > 0 x 1 1 Passing to the limit t 0 we obtain C 1- x . Passing to the limit 1 1 x1 1 we obtain C 1- . To show that B 2 we assume first that d 2. If we choose L(h, x) = h(1, 0, . . . , 0), f (x) = 2x1 - 1, x = 0, 1 , . . . , 1 and replace in (2) we obtain 2 2 (2q + d - 1) q , ()t > 0 2B+C t(d - 1) 21-d (we use K1,p, (f, t) f = 1). Passing to the limit t we obtain 1 0 B 2. If d = 1, then we choose L(h, x) = h(1), f = e2 - 2e1 on 0, 2 1 and f = 2e1 - 3e0 on 1 , 1 , x = 2 and replace in (2). We obtain 1 2 2 2-1 1 (B + C2 t ) · 1 , ()t > 0 (we use K1, (f, t) f = 2 ). Passing to the 2 limit t we obtain B 2. s For the estimate with K1,p, , 1 s < we use the estimate (1) and the following relation between the K-functionals (for the case of the Kfunctionals defined for an arbitrary couple of quasi-normed spaces see [1]): Lemma 1. Let 1 s < . Then for f C(S) and t > 0 we have (3) = inf 1 s K1,p, (f, u). 1 Proof. ":" Let g WC ,p (S) and u > 0. We have f -g 1 s 1 s max { f - g , u g p } . ts us 1 s s Since g is arbitrary this implie 1 + since u is arbitrary the above inequality implies K1,p, (f, u). Also inf 1 s K1,p, (f, u). s g p = 0 and K1,p, (f, t) + C f - g s s s where C = p 1 C(t). Indeed, by definition of the K-functional there exists g1 WC ,p (S) 1 "": Let > 0. We can choose g WC ,p (S) such that f - g = 0, s such that K1,p, (f, t) + f - g1 s 1 s s . So, if f - g1 = 0 p and g1 p = 0 we choose g = g1 . If f - g1 = 0 and g1 p = 0 we choose g = g1 + e0 . Then f - g = f - g1 - e0 = = 0, g p = g1 p = 0 and f -g 1 s = s 1 1 s s K1,p, (f, t) + 2. If f - g1 = 0 and g1 p = 0 we choose g = g1 + C1 1 , where C1 is a constant for which f - g = 0 and 1 is the projection on the first component. We can suppose that 0 < C1 < 1. Then f - g = f - g1 - C1 1 f - g1 + C1 1 , g p = C1 = 0 and f -g 1 s s f - g1 + C1 (1 + t) K1,p, (f, t) + (2 + t) . f - g + t g If f - g1 = 0 and g1 p = 0 we choose g = g1 + 1 . Then f - g = f - g1 - 1 = 1 = 0, g p = = 0 and f -g 1 s = (1 + t) f - g + t g p + (1 + t) . On the assumption above we have inf 1+t 1 s K1,p, (f, u) s 1 s 1+t g f -g s 1 s K1,p, f, f -g g p f -g g f -g · max 1 s f -g , f -g g g p s K1,p, (f, t) + C . Since is arbitrary this implies the inequality. Corollary 1. Under the conditions of theorem we have |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| 1 L ( e1 - xe0 q , x) + max 2L (e0 , x) , · 1- t(x) (4) 1 K1,p, (f, t) and |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| (5) L ( e1 - xe0 q , x)s 1 + 2 L (e0 , x) + · (1 - )s ts (x)s s s 1 s s for 1 < s < , s = s-1 . Conversely, · if exists A, B, C 0 such that |L(f, x) - f (x)| A · |f (x)| · |L (e0 , x) - 1| + max B · L (e0 , x) , C · L ( e1 - xe0 q , x) t(x) 1 K1,p, (f, t) holds for all positive linear operators, any f C(S), any x S\ vi |i = 0, d 1 and any t > 0, then A 1, B 2 and C 1- . · if exists A, B, C 0 such that |L(f, x) - f (x)| A · |f (x)| · |L (e0 , x) - 1| + L ( e1 - xe0 q , x)s B · L (e0 , x) + C · ts (x)s 1 s holds for all positive linear operators, any f C(S), any x S\ vi |i = 0, d 1 and any t > 0, then A 1, B 2s and C (1-)s . Proof. Let u > 0. For s = 1 we have: |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| + 2L (e0 , x) + L ( e1 - xe0 q , x) 1 · 1- u(x) |f (x)| · |L (e0 , x) - 1| K1,p, (f, u) + max 2L (e0 , x) , from whence (4). L ( e1 - xe0 q , x) 1 · 1- t(x) 1+ t u K1,p, (f, u), s s-1 For 1 < s < , we denote by s = have: |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| + 2L (e0 , x) + and by H¨lder's inequality we o 1 L ( e1 - xe0 q , x) · 1- u(x) |f (x)| · |L (e0 , x) - 1| s s · K1,p, (f, u) L ( e1 - xe0 q , x)s 1 · 2 L (e0 , x) + (1 - )s ts (x)s 1 s 1 s K1,p, (f, u), from whence (5). For the converse part we make the same choices like in Theorem 1. s Other estimates with K1,p, , 1 < s < are obtained as a consequence of Theorem 1 or directly using the H¨lder-type inequality for positive linear o operators. Theorem 2. Let L : C(S) - C(S) be a positive linear operator and f C(S). Then ()x S \ vi |i = 0, d , ()t > 0 |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| L 1 + 2s L (e0 , x) + (6) · (1 - )s s · L (e0 , x) s · K1,p, (f, t) 1 1 s e1 - xe0 s , x q ts (x)s s holds for 1 < s < , s = s-1 . Conversely, if exists A, B, C 0 such that (7) |L(f, x) - f (x)| A · |f (x)| |L (e0 , x) - 1| s1 L e1 - xe0 s , x q 1 s · L (e0 , x) s · K1,p, (f, t) + B · L (e0 , x) + C s (x)s t holds for all positive linear operators, any f C(S), any x S\ vi |i = 0, d 1 and any t > 0 then A 1, B 2s and C (1-)s . 1 Proof. Let g WC ,p [0, 1]. We have |f (y) - f (x)| 2 f - g + where 1 < s, s < : 1 s y-x q 1 · · t g p 1- t(x) y-x q 1 2, · s · ( f - g , t g p ) 1- t(x) 1 s = 1. Since g is arbitrary follows that 2, y-x 1 · 1- t(x) q s s · K1,p, (f, t). |f (y) - f (x)| Then |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| + L (|f - f (x)e0 | , x) |f (x)| · |L (e0 , x) - 1| 2e0 , +L e1 - xe0 1 · 1- t(x) |f (x)| · |L (e0 , x) - 1| q s , x s · K1,p, (f, t) which is (6). For the converse part we make the same choices like in Theorem 1. s 1 e1 - xe0 q 1 s s · +L 2e0 , · L (e0 , x) s · K1,p, (f, t) s , x 1- t(x) = |f (x)| · |L (e0 , x) - 1| s1 L e1 -xe0 s , x q 1 1 s · L (e0 , x) s + 2s L (e0 , x) + · (1-)s ts (x)s s Remark 1. From K1,p, (f, t) K1,p, (f, t) 2 it follows s that lim = K1,p, (f, t). Passing to the limit s (in this 1) in (6) we obtain (1). case s Example 1. The Bernstein's operators are defined by (8) Bn (f, x) = k1 +...+kd =0 kd k1 ,..., n n bn,k (x) 9 where bn,k (x) = n! k1 ! . . . kd ! · n - d i=1 ki n- d i=1 ki xk 1 1 . . . xk d d 1- xi i=1 1 with n N and k1 , . . . , kd N {0}. We consider p = s = 2 and = 2 . We have Bn e1 - xe0 2 , x 2 2 (x) (1 - xi ) n (x1 + . . . + xd ) (1 - x1 ) . . . (1 - xd ) d i=1 xi and from (6) we obtain |Bn (f, x) - f (x)| 2 1+ xi ) 1 2 K1,2, f, (x1 + . . . + xd ) (1 - x1 ) . . . (1 - xd ) n d i=1 xi (1 - 3. General estimates for smooth functions s s In the case d = 1 we use the notation K1, := K1,, , 1 s , ) , f C[0, 1], t > 0. The s (f, t) = inf 1 so K1, s gWC [0,1] ( f - g , t g quantitative estimate for the remainder in Taylor's formula using the least concave majorant of the modulus of continuity was established in [2]. In 1 thiaper the K-functional K1 (f, t) = inf gC1 [a,b] ( f - g , t g ) 1 , f C[a, b], t > 0, is used. Now we derive in the next lemma a result for the K-functional K1, . Lemma 2. If f Cr [0, 1], r N, x (0, 1) and y [0, 1] then for the remainder in Taylor's formula of order r we have the following estimate (9) |Rr,f,x (y)| |y - x|r r! 2+ |y-x| t(r-+1)(x) K1, f (r) , t , ()t>0. r+1 Proof. Let g WC [0, 1] and x (0, 1). Let y [0, 1], y > x. Using the integral form of the remainder we have |Rr,g,x (y)|= 1 r! y x g(r+1) (u)(y-u)r du 1 r! y x (u)g (r+1) (u) (y - u)r du (u) g(r+1) r! x g(r+1) (y - u)r 1 (x) y-x y-u du r!(r - + 1)(x) (y - x)r+1 . If y [0, 1], y < x then 1 - y > 1 - x and |Rr,g,x (y)| = 1 r! 1-y 1-x 1-y 1-x g(r+1) (1 - u)(y - 1 + u)r du (1 - u)g(r+1) (1 - u) 1-y 1 r! g(r+1) r! = 1-x g(r+1) (1 - y - u)r 1 (1 - x) (1 - y - u)r du (1 - u) x-y 1-y-u du r!(r - + 1)(x) (x - y)r+1 . g (r+1) r!(r-+1)(x) So for y [0, 1] arbitrary we have |Rr,g,x (y)| have |y - x|r+1 . We |Rr,f -g,x (y)| = (f - g)(y) - k=0 (f - g)(k) (x) (y - x)k k! (f - g)(r) (x) (y - x)r r! = Rr-1,f -g,x (y) - (f - g)(r) |y - x|r r |Rr-1,f -g,x (y)| + |y - x| 2 · f (r) -g(r) . r! r! Then |Rr,f,x (y)| |Rr,f -g,x (y)| + |Rr,g,x (y)| |y - x|r r! |y - x|r r! 2 f (r) - g (r) + 2+ |y - x| t(r - + 1)(x) |y - x| g(r+1) (r - + 1)(x) max f (r) - g(r) , t g (r+1) . Since g is arbitrary this implies (9). Theorem 3. Let r N, L : C[0, 1] - C[0, 1] a positive linear operator and f Cr [0, 1]. Then ()x (0, 1), ()t > 0 we have |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| (10) f (k)(x) · L (e1 - xe0 )k , x k! Conversely, if exists Ak , B, C 0, k = 0, r such that |L(f, x) - f (x)| A0 · |f (x)| |L (e0 , x) - 1| L |e1 - xe0 |r+1 , x 1 · K1, f (r) , t . + 2L (|e1 - xe0 |r , x) + r! t(r - + 1)(x) (11) Ak f (k)(x) · L (e1 - xe0 )k , x L |e1 - xe0 |r+1 , x t(x) · K1, (f (r) , t) holds for all positive linear operators L : C[0, 1] - C[0, 1], any f 1 Cr [0, 1], any x (0, 1) and any t > 0 then A0 1, Ak k! , k = 1, r, 1 1 2 B r! and for Ak = k! , k = 1, r we have C r!(r-+1) . Proof. We have f (y) - f (x) = where r f (k) (x) r k! + B · L (|e1 - xe0 |r , x) + C (y - x)k + Rr,f,x (y) from L (f - f (x)e0 , x) = Then f (k) (x) L (e1 - xe0 )k , x + L (Rr,f,x , x) . k! |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| + |L (f - f (x)e0 , x)| |f (x)| · |L (e0 , x) - 1| + |f (x)| · |L (e0 , x) - 1| + r f (k) (x) L (e1 - xe0 )k , x k! +L (|Rr,f,x | , x) f (k) (x) L (e1 - xe0 )k , x k! which is (10). If we choose L(h, x) = 0 and f = e0 and replace in (11) we obtain A0 1. If we choose L(h, x) = h(1), f = ek , k = 1, r and replace in (11) we obtain L |e1 - xe0 |r+1 , x 1 r · K1, f (r) , t , + 2L (|e1 - xe0 | , x) + r! t(r - + 1)(x) 1-x l=1 Al · f (l) (x) (1 - x)l . 1 Passing to the limit x 0 we obtain Ak k! . 2 To show that B r! we choose L(h, x) = h(1) and f (x) = 2xr+ with > 0. For g = (r + ) · (r + - 1) . . . ( + 1) e0 we have K1, f (r) , t max f (r) - g , t g = f (r) -g = (r + )·(r + - 1) . . . ( + 1) . We replace in (11) and passing to the limit t , x 0, 0 (in this order) 2 we obtain B r! . 1 1 To show that C r!(r-+1) if Ak = k! , k = 1, r we choose L(h, x) = h(0) and 1 r 1 u 1- - x du. f (x) = 1 - x1- We have f (0) = 1, r and K1, to the limit t 0 we obtain 1 1 (k) (x) = (-1)k r! 1- -x)r-k du, k = r-+1 , f 1- (r-k)! x1- (u f (r) , t t f (r+1) = tr!. We replace in (11) and passing - r-+1 1- ie 1- 1 r x1- x1- u 1- - x u 1- - x du xk du + C · r!xr+1- (1 - x) r k r-k 1 1 - u 1- - x r - + 1 1 - x1- 1 r 1 u 1- - u 1- - x 1 - x1- du du + C · r!xr+1- (1 - x) from where C 1 r-+1 1 r!(r-+1) . C· r! . (1-x) Passing to the limit x 0 we obtain Corollary 2. Under the conditions of theorem we have |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| and f (k) (x) · L (e1 - xe0 )k , x k! L |e1 - xe0 |r+1 , x 1 1 · K1, f (r) , t + max 2L (|e1 - xe0 |r , x) , r! t(r - + 1)(x) |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| + f (k) (x) · L (e1 - xe0 )k , x k! s for 1 < s < , s = s-1 . Conversely, · if exists Ak , B, C 0, k = 0, r such that r L |e1 - xe0 | ,x 1 s s r 2 L (|e1 - xe0 | , x) + s s (x)s r! t (r - + 1) r+1 s1 s · K1, f (r) , t |L(f, x) - f (x)| A0 · |f (x)| |L (e0 , x) - 1| + Ak f (k) (x) · L (e1 - xe0 )k , x holds for all positive linear operators L : C[0, 1] - C[0, 1], any f 1 Cr [0, 1], any x (0, 1) and any t > 0 then A0 1, Ak k! , k = 1, r, 1 1 2 B r! and for Ak = k! , k = 1, r we have C r!(r-+1) . · if exists Ak , B, C 0, k = 0, r such that |L(f, x) - f (x)| A0 · |f (x)| |L (e0 , x) - 1| L |e1 - xe0 |r+1 , x + max B · L (|e1 -xe0 |r , x) , C · K 1 (f (r) , t) 1, t(x) Ak f (k) (x) · L (e1 - xe0 )k , x 1 s B · L (|e1 - xe0 | , x) + C L |e1 - xe0 |r+1 , x ts s · K1, (f (r) , t) holds for all positive linear operators L : C[0, 1] - C[0, 1], any f 1 Cr [0, 1], any x (0, 1) and any t > 0 then A0 1, Ak k! , k = 1, r, B 2s r!s and for Ak = 1 k! , k = 1, r we have C 1 . (r!(r-+1))s Proof. The proof is like for the Corollary 1. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Annals of the Alexandru Ioan Cuza University - Mathematics de Gruyter

Estimates with Optimal Constants Using Weighted K -Functionals on Simplex

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Abstract

We present general estimates of the degree of approximation by positive s linear operators with the weighted K-functionals K1,,p , 1 s, p . Mathematics Subject Classification 2010: 41A36, 41A25. Key words: positive linear operators, Peetre's K-functional, degree of approximation. 1. Introduction Let S = x = (x1 , . . . , xd ) Rd |x1 , . . . , xd 0, x1 + . . . + xd 1 be the simplex in Rd , d N. Starting from the weight function used in [4], we consider the function (x) = [(x1 + . . . + xd ) (1 - x1 ) . . . (1 - xd )] , (0, 1). We denote by C (S) = 1 WC (S) = f C(S \ vi , i = 0, d )|() lim f (x)(x) R, i = 0, d xvi and f C(S) | f C (S), i = 1, d , xi where vi , i = 0, d are simplex vertices. We consider the K-functional s 1 K1,p, (f, t) = K s f, t; C(S), WC ,p (S) , t > 0, 1 s, p defined for the Banach space (C(S), · ) and the semi-Banach subspace 1 WC ,p (S), |·|W 1 C ,p , |f |W 1 C ,p = f f f ,..., x1 xd C ,p (S) s, s by K1,p, (f, t) = inf gW1 In Section 2 are given general estimates using the weighted K-functional s K1,p, , 1 s, p , and iroved the optimality of the constants (in the sense from [3]) appearing in these estimates. In Section 3 are established general estimates with optimal constants for the smooth functions in the case d = 1. These results are generalizations of those presented in [5] and [6]. We use the notation e0 for the function e0 : Rd R, e0 (x) = 1 and e1 for the function e1 : Rd Rd , e1 (x) = x. s 2. General estimates with K1,p, , 1 p, s ( f - g , t g p ) 1 s, p . Theorem 1. Let L : C(S) - C(S) a positive linear operator and f C(S). Then for every x S \ vi |i = 0, d and t > 0 we have |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| (1) + 2L (e0 , x) + 1 L ( e1 - xe0 q , x) · 1- t (x) 1 where q is such that p + 1 = 1. q Conversely, if there exists A, B, C 0 such that |L(f, x) - f (x)| A · |f (x)| |L (e0 , x) - 1| (2) + B · L (e0 , x) + C L ( e1 - xe0 q , x) t (x) holds for all positive linear operators, any f C(S), any x S\ vi |i = 0, d 1 and any t > 0, then A 1, C 1- and B 2. 1 Proof. Let g WC (S). For x, y S we denote by u(s) = (1 - s)x + sy, s [0, 1]. In [4], Lemma 2, iroved that the function s 1-s 1 is (u(s)) strictly decreasing on (0, 1). Then 1 1 ds (u(s)) (u(0)) (1 - s)- ds = 1 (1 - ) (x) and |f (y) - f (x)| 2 f - g + |g(y) - g(x)| = 2 f - g + |g(u(1))-g(u(0))| =2 f - g + (g (u(s))) ds 1 d i=1 d =2 f -g + 2 f -g + 2 f -g + g (u(s)) (yi - xi ) ds xi g |yi - xi | xi i=1 1 ds (u(s)) y-x q 1 · · t g p 1- t (x) y-x q 1 2+ · · max { f - g , t g p } . 1- t (x) Since g is arbitrary it follows that |f (y) - f (x)| Then |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| + |L (f - f (x)e0 , x)| |f (x)| · |L (e0 , x) - 1| + L 2e0 + 2+ 1 y-x q · 1- t (x) · K1,p, (f, t). |f (x)| · |L (e0 , x) - 1| + L (|f - f (x)e0 | , x) e1 - xe0 1 · 1- t (x) q , x · K1,p, (f, t) = |f (x)| · |L (e0 , x) - 1| + 2L (e0 , x) + 1 L ( e1 - xe0 q , x) · 1- t (x) which is (1). Now we prove the converse part. If we choose L(h, x) = 0 and f = e0 and replace in (2) we obtain A 1. We choose L(h, x) = h(1, 0, . . . , 0) and f (x) = (1 - x1 )1- . Since f f 1 WC (S) this implie t f p = t x1 = t (1 - ) . For x = (x1 , 0, . . . , 0), x1 (0, 1), we have L (f, x) = 0, L ( e1 - xe0 q , x) = 1 - x1 , f (x) = (1 - x1 )1- and (x) = x (1 - x1 ) . Replace in (2) and 1 we obtain (1 - x1 )1- Bt(1 - ) + C(1 - ) (1 - x1 )1- , ()t > 0 x 1 1 Passing to the limit t 0 we obtain C 1- x . Passing to the limit 1 1 x1 1 we obtain C 1- . To show that B 2 we assume first that d 2. If we choose L(h, x) = h(1, 0, . . . , 0), f (x) = 2x1 - 1, x = 0, 1 , . . . , 1 and replace in (2) we obtain 2 2 (2q + d - 1) q , ()t > 0 2B+C t(d - 1) 21-d (we use K1,p, (f, t) f = 1). Passing to the limit t we obtain 1 0 B 2. If d = 1, then we choose L(h, x) = h(1), f = e2 - 2e1 on 0, 2 1 and f = 2e1 - 3e0 on 1 , 1 , x = 2 and replace in (2). We obtain 1 2 2 2-1 1 (B + C2 t ) · 1 , ()t > 0 (we use K1, (f, t) f = 2 ). Passing to the 2 limit t we obtain B 2. s For the estimate with K1,p, , 1 s < we use the estimate (1) and the following relation between the K-functionals (for the case of the Kfunctionals defined for an arbitrary couple of quasi-normed spaces see [1]): Lemma 1. Let 1 s < . Then for f C(S) and t > 0 we have (3) = inf 1 s K1,p, (f, u). 1 Proof. ":" Let g WC ,p (S) and u > 0. We have f -g 1 s 1 s max { f - g , u g p } . ts us 1 s s Since g is arbitrary this implie 1 + since u is arbitrary the above inequality implies K1,p, (f, u). Also inf 1 s K1,p, (f, u). s g p = 0 and K1,p, (f, t) + C f - g s s s where C = p 1 C(t). Indeed, by definition of the K-functional there exists g1 WC ,p (S) 1 "": Let > 0. We can choose g WC ,p (S) such that f - g = 0, s such that K1,p, (f, t) + f - g1 s 1 s s . So, if f - g1 = 0 p and g1 p = 0 we choose g = g1 . If f - g1 = 0 and g1 p = 0 we choose g = g1 + e0 . Then f - g = f - g1 - e0 = = 0, g p = g1 p = 0 and f -g 1 s = s 1 1 s s K1,p, (f, t) + 2. If f - g1 = 0 and g1 p = 0 we choose g = g1 + C1 1 , where C1 is a constant for which f - g = 0 and 1 is the projection on the first component. We can suppose that 0 < C1 < 1. Then f - g = f - g1 - C1 1 f - g1 + C1 1 , g p = C1 = 0 and f -g 1 s s f - g1 + C1 (1 + t) K1,p, (f, t) + (2 + t) . f - g + t g If f - g1 = 0 and g1 p = 0 we choose g = g1 + 1 . Then f - g = f - g1 - 1 = 1 = 0, g p = = 0 and f -g 1 s = (1 + t) f - g + t g p + (1 + t) . On the assumption above we have inf 1+t 1 s K1,p, (f, u) s 1 s 1+t g f -g s 1 s K1,p, f, f -g g p f -g g f -g · max 1 s f -g , f -g g g p s K1,p, (f, t) + C . Since is arbitrary this implies the inequality. Corollary 1. Under the conditions of theorem we have |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| 1 L ( e1 - xe0 q , x) + max 2L (e0 , x) , · 1- t(x) (4) 1 K1,p, (f, t) and |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| (5) L ( e1 - xe0 q , x)s 1 + 2 L (e0 , x) + · (1 - )s ts (x)s s s 1 s s for 1 < s < , s = s-1 . Conversely, · if exists A, B, C 0 such that |L(f, x) - f (x)| A · |f (x)| · |L (e0 , x) - 1| + max B · L (e0 , x) , C · L ( e1 - xe0 q , x) t(x) 1 K1,p, (f, t) holds for all positive linear operators, any f C(S), any x S\ vi |i = 0, d 1 and any t > 0, then A 1, B 2 and C 1- . · if exists A, B, C 0 such that |L(f, x) - f (x)| A · |f (x)| · |L (e0 , x) - 1| + L ( e1 - xe0 q , x)s B · L (e0 , x) + C · ts (x)s 1 s holds for all positive linear operators, any f C(S), any x S\ vi |i = 0, d 1 and any t > 0, then A 1, B 2s and C (1-)s . Proof. Let u > 0. For s = 1 we have: |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| + 2L (e0 , x) + L ( e1 - xe0 q , x) 1 · 1- u(x) |f (x)| · |L (e0 , x) - 1| K1,p, (f, u) + max 2L (e0 , x) , from whence (4). L ( e1 - xe0 q , x) 1 · 1- t(x) 1+ t u K1,p, (f, u), s s-1 For 1 < s < , we denote by s = have: |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| + 2L (e0 , x) + and by H¨lder's inequality we o 1 L ( e1 - xe0 q , x) · 1- u(x) |f (x)| · |L (e0 , x) - 1| s s · K1,p, (f, u) L ( e1 - xe0 q , x)s 1 · 2 L (e0 , x) + (1 - )s ts (x)s 1 s 1 s K1,p, (f, u), from whence (5). For the converse part we make the same choices like in Theorem 1. s Other estimates with K1,p, , 1 < s < are obtained as a consequence of Theorem 1 or directly using the H¨lder-type inequality for positive linear o operators. Theorem 2. Let L : C(S) - C(S) be a positive linear operator and f C(S). Then ()x S \ vi |i = 0, d , ()t > 0 |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| L 1 + 2s L (e0 , x) + (6) · (1 - )s s · L (e0 , x) s · K1,p, (f, t) 1 1 s e1 - xe0 s , x q ts (x)s s holds for 1 < s < , s = s-1 . Conversely, if exists A, B, C 0 such that (7) |L(f, x) - f (x)| A · |f (x)| |L (e0 , x) - 1| s1 L e1 - xe0 s , x q 1 s · L (e0 , x) s · K1,p, (f, t) + B · L (e0 , x) + C s (x)s t holds for all positive linear operators, any f C(S), any x S\ vi |i = 0, d 1 and any t > 0 then A 1, B 2s and C (1-)s . 1 Proof. Let g WC ,p [0, 1]. We have |f (y) - f (x)| 2 f - g + where 1 < s, s < : 1 s y-x q 1 · · t g p 1- t(x) y-x q 1 2, · s · ( f - g , t g p ) 1- t(x) 1 s = 1. Since g is arbitrary follows that 2, y-x 1 · 1- t(x) q s s · K1,p, (f, t). |f (y) - f (x)| Then |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| + L (|f - f (x)e0 | , x) |f (x)| · |L (e0 , x) - 1| 2e0 , +L e1 - xe0 1 · 1- t(x) |f (x)| · |L (e0 , x) - 1| q s , x s · K1,p, (f, t) which is (6). For the converse part we make the same choices like in Theorem 1. s 1 e1 - xe0 q 1 s s · +L 2e0 , · L (e0 , x) s · K1,p, (f, t) s , x 1- t(x) = |f (x)| · |L (e0 , x) - 1| s1 L e1 -xe0 s , x q 1 1 s · L (e0 , x) s + 2s L (e0 , x) + · (1-)s ts (x)s s Remark 1. From K1,p, (f, t) K1,p, (f, t) 2 it follows s that lim = K1,p, (f, t). Passing to the limit s (in this 1) in (6) we obtain (1). case s Example 1. The Bernstein's operators are defined by (8) Bn (f, x) = k1 +...+kd =0 kd k1 ,..., n n bn,k (x) 9 where bn,k (x) = n! k1 ! . . . kd ! · n - d i=1 ki n- d i=1 ki xk 1 1 . . . xk d d 1- xi i=1 1 with n N and k1 , . . . , kd N {0}. We consider p = s = 2 and = 2 . We have Bn e1 - xe0 2 , x 2 2 (x) (1 - xi ) n (x1 + . . . + xd ) (1 - x1 ) . . . (1 - xd ) d i=1 xi and from (6) we obtain |Bn (f, x) - f (x)| 2 1+ xi ) 1 2 K1,2, f, (x1 + . . . + xd ) (1 - x1 ) . . . (1 - xd ) n d i=1 xi (1 - 3. General estimates for smooth functions s s In the case d = 1 we use the notation K1, := K1,, , 1 s , ) , f C[0, 1], t > 0. The s (f, t) = inf 1 so K1, s gWC [0,1] ( f - g , t g quantitative estimate for the remainder in Taylor's formula using the least concave majorant of the modulus of continuity was established in [2]. In 1 thiaper the K-functional K1 (f, t) = inf gC1 [a,b] ( f - g , t g ) 1 , f C[a, b], t > 0, is used. Now we derive in the next lemma a result for the K-functional K1, . Lemma 2. If f Cr [0, 1], r N, x (0, 1) and y [0, 1] then for the remainder in Taylor's formula of order r we have the following estimate (9) |Rr,f,x (y)| |y - x|r r! 2+ |y-x| t(r-+1)(x) K1, f (r) , t , ()t>0. r+1 Proof. Let g WC [0, 1] and x (0, 1). Let y [0, 1], y > x. Using the integral form of the remainder we have |Rr,g,x (y)|= 1 r! y x g(r+1) (u)(y-u)r du 1 r! y x (u)g (r+1) (u) (y - u)r du (u) g(r+1) r! x g(r+1) (y - u)r 1 (x) y-x y-u du r!(r - + 1)(x) (y - x)r+1 . If y [0, 1], y < x then 1 - y > 1 - x and |Rr,g,x (y)| = 1 r! 1-y 1-x 1-y 1-x g(r+1) (1 - u)(y - 1 + u)r du (1 - u)g(r+1) (1 - u) 1-y 1 r! g(r+1) r! = 1-x g(r+1) (1 - y - u)r 1 (1 - x) (1 - y - u)r du (1 - u) x-y 1-y-u du r!(r - + 1)(x) (x - y)r+1 . g (r+1) r!(r-+1)(x) So for y [0, 1] arbitrary we have |Rr,g,x (y)| have |y - x|r+1 . We |Rr,f -g,x (y)| = (f - g)(y) - k=0 (f - g)(k) (x) (y - x)k k! (f - g)(r) (x) (y - x)r r! = Rr-1,f -g,x (y) - (f - g)(r) |y - x|r r |Rr-1,f -g,x (y)| + |y - x| 2 · f (r) -g(r) . r! r! Then |Rr,f,x (y)| |Rr,f -g,x (y)| + |Rr,g,x (y)| |y - x|r r! |y - x|r r! 2 f (r) - g (r) + 2+ |y - x| t(r - + 1)(x) |y - x| g(r+1) (r - + 1)(x) max f (r) - g(r) , t g (r+1) . Since g is arbitrary this implies (9). Theorem 3. Let r N, L : C[0, 1] - C[0, 1] a positive linear operator and f Cr [0, 1]. Then ()x (0, 1), ()t > 0 we have |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| (10) f (k)(x) · L (e1 - xe0 )k , x k! Conversely, if exists Ak , B, C 0, k = 0, r such that |L(f, x) - f (x)| A0 · |f (x)| |L (e0 , x) - 1| L |e1 - xe0 |r+1 , x 1 · K1, f (r) , t . + 2L (|e1 - xe0 |r , x) + r! t(r - + 1)(x) (11) Ak f (k)(x) · L (e1 - xe0 )k , x L |e1 - xe0 |r+1 , x t(x) · K1, (f (r) , t) holds for all positive linear operators L : C[0, 1] - C[0, 1], any f 1 Cr [0, 1], any x (0, 1) and any t > 0 then A0 1, Ak k! , k = 1, r, 1 1 2 B r! and for Ak = k! , k = 1, r we have C r!(r-+1) . Proof. We have f (y) - f (x) = where r f (k) (x) r k! + B · L (|e1 - xe0 |r , x) + C (y - x)k + Rr,f,x (y) from L (f - f (x)e0 , x) = Then f (k) (x) L (e1 - xe0 )k , x + L (Rr,f,x , x) . k! |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| + |L (f - f (x)e0 , x)| |f (x)| · |L (e0 , x) - 1| + |f (x)| · |L (e0 , x) - 1| + r f (k) (x) L (e1 - xe0 )k , x k! +L (|Rr,f,x | , x) f (k) (x) L (e1 - xe0 )k , x k! which is (10). If we choose L(h, x) = 0 and f = e0 and replace in (11) we obtain A0 1. If we choose L(h, x) = h(1), f = ek , k = 1, r and replace in (11) we obtain L |e1 - xe0 |r+1 , x 1 r · K1, f (r) , t , + 2L (|e1 - xe0 | , x) + r! t(r - + 1)(x) 1-x l=1 Al · f (l) (x) (1 - x)l . 1 Passing to the limit x 0 we obtain Ak k! . 2 To show that B r! we choose L(h, x) = h(1) and f (x) = 2xr+ with > 0. For g = (r + ) · (r + - 1) . . . ( + 1) e0 we have K1, f (r) , t max f (r) - g , t g = f (r) -g = (r + )·(r + - 1) . . . ( + 1) . We replace in (11) and passing to the limit t , x 0, 0 (in this order) 2 we obtain B r! . 1 1 To show that C r!(r-+1) if Ak = k! , k = 1, r we choose L(h, x) = h(0) and 1 r 1 u 1- - x du. f (x) = 1 - x1- We have f (0) = 1, r and K1, to the limit t 0 we obtain 1 1 (k) (x) = (-1)k r! 1- -x)r-k du, k = r-+1 , f 1- (r-k)! x1- (u f (r) , t t f (r+1) = tr!. We replace in (11) and passing - r-+1 1- ie 1- 1 r x1- x1- u 1- - x u 1- - x du xk du + C · r!xr+1- (1 - x) r k r-k 1 1 - u 1- - x r - + 1 1 - x1- 1 r 1 u 1- - u 1- - x 1 - x1- du du + C · r!xr+1- (1 - x) from where C 1 r-+1 1 r!(r-+1) . C· r! . (1-x) Passing to the limit x 0 we obtain Corollary 2. Under the conditions of theorem we have |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| and f (k) (x) · L (e1 - xe0 )k , x k! L |e1 - xe0 |r+1 , x 1 1 · K1, f (r) , t + max 2L (|e1 - xe0 |r , x) , r! t(r - + 1)(x) |L(f, x) - f (x)| |f (x)| · |L (e0 , x) - 1| + f (k) (x) · L (e1 - xe0 )k , x k! s for 1 < s < , s = s-1 . Conversely, · if exists Ak , B, C 0, k = 0, r such that r L |e1 - xe0 | ,x 1 s s r 2 L (|e1 - xe0 | , x) + s s (x)s r! t (r - + 1) r+1 s1 s · K1, f (r) , t |L(f, x) - f (x)| A0 · |f (x)| |L (e0 , x) - 1| + Ak f (k) (x) · L (e1 - xe0 )k , x holds for all positive linear operators L : C[0, 1] - C[0, 1], any f 1 Cr [0, 1], any x (0, 1) and any t > 0 then A0 1, Ak k! , k = 1, r, 1 1 2 B r! and for Ak = k! , k = 1, r we have C r!(r-+1) . · if exists Ak , B, C 0, k = 0, r such that |L(f, x) - f (x)| A0 · |f (x)| |L (e0 , x) - 1| L |e1 - xe0 |r+1 , x + max B · L (|e1 -xe0 |r , x) , C · K 1 (f (r) , t) 1, t(x) Ak f (k) (x) · L (e1 - xe0 )k , x 1 s B · L (|e1 - xe0 | , x) + C L |e1 - xe0 |r+1 , x ts s · K1, (f (r) , t) holds for all positive linear operators L : C[0, 1] - C[0, 1], any f 1 Cr [0, 1], any x (0, 1) and any t > 0 then A0 1, Ak k! , k = 1, r, B 2s r!s and for Ak = 1 k! , k = 1, r we have C 1 . (r!(r-+1))s Proof. The proof is like for the Corollary 1.

Journal

Annals of the Alexandru Ioan Cuza University - Mathematics – de Gruyter

Published: Nov 24, 2014

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