Proceedings of the Estonian Academy of Sciences, 2011, 60, 4, 201–209 doi: 10.3176/proc.2011.4.01 Available online at www.eap.ee/proceedings
Approximation in variation by the Kantorovich operators Andi Kivinukk and Tarmo Metsm¨agi∗ Institute of Mathematics and Natural Sciences, Tallinn University, Narva mnt. 25, 10120 Tallinn, Estonia Received 1 November 2010, accepted 27 April 2011 Abstract. We discuss the rate of approximation of the Kantorovich operators. The rate of approximation is given with respect to the variation seminorm. Key words: approximation theory, Kantorovich operators, convergence in variation, order of approximation, absolutely continuous functions.
1. INTRODUCTION The paper deals with the convergence in variation of the Kantorovich operators. In [7] we proved similar results for the Meyer-K¨onig and Zeller operators. Let TV [0, 1], respectively AC[0, 1], denote the class of all functions of bounded variation, respectively the absolutely continuous functions on [0, 1]. Let Ln : TV [0, 1] → TV [0, 1] be an arbitrary positive operator, i.e. for f ≥ 0 we have Ln f ≥ 0. It is known that many linear positive operators have the variation detracting property (or the variational diminishing property, cf. [10]) in the following form: for all f ∈ TV [0, 1] with the total variation V[0,1] [ f ] we have Ln f ∈ TV [0, 1] and V[0,1] [Ln f ] ≤ V[0,1] [ f ]. For example, the variation detracting property is valid for the Bernstein, Meyer-K¨onig and Zeller, and Stancu operators (see [2], where the problem is posed and solved even for the ϕ -variation). The variation detracting property is needed to consider the convergence in variation, i.e. for all f ∈ TV [0, 1] there has to be V[0,1] [Ln f − f ] → 0. The convergence in ϕ -variation of many positive operators was considered in [2], but not of Kantorovichtype operators. The operators of Kantorovich n
(Kn f )(x) = (n + 1) ∑ pk,n (x) k=0
∗
Corresponding author,
[email protected]
Z
k+1 n+1 k n+1
f (u)du (x ∈ [0, 1]),
(1.1)
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Proceedings of the Estonian Academy of Sciences, 2011, 60, 4, 201–209
where
µ ¶ n k pk,n (x) := x (1 − x)n−k , k
were introduced in [6] and their asymptotic or approximation behaviour has been investigated in many works (see, for example, [1,4,5,8] and literature cited there). In the case of the Kantorovich operators the variation detracting property holds as follows. Theorem A. ([3], Proposition 3.3) If f ∈ TV [0, 1], then Kn f ∈ AC[0, 1] and V[0,1] [Kn f ] ≤ V[0,1] [ f ].
The convergence in variation for smooth functions is given in Section 2, Theorem 1. We prove not only the convergence in variation of the Kantorovich operators, but also give the rate of approximation in Theorem 2. For the following proof of Theorem 1 we calculate the derivative of Kn f , n+1 n (Kn f ) (x) = ∑ (k − nx)pk,n (x) X k=0 0
Z
k+1 n+1 k n+1
f (u)du (x ∈ (0, 1)),
(1.2)
where X = x(1 − x). In the same proof we need the sum moments for the operators (1.1). Let us define the sum moments as in [3]: n
Tr,n (x) :=
∑ [k − nx]r pk,n (x).
(1.3)
k=0
Then there hold the identities (see, e.g., [3] or original presentation in [9]) 1, r = 0, 0, r = 1, Tr,n (x) = nX, r = 2, nX(1 − 2x), r = 3.
(1.4)
2. APPROXIMATION IN VARIATION BY THE KANTOROVICH OPERATORS We start with studying the rate of approximation of smooth functions with respect to the variation seminorm. Theorem 1. If g00 ∈ AC[0, 1], then V[0,1] [Kn g − g] ≤
4 (V [g] +V[0,1] [g00 ]) n + 1 [0,1]
(n ≥ 3).
(2.1)
Proof. We represent f in (1.2) by Taylor’s formula with the integral remainder term g(t) = g(x) + (t − x)g0 (x) + (t − x)2
g00 (x) 1 + 2 2
Z t x
We have (Kn g)0 (x) = A0,n (x)g(x) + A1,n (x)g0 (x) + A2,n (x)
(t − v)2 g000 (v)dv.
g00 (x) + (Rn g)(x), 2
(2.2)
A. Kivinukk and T. Metsm¨agi: Kantorovich operators
203
where n+1 n A j,n (x) = ∑ (k − nx)pk,n (x) X k=0
Z
k+1 n+1 k n+1
and n+1 n (Rn g)(x) := ∑ (k − nx)pk,n (x) 2X k=0
Z
(t − x) j dt ( j = 0, 1, 2)
k+1 n+1 k n+1
dt
Z t x
(t − v)2 g000 (v)dv.
(2.3)
(2.4)
Calculating (2.3) by (1.3) and (1.4), we have A0,n (x) =
T1,n (x) = 0. X
Analogously, A1,n (x) =
T2,n (x) + ( 12 − x)T1,n (x) n = (n + 1)X n+1
and A2,n (x) =
T3,n (x) + (1 − 2x)T2,n (x) + ( 13 − x + x2 )T1,n (x) 2n(1 − 2x) = . (n + 1)2 X (n + 1)2
So, by (2.2), (2.3) and (2.4) we have for the derivative (Kn g)0 (x) =
n 0 n(1 − 2x) 00 g (x) + g (x) + (Rn g)(x). n+1 (n + 1)2
(2.5)
The integration domain of the double integral in the remainder (2.4) is ¯ n ¯ Dx,k = (t, v)¯
o k k+1 ≤t ≤ , v ∈ [x,t] . n+1 n+1
k We denote tk := n+1 . For fixed x ∈ (0, 1) we divide the summation indices k into three sets: tk+1 ≤ x, tk ≤ x < tk+1 or tk > x. Hence, for the remainder we get ([x] denotes the integer part of x)
(Rn g)(x) =
∑
... +
tk+1 ≤x
∑
[(n+1)x]−1
=
∑
... +
∑ ...
tk >x
tk ≤x n. In each summand we change the order of integration by splitting the double integration domain Dx,k in a suitable way. After that we get six different sums. So we have 6
6
i=1
i=1
(Rn g)(x) = ∑ Bi,n g(x) ≡ ∑ Bi,n g, where, denoting qk,n (x) :=
n+1 (k − nx)pk,n (x), 2X
(2.6)
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Proceedings of the Estonian Academy of Sciences, 2011, 60, 4, 201–209
we obtain Z
[(n+1)x]−1
∑
B1,n g := −
k+1 n+1
qk,n (x)
k n+1
k=0 [(n+1)x]−1
∑
B2,n g := −
qk,n (x)
∑
g000 (v)dv k+1
qk,n (x)
Z x k n+1
k=[(n+1)x]
Z
[(n+1)x]
∑
qk,n (x)
k+1 n+1
x
k=[(n+1)x]
Z
n
∑
B5,n g :=
qk,n (x)
∑
qk,n (x)
k n+1
k=[(n+1)x]+1
Let us estimate B1,n g in (2.7). Since for Z v k n+1
k n+1
≤v≤
k+1 n+1
k+1 n+1
(t − v)2 dt ≤
k+1 n+1 k n+1
Z v k n+1
Z
g000 (v)dv
k n+1
Z
n
g000 (v)dv
x
k=[(n+1)x]+1
B6,n g :=
k n+1
n+1
[(n+1)x]
B4,n g :=
Z v
Z
Z x
k=0
B3,n g := −
g000 (v)dv
k+1 n+1
v
Z
g000 (v)dv
(t − v)2 dt,
(2.7)
(t − v)2 dt,
(2.8)
(t − v)2 dt,
(2.9)
(t − v)2 dt,
(2.10)
k+1 n+1 k n+1
Z
g000 (v)dv
k+1 n+1
v
(t − v)2 dt,
(2.11)
(t − v)2 dt.
(2.12)
we have
1 , 3(n + 1)3
by using Cauchy’s inequality and (1.3), we get for (2.7) the estimate |(B1,n g)(x)| ≤ ≤
kg000 k 6X(n + 1)2
Ã
! 12 Ã
n
2
∑ (k − nx)
! 21
n
∑ pk,n (x)
pk,n (x)
k=0
k=0
1 1 1 (T2,n (x)) 2 (T0,n (x)) 2 kg000 k, 2 6X(n + 1)
where here and later on the norm is taken in L1 (0, 1), i.e. k f k := k f kL1 (0,1) . Finally, by (1.4) and
R1 1 √ dx = π we obtain for norms 0 X
||B1,n g|| ≤
π 6(n + 1)
3 2
||g000 ||.
(2.13)
We get in a similar way the same estimate for B6,n g in (2.12). In the following we estimate the sum B2,n g + B5,n g of expressions in (2.8) and (2.11). Let by (2.8) B2,n g = B12,n g + B22,n g :=
Z
[(n+1)x]−1
∑
...
k=0
k n
x
Z
[(n+1)x]−1
...dv... +
∑
...
k=0
k+1 n+1 k n
...dv...
(2.14)
and by (2.11) B5,n g = B15,n g − B25,n g :=
Z
n
∑
k=[(n+1)x]+1
...
x
k n
Z
n
...dv... −
∑
k=[(n+1)x]+1
...
k n k n+1
...dv... .
(2.15)
A. Kivinukk and T. Metsm¨agi: Kantorovich operators
205
k k+1 In expressions of B22,n g and B25,n g the variable v is on [ n+1 , n+1 ], therefore
Z
k+1 n+1 k n+1
·
¸ k k+1 v∈ , . n+1 n+1
1 (t − v) dt ≤ , 3(n + 1)3 2
(2.16)
Hence, the quantities B22,n g and B25,n g can be estimated in the same way as we did before for B1,n g. So we can state that the estimate (2.13) is valid also for B22,n g and B25,n g. Now consider the sum B12,n g + B15,n g :=
Z
[(n+1)x]−1
∑
qk,n (x)
k=0
Z
k n
g000 (v)dv
x
Z
n
∑
+
qk,n (x) Z
n
∑ qk,n (x)
=
k n
x
k=0
Z
∑
−
g (v)dv
Z
[(n+1)x]
k+1 n+1
qk,n (x)
k=[(n+1)x]
k n+1
k n+1
k+1 n+1
000
g (v)dv
x
k+1 n+1
(t − v)2 dt
(t − v)2 dt Z
k n
(t − v)2 dt
000
000
g (v)dv
k n+1
Z
k n
x
k=[(n+1)x]+1
k+1 n+1
k n+1
(t − v)2 dt
=: C1,n g −C2,n g. For C2,n g we can prove that (2.13) is valid. Indeed, for fixed x ∈ (0, 1) and n ∈ IN the quantity C2,n g consists of one summand with the index k = [(n + 1)x], i.e. Z
C2,n g(x) = qk,n (x)
Z
k n
k+1 n+1
000
g (v)dv
x
k n+1
(t − v)2 dt.
(2.17)
To estimate the norm kC2,n gk, we decompose the interval (0, 1) into n + 1 equal parts, n+1 n ∑ 2 i=0
Z
kC2,n gk =
n+1 n ∑ 2 i=0
Z
≤
i+1 n+1 i n+1 i+1 n+1 i n+1
Here the second relation holds because for Cauchy’s inequality we get kC2,n gk ≤ ≤
Z k+1 ¯ Z nk ¯ n+1 |k − nx| ¯ ¯ pk,n (x)¯ g000 (v)dv k (t − v)2 dt ¯dx x(1 − x) x n+1
|i − nx| pi,n (x) x(1 − x) i n+1
kg000 k n ∑ 6(n + 1)2 i=0 kg000 k 6(n + 1)2
≤x
0 such that V[0,1] [Kn f − f ] ≤ c1
2
sup V[h,1−h] [4h f ] + c2 1
0 0 such that c3 V[0,1] [Kn f − f ] ≤ √ n
1
sup V[ h ,1− h ] [4h f 0 ] + c2 1
0
