Abstract

We characterize the errors of the algebraic version of trigonometric Jackson integrals in weighted integral metric. We prove direct and strong converse theorem in terms of the weighted -functional.

1. Introduction

We study linear approximation process together with the characterization of the rate of convergence of the algebraic version of the trigonometric Jackson integrals defined by where We established the equivalence in uniform norm for the approximation error of the operator , the values of proper -functional, and modulus of smoothness (see [1, 2]): For every and the -functional is defined by where the differential operator The modulus The expression ~ means that there exists a positive constant independent of such that . The equivalence ~ consists of a direct inequality and strong converse inequality of type in the sense of [3]. Ditzian and Ivanov show that a converse inequality follows from several inequalities of Bernstein and Voronovskaya type. We apply their method.

Let be the weighted space with the norm The approximation error of in will be compared with the -functional, which for every and is defined by Our main result states the following.

Theorem 1. For every , and , one has

In Section 2 we state and prove some auxiliary lemmas. The proof of Theorem 1 is given in Section 3.

2. Auxiliary Lemmas

Lemma 2. Let be summable on and —periodic function, . Then the following holds true:

Proof. Using the Minkowski inequality we have

Lemma 3. For periodic integrable on function and every one has

Proof. We consider the function . By the Fubini theorem we have
As we obtain that the integral on the left-hand side of (13) is equal to The integral on the right-hand side of (13) is From (13), (14), and (15) we obtain . If then the sign of is constant and coincides with the sign of . Therefore which proves the lemma.

Lemma 4. For periodic function and every one has

Proof. For Lemma 4 is fulfilled as equality. Let . Using the Hölder inequality we get Therefore Then We apply Lemma 3 to to obtain Hence Thus

Definition 5. Set

Lemma 6 ([2, page 402]). Let be the space from Definition 5, , and . Then and for .

Lemma 7. Let be the space from Definition 5. Then for every and , one has

Proof. From we see that . In order to prove it is sufficient to show (see Lemma  2 in [4, page 116]) that for every and there exists such that and . Let . We put and consider . Since (see Lemma 6). We use the Jackson integrals of the following type: where , .
Since , it follows that is an even nonnegative trigonometric polynomial of degree is a trigonometric polynomial of degree , which is even as is even. From Jackson theorem (see [5, Chapter 7, Theorem 2.2]) we get By the substitution in we obtain an algebraic polynomial, which is desired function from . We set From and (27) we get From (26) we get Using the Jackson theorem we get Since and (31) implies For given we choose such that and to obtain This completes the proof of the lemma.

3. Proof of Theorem 1

The theorem will be proved if we show that ~ (see Lemma 7). First we prove the converse result , which is a strong converse inequality of type in the terminology of [3]. We utilize Theorems 3.1 and   4.1 in [3] with , , , and for , for . The result needed for inequality from Theorem  3.1 in [3] with is given by

We set . To prove (34) we recall the representation where And the convolution is defined by

We have

Using Lemma 2 as and are even, periodic functions, we estimate to complete the proof of (34).