%0 Journal Article
%T On Approximation by Algebraic Version of the Trigonometric Jackson Integrals in Weighted Integral Metric
%A Teodora Zapryanova
%J Journal of Operators
%D 2013
%I Hindawi Publishing Corporation
%R 10.1155/2013/324630
%X 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 ¡ªperiodic 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 H£¿lder 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 ,
%U http://www.hindawi.com/journals/joper/2013/324630/