ON THE L p -SATURATION OF THE YE-ZHOU OPERATOR

. We solve the saturation problem for a class of Ye-Zhou operator T n ( f, x ) = P n ( x ) A n L n ( f ) with suitable sequence of matrices { A n } n ≥ 1 . The solu-tion is based on the saturation theorem for the Kantorovich operator established by V. Maier and S. D. Riemenschneider.


INTRODUCTION
For f ∈ C[0, 1] the nth Bernstein polynomial is defined by D. X. Zhou showed in [6] for the Kantorovich operator (1) p n,k (x)(n + 1) He also showed [6] that if 1 < α < 2 then there exist no functions {ϕ n,α (x)} n≥1 such that Thus we cannot characterize the second orders of Lipschitz functions by the rate of convergence for the Kantorovich operators.To overcome this difficulty, M. D. Ye and D. X. Zhou [5] introduced a new method of linear approximation by means of matrices: let P n (x) ≡ (p n,0 (x), . . ., p n,n (x)) and k=0 is a bases of the linear space span{1, x, . . ., x n } and are functionals on C[0, 1], respectively.Then, for any (n + 1) × (n + 1) matrix A we get the so-called Ye-Zhou operator defined by The aim of the present paper is to solve the saturation problem for the class of operator {T n (f, x)} n≥1 for a suitable sequence of matrices {A n } n≥1 .By reason of (2), the saturation problem of the Ye-Zhou operator will be in connection with the saturation condition of the Kantorovich operator (see [2], [3], [4]).
The existence of a matrix A n with the properties (3)-( 7) is guaranteed by the following numerical example: Then the explicit expression of the Ye-Zhou operator will be the following: Theorem 2.1.Let A n and T n be defined as above.Then we have , where the kernel with I i = (i/(n + 1), (i + 1)/(n + 1)) and χ I i the characteristic function on I i , respectively.
Proof.(i) it is a direct consequence of ( 8) and ( 3); (ii) a simple calculation shows that Then, by ( 5),

, n, and
in view of (4).
Proof.In view of (8), ( 4) and ( 1) we obtain Thus we have for x ∈ [0, 1] Again, by (7), we have for x ∈ [0, 1] In conclusion By the interpolation theorem of Riesz-Thorin we get for 1 in view of (9) and (10).Hence we obtain the assertion of the theorem, because

THE SATURATION RESULT
The matrix A n and the corresponding operators T n are determined in Section 2. We can now state our main result: Theorem 3.1.Let {T n } n≥1 be defined as above.
Here we mention that we can deduce for T n similar statements to the lemmas established in [1, pp.316-318, Lemmas 6.2, 6.4 and 6.5] for K n from the application of Theorem 1.