Complex Shepard Operators and Their Summability

In this paper, we construct the complex Shepard operators to approximate continuous and complex-valued functions on the unit square. We also examine the effects of regular summability methods on the approximation by these operators. Some applications verifying our results are provided. To illustrate the approximation theorems graphically we consider the real or imaginary part of the complex function being approximated and also use the contour lines of the modulus of the function.


Introduction
In 1968, Donald Shepard introduced a sequence of operators that was highly effective in scattered data interpolation problems, which are known in the literature as Shepard operators [29]. Since these operators are of interpolatory-type, linear and positive, they are also quite useful in the classical approximation theory. Recall that the Shepard interpolatory operators are defined by for an arbitrary continuous real-valued function f on the unit interval [0, 1]. These operators defined on real-valued functions have been studied extensively. For example, error estimates, direct and converse theorems, saturation results, rational approximation, simultaneous approximation may be found in  [14][15][16]31,[34][35][36]. There are also several modifications of the original Shepard operators in order to increase the accuracy of approximation or to solve specific interpolation problems in CAGD such as scattered data and image compression [2,6,7,9,10,[17][18][19][20][21][22]24]. However, to the best of our knowledge, there is a limited information in the literature about a complex version of Shepard operators. By using the fundamental polynomials of Lagrange interpolation at the roots of unity, Hermann studied Shepard-type operators for functions on the unit circle [27].
In the present paper we mainly introduce the complex Shepard operators and systematically investigate their approximation properties. We also examine the effects of regular summability methods on the approximation by these operators. To visualize the corresponding approximation we consider the real or imaginary part of the complex function being approximated and also use the contour lines of the modulus of the complex function. Fig. 1). Let f be a complex-valued function defined on K. Then, for a positive real number λ, we define the complex Shepard operators by where we write n k,m=0 for the double summation n k=0 n m=0 . Observe that S n,λ (f ) interpolates at the sample points z k,m , i.e., We denote the space of all continuous and complex-valued functions on K by C(K, C). Then, for the complex Shepard operators defined by (1) we get the following approximation result. Theorem 1. For every f ∈ C (K, C) and λ ≥ 3, we have where, as usual, the symbol ⇒ denotes the uniform convergence.
To prove Theorem 1, we first need the following lemma. Lemma 1. Let z ∈ K with z = z k,m . Then, for every n ∈ N and λ > 0, ⎛ ⎝ n k,m=0 Proof. Define the indices k * and m * by Then, we observe that the proof is completed. Now for each fixed z ∈ K define the function ϕ z on K by ϕ z (w) := |w − z|. Then, we also need the next lemma.
Proof. Using the indices k * and m * given by (2) we may write from Lemma 1 that Hence we get Then, combining the above results, for λ > 3, we obtain Also, if λ = 3, then we have Therefore, the proof follows from (3) and (4), immediately.
Now we recall some terminology in the recent paper by Anastassiou [3].
Let K ⊂ C be a compact convex set and L : C(K, C) → C(K, C) be a linear operator. Then, consider a positive linear operatorL : In this case,L is called the companion operator of L. Then, using the test function e 0 (w) = 1 and ϕ z (w) = |w − z|, Anastassiou proved the next approximation result.

Theorem 2.
(see Corollary 19 in [3]) Let (L n ) be a sequence of linear operators from C(K, C) into itself and L n be a sequence of their companion operators from C(K, R) into itself. Additionally, assume that, for every g ∈ C(K, R), Then, for every f ∈ C(K, C), we have where · denotes the usual supremum norm on K and ω (f, δ) (δ > 0) is the modulus of continuity of f on K. In this case, if L n (e 0 ) ⇒ e 0 andL n (ϕ z ; z) ⇒ 0 on K, then, for every f ∈ C(K, C), we have Now we are ready to prove Theorem 1.
Then, it is easy to check from (5) that, for each n ∈ N and λ > 0,S n,λ is the companion operator of S n,λ . Also, these operators satisfy the assumption (6). We observe thatS n,λ (ϕ z ) = S n,λ (ϕ z ) since ϕ z is a real-valued function. Therefore, using the fact thatS n,λ (e 0 ) = e 0 = 1 on K and taking into account Lemma 2, all conditions of Theorem 2 hold for λ ≥ 3, which immediately gives the proof of Theorem 1.
wherez denotes the conjugate of z and exp(·) is the complex exponential function. Since f ∈ C(K, C), it follows from Theorem 1 that, for every λ ≥ 3, To visualize this uniform approximation we may consider the real or imaginary part of the function f. Then, for example, it is easy to see that which is indicated in Fig. 2 for the values λ = 3.7 and n = 8, 15, 30.
Example 2. Define the function g by Then, Theorem 1 implies that, for every λ ≥ 3, S n,λ (g) ⇒ g on K holds, which also gives This approximation is indicated in Fig. 3 via the contour lines, where larger values are colored lighter. In Fig. 3 we use the parameter values λ = 5 and n = 10, 20, 50.

Summability by the Complex Shepard Operators
Now let (u n ) be a sequence of complex functions from K into itself. Then we consider the following slight modification of the complex Shepard operators: for z ∈ K, n ∈ N, λ > 0 and f ∈ C(K, C). Defining the function e 1 (z) = z on K, we get the next approximation theorem.
Theorem 3. If u n ⇒ e 1 on K, then, for every f ∈ C(K, C) and λ ≥ 3, we have S * n,λ (f ) ⇒ f on K. Proof. From the definitions of (1) and (10) , we may write that | for any f ∈ C(K, C). Then, using (3) , (4) and (7) we get for λ = 3. Since u n ⇒ e 1 on K and f is uniformly continuous on K, we immediately observe that, for every λ ≥ 3, Then, it is natural to arise the following problem.   (9) To find an affirmative answer to Problem 1, we will mainly consider nonnegative regular summability matrices. It is known that a summability method is a common and useful way to overcome the lack of the usual convergence [8,26]. Recent studies demonstrate that it is also quite effective in the classical approximation theory (see, for instance, [1,4,5,23,25,28,30,32]). We now use it in the approximation by complex Shepard operators.
Recall that, for a given infinite matrix A := [a jn ] (j, n ∈ N) and a sequence x := (x n ) , the A-transformed sequence of (x n ) is given by Ax := ((Ax) j ) = ∞ n=1 a jn x n provided that the series is convergent for every j ∈ N. In this case, we say that A is a summability matrix method. If the Atransformed sequence of (x n ) converges to a number L, then the sequence (x n ) is said to be A-summable (or A-convergent) to L, which is denoted by which is denoted by f n A ⇒ f on D. A regular matrix summability method is a matrix transformation of a convergent sequence which preserves the limit. It is well-known that the Silverman-Toeplitz theorem characterizes the regularity of a matrix summability method (see [8,26]).
Throughout the paper, we will consider A = [a jn ] as a nonnegative regular summability matrix. By the nonnegativity of a matrix method, we mean all entries of the matrix are nonnegative.
Then, the first idea that comes to mind regarding the solution of Problem 1 is as follows.
holds for every f ∈ C(K, C) and λ ≥ 3. However, the next example shows that the claim in (13) is not satisfied for the function e 2 (z) = z 2 on K.
Example 3. Define the function sequence (u n ) on the set K by u n (z) = 2z, if n is odd 0, if n is even.
Assume that A = [a jn ] is any nonnegative regular summability matrix such that u n A ⇒ e 1 on K.
It follows from (14) that Taking supremum over z ∈ K, we get Since u n A ⇒ e 1 on K, we see that On the other hand, we may write from (14) that e 2 (u n (z)) = 4z 2 , if n is odd 0, if n is even.
Then, defining the function h(z) = 2z 2 on K, we see that a jn e 2 (u n (z)) a jn |S n,λ (e 2 ; u n (z)) − e 2 (u n (z))| Therefore, we get from (3), (4) and (7) that for λ = 3. Using (15) in the last two inequalities and also considering the regularity of A, we obtain S * n,λ (e 2 ) A ⇒ h = e 2 on K for every λ ≥ 3. Hence, (13) does not work for the test function e 2 .
We understand from Example 3 that in order to find a solution to Problem 1, we need a slightly stronger concept than the A-summability. In fact, there is such a concept in the summability theory, the so-called "strong summability".
Recall that, for a given summability matrix A = [a jn ], a sequence x = (x n ) is said to be strongly A-summable to a value L if all sums ∞ n=1 a jn |x n −L| (j = 1, 2, . . .) exist and converges to zero as j → ∞. We denote this by x n |A| → L. The concept of strong summability can be also given for a sequence of functions. Let (f n ) be a sequence of complex-valued functions defined on a set D ⊂ C. Then, we say that (f n ) is strongly uniform A-summable to a function f on D provided that