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Abstract and Applied Analysis

Volume 2014 (2014), Article ID 781068, 10 pages

http://dx.doi.org/10.1155/2014/781068

## Strong Inequalities for Hermite-Fejér Interpolations and Characterization of -Functionals

Department of Mathematics, Shaoxing College of Arts and Sciences, Shaoxing, Zhejiang 312000, China

Received 10 October 2013; Accepted 6 December 2013; Published 16 January 2014

Academic Editor: Ding-Xuan Zhou

Copyright © 2014 Gongqiang You. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

The works of Smale and Zhou (2003, 2007), Cucker and Smale (2002), and Cucker and Zhou (2007) indicate that approximation operators serve as cores of many machine learning algorithms. In this paper we study the Hermite-Fejér interpolation operator which has this potential of applications. The interpolation is defined by zeros of the Jacobi polynomials with parameters , . Approximation rate is obtained for continuous functions. Asymptotic expression of the -functional associated with the interpolation operators is given.

#### 1. Introduction

Zhou and Jetter [1] used Bernstein-Durrmeyer operators for studying support vector machine classification algorithms. This work initiates the direction of applying more linear operators from approximation theory to learning theory. We will follow this direction and study Hermite-Fejér interpolation operator. It would be interesting to derive explicit learning rates by means of these operators for some specific learning algorithms.

Let denote the Jacobi polynomial of order . Let be the zeros of . We assume that . For any continuous function on , the Hermite-Fejér interpolation is a polynomial of order that satisfies for any . Let be the norm on (, ). Without introducing ambiguity we also use to denote the norm on (which is the totality of the continuous functions on with period , and in this case ).

One has (see e.g., [2]) if and only if .

Define Denote by the conjugate function of , and write . When , , one has (see [3, 4]) So when , , is saturation with and the saturation class is

Note that for all , the associated classes are identical ([4, Theorem 6]). See [2, 5–7] for related works.

Denote that and recursively with . Write . Define For any , one has [8] Let and . We use the following definition of -functional from [9]:

We cite the following three Theorems from [9].

Theorem 1. *Let , be fixed. Then there is a constant such that for all and all , one has
**
in which the symbol does not rely on and .*

*Theorem 2. Let , be fixed. Then the following relation holds:
*

*Theorem 3. Let , be fixed. Then there is a constant such that, for all and all , one has
*

Here the symbol does not rely on and .

*The -functional for all , is characterized by the following*

*Theorem 4. Let , and , be fixed. Then, for all and all , the following holds:
*

in which

*
Moreover, if , , then, due to (see (2.6) of [9]), we have
*

*Theorem 4 will be proved in Section 3. In Section 2 we will discuss some properties of Jacobi polynomials and make some remarks concerning the conjugate function .*

*2. Estimates for Jacobi Polynomials and Conjugate Functions*

*2. Estimates for Jacobi Polynomials and Conjugate Functions**Chapter 8 of [10] gave the following.*

*Lemma 5. Let be fixed. For all and , one has
*

in which and . If , then there exists a making and . Moreover, let . Then

and, for , one has with .

*Define
*

*
and we have
*

*
For we have a conclusion similar to (17) (see [11]) as follows.*

*Lemma 6. For all fixed and , there is such that
*

*Denote by the set of -order algebraic polynomials and by the set of -order trigonometric polynomials. Denote further that
*

*
The following identity can be found in [9, 12] (see Lemma 5 of [12] and its proof).*

*Lemma 7. Let , be fixed. Then for all *

*For our purpose, we need the following result, which improves the estimate of [4] (see Lemma 4 of [4]).*

*Lemma 8. Let , be fixed. If and , then for all , the following holds:
*

*Proof. *Write . So and for

On the other hand,

Through integration by parts,

Moreover, we have

but

So we have

Thus, if , then

If , then and

In this case, for , we have

therefore,

Let in (35) and, with (32), we obtain

For , it is clear that if

then (31) and (37) imply that

On the other hand, it is easy to see that the above estimate also holds for . Thus, the desired inequality is obtained.

Next, we are going to prove (37). This time we define

Simple calculation shows that

With (36), we obtain

Hence,

Similar to the case of , we have

which obviously implies (37).

*In what follows, we will give an estimate of the conjugate function defined in the saturation class .*

*Lemma 9. Let , be fixed. Then, for all and even , one has
*

in which does not depend on and . Moreover, let be the best approximation of . Then, for all , one gets

*Proof. *Since

we have

Integrating by parts, we obtain

Therefore,

To deal with the second term of the above estimate, we note that, if , then , and

Thus,

If , rewrite the previous term as

Obviously, we get

On the other hand, we have

Since

we obtain

Moreover, the following estimates hold:

Consequently, for all , we get

which proves the first assertion of the lemma.

The second estimate can be obtained from Lemma 8, the first estimate, and integration by parts.

*Lemma 10. There exists an absolute constant such that, for all even ,
*

*Proof. *We may assume that and , . Thus, by [13, 14], for Fejér mean of , we have

therefore,

For , we have

Consequently, for with and , we obtain

We may assume that

Otherwise, choose and to make even and . Then

We conclude that

which gives the desired inequality.

*3. Proof of Theorem 4*

*3. Proof of Theorem 4**We need to prove the following Lemma before Theorem 4.*

*Lemma 11. Given , , there is such that, for all and all , one has
*

*Proof. *Denote that . Then from Theorem 3, we have

Next, let us estimate .

We know that with and . But (1) tells that . Hence, for . Lemma 5 tells that, for each , there is satisfying . Assume that . Then and further

We may assume that . Thus, the Bernstein inequality for trigonometric polynomials yields

But

So, we have

Combining this inequality with (68), we get

Now we need only to prove

Obviously,

Lemma 5 tells the following. Let be fixed, then for ,

Thus, for those ,

Since , following (23), (18), and Lemma 6, we have, for those ,

If and , then there is a satisfying (see Lemma 5). Hence, from Bernstein inequality , we have

Moreover,

Consequently, (78) obtained from Lemma 9 holds for all . Finally, from (78) and (75) we obtain (74).

*Proof of Theorem 4. *Firstly, we prove that

Let , , and . From Theorems 1 and 2 we conclude that, for some ,

We know that , so (see [14, page 43])

Thus, (18) and Lemmas 6 and 7 imply that

Consequently, by Lemma 11, we get

Obviously,

If , then Lemma 9 implies that

Since