Pólya-type polynomial inequalities in Orlicz spaces and best local approximation

Abstract

We obtain an extension of Pólya-type inequalities for univariate real polynomials in Orlicz spaces. We also give an application to a best local approximation problem.

MSC 2010: 41A10; 41A17.

1 Introduction

Let X be a bounded open subset of ℝ. Consider the measure space (X, $ℬ$, μ), where μ is the Lebesgue measure, and denote $ℳ=ℳ\left(X\right)$ the system of all equivalence classes of Lebesgue measurable real valued functions on X. Let Φ be the set of convex functions ϕ : ℝ₊ → ℝ₊, with ϕ(x) > 0 for x > 0, and ϕ(0) = 0.

Given ϕ ∈ Φ, we define

$$L^{\varphi }=L^{\varphi }\left(X\right):=\left\{f\in ℳ:\underset{X}{\int }\varphi \left(\alpha \left|f\left(x\right)\right|\right)dx<\infty ,\text{for}\text{some}\alpha >0\right\}.$$

The space L^ϕis called the Orlicz space determined by ϕ. This space is endowed with the Luxemburg norm,

$${∥f∥}_{\varphi ,X}=\text{inf}\left\{\lambda >0:\underset{X}{\int }\varphi \left(\frac{\left|f\left(x\right)\right|}{\lambda }\right)\frac{dx}{\mu \left(X\right)}\le 1\right\}.$$

The space L^ϕwith this norm is a Banach space (see [1]). If $E\in ℬ$ and μ(E) > 0, then ∥ · ∥_ϕ,Eis a seminorm on L^ϕ(X). In the particular case, ϕ (t) = t^p, we will use the notation ∥ · ∥_p,Einstead of ∥ · ∥_ϕ,E.

Let ${\Pi }^N\subset ℳ$, N ∈ ℕ, be the class of all algebraic polynomials of degree at most N, with real coefficients.

Given $E\in ℬ$, we recall that a polynomial g _E ∈ Π^Nis a best approximation of f ∈ L^ϕ(X) from Π^Nrespect to ∥ · ∥_ϕ,E, if

$${∥f-g_E∥}_{\varphi ,E}=\text{inf}\left\{{∥f-P∥}_{\varphi ,E}:P\in {\prod }^N\right\}.$$

Let x _k , 1 ≤ k ≤ n, be n points in X. We consider a net of measurable sets $\left\{E\right\}\subset ℬ$ such that $E={\bigcup }_{k=1}^n{E}_k$, with μ(E _k ) > 0 and

$$\underset{1\le k\le n}{\text{sup}}\underset{y\in E_k}{\text{sup}}\left|x_k-y\right|\to 0,\text{as}\mu \left(E\right)\to 0.$$

Given f ∈ L^ϕ(X) and Π^N, we consider a net of best approximation functions {g _E }. If it has a limit in Π^Nas μ(E) → 0, this limit is called the best local approximation of f from Π^Non {x₁,..., x _n }. If the points in our approximation problem have not the same importance the neighborhoods E _k can be adjusted to reflect it. In [2], Chui et al. introduced the balanced neighborhood concept and they studied existence and characterization of best local approximation in L^p-spaces for several points with different size neighborhoods. In [3, 4], the last problem was considered for ϕ-approximation and ∥ · ∥ _ϕ -approximation, respectively, in Orlicz spaces. Other results in these spaces about best local approximation with non balanced neighborhoods were considered in [5].

Polynomial inequalities on measurable sets have been studied extensively in the literature (see [6–8]). In [9], the authors proved the following extension of the Pólya inequality in L^p-spaces, 0 < p ≤ ∞.

Theorem 1.1. Let 0 < p ≤ ∞ and n, N ∈ ℕ. Let i _k , 1 ≤ k ≤ n, be n positive integers such that${\sum }_{k=1}^n{i}_k=N+1$. Let B _k , 1 ≤ k ≤ n, be disjoint pairwise compact intervals in ℝ with 0 < μ(B _k ) ≤ 1. Then there exists a constant K depending on p, i _k and B _k , for 1 ≤ k ≤ n, such that

$$\left|c_j\right|\le \frac{K}{\underset{1\le k\le n}{\text{min}}\mu {\left(E\cap B_k\right)}^{i_k-1+1/p}}{∥P∥}_{p,E,}0\le j\le N,$$

for all$P\left(x\right)={\sum }_{j=0}^N{c}_j{x}^j,E\subset {\bigcup }_{k=1}^n{B}_k,\mu \left(E\cap B_k\right)>0,1\le k\le n.$.

They gave an application of this theorem to the existence of the best multipoint local approximation in L^pspaces, with balanced neighborhoods.

In this article, we generalize Theorem 1.1 and the balanced neighborhood concept to L^ϕ. As a consequence of this extension we prove the existence of the best local approximation of a function from Π^Non {x₁,..., x _n }, with balanced neighborhoods, following the pattern used in [9]. Moreover, we prove that the best local approximation polynomial is the Hermite interpolating polynomial.

We say that a function ϕ ∈ Φ satisfies the Δ₂-condition if there exists a constant k > 0 such that ϕ(2x) ≤ kϕ(x), for x ≥ 0, and we say that ϕ satisfies the Δ'-condition if there exists a constant c > 0 such that ϕ(xy) ≤ cϕ(x)ϕ (y) for x, y ≥ 0. We point out that the Δ'-condition implies the Δ₂-condition. A detailed treatment about these subjects may be found in [1].

If ϕ satisfies the Δ'-condition, it is easy to see that there exists a constant K > 0 such that

$${\varphi }^{-1}\left(x\right){\varphi }^{-1}\left(y\right)\le K{\varphi }^{-1}\left(xy\right),\text{for}\text{all}x,y\ge 0.$$

(1)

We assume in this article that ϕ ∈ Φ and it satisfies the Δ'-condition.

2 Preliminary results

Let ${\mathcal{X}}_A$ denotes the characteristic function on the measurable set A ⊂ X.

Proposition 2.1. The family of all seminorms ∥ · ∥_ϕ.Ewith μ(E) > 0, has the following properties:

(a)${∥{\mathcal{X}}_E∥}_{\varphi ,E}=\frac{1}{{\varphi }^{-1}\left(1\right)}$.

(b) if f, g ∈ L^ϕ(X) satisfy |f| ≤ |g| on E, then ∥f∥_ϕ,E≤ ∥g∥_ϕ,E. The inequality is strict if |f| < |g| on some subset of E with positive measure.

(c) There exists a constant M > 0 such that

$${∥f∥}_{\varphi ,G}\le \frac{M}{{\varphi }^{-1}\left(\frac{\mu \left(G\right)}{\mu \left(D\right)}\right)}{∥f∥}_{\varphi ,D,}f\in L^{\varphi }\left(X\right),$$

(2)

for all pair of measurable sets G, D, with G ⊂ D and μ(G) > 0.

Proof (a) For λ := 1/ϕ^-1(1) we have

$$\underset{E}{\int }\varphi \left(\frac{\left|{\mathcal{X}}_E\right|}{\lambda }\right)\frac{dx}{\mu \left(E\right)}=\underset{E}{\int }\frac{dx}{\mu \left(E\right)}=1.$$

Now, the Δ₂- condition implies ${∥{\mathcal{X}}_E∥}_{\varphi ,E}=1/{\varphi }^{-1}\left(1\right)$.

• (b) If |f| ≤ |g| on E, then

  $$\underset{E}{\int }\varphi \left(\frac{\left|f\right|}{\lambda }\right)\frac{dx}{\mu \left(E\right)}\le \underset{E}{\int }\varphi \left(\frac{\left|g\right|}{\lambda }\right)\frac{dx}{\mu \left(E\right)},\lambda >0,$$

and so ∥f∥_ϕ,E≤ ∥g∥_ϕ,E. In addition, if |f| < |g| on some subset of E with positive measure, the above inequality is strict. So, the Δ₂-condition implies the assertion.

• (c) Given G ⊂ D, μ(G) > 0, and f ∈ L ^ϕ(X), for each λ > 0, we denote

  $$\mathfrak{U}\left(\lambda \right):=\underset{G}{\int }\varphi \left(\frac{\left|f\right|}{\lambda }\right)\frac{dx}{\mu \left(G\right)}\text{and}\mathfrak{B}\left(\lambda \right):=\underset{D}{\int }\varphi \left(\frac{\left|f\right|}{\lambda }\right)\frac{dx}{\mu \left(D\right)}.$$

We consider λ > 0 such that $\mathfrak{B}\left(\lambda \right)\le 1$. By the Δ'-condition we obtain

$$\mathfrak{U}\left(\frac{\lambda }{{\varphi }^{-1}\left(\frac{\mu \left(G\right)}{c\mu \left(D\right)}\right)}\right)\le \underset{D}{\int }c\frac{\mu \left(G\right)}{c\mu \left(D\right)}\varphi \left(\frac{\left|f\right|}{\lambda }\right)\frac{dx}{\mu \left(G\right)}=\mathfrak{B}\left(\lambda \right)\le 1.$$

Then ${∥f∥}_{\varphi ,G}\le \frac{\lambda }{{\varphi }^{-1}\left(\frac{\mu \left(G\right)}{c\mu \left(D\right)|}\right)}$, for all λ > 0 with $\mathfrak{B}\left(\lambda \right)\le 1$. So, the definition of ∥f∥_ϕ,Dand (1) imply ${∥f∥}_{\varphi ,G}\le \frac{M}{{\varphi }^{-1}\left(\frac{\mu \left(G\right)}{\mu \left(D\right)}\right)}{∥f∥}_{\varphi ,D}$ with $M=\frac{K}{{\varphi }^{-1}\left(c^{-1}\right)}$.

Lemma 2.2. There exists a constant M > 0 such that

$$\left|P^{\left(j\right)}\left(a\right)\right|\le \frac{M}{{\epsilon }^j}{∥P∥}_{\varphi ,\left[a-\epsilon ,a+\epsilon \right],}$$

for all P ∈ Π^N, [a - ϵ, a + ϵ] ⊂ X, and 0 ≤ j ≤ N.

Proof. Given P ∈ Π^Nand [a - ϵ, a + ϵ] ⊂ X, we divide that interval in 2(N + 1) close subintervals with the same size. Let J_ϵ be one of them. From Proposition 2.1 (c), we get ${∥P∥}_{\varphi ,J_{\epsilon }}\le M{∥P∥}_{\varphi ,\left[a-\epsilon ,a+\epsilon \right]}$, where M is independent on P, a, and ϵ. In addition, there exists y_ϵ ∈ J_ϵ such that $\left|P\left(y_{\epsilon }\right)\right|\le {\varphi }^{-1}\left(1\right){∥P∥}_{\varphi ,J_{\epsilon }}$. In fact, if ${\varphi }^{-1}\left(1\right){∥P∥}_{\varphi ,J_{\epsilon }}<\left|P\left(y\right)\right|$, for all y ∈ J_ϵ, then Proposition 2.1 (a) and (b) yield ${∥P∥}_{\varphi ,J_{\epsilon }}>{∥P∥}_{\varphi ,J_{\epsilon }}$. A contradiction.

From the family of intervals J_ϵ, we choose pairwise disjoint (N + 1) intervals, and we denote them with J_i,ϵ, 1 ≤ i ≤ N + 1. Let y_i,ϵ∈ J_i,ϵbe such that

$$\left|P\left(y_{i,\epsilon }\right)\right|\le M{\varphi }^{-1}\left(1\right){∥P∥}_{\varphi ,\left[a-\epsilon ,a+\epsilon \right],}1\le i\le N+1.$$

(3)

If $t_{i,\epsilon }:=\frac{y_{i,\epsilon }-a}{\epsilon }\in \left[-1,1\right]$, we have

$$P\left(y_{i,\epsilon }\right)=\sum _{j=0}^N\frac{P^{\left(j\right)}\left(a\right)}{j!}{\left(y_{i,\epsilon }-a\right)}^j=\sum _{j=0}^N\frac{P^{\left(j\right)}\left(a\right)}{j!}{\epsilon }^j{t}_{i,\epsilon }^j,1\le i\le N+1.$$

(4)

The matrix of the linear system (4), $\left(t_{i,\epsilon }^j\right)$, is a Vandermonde matrix whose determinant has a positive lower bound, because t_i,ϵ- t_i',ϵ≥ 1/N + 1 for i > i'. Using Cramer's rule and (3), there is a constant which we again denote by M such that

$$\left|P^{\left(j\right)}\left(a\right){\epsilon }^j\right|\le M{∥P∥}_{\varphi ,\left[a,\epsilon ,a+\epsilon \right]}0\le j\le N.$$

The proof of the following lemma is analogous to the one of Lemma 2.3 in [9], however we give it for sake of completeness.

Lemma 2.3. Let C ⊂ X be an interval, E ⊂ C, μ(E) > 0. For all P ∈ Π^N, there exists an interval F := F(E,P) ⊂ C such that

a)$\mu \left(F\right)\ge \frac{\mu \left(E\right)}{2N}$,

b) ∥P∥_ϕ,F≤ 2N∥P∥_ϕ,E.

Proof. Let P ∈ Π^N, S = 2N, and let D _a := {x ∈ C : |P(x)| < a}. It easy to see that the function G(a): = μ(D _a ) is continuous, G(0) = 0 and $\underset{a\to \infty }{\text{lim}}G\left(a\right)=\mu \left(C\right)$. Therefore, there exists a constant a_* ∈ ℝ⁺ such that $\mu \left(D_{a_{*}}\right)=\mu \left(E\right)/2$. Since {x ∈ C : |P(x)| = a_*} has at most 2N elements, there exists k, 1 ≤ k ≤ N, and pairwise disjoint intervals E _j , 1 ≤ j ≤ k, such that $D_{a_{*}}={\bigcup }_{j=1}^k{E}_j$.

We denote $\overline{A}=C\backslash A$, for any set A. Then

$$\mu \left(E\cap {\overline{D}}_{a_{*}}\right)=\mu \left(E\right)-\mu \left(E\cap D_{a_{*}}\right)\ge \mu \left(E\right)-\mu \left(D_{a_{*}}\right)=\frac{\mu \left(E\right)}{2}.$$

(5)

There exists j, 1 ≤ j ≤ k, such that μ(E _j ) ≥ μ(E)/S. In fact, if μ(E _j ) < μ(E)/S for all j , 1 ≤ j ≤ k, we obtain $\mu \left(D_{a_{*}}\right)<k/S\mu \left(E\right)\le \mu \left(E\right)/2$, which is a contradiction. So, we have proved a) with F := E _j .

Using (5), we obtain

$$\mu \left(E\cap {\overline{D}}_{a_{*}}\right)\ge \frac{\mu \left(E\right)}{2}=\mu \left(D_{a_{*}}\right)\ge \mu \left(F\right)\mu \left(F\cap \overline{E}\right).$$

Therefore

$$\begin{array}{ll}\hfill \underset{F}{\int }\varphi \left(\frac{\left|P\right|}{\lambda }\right)\frac{dx}{\mu \left(F\right)}& \le \underset{F\cap E}{\int }\varphi \left(\frac{\left|P\right|}{\lambda }\right)\frac{dx}{\mu \left(F\right)}+\varphi \left(\frac{a_{*}}{\lambda }\right)\frac{\mu \left(E\cap {\overline{D}}_{a_{*}}\right)}{\mu \left(F\right)}\\ \le \underset{E\cap D_{a_{*}}}{\int }\varphi \left(\frac{\left|P\right|}{\lambda }\right)\frac{dx}{\mu \left(F\right)}+\underset{E\cap {\overline{D}}_{a_{*}}}{\int }\varphi \left(\frac{\left|P\right|}{\lambda }\right)\frac{dx}{\mu \left(F\right)}\\ =\underset{E}{\int }\varphi \left(\frac{\left|P\right|}{\lambda }\right)\frac{dx}{\mu \left(F\right)}.\end{array}$$

So, (a) implies

$${\mathcal{A}}_F\left(\lambda \right):=\underset{F}{\int }\varphi \left(\frac{\left|P\right|}{\lambda }\right)\frac{dx}{\mu \left(F\right)}\le S\underset{E}{\int }\varphi \left(\frac{\left|P\right|}{\lambda }\right)\frac{dx}{\mu \left(E\right)}=:S{\mathcal{A}}_E\left(\lambda \right).$$

Let λ be such that ${\mathcal{A}}_E\left(\lambda \right)=1$. The convexity of ϕ implies ${\mathcal{A}}_F\left(S\lambda \right)\le 1$. So, ∥P∥_ϕ,F≤ S∥P∥_ϕ,E.

3 Pólya inequality

Now, we present the main result concerning to Pólya inequality in L^ϕ.

Theorem 3.1. Let ϕ ∈ Φ, and n, N ∈ ℕ. Let i _k , 1 ≤ k ≤ n, be n positive integers such that${\sum }_{k=1}^n{i}_k=N+1$. Let B _k , 1 ≤ k ≤ n, be disjoint pairwise compact intervals in ℝ, with 0 < μ(B _k ) ≤ 1. Then there exists a positive constant M depending on ϕ, i _k , and B _k , 1 ≤ k ≤ n, such that

$$\left|c_j\right|\le \frac{M}{{\text{min}}_{1\le k\le n}\left\{\mu {\left(E\cap B_k\right)}^{i_k-1}{\varphi }^{-1}\left(\frac{\mu \left(E\cap B_k\right)}{\mu \left(E\right)}\right)\right\}}{∥P∥}_{\varphi ,E,}0\le j\le N,$$

(6)

for all$P\left(x\right)={\sum }_{j=0}^N{c}_j{x}^j,E\subset {\bigcup }_{k=1}^n{B}_k$with μ(E ∩ B _k ) > 0, 1 ≤ k ≤ n.

Proof. In the following proof, the constant M can be different in each occurrence. Let $P\left(x\right)={\sum }_{j=0}^N{c}_j{x}^j\in {\Pi }^N$, and let $E\subset {\bigcup }_{k=1}^n{B}_k$ be a measurable set with μ(E ∩ B _k ) > 0, 1 ≤ k ≤ n. By Lemma 2.3 for C = B _k , there exist n intervals F _k = [a _k - r _k , a _k + r _k ] ⊂ B _k , 1 ≤ k ≤ n, such that μ(F _k ) ≥ μ(E ∩ B _k )/2N and ${∥P∥}_{\varphi ,F_k}\le 2N{∥P∥}_{\varphi ,E\cap B_k}$. From Lemma 2.2, there exists a positive constant M depending on p, i _k , and B _k , 1 ≤ k ≤ n, such that for all j, 0 ≤ j ≤ i _k - 1, 1 ≤ k ≤ n, it verifies

$$\left|P^{\left(j\right)}\left(a_k\right)\right|\le \frac{M}{\mu {\left(F_k\right)}^j}{∥P∥}_{\varphi ,F_k}\le \frac{M}{\mu {\left(F_k\right)}^{i_k-1}}{∥P∥}_{\varphi ,F_k}\le \frac{M}{\mu {\left(E\cap B_k\right)}^{i_k-1}}{∥P∥}_{\varphi ,E\cap B_k}.$$

(7)

From (7) and (2), there is a constant M such that

$$\left|P^{\left(j\right)}\left(a_k\right)\right|\le \frac{M}{\mu {\left(E\cap B_k\right)}^{i_k-1}{\varphi }^{-1}\left(\frac{\mu \left(E\cap B_k\right)}{\mu \left(E\right)}\right)}{∥P∥}_{\varphi ,E}$$

for 0 ≤ j ≤ i _k - 1, 1 ≤ k ≤ n. So

$$\left|P^{\left(j\right)}\left(a_k\right)\right|\le \frac{M}{{\text{min}}_{1\le s\le n}\left\{\mu {\left(E\cap B_s\right)}^{i_s-1}{\varphi }^{-1}\left(\frac{\mu \left(E\cap B_s\right)}{\mu \left(E\right)}\right)\right\}}{∥P∥}_{\varphi ,E},$$

for 0 ≤ j ≤ i _k - 1, 1 ≤ k ≤ n. From the equivalence of the norms ∥ ⋅ ∥₁ and ∥ ⋅ ∥₂ on Π^N,

$${∥P∥}_1=\underset{1\le k\le n}{\text{max}}\underset{a_k\in B_k}{\text{sup}}\underset{0\le j\le i_k-1}{\text{max}}\left|P^{\left(j\right)}\left(a_k\right)\right|\text{and}{∥P∥}_2=\underset{0\le j\le N}{\text{max}}\left|c_j\right|,$$

we obtain (6).

4 Best local approximation

In this section, we introduce a concept of balanced neighborhood in L^ϕand we prove the existence of the best local approximation using the neighborhoods E _k , 1 ≤ k ≤ n, mentioned in the Section 1.

It is easy to see that E _k = x _k + μ(E _k )A _k , where A _k is a measurable set with measure 1. Henceforward, we assume the sets A _k are uniformly bounded.

For each α ∈ ℝ and k, 1 ≤ k ≤ n, we denote

$${\mathcal{A}}_k\left(\alpha \right):=\frac{\mu {\left(E_k\right)}^{\alpha }}{{\varphi }^{-1}\left(\frac{\mu \left(E\right)}{\mu \left(E_k\right)}\right)}.$$

We assume the following condition, which allows us that ${\mathcal{A}}_k\left(\alpha \right)$ can be compared with each other as functions of α when μ(E) → 0.

For any nonnegative integers α and β, and any pair j, k, 1 ≤ j, k ≤ n,

$$\text{either}{\mathcal{A}}_k\left(\alpha \right)=O\left({\mathcal{A}}_j\left(\beta \right)\right)\text{or}{\mathcal{A}}_j\left(\beta \right)=o\left({\mathcal{A}}_k\left(\alpha \right)\right),\text{as}\mu \left(E\right)\to 0.$$

(8)

Let < i _k > be an ordered n-tuple of nonnegative integers. We say that ${\mathcal{A}}_j\left(i_j\right)$ is a maximal element of $<{\mathcal{A}}_k\left(i_k\right)>$ if ${\mathcal{A}}_k\left(i_k\right)=O\left({\mathcal{A}}_j\left(i_j\right)\right)$ for all 1 ≤ k ≤ n. We denote it by

$${\mathcal{A}}_j\left(i_j\right)=\text{max}\left\{{\mathcal{A}}_k\left(i_k\right)\right\}.$$

Observe that ${\sum }_{k=1}^n{\mathcal{A}}_k\left(i_k\right)=O\left(\text{max}\left\{{\mathcal{A}}_k\left(i_k\right)\right\}\right).$.

Definition 4.1. An n-tuple < i _k > of nonnegative integers is balanced if

$$\sum _{k=1}^n{\mathcal{A}}_k\left(i_k\right)=o\left(\underset{1\le k\le n}{\text{min}}\left\{\mu {\left(E_k\right)}^{i_k-1}{\varphi }^{-1}\left(\frac{\mu \left(E_k\right)}{\mu \left(E\right)}\right)\right\}\right).$$

In this case, we say that${\sum }_{k=1}^n{i}_k$is a balanced integer, and < E _k > are balanced neighborhoods.

Lemma 4.2. To each balanced integer there corresponds exactly one balanced n-tuple.

Proof. Let < i _k > be a balanced n-tuple. If $<i_k^{\prime }>$ is distinct from < i _k > and ${\sum }_{k=1}^n{i}_k={\sum }_{k=1}^n{i^{\prime }}_k$, there exist indices j and s such that $i_j\ge i_j^{\prime }+1$ and $i_s^{\prime }\ge i_s+1$. From definition of balanced neighborhood, we have

$$\mathcal{A}:=\sum _{k=1}^n{\mathcal{A}}_k\left(i_k\right)=o\left(\mu {\left(E_j\right)}^{i_j-1}{\varphi }^{-1}\left(\frac{\mu \left(E_j\right)}{\mu \left(E\right)}\right)\right).$$

In addition, by (1) we get $\mu {\left(E_j\right)}^{i_j-1}{\varphi }^{-1}\left(\frac{\mu \left(E_j\right)}{\mu \left(E\right)}\right)\le \mu {\left(E_j\right)}^{{i^{\prime }}_j}{\varphi }^{-1}\left(\frac{\mu \left(E_j\right)}{\mu \left(E\right)}\right)\le K{\varphi }^{-1}\left(1\right){\mathcal{A}}_j\left(i_j^{\prime }\right)$. So, $\mathcal{A}=o\left(\sum _{k=1}^n{\mathcal{A}}_k\left({i^{\prime }}_k\right)\right)$. Again, by (1) we get

$$\frac{\sum _{k=1}^n{\mathcal{A}}_k\left({i^{\prime }}_k\right)}{\mu {\left(E_s\right)}^{{i^{\prime }}_s-1}{\varphi }^{-1}\left(\frac{\mu \left(E_s\right)}{\mu \left(E\right)}\right)}\ge \frac{\sum _{k=1}^n{\mathcal{A}}_k\left({i^{\prime }}_k\right)}{\mu {\left(E_s\right)}^{i_s}{\varphi }^{-1}\left(\frac{\mu \left(E_s\right)}{\mu \left(E\right)}\right)}\ge \frac{\sum _{k=1}^n{\mathcal{A}}_k\left({i^{\prime }}_k\right)}{K{\varphi }^{-1}\left(1\right){\mathcal{A}}_s\left(i_s\right)}\to \infty .$$

Then $<i_k^{\prime }>$ cannot be balanced.

The following lemma allows us to state an algorithm to compute all the balanced integers greater than a given balanced integer.

Lemma 4.3. Let < i _k > and$<i_k^{\prime }>$be two balanced n-tuples with${\sum }_{k=1}^n{i}_k<{\sum }_{k=1}^n{i^{\prime }}_k$. Let$A=A\left(<i_k>\right):=\left\{j:{\mathcal{A}}_j\left(i_j\right)=\text{max}\left\{{\mathcal{A}}_k\left(i_k\right)\right\}\right\}$and B = B(< i _k >):= {1,2,...,n}\ A. Then

(a) for j ∈ A$i_j^{\prime }\ge i_j+1$.

(b) for j ∈ A$i_j^{\prime }\ge i_j$.

Proof. (a) Suppose $i_j^{\prime }\le i_j$ for some j ∈ A. For any l ∈ B, from (8) we get ${\mathcal{A}}_l\left(i_l\right)=o\left({\mathcal{A}}_j\left(i_j\right)\right)$. Assume now $i_l^{\prime }\ge i_l+1$ for some l ∈ B. By (1), there exists a constant M > 0 such that

$$\frac{{\mathcal{A}}_j\left({i^{\prime }}_j\right)}{\mu {\left(E_l\right)}^{{i^{\prime }}_l-1}{\varphi }^{-1}\left(\frac{\mu \left(E_l\right)}{\mu \left(E\right)}\right)}\ge \frac{{\mathcal{A}}_j\left(i_j\right)}{\mu {\left(E_l\right)}^{i_l}{\varphi }^{-1}\left(\frac{\mu \left(E_l\right)}{\mu \left(E\right)}\right)}\ge \frac{{\mathcal{A}}_j\left(i_j\right)}{M{\mathcal{A}}_l\left(i_l\right)}\to \infty ,$$

as μ(E) → 0. Thus $<i_k^{\prime }>$ cannot be balanced, a contradiction. Therefore, either $B=\varnothing$ or $i_l^{\prime }\le i_l$, for all l ∈ B. On the other hand, since ${\sum }_{k=1}^n{i}_k<{\sum }_{k=1}^n{i^{\prime }}_k$, there is s ∈ A such that $i_s^{\prime }\ge i_s+1$. According to (1) and the definition of A we obtain

$$\frac{{\mathcal{A}}_j\left({i^{\prime }}_j\right)}{\mu {\left(E_s\right)}^{{i^{\prime }}_s-1}{\varphi }^{-1}\left(\frac{\mu \left(E_s\right)}{\mu \left(E\right)}\right)}\ge \frac{{\mathcal{A}}_j\left(i_j\right)}{\mu {\left(E_s\right)}^{i_s}{\varphi }^{-1}\left(\frac{\mu \left(E_s\right)}{\mu \left(E\right)}\right)}\ge \frac{{\mathcal{A}}_j\left(i_j\right)}{M{\mathcal{A}}_s\left(i_s\right)}\ge M^{\prime },$$

as μ(E) → 0, for some constant M' > 0. Therefore, $<i_k^{\prime }>$ cannot be balanced.

• (b) Suppose $i_j^{\prime }<i_j$ for some j ∈ B. From (a), (1) and the definition of balanced n-tuple, we obtain for each l ∈ A,

  $$\frac{{\mathcal{A}}_j\left({i^{\prime }}_j\right)}{\mu {\left(E_l\right)}^{{i^{\prime }}_l-1}{\varphi }^{-1}\left(\frac{\mu \left(E_l\right)}{\mu \left(E\right)}\right)}\ge \frac{{\mathcal{A}}_j\left(i_j-1\right)}{M{\mathcal{A}}_l\left(i_l\right)}\ge M^{\prime }\frac{\mu {\left(E_j\right)}^{i_l-1}{\varphi }^{-1}\left(\frac{\mu \left(E_j\right)}{\mu \left(E\right)}\right)}{{\mathcal{A}}_l\left(i_l\right)}\to \infty ,$$

as μ(E) → 0. Therefore $<i_k^{\prime }>$ cannot be balanced.

Given a balanced integer, the above lemma gives us a necessary condition which must satisfy the next balanced integer. The following example shows that the conditions of Lemma 4.3 are not sufficient to get a balanced n-tuple.

Example 4.4. Define ϕ (x) = x³(1 + |ln x|), x > 0, and ϕ(0) = 0. Consider two points x₁, x₂ with μ(E₁) = δ^4/3, μ (E₂) = δ^1/3, and A₁ = A₂ = [0,1]. The 2-tuple < 0,1 > is balanced. Here, the set A(< 0,1 >) = {0}, however < 1,1 > is not a balanced 2-tuple. In fact, if < i _k >=< 0,1 > we obtain

$$\underset{1\le k\le 2}{\text{min}}\left\{\mu {\left(E_k\right)}^{i_k-1}\left(\frac{\mu \left(E_k\right)}{\mu \left(E\right)}\right)\right\}=\text{min}\left\{\frac{{\varphi }^{-1}\left(\delta \right)}{{\delta }^{4/3}},{\varphi }^{-1}\left(1\right)\right\}+o\left(1\right)\to {\varphi }^{-1}$$

as δ → 0. Since ${\mathcal{A}}_2\left(i_2\right)=o\left({\mathcal{A}}_1\left(i_1\right)\right)$ and ${\mathcal{A}}_1\left(i_1\right)=o\left(1\right)$, as δ → 0, we have

$$\frac{{\sum }_{k=1}^2{\mathcal{A}}_k\left(i_k\right)}{\underset{1\le k\le 2}{\text{min}}\left\{\mu {\left(E_k\right)}^{i_k-1}{\varphi }^{-1}\left(\frac{\mu \left(E_k\right)}{\mu \left(E\right)}\right)\right\}}=o\left(1\right),\text{as}\delta -0.$$

So < 0,1 > is a balanced 2-tuple, A(< 0,1 >) = {0}, and < 1,1 > is the next 2-tuple generated by the algorithm. For < i _k > = < 1,1 > we have

$$\frac{{\mathcal{A}}_2\left(i_2\right)}{\underset{1\le k\le 2}{\text{min}}\left\{\mu {\left(E_k\right)}^{i_k-1}{\varphi }^{-1}\left(\frac{\mu \left(E_k\right)}{\mu \left(E\right)}\right)\right\}}\ge \frac{{\mathcal{A}}_2\left(i_2\right)}{{\varphi }^{-1}\left(\frac{\mu \left(E_1\right)}{\mu \left(E\right)}\right)}\to \infty ,\text{as}\delta \to 0.$$

Thus < 1,1 > is not a balanced 2-tuple.

Next, we establish an algorithm which gives all balanced n-tuples. First, we observe that < 0 > is a balanced n-tuple. In fact, since ϕ^-1 is a concave positive function on ℝ₊ with ϕ^-1(0) = 0, we have ϕ^-1(x) ≥ ϕ^-1(1)x, for x ≤ 1. This yields

$$\frac{\mu \left(E_j\right)}{{\varphi }^{-1}\left(\frac{\mu \left(E\right)}{\mu \left(E_k\right)}\right){\varphi }^{-1}\left(\frac{\mu \left(E_j\right)}{\mu \left(E\right)}\right)}\le \frac{\mu \left(E\right)}{{\left({\varphi }^{-1}\left(1\right)\right)}^2},1\le j,k\le n.$$

Algorithm. Let v _q be a balanced integer and let $<i_k^{\left(vq\right)}>$ be the corresponding balanced n-tuple. To build the next n-tuple, $<i_k^{\left(vq+1\right)}>$, put $i_k^{\left(vq+1\right)}=i_k^{\left(vq\right)}+1$ for $k\in A\left(<i_k^{\left(vq\right)}>\right)$ and $i_k^{\left(v_q+1\right)}=i_k^{\left(v_q\right)}$ for $k\in B\left(<i_k^{\left(vq\right)}>\right)$.

The following lemma shows that all balanced n-tuples are contained in the set of n-tuples generated by the algorithm.

Lemma 4.5. if < i _k > is a balanced n-tuple with${\sum }_{k=1}^n{i}_k=q$, then the algorithm generates all the balanced n-tuple$<i_k^{*}>$with${\sum }_{k=1}^n{i}_k^{*}>q$.

Proof. Suppose $<i_k^{*}>$ is a balanced n-tuple with ${\sum }_{k=1}^n{i}_k^{*}=m>q$, and the n-tuple $<i_k^{\left(m\right)}>$ is not balanced. Since $\sum _{k=1}^n{i}_k^{*}=\sum _{k=1}^n{i}_k^{\left(m\right)}$, there exist r and s such that $i_r^{\left(m\right)}>i_r^{*}$ and $i_s^{*}>i_s^{\left(m\right)}$. By definition of balanced integer we have

$${\mathcal{A}}_r\left(i_r^{\left(m\right)}-1\right)=O\left({\mathcal{A}}_r\left(i_r^{*}\right)\right)=o\left(\mu {\left(E_s\right)}^{i_s^{*}-1}{\varphi }^{-1}\left(\frac{\mu \left(E_s\right)}{\mu \left(E\right)}\right)\right),$$

(9)

and (1) implies $\mu {\left(E_s\right)}^{i_s^{*}-1}{\varphi }^{-1}\left(\frac{\mu \left(E_s\right)}{\mu \left(E\right)}\right)\le K{\varphi }^{-1}\left(1\right){\mathcal{A}}_s\left(i_s^{*}-1\right)$. So, ${\mathcal{A}}_r\left(i_r^{\left(m\right)}-1\right)=o\left({\mathcal{A}}_s\left(i_s^{\left(m\right)}\right)\right)$.

On the other hand, since m > q, Lemma 4.3 implies $i_r^{*}\ge i_r$, so $i_r^{\left(m\right)}>i_r$. Therefore ${\mathcal{A}}_r\left(i_r^{\left(m\right)}-1\right)$ is maximal in a previous step of the algorithm, i.e., there exists m', q ≤ m' < m, such that ${\mathcal{A}}_r\left(i_r^{\left(m\right)}-1\right)$ is maximal of $<{\mathcal{A}}_k\left(i_k^{\left(m^{\prime }\right)}\right)>$. Since the exponents $i_k^{\left(m\right)}$ are nondecreasing,

$${\mathcal{A}}_s\left(i_s^{\left(m\right)}\right)=O\left({\mathcal{A}}_s\left(i_s^{\left(m^{\prime }\right)}\right)\right)=O\left({\mathcal{A}}_r\left(i_r^{\left(m\right)}-1\right)\right),$$

which contradicts (9).

Remark 4.6. If we assume the additional condition ϕ^-1(x)ϕ^-1 (1/x) ≥ c > 0 for x > 0, given a balanced n-tuple < i _k >, it is easy to see that the n-tuple $<i_k^{\prime }>$ defined by $i_k^{\prime }=i_k+1$ for $k\in A\left(<i_k^{\left(vq\right)}>\right)$, and $i_k^{\prime }=i_k$ for $k\in B\left(<i_k^{\left(vq\right)}>\right)$, is balanced. It give us an algorithm that generates the infinite sequences of all balanced n-tuples.

Let PC^m(X) be the class of functions with derivatives up to order m - 1 and with bounded piecewise continuous m^th derivative on X.

Next, we prove the following auxiliary lemma.

Lemma 4.7. Let < i _k > be an ordered n-tuple of nonnegative integers. Suppose h ∈ PC^m(X), where m = max{i _k } and h^(j)(x _k ) = 0, 0 ≤ j ≤ i _k - 1, 1 ≤ k ≤ n. Then

$${∥h∥}_{\varphi ,E}=O\left(\text{max}\left\{{\mathcal{A}}_k\left(i_k\right)\right\}\right).$$

Proof. Expanding h by the Taylor polynomial at x _k up to the order n, we obtain

$$h\left(x\right)=\sum _{k=1}^n{h}^{\left(i_k\right)}\left({\xi }_k\right)\frac{{\left(x-x_k\right)}^{i_k}}{i_k!}{\chi }_{E_k}\left(x\right),x\in E,$$

where ξ _k is between x and x _k . The change of variable x - x _k = ϵy, y ∈ A _k , yields

$${∥h∥}_{\varphi ,E}=\text{inf}\left\{\lambda >0:\sum _{k=1}^n\underset{A_k}{\int }\mu \left(E_k\right)\varphi \left(\frac{\left|h^{\left(i_k\right)}\left({\xi }_k\right)\right|\frac{\mu {\left(E_k\right)}^{i_k}\left|y^{i_k}\right|}{i_k!}}{\lambda }\right)\frac{dy}{\mu \left(E\right)}\le 1\right\}.$$

For

$$\lambda :=M\sum _{j=1}^n\frac{\mu {\left(E_j\right)}^{i_j}}{{\varphi }^{-1}\left(\frac{\mu \left(E\right)}{n\mu \left(E_j\right)}\right)},$$

where $M=\underset{1\le k\le n}{\text{max}}\left\{\frac{1}{i_k!}\underset{x\in X}{\text{max}}\left\{\left|h^{\left(i_k\right)}\left(x\right)\right|\right\}\underset{y\in A_k}{\text{max}}\left\{{\left|y\right|}^{i_k}\right\}\right\}$, we obtain

$$\sum _{k=1}^n\underset{A_k}{\int }\mu \left(E_k\right)\varphi \left(\frac{\left|h^{\left(i_k\right)}\left({\xi }_k\right)\right|\frac{\mu {\left(E_k\right)}^{i_k}\left|y^{i_k}\right|}{i_k!}}{\lambda }\right)\frac{dy}{\mu \left(E\right)}\le 1.$$

Therefore ${∥h∥}_{\varphi ,E}=O\left({\sum }_{k=1}^n\frac{\mu {\left(E_k\right)}^{i_k}}{{\varphi }^{-1}\left(\frac{\mu \left(E\right)}{n\mu \left(E_k\right)}\right)}\right)$. Using the convexity of ϕ, we have $\frac{{\varphi }^{-1}\left(x\right)}{n}\le {\varphi }^{-1}\left(\frac{x}{n}\right)$, x ≥ 0. So, ${∥h∥}_{\varphi ,E}=O\left(\text{max}\left\{{\mathcal{A}}_k\left(i_k\right)\right\}\right)$.

If a polynomial P ∈ Π^N, $N+1={\sum }_{k=1}^n{i}_k$, satisfies P^(j)(x _k ) = f^(j)(x _k ), 1 ≤ j ≤ i _k - 1, 1 ≤ k ≤ n, we call it the Hermite interpolating polynomial of the function f on {x₁,...,x _n }.

Now, we are in condition to prove the main result in this Section.

Theorem 4.8. Let < i _k > be a balanced n-tuple and$N+1={\sum }_{k=1}^n{i}_k$. If m = max{i _k } and f ∈ PC^m(X), then the best local approximation of f from Π^Non {x₁,...,x _k } is the Hermite interpolating polynomial of f on {x₁,...,x _n }.

Proof Let H ∈ Π^Nbe the Hermite interpolating polynomial and let {g _E } be a net of best approximations of f from Π^Nrespect to ∥ · ∥_ϕ,E. From Lemma 4.7,

$${∥g_E-H∥}_{\varphi ,E}=O\left(\text{max}\left\{{\mathcal{A}}_k\left(i_k\right)\right\}\right).$$

Using Theorem 3.1 and the equivalence of the norms in Π^N, we get

$${∥g_E-H∥}_{\infty }\le \frac{K}{\underset{1\le k\le n}{\text{min}}\left\{\mu {\left(E_k\right)}^{i_k-1}{\varphi }^{-1}\left(\frac{\mu \left(E_k\right)}{\mu \left(E\right)}\right)\right\}}{∥g_E-H∥}_{\varphi ,E.}$$

So, the definition of balanced n-tuple implies g _E → H, as μ(E) → 0.