A Wavelet Characterization for the Upper Global Hölder Index

Abstract

The upper Hölder index has been introduced to describe smoothness properties of a continuous function. It can be seen as the irregular counterpart of the usual Hölder index and has been used to investigate the behavior at the origin of the modulus of smoothness in many classical cases.

In this paper, we prove a characterization of the upper Hölder index in terms of wavelet coefficients. This result is a first step in the estimation of this exponent using wavelet methods.

Appendix: Optimality of the Assumptions of Theorem 3

We prove here the optimality of the assumptions of Proposition 5 and thus of Theorem 3. To this end we use two counter-examples already introduced in [19].

1.1 A.1 A Uniform Irregular Function Satisfying Property (8)

Let α∈(0,1), ℓ ₀∈N and define the two following sequences of integers (j _n ) _n∈N and (j _n,α ) _n∈N as

$$\left\{\begin{array}{l@{\quad}l}j_1=\ell_0,\\[3pt]j_{n+1}=[\frac{1}{1-\alpha}2^{j_n \alpha}-j_n \alpha],& \forall n\geq1,\\[6pt]j_{n,\alpha}={[2^{j_n \alpha}]},& \forall n\geq1.\end{array}\right.$$

We aim at proving the following result.

Proposition 12

Let us assume that the multiresolution analysis is compactly supported. Let ε∈(0,1) and ℓ ₀ be such that $\mathrm {supp}(\psi)\subset[-2^{\ell_{0}},2^{\ell_{0}}]$. Furthermore, let us assume that ψ(0)≠0. The function f defined as

satisfies the following properties:

• 1. f is not a uniformly Hölderian function,

• 2. the wavelet coefficients of f satisfy property (8),

• 3. f is uniformly irregular with exponent β, where

  $$ \beta=\max\biggl(\alpha\varepsilon,\frac{\alpha\varepsilon}{(1-\alpha )+\alpha\varepsilon}\biggr)<\alpha.$$

  (27)

Proof

The two first properties being straightforward, we just have to prove that f is uniformly irregular with exponent β. Let n∈N and define

$$f_j(x)=\sum_{\ell=j+2}^{j_{n,\alpha}} \ell^{-\varepsilon} \psi \big(2^\ell(x-2^{-(j-\ell_0)})\big),$$

for j∈{j _n ,…,j _n,α } and

$$f_j(x)=\sum_{\ell=j+2}^{j_{n+1}} 2^{-\ell} \ell^{-\varepsilon}\psi\big( 2^\ell(x-2^{-(j-\ell_0)})\big)+\sum_{\ell=j_{n+1}}^{j_{n+1,\alpha}}\ell^{-\varepsilon}\psi\big (2^\ell(x-2^{-(j-\ell_0)})\big),$$

for j∈{j _n,α ,…,j _n+1−1}. We need to estimate

$$f(2^{-(j-\ell_0)})-f(0) = f(2^{-(j-\ell_0)})$$

for any j∈N. First, observe that for j≠j′, supp(f _j )∩supp(f _j′)=∅. Indeed for any j, we have

$$\mathrm{supp}(f_j)\subset[3.2^{-(j+2-\ell_0)},5.2^{-(j+2-\ell_0)}]\;$$

and hence $f(2^{-(j-\ell_{0})})-f(0)=f_{j}(2^{-(j-\ell_{0})})$ for any j∈N.

We now distinguish two cases. Let us first assume that j∈{j _n ,…,j _n,α }; we have

$$f(2^{-(j-\ell_0)})=2^{-j_n \alpha} \sum_{\ell=j+2}^{j_{n,\alpha}} \ell^{-\varepsilon } \psi(0)\ge2^{-j_n \alpha}((j_{n,\alpha}+1)^{1-\varepsilon }-(j+2)^{1-\varepsilon})$$

Therefore, if j _n ≤j≤j _n,α /2,

$$f(2^{-(j-\ell_{0})})\ge2^{-j_n \alpha}(j_{n,\alpha}+1)^{1-\varepsilon }(1-2^{-(1-\varepsilon)})\ge C' 2^{-j\alpha\varepsilon},$$

whereas if j _n,α /2≤j≤j _n,α ,

$$f(2^{-(j-\ell_{0})})\ge2^{-j_n \alpha}j_{n,\alpha}^{-\varepsilon}\ge j^{-1-\varepsilon}.$$

Gathering these inequalities, we have, for any j∈{j _n ,…,j _n,α },

$$ f(2^{-(j-\ell_0)})\ge C' 2^{-j\alpha\varepsilon}.$$

(28)

Let us now consider the second case, where j∈{j _n,α +1,…,j _n+1−1} for some n∈N. We have

$$f(2^{-(j-\ell_0)})= \Biggl(2^{j_{n+1}(1-\alpha)}\sum_{\ell =j+2}^{j_{n+1}}2^{-\ell}\ell^{-\varepsilon}+ 2^{-j_{n+1}\alpha}\sum _{\ell=j_{n+1}}^{j_{n+1,\alpha}}\ell^{-\varepsilon}\Biggr)\psi(0).$$

If one remarks that

then for any j _n,α +1≤j≤((1−α)+αε)j _n+1, we get

$$ f(2^{-(j-\ell_{0})})\ge C' 2^{j\frac{1-\alpha}{(1-\alpha)+\alpha\varepsilon}}2^{-j}j^{-\varepsilon}= C' 2^{-j\frac{\alpha\varepsilon}{(1-\alpha)+\alpha\varepsilon }}j^{-\varepsilon},$$

(29)

whereas if ((1−α)+αε)j _n+1≤j≤j _n+1−1,

$$ f(2^{-(j-\ell_0)})\ge C' 2^{-j_{n+1}\alpha\varepsilon}\ge C' 2^{-j\frac{\alpha\varepsilon}{(1-\alpha)+\alpha\varepsilon}}.$$

(30)

Inequalities (28), (29) and (30) together imply f∈UI ^β(R ^d). □

1.2 A.2 Necessity of the Logarithmic Correction in the Wavelet Criteria

Let ε,α∈(0,1), β>1 and define (j _n ) _n∈N as

$$j_n={[\beta^n]},$$

for any n∈N. Let us also define the function f _α,β,ε on R as follows,

$$ f_{\alpha,\beta,\varepsilon}(x)=\sum_{n=0}^{\infty} \sum_{j=j_n+1}^{j_{n+1}} \frac{\inf (2^{-j_{n}\alpha},2^{j_{n+1}(1-\alpha)}2^{-j})}{j^\varepsilon} \sin (2^j \pi x).$$

(31)

We first give an estimation of the wavelet coefficients (c _j,k ) of f _α,β,ε .

Proposition 13

Assume that the multiresolution analysis is the Meyer multiresolution analysis. Then for n≥1, any j∈{j _n ,…,j _n+1−1} and any C>0,

$$ \sup_{k\in\mathbf{Z}} |c_{j,k}|\le C\inf (2^{-j_n\alpha },2^{j_{n+1}(1-\alpha)}2^{-j}),$$

(32)

for n sufficiently large.

Proof

Let n∈N and ℓ∈{j _n ,…,j _n+1−1}. By definition of the wavelet coefficients of a bounded function, we have

$$c_{\ell,k}= 2^{\ell}\int_{\mathbf{R}^d} f_{\alpha ,\beta,\varepsilon}(x)\psi(2^{\ell}x-k) \,dx.$$

Since the trigonometric series f _α,β,ε is uniformly converging on any compact,

$$c_{\ell,k}=2^{\ell} \sum_{n=0}^{\infty} \sum_{j=j_n+1}^{j_{n+1}}\frac{\inf(2^{-j_n \alpha},2^{j_{n+1}(1-\alpha )}2^{-j})}{j^\varepsilon} \int_{\mathbf{R}^d} \sin (2^j \pi x)\psi(2^\ell x-k) \,dx,$$

or

that is,

(33)

Since the Meyer wavelet belongs to the Schwartz class, its Fourier transform is symmetric and compactly supported with

$$\mathrm{supp}(\hat{\psi})\subset\biggl[-\frac{8\pi}{3},-\frac{2\pi }{3}\biggr]\cup\biggl[\frac{2\pi}{3},\frac{8\pi}{3}\biggr],$$

the sum in equality (33) contains at most five terms corresponding to

One directly checks that for any n∈N, j∈{j _n ,…,j _n+1−1}, this implies inequality (32). □

Let us now prove the uniform irregularity properties of the functions f _α,β,ε .

Proposition 14

For any β>1 and any (α,ε)∈(0,1)², $f_{\alpha,\beta,\varepsilon}\in \mathit{UI}^{\alpha}_{1-\varepsilon }(\mathbf{R})$.

Proof

Let us remark that it is sufficient to prove that for any ℓ∈N,

$$ f_{\alpha,\beta,\varepsilon} (2^{-\ell})\ge2^{-\alpha\ell} \ell ^{1-\varepsilon}.$$

(34)

Let n ₀∈N and $\ell\in\{j_{n_{0}+1},\ldots,j_{n_{0}+1}\}$. By definition, we have

The classical inequality sin(x)≥(2/π)x valid for any x∈[0,π/2] leads to the following inequality if $j_{n_{0}}+1\le\ell\le j_{n_{0}+1}$,

Let t∈(1,β) such that $\ell=tj_{n_{0}}$, that is $j_{n_{0}}=\ell/t$. We get

$$f_{\alpha,\beta,\varepsilon}(2^{-\ell})\ge 2\inf(\ell^{1-\varepsilon}2^{-\ell\frac{\alpha}{t}},\ell ^{1-\varepsilon} 2^{-\ell(1-\frac{\beta\ell}{t}+\frac{\alpha \beta\ell}{t})}).$$

Since

$$\sup_{t\in[1,\beta]} \max(\alpha/t,1-\beta\ell/t+\alpha\beta \ell/t)\le\alpha,$$

inequality (34) is satisfied for any ℓ∈N. □

Propositions 13 and 14 together imply the following proposition.

Proposition 15

For any (α,ε,β)∈(0,1)²×(1,+∞), the functions f _α,β,ε defined by the relation (31) are uniformly Hölderian, satisfy (8) and belong to $\mathit{UI}^{\alpha}_{1-\varepsilon }(\mathbf{R})$.