Wavelet-Type Expansion of the Generalized Rosenblatt Process and Its Rate of Convergence

Abstract

Pipiras introduced in the early 2000s an almost surely and uniformly convergent (on compact intervals) wavelet-type expansion of the classical Rosenblatt process. Yet, the issue of estimating, almost surely, its uniform rate of convergence remained an open question. The main goal of our present article is to provide an answer to it in the more general framework of the generalized Rosenblatt process, under the assumption that the underlying wavelet basis belongs to the class due to Meyer. The main ingredient of our strategy consists in expressing in a non-classical (new) way the approximation errors related with the approximation spaces of a multiresolution analysis of $L^2({{\mathbb {R}}}^2)$. Such a non-classical expression may also be of interest in its own right.

Appendix

Lemma 4.1

For every $(x,y)\in {{\mathbb {R}}}_+^2$, one has

$$\begin{aligned} \log (3+x+y)\le \log (3+x)\log (3+y). \end{aligned}$$

(4.1)

Moreover, for any fixed positive real number T there exists a constant $c>0$ such that, for all $x\in {\mathbb {R}}_+$, the following inequality holds:

$$\begin{aligned} \log (3+x+2^xT)\le c (1+x). \end{aligned}$$

(4.2)

The proof of Lemma 4.1 is standard and easy this is why it has been omitted.

Lemma 4.2

For any fixed real number $L> 1$, there exists a constant $c>0$ such that, for all $j\in {{\mathbb {Z}}}$ and for each $s\in {{\mathbb {R}}}$, one has:

$$\begin{aligned} \sum _{k\in {{\mathbb {Z}}}}\frac{\sqrt{\log (3+|j|+|k|)}}{(3+|2^js-k|)^L}\le c\,\sqrt{\log \big (3+|j|+ 2^j |s| \big )}. \end{aligned}$$

(4.3)

Proof of Lemma  4.2

Setting $m=k-\lfloor 2^js\rfloor$, where $\lfloor 2^js \rfloor$ denotes the integer part of $2^j s$, and using the triangle inequality and (4.1), one obtains that

$$\begin{aligned} \sum _{k\in {{\mathbb {Z}}}}\frac{\sqrt{\log (3+|j|+|k|)}}{(3+|2^js-k|)^L}&=\sum _{m\in {{\mathbb {Z}}}}\frac{\sqrt{\log \big (3+|j|+|m+\lfloor 2^js\rfloor |)}}{(3+|2^js-\lfloor 2^js\rfloor -m|)^L}\\&\le \sqrt{\log \big (3+|j|+ 2^j |s| \big )}\sum _{m\in {{\mathbb {Z}}}}\frac{\sqrt{\log (4+|m|)}}{(3+|2^js-\lfloor 2^js\rfloor -m|)^L}. \end{aligned}$$

Then, noticing that

$$\begin{aligned} 3+|2^js-\lfloor 2^js\rfloor -m|\ge 2+|m|, \end{aligned}$$

one gets that

$$\begin{aligned} \sum _{k\in {{\mathbb {Z}}}}\frac{\sqrt{\log (3+|j|+|k|)}}{(3+|2^js-k|)^L}\le c\,\sqrt{\log (3+|j|+2^j |s| \big )}, \end{aligned}$$

where the constant

$$\begin{aligned} c:=\sum _{m\in {{\mathbb {Z}}}}\frac{\sqrt{\log (4+|m|)}}{(2+|m|)^L}<+\infty . \end{aligned}$$

$\square$

Lemma 4.3

For each fixed real number $L> 1$, there exists a constant $c>0$ such that, for every $t\in {{\mathbb {R}}}_+$, for all $s\in [0,t]$ and for any $j\in {{\mathbb {Z}}}_+$, one has

$$\begin{aligned} \sum _{k\in D_j^3(t)}\frac{\sqrt{\log (3+j+|k|)}}{(3+|2^js-k|)^L}\le c (j+1)2^{-j (L-1)(1-a)}\sqrt{\log (3+t)}\,, \end{aligned}$$

(4.4)

where $D_j^3(t)$ is defined through (2.14) and (2.11).

Proof of Lemma 4.3

In view of (2.14) and (2.11), one has $D_j^3(t)=D_j^{3,+}(t)\cup D_j^{3,-}(t),$ where $D_j^{3,+}(t)$ and $D_j^{3,-}(t)$ are the two disjoint sets defined as:

$$\begin{aligned} D_j^{3,+}(t)=\left\{ k\in {{\mathbb {Z}}}, \ k>2^jt+2^{j(1-a)}\right\} \end{aligned}$$

and

$$\begin{aligned} D_j^{3,-}(t)=\left\{ k\in {{\mathbb {Z}}}, \ k<-2^{j(1-a)}\right\} . \end{aligned}$$

Thus, one gets that

$$\begin{aligned} \sum _{k\in D_j^3(t)}\frac{\sqrt{\log (3+j+|k|)}}{(3+|2^js-k|)^L}= \sum _{k>2^jt+2^{j(1-a)}}\frac{\sqrt{\log (3+j+k)}}{(3+|2^js-k|)^L} +\sum _{k<-2^{j(1-a)}}\frac{\sqrt{\log (3+j+|k|)}}{(3+|2^js-k|)^L}.\nonumber \\ \end{aligned}$$

(4.5)

Let us now provide an appropriate upper bound for the first term in the right-hand side of (4.5). One denotes by $\lfloor \cdot \rfloor$ the integer part function. Using the change of variable $m=k-\lfloor 2^jt\rfloor$, the triangle inequality, (4.1), the inequality $\lfloor 2^jt\rfloor -2^js >-1$, the inequality $\log (3+m)\le 2+m$, and the fact that $x\mapsto (1+x)^{-L}\sqrt{\log (2+x)}$ is a decreasing function on ${{\mathbb {R}}}_+$, one obtains that

$$\begin{aligned} \sum _{k>2^jt+2^{j(1-a)}}\frac{\sqrt{\log (3+j+k)}}{(3+|2^js-k|)^L}&=\sum _{k>2^jt+2^{j(1-a)}}\frac{\sqrt{\log (3+j+k)}}{(3+k-2^js)^L} \nonumber \\&=\sum _{m>2^jt-\lfloor 2^jt\rfloor +2^{j(1-a)}}\frac{\sqrt{\log (3+j+\lfloor 2^jt\rfloor +|m|)}}{(3+\lfloor 2^jt\rfloor -2^js+m)^L} \nonumber \\&\le \sqrt{\log (3+j+2^j t)}\sum _{m>2^{j(1-a)}}\frac{\sqrt{\log (3+m)}}{(2+m)^L}\nonumber \\&\le \sqrt{\log (3+j+2^j t)}\int _{2^{j(1-a)}}^{+\infty }\frac{\sqrt{\log (2+x)}}{(1+x)^{L}}\,dx\nonumber \\&\le c_1\sqrt{(j+1)\log (3+t)}\,\,2^{-j(L-1)(1-a)}\sqrt{\log \big (2+2^{j(1-a)}\big )}\nonumber \\&\le c_2 (j+1) 2^{-j(L-1)(1-a)}\sqrt{\log (3+t)}\, , \end{aligned}$$

(4.6)

where $c_1$ and $c_2$ are two positive finite constants not depending on j, t, s and a. Similarly to (4.6), it can be shown that

$$\begin{aligned} \sum _{k<-2^{j(1-a)}}\frac{\sqrt{\log (3+j+|k|)}}{(3+|2^js-k|)^L}\le c_3 2^{-j(L-1)(1-a)}\sqrt{(j+1)\log (3+j)}\,, \end{aligned}$$

(4.7)

where $c_3$ is a positive finite constant not depending on j, t, s and a. Finally, putting together (4.5), (4.6), (4.7) and (4.2), it follows that (4.4) holds. $\square$