Lévy–Ciesielski construction of a Brownian motion

Question

The family $\left\{h_k(\cdot)\right\}_{k=0}^{\infty}$ of Haar functions are defined for $0 \leq t \leq 1$ as follows: $$ \begin{gathered} h_0(t):=1 \quad \text { for } 0 \leq t \leq 1 \\ h_1(t):=\left\{\begin{array}{lr} 1 & \text { for } 0 \leq t \leq \frac{1}{2} \\ -1 & \text { for } \frac{1}{2}<t \leq 1 \end{array}\right. \end{gathered} $$ If $2^n \leq k<2^{n+1}, n=1,2, \ldots$, we set $$ h_k(t):=\left\{\begin{array}{l} 2^{n / 2} \text { for } \frac{k-2^n}{2^n} \leq t \leq \frac{k-2^n+1 / 2}{2^n} \\ -2^{n / 2} \text { for } \frac{k-2^n+1 / 2}{2^n}<t \leq \frac{k-2^n+1}{2^n} \\ 0 \text { otherwise. } \end{array}\right. $$ For $k=0,1,2, \ldots$, $$ s_k(t):=\int_0^t h_k(s) d s \quad(0 \leq t \leq 1) $$ is the $k$-th Schauder function.

This offers the premises for the Lévy–Ciesielski construction of a Brownian motion. In fact, we define $$ W(t):=\sum_{k=0}^{\infty} A_k s_k(t) \quad \quad \quad\quad \quad \quad\quad \quad \quad (1) $$ for times $0 \leq t \leq 1$, where the coefficients $\left\{A_k\right\}_{k=0}^{\infty}$ are independent, $N(0,1)$ random variables defined on some probability space. (I took this part from "Evans, Lawrence C. An introduction to stochastic differential equations. Vol. 82. American Mathematical Soc., 2012.")

Anyway, I have some doubts.

1. The first is, if $2^n \leq k<2^{n+1}, n=1,2, \ldots$ why the sum in (1) is from $0$ to $\infty$?
2. The system $\{h_k\}_{k\in\mathbb Z}=\{h_{n,k}\}_{n,k\in\mathbb Z}$ should be the Haar system (Haar wavelets), where $h_1$ is the Haar mother wavelet. Why don't we have also the sum over n in (1)?

Answer 1

In the usual notations for wavelets, were we have two independent indexes, $n$ for dilatations and $k$ for translations, and we set $\varphi_{n,k}(x) := 2^{n/2}\phi(2^{-n}x-k)$ for every $n \in \mathbb{Z}$, $k \in \mathbb{Z}$ and $x \in \mathbb{R}$.

Here, we use an orthonormal basis of $L^2([0,1]$, not $L^2(\mathbb{R})$. Given $n \ge 0$, we have only finitely many values of $k$ to consider. The choice which has be made is to link the two indexes by setting $2^n \le k < 2^{n+1}$ instead of $0 \le k < 2^n$. Therefore, when $k \ne 0$, $n$ is a function of $k$, namely $n = \lfloor \log_2 k \rfloor$. Therefore, a summation over all $k$ in $\mathbb{N}$ is sufficient.