Show that $\{f_n\}({x}) = \text{sgn}(\sin(2^n\pi{x}))$ is orthonormal on a given interval

Question

This is from the following question :

Show that the set $\{f_n\} \mathbb{_n} _\in \mathbb{_N}$ where $\{f_n\}({x}) = \text{sgn}(\sin(2^n\pi{x}))$ is orthonormal on the interval $[0,1]$

Honestly stuck on figuring out how to compute the following integral:

$$\int_{0}^{1}\text{sgn}(\sin(2^j\pi {x}))\text{sgn}(\sin(2^k\pi {x}))dx = 0$$

I'm fairly sure that when $j = k$, we are left with $\text{sgn}(-)^2$ which we can just assume is going to be $1$ for all cases, but for cases where $j \neq k$ I'm a bit stuck.

Answer 1

Hint : Use the fact that:

$$\text{sgn}\sin (x) = \sum_{l=0}^{\infty}\left(\mathbf 1_{\left[2l\pi, (2l+1)\pi\right]}(x)-\mathbf 1_{\left[(2l+1)\pi, 2(l+1)\pi\right]}(x)\right)$$

Since

$$ 1_{\left[p\pi, (p+1)\pi\right]}(2^j\pi x) = \mathbf 1_{\left[\frac{p}{2^{j}}, \frac{p}{2^{j}} + \frac{1}{2^j}\right]}(x) $$

then for $k < j$

\begin{align} \int_{0}^1 1_{\left[p\pi, (p+1)\pi\right]}(2^j\pi x)1_{\left[m\pi, (m+1)\pi\right]}(2^k \pi x)\mathrm d x &= \int_{0}^1 1_{\left[\frac{p}{2^{j}}, \frac{p}{2^{j}} + \frac{1}{2^j}\right]}(x)1_{\left[\frac{2^{j-k}m}{2^{j}}, \frac{2^{j-k}m}{2^{j}} + \frac{2^{j-k}}{2^j}\right]}(x)\mathrm d x\\ &=\left\{\begin{array}{ll} \frac1{2^j} & \text{if $2^{j-k}m \le p\le 2^{j-k}m + 2^{j-k}$}\\ 0 & \text{otherwise} \end{array}\right. \end{align} \begin{align} \implies \quad \int_{0}^1 \left(1_{\left[2p\pi, (2p+1)\pi\right]}(2^j\pi x)-1_{\left[(2p+1)\pi, 2(p+1)\pi\right]}(2^j\pi x)\right)1_{\left[m\pi, (m+1)\pi\right]}(2^k \pi x)\mathrm d x &= 0&&\forall p \in \mathbb N \end{align}

After summing you will find what you are looking for.

Answer 2

Note that $f_n$ is periodic with period $2^{-n}$, i.e., $f_n(x)=f_n(x+T)$ if $T$ is a multiple of $2^{-n}$. Even more: $f_n(x)=-f_n(x+T)$ if $T$ is an odd multiple of $2^{-n-1}$. If $1\le j<k$ and $T=2^{-k-1}$, conclude that $$ \int_0^af_j(x)f_k(x)\,\mathrm dx=-\int_T^{a+T}f_j(x)f_k(x)\,\mathrm dx$$

Using this and that $1$ is an even multiple of $T$, $$\int_0^1=\int_0^T+\int_T^1 = \int_1^{1+T}+\int_T^1=\int_T^{1+T}=-\int_0^1.$$