I'm trying to solve the part (b) of the following exercise:
Let consider the space $L^2(0,1)$ and define $r_0(t)=1$ and $$r_n(t)=\sum_{k=1}^{2^n}(-1)^{k-1}\xi_{\left[\frac{k-1}{2^n},\frac{k}{2^n}\right]},\quad\forall n\in\Bbb{N},$$ where $\xi_E$ denotes the characteristic function of a set $E$.
(a) Show that $r_n(t)=\text{sgn}(\sin(2^n\pi t))$, for all $t\in[0,1]$.
(b) Show that $\{r_n(t)\}_{n=0}^{\infty}$ is an orthonormal set in $L^2(0,1)$ but that it is not complete.
The part (a) is ok, also that $\langle r_n,r_n\rangle=1$, for all $n$, but I don't see how to conclude that $\langle r_n,r_m\rangle=0$, when $m\ne n$.
Thanks for any help.
These are the Rademacher functions.
For convenience, extend the $r_n$ so they are ${1\over 2^n}$-periodic on the real line. (The values at a countable number of points can be ignored, if the slight overlap in definition is bothersome.)
Note that if $m >n$, $\int_{[{k-1 \over 2^n}, {k \over 2^n}]} r_m(t) dt = \int_{[{k-1 \over 2^n}, {k \over 2^n}]} r_m(t) r_n(t) dt = 0$. Sum these to get the desired result, $\langle r_m, r_n \rangle = 0$.
Take $f=1_{[{1 \over 4}, {3 \over 4}]}-{1 \over 2}$. Note that $\langle r_n, f \rangle = 0 $ for all $n$.