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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.

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Mathecm
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1 Answers1

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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 $n$.

copper.hat
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  • The symbol $1_E$ is the characteristic function of $E$? I never used this symbol before... – Mathecm Oct 17 '19 at 10:41
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    Yes. It it a fairly standard symbol. – copper.hat Oct 17 '19 at 11:38
  • I was doing the math here, but I can’t see how $r_n$ converges to $f$. Because $|r_n-f|^2=\langle r_n,r_n\rangle-\langle r_n,f\rangle-\langle f,r_n\rangle+\langle f,f\rangle$, wich is not zero... – Mathecm Oct 19 '19 at 12:47
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    The collection is not complete. F Isa non zero function that is orthogonal to all rn so no combination of rn can ever converge to f. That is the whole point. – copper.hat Oct 19 '19 at 13:00
  • Now I understand what you tell me, you use the fact that $\langle f,r_n\rangle=0$, with $f≠0$. This is a way to define complete orthonormal that I didn't know, I just learn right now ;D. I was thinking about "complete" in the sense of "Cauchy implying convergence". Thanks a lot! – Mathecm Oct 20 '19 at 02:50