Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I would like to know how to show that the function

$r_n(t)=\mbox{sgn}\big(\sin(2^n \pi t)\big)$, sgn = sign function

is an orthonormal system but not an orthonormal basis from $L_2([0,1])$

share|improve this question
What have you tried? –  Chris Eagle Nov 9 '12 at 17:29
I pick to different variables i,k and show that $<r_i (t),r_k (t)>=0$, so I can take the integral from 0 to 1 over $r_i (t)*r_k (t)$, but I do not know how to show that this is 0. –  Alexander Nov 9 '12 at 17:51

1 Answer 1

up vote 0 down vote accepted

To compute $\langle r_n, r_m\rangle$ with $n<m$, say, observe that $$\langle r_n, r_m\rangle=\int_0^1 \operatorname{sgn}(\sin(2^n\pi t))\operatorname{sgn}(\sin(2^m\pi t))\,dt\\=\sum_{k=0}^{2^n-1}\int_{k2^{-n}}^{(k+1)2^{-n}} \operatorname{sgn}(\sin(2^n\pi t))\operatorname{sgn}(\sin(2^m\pi t))\,dt$$ In each summand, $\operatorname{sgn}(\sin(2^n\pi t))$ is constant and $\operatorname{sgn}(\sin(2^m\pi t))$ runs over $2^{m-n}$ full periods. Thus each summand is zero.

Remark: What is still left for you to show that this orthnormal system fails to be a basis?

share|improve this answer
Thanks for your answer, but I am not sure if I understand the last part. May you could explain a little bit better why $sgn(sin(2^n \pi t))$ is constant and $sgn(sin(2^m \pi t))$ runs over $2^{m-n}$ full periods. To show that it is a basis, it is enough to show that the orthogonal complement is 0, so I need to find an element of L2[0,1] such that the inner product of this element with $r_n$ is not 0. –  Alexander Nov 10 '12 at 16:05
If $t$ runs from $k2^{-n}$ to $(k+1)2^{-n}$, then $2^n\pi t$ runs from $k\pi$ to $(k+1)\pi$, that is from one zero of the sine to the next. Between these zeroes (i.e. ignoring the interval end points) i i therefore postive throughout or negative thoughout, hence th signum is constantly $+1$ or conantly $-1$. –  Hagen von Eitzen Nov 11 '12 at 10:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.