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I would like to know how to show that the functions $$r_n(t)=\operatorname{sgn}\big(\sin(2^n \pi t)\big)$$ (where $\operatorname{sgn}$ is the sign function) form an orthonormal system but not an orthonormal basis from $L_2([0,1])$.

Progress: I pick two different variables $i,k$ and show that $\langle r_i (t),r_k (t)\rangle =0$, so I can take the integral $\int_0^1 r_i (t) r_k (t)\,dt$, but I do not know how to show that this is 0.

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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
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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?

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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

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