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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 $L_2([0,1])$

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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
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1 Answer

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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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