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Let $$\psi_{j,k}(t)=\begin{cases} 2^{j/2}, & 2^{-j}k < t < 2^{-j}(k+1/2)\\-2^{j/2} ,& 2^{-j}(k+1/2) < t < 2^{-j}(k+1) \\ 0, & \textrm{otherwise.} \end{cases}$$ How to prove that it is orthogonal? In other words how to prove that $\langle \psi_{j_1,k_1}(t), \psi_{j_2, k_2}(t) \rangle = 0$, $j_1<j_2$ and $k_1<k_2$. I have already proved that it's norm is one $|| \psi_{j,k} ||_2 = 1$, but now I'm stuck in this orthogonality issue, because it involves many nontrivial computations.

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$(j_1,k_1)\ne(j_2,k_2)$ means (a) $j_1\ne j_2$ or (b) $j_1=j_2$ and $k_1\ne k_2$. In each of these cases the statement follows immediately by looking at the graphs of two such $\psi$'s.

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But now is one problem. if it is that easy why then you have not any proofs on this in wikipedia for example. I have found none proofs on the internet. Does anybody know where to find one? –  laovultai Dec 14 '10 at 13:08
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(a) If $j_1<j_2$ then $\psi_{j_1,k_1}$ is constant on the support of $\psi_{j_2,k_2}$, and $\psi_{j_2,k_2}$ integrates to 0. (b) The supports of $\psi_{j_1,k_1}$ and $\psi_{j_1,k_2}$ are disjoint. –  Christian Blatter Dec 14 '10 at 15:23
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If you are working in $L_2(0,1)$ space it is then quite easy. The scalar product then is the integral \begin{align*} \int_0^1\psi_{j_1,k_1}(t)\psi_{j_2,k_2}(t)dt \end{align*}

Rewrite $\psi_{j,k}(t)$ as:

\begin{align*} \psi_{j,k}(t)=2^j1\left\{t\in\left(\frac{k}{2^j},\frac{k+1/2}{2^j}\right)\right\}- 2^{-j}1\left\{t\in\left(\frac{k+1/2}{2^j},\frac{k+1}{2^j}\right)\right\} \end{align*} then \begin{align*} \psi_{j,k}(t)\psi_{l,m}(t)&=2^{j+l}1\left\{t\in\left(\frac{k}{2^j},\frac{k+1/2}{2^j}\right)\cap\left(\frac{m}{2^l},\frac{m+1/2}{2^l}\right)\right\}\\ &-2^{j+l}1\left\{t\in\left(\frac{k}{2^j},\frac{k+1/2}{2^j}\right)\cap\left(\frac{m+1/2}{2^l},\frac{m+1}{2^l}\right)\right\}\\ &-2^{j+l}1\left\{t\in\left(\frac{k+1/2}{2^j},\frac{k+1}{2^j}\right)\cap\left(\frac{m}{2^l},\frac{m+1/2}{2^l}\right)\right\}\\ &+2^{j+l}1\left\{t\in\left(\frac{k+1/2}{2^j},\frac{k+1}{2^j}\right)\cap\left(\frac{m+1/2}{2^l},\frac{m+1}{2^l}\right)\right\} \end{align*}

Now if $j<l$, we have that $2^{l-j-1}\ge 1$ and we need to investigate the following cases:

  1. $m<2^{l-j}k$. The product is immediately zero, since the intervals do not overlap
  2. $2^{l-j}k\le m <2^{l-j}k+2^{l-j-1}$. Then the 3rd and 4th terms in the sum are zero and the product is equal to $2^j\psi_{l,m}$, whose integral is zero.
  3. $2^{l-j}k+2^{l-j-1}\le m<2^{l-j}(k+1)$. Then the 1st and 2nd terms in the sum are zero and the product is equal to $-2^{j}\psi_{l,m}$, whose integral is again zero.
  4. $m\ge 2^{l-j}(k+1)$. The product is zero, since the intervals do not overlap.
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What does this notion means $2^j 1{(\frac{k}{2^j},\frac{k+1/2}{2^j})}−2^{−j} 1{(\frac{k+1/2}{2^j} ,\frac{k+1}{2^j} )}$? It is quite new to me. –  laovultai Dec 14 '10 at 13:34
    
The indicator function. If $A$ is the interval $1_{A}(t)=1,$ if $t\in A$ and zero otherwise. –  mpiktas Dec 14 '10 at 13:35
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