Let $\{ \psi_{j,k}(t)\}$ haar system. How to prove that it is orthogonal? 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. 
 A: 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:


*

*$m<2^{l-j}k$. The product is immediately zero, since the intervals do not overlap

*$2^{l-j}k\le m <2^{l-j}k+2^{l-j-1}$. 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.

*$2^{l-j}k+2^{l-j-1}\le m<2^{l-j}(k+1)$. 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.

*$m\ge 2^{l-j}(k+1)$. The product is zero, since the intervals do not overlap.

A: $(j_1,k_1)\ne(j_2,k_2)$ 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.
