# How to show that $u_l(0)= \frac{1}{\sqrt{2}}$, $u_l(1)= \frac{1}{\sqrt{2}}$ and $u_l(n)= 0$ for $2 \leq n \leq (\frac{N}{2^{l-1}}-1$ $\dots$

Suppose that $$u_1 = \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,0,\ldots,0\right)\quad\text{and}\quad v_1 = \left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0,0,\ldots,0\right)$$ and $$u_l(n) = \sum_{k=0}^{2^{l-1}-1} u_1\left(n+ \frac{kN}{2^{l-1}}\right)\quad\text{and}\quad v_l(n) = \sum_{k=0}^{2^{l-1}-1} v_1\left(n+ \frac{kN}{2^{l-1}}\right).$$ How to show that \begin{align*} u_l(0)&= \frac{1}{\sqrt{2}},\\ u_l(1)&= \frac{1}{\sqrt{2}},\\ u_l(n)&= 0 &\quad&\text{for $2 \leq n \leq \frac{N}{2^{l-1}}-1$,}\\ v_l(0)&= \frac{1}{\sqrt{2}},\\ v_l(1)&= -\frac{1}{\sqrt{2}},\\ v_l(n)&= 0&&\text{for $2 \leq n \leq \frac{N}{2^{l-1}}-1$.} \end{align*}

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Where are you stuck? –  Aryabhata Dec 30 '10 at 18:46
Are $u_1,v_1$ in $\mathbb{R}^N$? And apparently $2^{l-1}$ has to divide $kN$ for $u_1(n+\frac{kN}{2^{l-1}})$ to make sense. The problem is not clear at all. –  TCL Dec 30 '10 at 18:49
I’m stuck at following $u_l(0)= \sum_{k=0}^{2^{l-1}-1} u_1(0+ \frac{kN}{2^{l-1}}) = ? = \frac{1}{\sqrt{2}}$ –  alvoutila Dec 30 '10 at 19:09
What is $N$? What does $u_1(\cdot)$ mean? Does $u_1(k)$ mean the $k$th coordinate of $u_1$? Please clarify. –  William Dec 30 '10 at 20:54
$N$ means that “it” is $\frac{N}{2^{l-1}}$-periodic( one must assume that $N$ is divisible by $2^p$. $u_1$ and $v_1$ form first stage Haar basis. Yes, $u_1(k)$ mean the $k$:th coordinate of $u_1$. Thus this is my clarification. –  alvoutila Dec 30 '10 at 22:27