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Suppose $ \phi \in C_0( \mathbb{R}) $ (compact support) satisfy the scaling relation $$ \phi(x) = \sum_{k \in \mathbb{Z}} p_k \phi(2x-k) , $$ with $$ p_k = 2^{1/2} \int_{- \infty}^\infty \phi(x) \overline{\phi(2x-k)} dx.$$ Let $ a = \inf\{x | \phi(x) \neq 0 \}$ and $ b = \sup\{x | \phi(x) \neq 0 \}$, show that $ a,b \in \mathbb{Z}.$

OK, so what a need to prove is that if $ \overline{\{ x| \phi(x) \neq 0\}} = [a,b]$, show that $a, b \in \mathbb{Z}$.

My toughts so far:

For fixed $x$, let $u = \lceil 2x-a \rceil$ and $l = \lfloor 2x - b \rfloor$, then $$ \phi(x) = \sum_{k = l}^u p_k \phi(2x-k) $$

Especially in $x=n \in \mathbb{Z}$ we would like to show that $u = 2x-a $ and $l = 2x - b $, but I'm pretty blank on where to start. Any hint?

Thanks

Henrik Finsberg
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1 Answers1

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Each rescaled function $\phi_k(x)=\phi(2x-k)$ satisfies $$\inf\{x|\phi_k(x)\ne 0\}=\frac{a+k}{2},\quad \sup\{x|\phi_k(x)\ne 0\}=\frac{b+k}{2} \tag1$$ Let $m=\min\{k:p_k\ne 0\}$ and $n=\max\{k:p_k\ne 0\}$. Using (1), you should be able to prove that $$\inf\left\{x:\sum_{k \in \mathbb{Z}} p_k \phi_k(x)\ne 0\right\} = \frac{a+m}{2}$$ Hence $a=\frac{a+m}{2}$, from where the conclusion about $a$ follows. Similarly for $b$.

40 votes
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