Suppose $\left \{ V_{j} ; j\in \mathbb{Z} \right \}$ is a multiresolution analysis with scaling function $\varphi$ . then the following scaling relation hold: $ \varphi (x)=\sum_{k\in \mathbb{Z}} p_{k}\varphi (2x-k) $ Where $P_{k}= 2\int_{-\infty }^{\infty}\varphi (x)\overline{\varphi (2x-k)}dx $ Moreover, we also have (*) $\varphi (2^{j-1}x-l)=\sum_{k\in \mathbb{Z}}p_{k-2l}\varphi (2^jx-k) $ And (**) $\sum p_{k-2l} \overline{p}_{k}= 2\delta_{l0}$
Now, how can I prove (٭) and (**)?
Let be $$ \varphi (x)=\sum_{k\in \mathbb{Z}} p_{k}\varphi (2x-k)\tag 1 $$ where $$ p_{k}= 2\int_{-\infty }^{\infty}\varphi (x)\overline{\varphi (2x-k)}\operatorname{d}!x=2\langle \varphi (x), \varphi (2x-k)\rangle\tag 2 $$ being $\langle\cdot, \cdot\rangle$ the scalar product in $L^2$.
Substituting $x$ with $2^{j-1}x-l$ in (1), we have $$ \begin{align} \varphi (2^{j-1}x-l)&=\sum_{k\in \mathbb{Z}} p_{k}\varphi \left(2(2^{j-1}x-l)-k\right)\\ &=\sum_{k\in \mathbb{Z}} p_{k}\varphi (2^{j}x-(k+2l))\\ &=\sum_{\nu\in \mathbb{Z}} p_{\nu-2l}\varphi (2^{j}x-\nu)\qquad (k+2l=\nu) \end{align} $$ and finally, changing the dummy index $\nu\to k$, $$\color{red}{(\star)}\qquad\color{blue}{ \varphi (2^{j-1}x-l)=\sum_{k\in \mathbb{Z}} p_{k-2l}\varphi (2^{j}x-k)}. $$
The function $\varphi$ belongs to $V_0$ and the set $\{\varphi(x-2k),k\in\Bbb Z\}$ is an orthonormal basis for $V_0$. So we have $$ \langle \varphi (x-m), \varphi (x-l)\rangle=\delta_{lm}\tag 3 $$ From the (1), substituting $x\to x-l$, we have $$ \varphi (x-l)=\sum_{k\in \mathbb{Z}} p_{k}\varphi \left(2x-(k+2l)\right) $$ Letting $m=0$, we have $$\begin{align} \delta_{l0}&= \langle \varphi (x), \varphi (x-l)\rangle\\ &=\left\langle \varphi (x),\sum_{k\in \mathbb{Z}} p_{k}\varphi\left(2x-(k+2l)\right)\right\rangle\\ &=\sum_{k\in \mathbb{Z}} p_{k}\underbrace{\left\langle \varphi (x),\varphi\left(2x-(k+2l)\right)\right\rangle}_{=\frac{1}{2}\bar{p}_{k+2l}\;\text{using the (2)}}\\ &=\sum_{k\in \mathbb{Z}} p_{k}\frac{1}{2}\bar{p}_{k+2l}\qquad\text{(put }\nu=k+2l)\\ &=\frac{1}{2}\sum_{k\in \mathbb{Z}} p_{\nu-2l}\bar{p}_{\nu} \end{align} $$ and finally replacing the dummy variable $\nu\to k$ $$\color{red}{(\star\star)}\qquad\color{blue}{ \sum_{k\in \mathbb{Z}} p_{k-2l}\bar{p}_{k}=2\delta_{l0}.} $$
