Let $\phi$ be a compactly supported continuous function such that $\phi (x)=0$ outside of some finite interval.
$\phi$ satisfies the following refinable equation: $$\phi (x)=\sum_{k=0}^{M}c_k\phi (2x-k)$$ and $\hat{\phi}(0)=1$ . $\hat{\phi}(\omega)$ is the Fourier transform of $\phi (x)$.
I've shown that $\sum_{k=0}^{M}c_k=2$
Show that {$\phi (x-n), n\in \Bbb{Z}$} is an orthonormal sequence in $L^2(\Bbb{R})$ only if $$\sum_{k\in \Bbb{Z}} |\hat{\phi}(\omega + 2k\pi)|^2 = 1, \forall \omega \in \Bbb{R}$$
The question is asking for a biconditional proof so we have to prove both directions. For the forward direction, Since {$\phi (x-n), n\in \Bbb{Z}$} is an orthonormal sequence, we know that $$ \int_{-\infty}^{\infty} \phi (x-n)\phi (x-m) dx = \left\{\begin{array}{ll} 1 & : n=m\\ 0 & : n\neq m \end{array} \right.\ \ \ (*)$$
We have that $$\int_{-\infty}^{\infty} \phi (x-n) e^{-i\omega (x-n)} dx = \hat{\phi}(\omega)\\\Rightarrow \int_{-\infty}^{\infty} \phi (x-n) e^{-i\omega x} dx = e^{-i\omega n}\hat{\phi}(\omega)$$
I'm not sure what to do next. Can I square both sides to try and get to the form of $(*)$ for the case $n=m$? Or am I not doing this correctly? Also I replaced $\omega$ with $\omega + 2k\pi$ but still nothing comes to mind about how to get to the required identity. Will greatly appreciate if someone can work out the reverse direction too, ie given $$ \sum_{k\in \Bbb{Z}} |\hat{\phi}(\omega + 2k\pi)|^2 = 1, \forall \omega \in \Bbb{R}$$ show that {$\phi (x-n), n\in \Bbb{Z}$} is an orthonormal sequence.
You essentially want to show $$ g[m] = \int_{\mathbb{R}} \phi(t)\phi(t-m) dt = \delta[m] $$ Note that the identity is defined only for $m \in \mathbb{Z}$. Consider taking the discrete-time Fourier transform of both sides:
$$ \sum_{n\in\mathbb{Z}} g[n] e^{-inw} = 1 $$
Now, write $g(t) = (\phi * \bar{\phi})(t)$ where $\bar{\phi} = \phi(-t)^*$ (the time-reversed conjugate). Using the Poisson summation formula,
\begin{align*} \sum_{n\in\mathbb{Z}} g[n] e^{-inw} &= \sum_{n\in\mathbb{Z}} \hat{g}(w+2\pi n) \\ &= \sum_{n \in \mathbb{Z}} |\hat{\phi}(w+2\pi n)|^2 \end{align*} Note that I'm using the Fourier transform definition: $$ \hat{f}(w) = \int_{\mathbb{R}} f(t) e^{-iwt} dt $$