Can I make this deduction involving inner product, discrete/continuous convolution and Fourier transform? Let $(\varphi_t)_{t\in\mathbb{Z}}$, where $\varphi_t \in L^2(\mathbb{R})$ for all $t\in \mathbb{Z}$ and
$$ \varphi_t (x) := \varphi(x-t).$$
Furthermore assume that I can write a function $h\in L^2(\mathbb{R})$ in terms of
$$ h(x) = \sum_{t\in\mathbb{Z}} \left\langle f, \varphi_t \right\rangle \varphi_t(x). $$
where $f\in L^2(\mathbb{R})$. Now I do the following deduction:
$$ \left\langle f, \varphi_t \right\rangle = \int_\mathbb{R} f(x) \varphi(x-t) \text{d}x = (f \ast \varphi^\ast)(t), $$
where $\varphi^\ast(x) := \varphi(-x)$. Thus
$$ h(x) = \sum_{t\in\mathbb{Z}} (f \ast \varphi^\ast)(t) \varphi_t(x) = \sum_{t\in\mathbb{Z}} (f \ast \varphi^\ast)(t) \varphi(x-t)= f \ast \varphi^\ast \circledast \varphi(x), $$
where $\circledast$ denots the discrete convolution. Now the Fourier transform turns convolution into multiplication, hence applying on both sides gives me
$$ \widehat h (\xi) = \widehat f(\xi) \cdot |\widehat \varphi (\xi)|^2.$$
I am not sure whether this is correct, because there is this mix between continuous convolution $\ast$ and the discrete convolution $\circledast$. Is the last equality well-defined?
Thanks for any comments!
 A: Let me show the difference between the case in the paper and your case.
For the ordinary convolution, defined by
$$
(f*g)(x) = \int f(y) g(x-y) \, dy
$$
we get
$$
\widehat{f*g}(\xi)
= \int (f*g)(x) e^{-ix\xi} dx
= \int \left( \int f(y) g(x-y) \, dy \right) e^{-ix\xi} dx
= \iint f(y) g(x-y) e^{-ix\xi} \, dx \, dy
\\
= \int f(y) \left( \int g(x-y) e^{-ix\xi} \, dx \right) \, dy
= \int f(y) \left( \int g(z) e^{-i(y+z)\xi} \, dz \right) \, dy
\\
= \int f(y) e^{-iy\xi} \left( \int g(z) e^{-iz\xi} \, dz \right) \, dy
= \hat{f}(\xi) \, \hat{g}(\xi)
.
$$
For the discrete convolution, defined by
$$
(f \circledast g)(x) = \sum_{k\in\mathbb{Z}} f(k) g(x-k)
$$
we instead get
$$
\widehat{f \circledast g}(\xi)
= \int (f \circledast g)(x) e^{-ix\xi} \, dx
= \int \left( \sum_{k\in\mathbb{Z}} f(k) g(x-k) \right) e^{-ix\xi} \, dx
\\
= \sum_{k\in\mathbb{Z}} f(k) \int g(x-k) e^{-ix\xi} \, dx
= \sum_{k\in\mathbb{Z}} f(k) \int g(z) e^{-i(z+k)\xi} \, dz
\\
= \sum_{k\in\mathbb{Z}} f(k) e^{-ik\xi} \int g(z) e^{-iz\xi} \, dz
= \left( \sum_{k\in\mathbb{Z}} f(k) e^{-ik\xi} \right) \hat{g}(\xi)
.
$$
Thus, only one of the factors becomes an ordinary Fourier transform. The other becomes a kind of discrete Fourier transform.
