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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!

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    $\begingroup$ Did you turn the discrete convolution into a product when taking the transform? $\endgroup$
    – md2perpe
    Nov 25, 2021 at 16:55
  • $\begingroup$ @md2perpe yes, and the Fourier transform of $\varphi^\ast$ is the complex conjugate, so $\widehat{\varphi^\ast \circledast \varphi} = | \varphi |^2$ $\endgroup$
    – stish
    Nov 26, 2021 at 7:40
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    $\begingroup$ But the transform is taken in the variable $x$ and the factor $(f \ast \varphi^\ast)(t)$ doesn't even contain $x$. $\endgroup$
    – md2perpe
    Nov 26, 2021 at 8:27
  • $\begingroup$ Thanks for pointing that out! I got the idea from this paper from proof of Theorem 1, where I basically replaced the integral with the sum. Is it false in here, too? $\endgroup$
    – stish
    Nov 26, 2021 at 9:24
  • $\begingroup$ The formula in the paper is correct. But in you case you have a discrete sum, not a continuous sum (integral), and you will then not get an ordinary continuous Fourier transform but a kind of discrete transform. $\endgroup$
    – md2perpe
    Nov 26, 2021 at 10:47

1 Answer 1

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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.

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