# 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?

• Did you turn the discrete convolution into a product when taking the transform? Nov 25, 2021 at 16:55
• @md2perpe yes, and the Fourier transform of $\varphi^\ast$ is the complex conjugate, so $\widehat{\varphi^\ast \circledast \varphi} = | \varphi |^2$ Nov 26, 2021 at 7:40
• But the transform is taken in the variable $x$ and the factor $(f \ast \varphi^\ast)(t)$ doesn't even contain $x$. Nov 26, 2021 at 8:27
• 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? Nov 26, 2021 at 9:24
• 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. Nov 26, 2021 at 10:47

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.