The following section concerns the existence of the mollifier used in the proof of the [[Fourier Inversion Theorem]]. > [!theorem] > > Let $d \in \nat$, then there exists $\phi \in L^1(\real^d)$ such that > 1. $\phi$ is bounded, continuous at $0$, with $\phi(0) = 1$. > 2. $\wh \phi$ is bounded, continuous at $0$, with $\wh \phi(0) = 1$. > 3. $\phi, \wh \phi \in L^1$ with $\int \phi = \int \wh \phi = 1$. > > *Proof*. Let $\phi(x) = e^{-\pi \abs{x}^2}$, and $1 \le j \le d$, then > $ > \partial^{e_j}_\xi(\wh \phi)(\xi) = \braks{-2\pi i x^{e_j}e^{-\pi \abs x^2}}^\wedge(\xi) = i (\wh{\partial^{e_j}\phi})(\xi) = -2\pi \xi_j \wh \phi(\xi) > $ > where by the product rule, > $ > \partial^{e_j}_\xi\braks{e^{\pi \abs \xi^2}\wh \phi(\xi)} = 2\pi \xi_je^{\pi \abs \xi ^2}\wh \phi(\xi) - 2 \pi \xi_j e^{\pi \abs \xi^2}\wh \phi(\xi) = 0 > $ > Hence $e^{\pi \abs \xi^2} \wh \phi(\xi)$ is a constant. Given the Gaussian integral, > $ > \wh f(\xi) = \prod_{j = 1}^d \braks{\int_{\real} e^{-\pi x_j^2 - 2 \pi i \xi_jx_j}dx_j} = \prod_{j = 1}^d \braks{e^{-\pi \xi_j^2}} = e^{-\pi \abs \xi^2} > $