> [!theoremb] Fourier Inversion Theorem > > Let $f \in L^1(\real^d)$ be an [[Integrable Function|integrable function]]. Define the **inverse Fourier transform** > $ > \check f(x) = \int_{\real^d} f(\xi)e^{2\pi i \angles{\xi, x}}d\xi > $ > then for any $g \in L^1(\real^d)$ such that the [[Fourier Transform|Fourier transform]] $\wh g \in L^1(\real^d)$ is also integrable, > - There exists a [[Continuity|continuous]] function $g' \in C(\real^d)$ such that $g = g'$ [[Almost Everywhere|a.e.]] > - $(\wh g)^\vee = (\check g)^\wedge = g'$. > > *Proof*. Let $\phi \in L^1(\real^d)$ such that[^1] > > 1. $\phi$ is bounded, continuous at $0$, with $\phi(0) = 1$. > 2. $\wh \phi \in L^1(\real^d)$ with $\int \wh \phi = 1$. > > Let $x \in \real^d$, and $t > 0$. Define > $ > \phi^x_t(\xi) = e^{2\pi i \angles{\xi, x}} \cdot \phi(t\xi) > $ > then $\wh{\phi_t^x} = \tau_{x} \wh{\phi_t^0} = \tau_x(\wh \phi)_t$, where > $ > \wh \phi_t^0(y) = \int_{\real^d} \phi(t\xi)e^{-2\pi i \angles{\xi, y}}d\xi = \frac{1}{t^d}\int_{\real^d}\phi(\xi)e^{-2\pi i \angles{\xi, t^{-1}y}}d\xi = (\wh \phi)_t(y) > $ > so > $ > \anglesn{\phi^x_t, \wh g} = \int_{\real^d}\phi^x_t \wh g = \int_{\real^d} \wh{\phi^x_t} g = \int_{\real^d}\tau_x (\wh \phi)_t \cdot g = [g * (\wh \phi)_t](x) > $ > As $\wh \phi$ is a [[Mollifier|mollifier]] by assumption $(2)$, $\anglesn{\phi^x_t, \wh g} = g * (\wh \phi)_t(x) \to g(x)$ in $L^1$ as $t \to 0$. On the other hand, > $ > \anglesn{\phi_t^x, \wh g} = \int_{\real^d} \phi_t^x(\xi) \wh g(\xi)d\xi = \int_{\real^d} \phi(t\xi)e^{2\pi i \angles{\xi, x}} \wh g(\xi)d\xi > $ > By assumption $(1)$ and the [[Dominated Convergence Theorem]], $\anglesn{\phi^x_t, \wh g} \to (\wh g)^\vee(x)$ [[Pointwise Convergence|pointwise]]. Therefore $g = (\wh g)^\vee$ [[Almost Everywhere|a.e.]], with $(\wh g)^\vee$ being continuous by the Dominated Convergence Theorem again. [^1]: See [[The Fourier Mollifier]].