> [!theoremb] Plancherel's Theorem > > Let $f \in L^1(\real^d) \cap L^2(\real^d)$, then $\wh f \in L^2(\real^d)$. The [[Fourier Transform|Fourier transform]] restricted to $L^1 \cap L^2$ [[Linear Extension Theorem|extends uniquely]] to a [[Unitary Map|unitary]] [[Space of Toplinear Isomorphisms|isomorphism]] on $L^2$. > > *Proof*. Let $\cm = \bracsn{f \in L^1: \wh f \in L^1}$, then any $f \in \cm$ is bounded. By the [[Intermediate Interpolation of Lp Spaces|intermediate interpolation]] of $L^p$ spaces, $\cm \subset L^1 \cap L^\infty \subset L^2$. Since the [[Schwartz Space|Schwartz space]] $\cs \subset \cm$, $\cm$ is [[Dense|dense]] in $L^2$. > > Let $f, g \in \cm$ and $h = \ol{\wh g}$, then by the [[Fourier Inversion Theorem]], > $ > \wh h(\xi) = \int_{\real^d} \ol{\wh g(x)}e^{-2\pi i \angles{\xi, x}}dx = \int_{\real^d}\ol{\wh g(x)e^{2\pi i \angles{\xi, x}}} = \ol{g(\xi)} > $ > so > $ > \anglesn{f, g}_{L^2} = \angles{f, \ol g} = \anglesn{f, \wh h} = \anglesn{\wh f, h} = \anglesn{\wh f, \ol{\wh g}} = \anglesn{\wh f, \wh g}_{L^2} > $ > where $\angles{\cdot, \cdot}_{L^2}$ conjugates the second function, and $\angles{\cdot, \cdot}$ does not. By the [[Linear Extension Theorem]] on both the Fourier transform and the inverse Fourier transform, the Fourier transform restricted to $\cm$ extends uniquely into a unitary isomorphism on $L^2$. > > It remains to show that the extension agrees with the original Fourier transform on $L^1 \cap L^2$. Let $\phi \in L^1$ such that[^1] > 1. $\phi, \wh \phi \in L^1$ with $\int \phi = 1$. > 2. $\wh \phi$ is bounded, continuous at $0$, with $\wh \phi(0) = 1$. > > Let $f \in L^1 \cap L^2$. By [[Young's Inequality]], $f * \phi_t \in L^1$. Moreover, $(f * \phi_t)^\wedge = \wh{(\phi_t)} \wh f$ with $\wh f \in L^\infty$ and $\wh {\phi_t} \in L^1$, so $(f * \phi_t)^\wedge \in L^1$ as well. Therefore $f * \phi_t \in \cm$ for all $t > 0$. > > Since $\phi$ is a [[Mollifier|mollifier]], $f * \phi_t \to f$ in $L^1$ and $L^2$, and $(f * \phi_t)^\wedge \to \wh f$ uniformly. In addition, > $ > \normn{(f * \phi_t)^\wedge - \wh f}_2^2 \le \int_{\real^d} |\wh f(\xi)|^2 |\wh \phi(t\xi) - 1|^2 d\xi > $ > where $\wh{\phi_t}(\xi) = \wh \phi(t\xi)$. By the [[Dominated Convergence Theorem]], $(f * \phi_t)^\wedge \to \wh f$ in $L^2$ as well. Therefore the extension agrees with the Fourier transform on $L^1 \cap L^2$. [^1]: [[The Fourier Mollifier]]