PDF: schuller-QM18-parseval.pdf.
Commented proof for Parseval’s theorem, used in F.
Schuller - QM18 - Fourier Operator1.
Note that this theorem seems to also be referred to as Plancherel
identity/theorem, for example in Teschl.
Definition 1. The Fourier operator is the linear map \(\mathcal{F} : S(\mathbb{R}^d) \rightarrow S(\mathbb{R}^d)\) defined for \(x\in\mathbb{R}^d\) by: \[\boxed{( \mathcal{F}(f))(x) := \frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \exp{(-ixy)}f(y)d^dy }\]
Theorem 1. \(\mathcal{F} : S(\mathbb{R}^d) \rightarrow S(\mathbb{R}^d)\) is invertible and: \[\boxed{ (\mathcal{F}^{-1}(g))(x) = \frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d} \exp{(ipx)}g(p)d^dp }\]
Theorem 2 (Parseval’s theorem). Let \(f\in S(\mathbb{R}^d)\). \[\boxed{ \int_{\mathbb{R}^d}\left|\mathcal{F}(f)(p)\right|^2d^dp = \int_{\mathbb{R}^d}\left|f(x)\right|^2d^dx }\] Saying it otherwise, the Fourier operator preserves the \(L^2\) norm.
Proof. We know from \(\mathbb{C}\)-analysis2 that, with \(\alpha^*\) the \(\mathbb{C}\)-conjugate of \(\alpha\in\mathbb{C}\): \[\alpha\alpha^* = |\alpha|^2\] Hence: \[f(x)(f(x))^* = \left|f(x)\right|^2;\qquad \mathcal{F}(f)(p)(\mathcal{F}(f)(p))^* = \left|\mathcal{F}(f)(p)\right|^2\]
From the invertibility of the Fourier transform, we have: \[f(x) = (\mathcal{F}^{-1}(\mathcal{F}(f)))(x) = \frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d}\exp{(ipx)}(\mathcal{F}(f))(p)d^dp\] And so: \[(f(x))^* = \frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d}\exp{(-ipx)}((\mathcal{F}(f))(p))^*d^dp\] Indeed, we’re integrating on \(\mathbb{R}^d\), and not on \(\mathbb{C}\)3, so for \(\phi : \mathbb{R}^d \rightarrow\mathbb{C}\), using the linearity of the Lebesgue integral: \[\begin{aligned} \left(\int_{\mathbb{R}^d}\phi(p)d^dp\right)^* &:=&& \left( \int_{\mathbb{R}^d}\Re(\phi(p))d^dp +i\int_{\mathbb{R}^d}\Im(\phi(p))d^dp \right)^* \\ ~ &=&& \int_{\mathbb{R}^d}\Re(\phi(p))d^dp -i\int_{\mathbb{R}^d}\Im(\phi(p))d^dp \\ ~ &=&& \int_{\mathbb{R}^d}\left( \underbrace{\Re(\phi(p))-i\Im(\phi(p))}_{=:(\phi(p))^*} \right)d^dp \\ ~ &=&& \int_{\mathbb{R}^d}(\phi(p))^*d^dp \\ \end{aligned}\]
And this holds in particular for \(\phi(p) = \exp{(ipx)}(\mathcal{F}(f))(p)\). We can then expand the right-hand-side of the theorem:
\[\begin{aligned} \int_{\mathbb{R}^d}\left|f(x)\right|^2d^dx &=&& \int_{\mathbb{R}^d}f(x)(f(x))^*d^dx \\ ~ &~&& \textcolor{blue}{ (\left|f(x)\right|^2 = f(x)(f(x))^*)} \\ ~ &=&& \int_{\mathbb{R}^d}f(x) \left( \frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d}\exp{(-ipx)}((\mathcal{F}(f))(p))^*d^dp \right) d^dx \\ ~ &~&& \textcolor{blue}{\text{ (inserting our previous expression for $(f(x))^*$})} \\ ~ &=&& \int_{\mathbb{R}^d}((\mathcal{F}(f))(p))^* \left( \underbrace{\frac1{(2\pi)^{d/2}}\int_{\mathbb{R}^d}\exp{(-ipx)}f(x)d^dx}_{ =: \mathcal{F}(f)(p) } \right) d^dp \\ ~ &~&& \textcolor{blue}{\text{ (Fubini)}} \\ ~ &=&& \int_{\mathbb{R}^d}\underbrace{ ((\mathcal{F}(f))(p))^*\mathcal{F}(f)(p) }_{= \left|\mathcal{F}(f)(p)\right|^2} d^dp \\ ~ &~&& \textcolor{blue}{\text{ (Fourier transform identification)}} \\ ~ &=&& \int_{\mathbb{R}^d}\left|\mathcal{F}(f)(p)\right|^2 d^dp \\ \end{aligned}\] ◻
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