Plancherel's Theorem

Theorem

For all $f,g \in L^{2}$ , the following equation holds.

$\begin{align} \langle \hat{f},\hat{g} \rangle &= 2\pi \left\langle f,g \right\rangle \\[1em] \| \hat{f} \|_{2}^{2} &= 2\pi \| f \|_{2}^{2} \end{align}$

Here $\hat{f}$ is the Fourier transform of $f$ .

Explanation

If expressed in integral form, it is as follows.

$\begin{align} \int \overline{f(x)}g(x)dx &= \dfrac{1}{2\pi} \int \overline{\hat{f}(\xi)} \hat{g}(\xi) d\xi \tag{1} \\[1em] \int \left| f(x) \right|^{2} dx &= \dfrac{1}{2\pi} \int | \hat{f}(\xi) |^{2} d\xi \end{align}$

Looking at the process of defining the Fourier transform of $f$ , $f$ has to be a $L^{1}$ function, and it only needs to be a $L^{1}$ function. However, we want to freely use the Fourier transform not only in $L^{1}$ space but also in $L^{2}$ space. $L^{2}$ space is the only Lebesgue space that is also a Hilbert space, so the importance of this issue goes without saying. The Plancherel theorem tells us that this is actually possible, and that the operator $\mathcal{F}$ known as the Fourier transform can be treated as follows.

$\mathcal{F} : L^{2} \to L^{2}$

Also, depending on how the Fourier transform is defined, the constant in front of $(1)$ and $(2)$ may disappear or a $\sqrt{2\pi}$ may be added instead. Equation $(2)$ is also known as the Parseval’s theorem concerning the Fourier transform.