What is the strategy to prove Parseval's formula?

Question

Parseval's theorem says that:

If $f(x)\in\mathcal{L}_2(-\infty,\infty)$

$\int_{-\infty}^{\infty} dx |f(x)|^2=\frac{1}{2\pi}\int_{-\infty}^{\infty} dk |\tilde f(k)|^2$

where $\tilde f(k)$ is the Fourier transform of $f(x)$.

I looked up Parseval's theorem on Wikipedia, but there the proof is not given. Some articles just say that it proves that total energy is same in position space and momentum space (i.e. the physics explanation).

Could some provide any sources or suggestions about the proof of the theorem?

Answer 1

I would say it usually comes with the 'package' that shows that Fourier transforms extend to $L^2$ functions.

One starting point is to consider smooth functions and knowing one particular Fourier transform. For $\sigma>0$, let $$ \phi_\sigma (t) = \exp( -\sigma^2 t^2 /2) $$ Then somehow (you would typically need some complex analysis to do so) show that: $$ \widehat{\phi}_\sigma(x) = \frac{1}{\sqrt{2\pi} \sigma} \exp(-x^2/ (2\sigma^2))$$ When $f,g$ are Schwarz functions (smooth, rapid decays), the triple integral: $$ \int\int\int e^{it(x-y)} \phi_\sigma(t) f(x)g(y) \,dt \,dx\, dy $$ converges absolutely and Fubini applies. Carrying out the $t$ integration you get $$ \int \int \widehat{\phi}_\sigma(x-y) f(x)g(y)\,dx \,dy$$ and you show that carrying out the $x$ integral and letting $\sigma\rightarrow 0$ this converges to $\int f(y) g(y) \, dy$. While if you first carry out the $x$ and $y$ integrals you get: $$ \int \int \phi_\sigma(t) \hat{f}(t)\hat{g}(-t) \, dt$$ which converges to $\int \hat{f}(t) \hat{g}(-t) \, dt$. This looks nicer if you introduce the complex $L^2$ scalar product $(f,g) = \int f(x)\overline{g(x)} dx$ as it takes the simple form of an isometry. Writing ${\cal F}$ for the Fourier transform (including a factor $1/\sqrt{2\pi}$): $$ ({\cal F} f, {\cal F} g) = (f,g) $$ This shows that ${\cal F}$ is uniformly continuous on Schwarz functions and as they are dense in $L^2$, the identity extends by continuity to all $L^2$ functions (with a more abstract interpretation).

From the above identity all the standard properties of ${\cal F}$ follows: Parseval/Plancherel (let $f=g$) as well as the Fourier inversion formulas, noting e.g. that $ ({\cal F} f, {\cal F} g) = ({\cal F}^* {\cal F}f, g) = (f,g) $

Answer 2

Suppose $f(x,y)$ is a bounded harmonic function in the upper half plane where $y > 0$. And further suppose $f$ is continuous on the closed half plane where $y \ge 0$. Then $f$ is the Poisson integral of its boundary function: $$ f(x,y) = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{yf(x',0)}{(x-x')^2+y^2}dx. $$ Harmonic functions in the upper half plane of this type obey an ergodic type of conservation law: $$ \lim_{y\uparrow\infty}yf(x,y)=\frac{1}{\pi}\int_{-\infty}^{\infty}f(x',0)dx'. $$ There are unusual consequences of this conservation law. Generalized Parseval identities can be proved from this for a large class of eigenfunction expansions associated with selfadjoint Math-Physics equations.

This can be used to prove what you want by using the function $$ f(\alpha,\beta)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-\beta|r|}e^{-i\alpha r}\left(\int_{-\infty}^{\infty}g(s)\overline{g(s-r)}ds\right)dr $$ This $f$ is harmonic in the upper half plane and is continuous on the closed half plane, assuming that $g$ is in $L^1\cap L^2$ on $\mathbb{R}$. The boundary function of $f$ is $$ f(\alpha,0)=|\hat{g}(\alpha)|^2. $$ So $f$ is non-negative and harmonic, which allows the ergodic law to be applied: $$ \lim_{\beta\uparrow\infty}\beta f(\alpha,\beta)=\frac{1}{\pi}\int_{-\infty}^{\infty}|\hat{g}(\alpha)|^2d\alpha. $$ Starting with the definition of $f(\alpha,\beta)$ and noting that $\beta e^{-\beta|r|}$ behaves like $2\delta_0(r)$ as $\beta\uparrow\infty$ gives the result that $$ \frac{1}{\pi}\int_{-\infty}^{\infty}|\hat{g}(\alpha)|^2d\alpha=\left.\frac{1}{\pi}\int_{-\infty}^{\infty}g(s)\overline{g(s-r)}ds\right|_{r=0}=\frac{1}{\pi}\int_{-\infty}^{\infty}|g(s)|^2ds. $$ And that's the Parseval identity.