Prove $\int_{-\infty}^{\infty} |F(k)|^2 \ dk=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} |f(x)|^2 \ dx$ (Parseval's theorem)

Question

I would like to know how to prove Parseval's theorem, $$\int_{-\infty}^{\infty} |F(k)|^2 \ dk=\int_{-\infty}^{\infty} |f(x)|^2 \ dx.$$ The definition of the Fourier transformed used is $$F(k)=\mathcal{F}(f(x))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-ikx} f(x) \ dx.$$

I don't quite know how to start.

Answer 1

Assuming that we may interchange the order of integration we have (all integrals go from $-\infty$ to $+\infty$): $$ \int F(k) \, \overline{G(k)} \, dk = \int \left( \frac{1}{\sqrt{2\pi}} \int f(x) \, e^{-ikx} \, dx \right) \, \overline{G(k)} \, dk = \int f(x) \left( \frac{1}{\sqrt{2\pi}} \int e^{-ikx} \, \overline{G(k)} \, dk \right) dx \\ = \int f(x) \overline{\left( \frac{1}{\sqrt{2\pi}} \int e^{ikx} \, G(k) \, dk \right)} dx = \int f(x) \, \overline{g(x)} \, dx . $$ Taking $g=f$ and thus $G=F$ gives us $$\int |F(k)|^2 \, dk = \int |f(x)|^2 \, dx.$$

What are the requirements on $f$ and $g$ for the above to work? For the Fourier transform to be defined through a pure integral (first step), we must have $f \in L^1(\mathbb{R}).$ Likewise, for the inverse Fourier transform to be defined through a pure integral (last step), we must have $G \in L^1(\mathbb{R}).$ These two conditions also make the interchange of order of integration valid. On the other hand, for $\int F(k) \, \overline{G(k)} \, dk$ and $\int f(x) \, \overline{g(x)} \, dx$ to be defined we should have $f,g,F,G \in L^2(\mathbb{R}).$ Thus, assuming $f,g,F,G \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$ make the calculations valid.