Prove Parseval for the Fourier transform Can you please show me how to prove $$\int_{-\infty}^\infty f(x)^2 dx = \frac{1}{2 \pi} \int_{-\infty}^\infty [Ff(x)]^2 dx$$ where $Ff(t) = \displaystyle\int_{-\infty}^\infty f(x) e^{-itx}dx$.  
 A: We have that
$$\hat{f}(t) = F[f(x)] = \int_{-\infty}^\infty f(x) e^{-itx} \, dx,$$
and by the Fourier inversion formula
$$F^{-1}[g(t)] = \frac{1}{2\pi}\int_{-\infty}^\infty g(t) e^{itx} \, dt.$$
Now
$$F[\delta(x)] = \int_{-\infty}^\infty \delta(x) e^{-itx} \, dx = 1,$$
and hence applying the Fourier inversion formula gives
$$\delta(x) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{itx} \, dt. \tag{$\ast$}$$
Now we simply calculate the following:
\begin{align*}
\int_{-\infty}^\infty f(x)^2 \, dx & = \int_{-\infty}^\infty \left( \frac{1}{2\pi} \int_{-\infty}^\infty \hat{f}(t) e^{itx} \, dt \right) \left(\frac{1}{2\pi} \int_{-\infty}^\infty \hat{f}(\tau)e^{i\tau x} \, d\tau \right) \, dx \\
 & = \int_{-\infty}^\infty \hat{f}(t) \left( \frac{1}{2\pi} \int_{-\infty}^\infty \hat{f}(t) \left( \frac{1}{2\pi} \int_{-\infty}^\infty e^{i(t - \tau)x} \, dx \right) \, d\tau \right) \, dt \\
 & = \int_{-\infty}^\infty \hat{f}(t) \left( \frac{1}{2\pi} \int_{-\infty}^\infty \hat{f}(\tau) \delta(t - \tau) \, d\tau \right) \, dt \tag{by $(\ast)$}\\
 & = \frac{1}{2\pi} \int_{-\infty}^\infty \hat{f}(t)^2 \, dt.
\end{align*}
This establishes Parseval's identity.
A: Expanding on my hint on the main question, for real-valued square-integrable functions $f(x)$, the Fourier transform is
$$\hat{f}(t) = \int_{-\infty}^\infty f(x)e^{-ixt}\,\mathrm dx$$
and $f(x)$ can be obtained from $\hat{f}(t)$ via
the inverse Fourier transform
$$f(x) = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{f}(t)e^{ixt}\,\mathrm dt. $$
Thus,
$$\begin{align}
\int_{-\infty}^\infty (f(x))^2\,\mathrm dx
&=  \int_{-\infty}^\infty f(x)
\left [\frac{1}{2\pi}\int_{-\infty}^\infty \hat{f}(t)e^{ixt}\,\mathrm dt \right ]
\,\mathrm dx\\
&= \frac{1}{2\pi}\int_{-\infty}^\infty \hat{f}(t)
\left [\int_{-\infty}^\infty f(x)e^{ixt}\,\mathrm dx \right ]
\,\mathrm dt &\scriptstyle{\text{be sure to justify the interchange of order of integration}}\\
&= \frac{1}{2\pi}\int_{-\infty}^\infty \hat{f}(t)(\hat{f}(t))^*\,\mathrm dt\\
&= \frac{1}{2\pi}\int_{-\infty}^\infty \left|\hat{f}(t)\right|^2\,\mathrm dt
\end{align}$$
The definitions of Fourier transform given above apply to complex-valued
functions $f(x)$ as well, and the method described above can be used to
show that
$$\int_{-\infty}^\infty |f(x)|^2\,\mathrm dx
=\frac{1}{2\pi}\int_{-\infty}^\infty |\hat{f}(t)|^2 \,\mathrm dt$$
