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 $Ff(t) = \displaystyle\int_{-\infty}^\infty f(x) e^{-itx}dx$.
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$\begingroup$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 $$\int_{-\infty}^\infty |f(x)|^2\,\mathrm dx =\frac{1}{2\pi}\int_{-\infty}^\infty |\hat{f}(t)|^2 \,\mathrm dt$$
$\endgroup$ 0 $\begingroup$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.
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