For \(f\in L^1(\RR^n)\) let \(\hat f(\xi)\coloneqq\int_{\RR^n}f(x)e^{-2\pi i x\cdot\xi}\dx\). Then \(\hat f\) is continuous, bounded by \(\norm f_1\), and vanishes at infinity by the Riemann–Lebesgue lemma; two functions in \(L^1\) with the same transform agree almost everywhere. Fourier inversion recovers \(f\) from \(\hat f\) under the additional hypothesis \(\hat f\in L^1\).
Fourier transform is injective on $L^1$ (Tag PR-47TTS)
Let \(f,g\in L^1(\RR^n)\). If their Fourier transforms satisfy \(\widehat f(\xi) = \widehat g(\xi)\) for all \(\xi\in\RR^n\), then \(f=g\) almost everywhere.
Riemann--Lebesgue lemma: $\widehat f$ is bounded, continuous, and vanishes at infinity (Tag PR-IGMH4)
Let \(f\in L^1(\RR^n)\). Then its Fourier transform \(\widehat f\) is continuous on \(\RR^n\), satisfies \(\sup_{\xi\in\RR^n}\abs{\widehat f(\xi)}\leq\norm{f}_1\), and \[ \widehat f(\xi)\to0 \quad\text{as } \abs{\xi}\to\infty . \]
Fourier inversion (Tag T-DTXIA)
Let \(f\in L^1(\RR^n)\), and suppose its Fourier transform \(\widehat f(\xi)=\int_{\RR^n}f(x)e^{-2\pi i x\cdot\xi}\,dx\) also lies in \(L^1(\RR^n)\). Then \[ F(x)\coloneqq\int_{\RR^n}\widehat f(\xi)e^{2\pi i x\cdot\xi}\,d\xi \] defines a bounded continuous function \(F\colon\RR^n\to\CC\), and \(F(x)=f(x)\) for almost every \(x\in\RR^n\). More precisely, \(F(x)=f(x)\) at every Lebesgue point \(x\) of \(f\), that is, every \(x\) with \[ \lim_{r\to0^+}\frac{1}{m(B_r(x))}\int_{B_r(x)}\abs{f(y)-f(x)}\,dy=0 ; \] in particular, \(F(x)=f(x)\) at every point \(x\) at which \(f\) is continuous.
Since \(\abs{\widehat f(\xi)e^{2\pi ix\cdot\xi}}=\abs{\widehat f(\xi)}\) and \(\widehat f\in L^1\), the bound \(\abs{F(x)}\leq\norm{\widehat f}_1\) holds and dominated convergence shows that \(F\) is continuous.
For \(t>0\), set \[ G_t(x)\coloneqq t^{-n}e^{-\pi \abs{x}^2/t^2}. \] Its Fourier transform is \[ \widehat{G_t}(\xi)=e^{-\pi t^2\abs{\xi}^2}. \] Indeed, in one dimension the function \[ H(\xi)\coloneqq\int_{\RR}e^{-\pi x^2}e^{-2\pi i x\xi}\,dx \] satisfies, by differentiation under the integral sign and integration by parts, \[ H'(\xi)=-2\pi\xi H(\xi), \qquad H(0)=1, \] so \(H(\xi)=e^{-\pi\xi^2}\). The \(n\)-dimensional identity follows by Fubini’s theorem, since \(e^{-\pi\abs{x}^2}=\prod_{j=1}^ne^{-\pi x_j^2}\), and the formula for \(G_t\) follows by the substitution \(x=ty\). Since \(G_t\) is even, the same computation gives \(\int_{\RR^n}e^{2\pi i w\cdot\xi}e^{-\pi t^2\abs{\xi}^2}\,d\xi=G_t(w)\) for \(w\in\RR^n\).
Fix \(x\in\RR^n\). By Fubini’s theorem T-4GPEF, \[ \begin{aligned} \int_{\RR^n}\widehat f(\xi) e^{2\pi i x\cdot\xi}e^{-\pi t^2\abs{\xi}^2}\,d\xi &=\int_{\RR^n}f(y) \qty{\int_{\RR^n}e^{2\pi i(x-y)\cdot\xi} e^{-\pi t^2\abs{\xi}^2}\,d\xi}dy\\ &=\int_{\RR^n}f(y)G_t(x-y)\,dy\\ &=(f*G_t)(x). \end{aligned} \] Fubini’s theorem applies because \[ \int_{\RR^n}\int_{\RR^n} \abs{f(y)}e^{-\pi t^2\abs{\xi}^2}\,d\xi\,dy=\norm{f}_1\,t^{-n}<\infty. \]
As \(t\to0^+\), dominated convergence on the left, with dominating function \(\abs{\widehat f}\), gives \[ \int_{\RR^n}\widehat f(\xi)e^{2\pi i x\cdot\xi}e^{-\pi t^2\abs{\xi}^2}\,d\xi \longrightarrow F(x). \] On the right, \(G_t(y)=t^{-n}G_1(y/t)\), where \(G_1\geq0\) has \(\int G_1=1\) and is radial and decreasing in \(\abs{y}\), so \((G_t)_{t>0}\) is an approximate identity with a radially decreasing integrable majorant. For such a family, \((f*G_t)(x)\to f(x)\) as \(t\to0^+\) at every Lebesgue point \(x\) of \(f\). Therefore \(F(x)=f(x)\) at every Lebesgue point of \(f\). By the Lebesgue differentiation theorem, almost every point of \(\RR^n\) is a Lebesgue point of \(f\), and every point of continuity of \(f\) is a Lebesgue point.
The function \(\hat f = \chi_{[-1,1]}\) is not the Fourier transform of any \(f\in L^1(\RR)\), since it is not continuous. For \(f=\chi_{[-1,1]}\in L^1(\RR)\), \(\hat f(\xi) = \frac{\sin 2\pi\xi}{\pi\xi}\) is not in \(L^1(\RR)\), so Fourier inversion does not apply to it.
The Gaussian is an eigenfunction of the Fourier transform (Tag PR-DPRY7)
Let \(g\colon\RR^d\to\RR\) be the Gaussian \(g(x) \coloneqq e^{-\pi \abs{x}^2}\). Then its Fourier transform is \(\widehat g = g\). Moreover, for \(t>0\) the dilation \(g_t(x)=t^{-d}g(t^{-1}x)\) satisfies \[ \widehat{g_t}(\xi) = g(t\xi) = e^{-\pi t^2 \abs{\xi}^2}, \qquad \xi\in\RR^d . \]
Fourier transforms of convolutions, translates, modulations, linear substitutions and derivatives (Tag PR-DY2B3)
Use the Fourier transform \(\widehat f(\xi)=\int_{\RR^n} f(x)e^{-2\pi i x\cdot\xi}\,dx\) on \(L^1(\RR^n)\). For \(h\in\RR^n\) and a function \(g\) on \(\RR^n\), let \(\tau_h g\) be the translate \((\tau_h g)(x)\coloneqq g(x+h)\). Let \(f,g\in L^1(\RR^n)\), \(h\in\RR^n\), and \(T\in\GL_n(\RR)\). Then for every \(\xi\in\RR^n\) [@Fol13]:
\(\widehat{f * g}(\xi) = \widehat f(\xi)\,\widehat g(\xi)\), where \(f*g\) is the convolution;
\(\widehat{\tau_h f}(\xi) = e^{2\pi i \xi\cdot h}\,\widehat f(\xi)\);
for \(m_h(x)\coloneqq e^{2\pi i x\cdot h}f(x)\), \(\widehat{m_h}(\xi) = \widehat f(\xi-h) = (\tau_{-h}\widehat f)(\xi)\);
\(\widehat{f\circ T}(\xi) = \abs{\det T}^{-1}\,\widehat f\big((T^{-1})^{t}\xi\big)\);
if \(x_jf\in L^1(\RR^n)\) for some \(j\), then \(\partial_{\xi_j}\widehat f\) exists and \(\partial_{\xi_j}\widehat f(\xi) = \widehat{(-2\pi i x_j f)}(\xi)\);
if \(f\in C^1(\RR^n)\), \(\partial_{x_j} f\in L^1(\RR^n)\), and \(f\) vanishes at infinity, then \(\widehat{\partial_{x_j} f}(\xi) = 2\pi i \xi_j\,\widehat f(\xi)\).
With the other common convention \((\tau_h g)(x)\coloneqq g(x-h)\), formula (b) reads \(\widehat{\tau_h f}(\xi)=e^{-2\pi i\xi\cdot h}\widehat f(\xi)\) and (c) reads \(\widehat{m_h}=\tau_h\widehat f\).
(Operations under the transform)
For \(f, g\in L^1(\RR)\), \(h\in\RR\), and \(a\neq 0\), with \(\hat f(\xi)=\int_\RR f(x)e^{-2\pi i x\xi}\dx\):
| Operation on \(f\) | Transform |
|---|---|
| \(f(x-h)\) | \(e^{-2\pi i h\xi}\hat f(\xi)\) |
| \(e^{2\pi i hx}f(x)\) | \(\hat f(\xi-h)\) |
| \(f(ax)\) | \(\abs a^{-1}\hat f(\xi/a)\) |
| \(f'\), for \(f\) absolutely continuous with \(f'\in L^1\) | \(2\pi i\xi\hat f(\xi)\) |
| \(f*g\) | \(\hat f(\xi)\hat g(\xi)\) |
The identity \(\widehat{f*g}=\hat f\hat g\) converts a convolution equation \(f*g = k\) into the pointwise equation \(\hat f\hat g = \hat k\).
Fourier coefficients
Bessel's inequality (Tag T-4BDE3)
Let \(\mch\) be a Hilbert space, let \((u_n)_{n\geq1}\) be an orthonormal sequence in \(\mch\), and let \(x\in\mch\). For every \(N\geq1\), \[ \norm{x-\sum_{n=1}^{N}\inner{x}{u_n}u_n}^{2}=\norm{x}^{2}-\sum_{n=1}^{N}\abs{\inner{x}{u_n}}^{2}, \] and consequently \[ \sum_{n=1}^{\infty}\abs{\inner{x}{u_n}}^{2} \leq\norm{x}^{2}. \]
Fix \(N\geq 1\) and put \(s_N\coloneqq\sum_{k=1}^{N}\inner{x}{u_k}u_k\). By orthonormality, \(\inner{x}{s_N} = \inner{s_N}{s_N} = \sum_{k=1}^{N}\abs{\inner{x}{u_k}}^2\), so \[ 0\leq\norm{x - s_N}^2 = \norm{x}^2 - 2\operatorname{Re}\inner{x}{s_N} + \norm{s_N}^2 = \norm{x}^2 - \sum_{k=1}^{N}\abs{\inner{x}{u_k}}^2. \] Letting \(N\to\infty\) gives the inequality.
The inequality extends to an orthonormal family \((u_i)_{i\in I}\) indexed by an arbitrary, possibly uncountable, set \(I\): for every \(x\in\mch\), \[ \sum_{i\in I}\abs{\inner{x}{u_i}}^2\coloneqq\sup_{\substack{F\subseteq I\\ F\text{ finite}}}\sum_{i\in F}\abs{\inner{x}{u_i}}^2\leq\norm{x}^2 . \] Consequently, for each \(x\) only countably many of the coefficients \(\inner{x}{u_i}\) are nonzero [@Fol13].
For an orthonormal set \(\theset{u_n}\) in a Hilbert space \(H\), Bessel’s inequality \(\sum_n\abs{\inner{x}{u_n}}^2 \leq \norm x^2\) holds for every \(x\in H\). Equality \(\sum_n\abs{\inner{x}{u_n}}^2 = \norm x^2\) for every \(x\) is Parseval’s identity, which holds if and only if \(\theset{u_n}\) is an orthonormal basis.