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 \(\widehat{m_h}=\tau_h\widehat f\).