= Hilbert-transform Fourier multiplier
{c}
{title2=$\widehat{\mathcal Hf}=-i\operatorname{sgn}(\xi)\widehat f$}
For $\mathcal Hf(x)=\pi^{-1}\operatorname{pv}\int f(y)/(x-y)\,dy$ and $\widehat f(\xi)=\int e^{-ix\xi}f(x)dx$, the <Fourier transform> multiplier is $-i\operatorname{sgn}\xi$. The <Heaviside function> transform and the double-transform reflection identity give the sign. <Plancherel theorem> proves the $L^2$ isometry, and multiplication by $i\xi$ proves commutation with differentiation. The map need not preserve the <Schwartz space>.
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