 Solution

Let $f=\hat{\phi }$, a Schwartz function. Splitting the inverse Fourier transform at the jump in the Hilbert-transform Fourier multiplier givesFor $x\ne 0$, integration by parts on each half-line showsBoth restricted derivatives are in $L^1$. The Riemann-Lebesgue lemma makes the bracket tend to zero as $|x|\to \infty$. Hence the two-sided tail isHere $\hat{\phi }(0)=\int \phi$. The large-distance tail of the Hilbert transform thus depends on the zeroth moment of the input. If that moment is nonzero, the $1/x$ tail proves that the output is not a Schwartz function, despite being smooth and square-integrable.