 Solution

The Hilbert-transform Fourier multiplier is bounded, and ${\xi }^r\hat{\phi }(\xi )$ is integrable for every nonnegative integer $r$. Therefore its inverse Fourier transform can be differentiated under the integral arbitrarily many times:These derivatives are continuous by dominated convergence theorem. Since $\widehat{{\phi }^{(r)}}=(i\xi {)}^r\hat{\phi }$, the Hilbert transform is smooth and commutes with differentiation:The Fourier proof avoids differentiating a singular kernel without preserving its Cauchy principal value prescription.