Smoothness of Fourier transform of a measure

Question

Is the Fourier transform of a finite Borel measure on $\mathbb{R}$ necessarily a smooth function?( $\widehat{\mu}(x)=\int_\mathbb{R}e^{-i\pi xy} d\mu(y)$)

Answer 1

No. It's continuous, but in general not smooth.

If we take for $\mu$ the measure given by the density

$$f(x) = \frac{1}{1+x^2}$$

with respect to the Lebesgue measure, we find that

$$\hat{\mu}(y) = \pi e^{-\lvert y\rvert},$$

which is not differentiable at $0$.