1

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

BigM
  • 3,936
  • 1
  • 26
  • 36

1 Answers1

3

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 $0$.

Daniel Fischer
  • 206,697
  • Thanks.i was certain that should be case but didnt know examples(otherwise all those propositions in Katznelson's book on characterizing measures whose their Fourier trans. Have certain regularity would have been void.) – BigM Nov 04 '14 at 16:29