9
$\begingroup$

This should really be well-known, but I was not able to find a definite answer to this question:

Is the Fourier transform of a bounded function always a borel measure (i.e. an order 0 distribution)?

In some sense, the distributional order corresponds to the order of a bounding polynomial and a bounded function can be bounded by an order zero polynomial. But I could not find any reference.

If the above statement is false, what is a counterexample?

If the statement is true: How does bounded variation correspond to all this? I.e. are there criterions for the Fourier transform to be of bounded variation?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
15
$\begingroup$

The answer is no : Fourier transform of the signum function is a constant times the distribution $\lim_{\epsilon\to 0} 1_{|x|>\epsilon}/x$, of order $1$, also known as the principal value of $1/x$. Another keyword is Hilbert transform.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I suppose, continuity of the function does not help either? $\endgroup$
    – Matthias Ludewig
    Commented May 9, 2013 at 16:04
  • 1
    $\begingroup$ No, because modifying the signum function, say on [-1,1], to render it continuous (or even smooth), would just add an $L^1$ function to it, hence a $C_0$ function to its Fourier transform. On the other hand continuous [positive definite][1] functions on $\mathbb{R}$ are exactly the Fourier transforms of finite positive measures, by Bochner's theorem. [1]: en.wikipedia.org/wiki/Positive-definite_function $\endgroup$
    – BS.
    Commented May 9, 2013 at 16:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .