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?



Fourier transform of a bounded function
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](http://en.wikipedia.org/wiki/Cauchy_principal_value) of $1/x$. Another keyword is [Hilbert transform](http://en.wikipedia.org/wiki/Hilbert_transform).


