most recent 30 from http://mathoverflow.net 2013-05-19T13:17:50Z http://mathoverflow.net/feeds/question/38308 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38308/fourier-transform-of-radon-measure Philipp 2010-09-10T14:01:20Z 2010-09-11T22:48:19Z

<p>hi, </p> <p>assume that I have a function $q$ which is a Fourier Multiplier of order zero, i.e. $$ \left|\left( \frac{d}{dx}\right)^nq(x)\right|\lesssim \left(\frac{1}{1+|x|}\right)^n\quad \mbox{for all }n\geq 0. $$ Can I conclude that $q$ is the Fourier Transform of a finite Radon Measure? If not, what are the conditions on $q$?</p> <p>Actually, what I really want to know is when is the Fourier Multiplication Operator defined by $q$ bounded on $L^p$? The Mihlin Multiplier theorem gives an affirmative answer for <code>$1&lt;p&lt;\infty$</code>. How about the boundary cases $1$ and $\infty$? </p>

http://mathoverflow.net/questions/38308/fourier-transform-of-radon-measure/38428#38428 fedja 2010-09-11T22:48:19Z 2010-09-11T22:48:19Z

<p>No, such function doesn't need to be a Fourier transform of a finite measure (and, thereby, doesn't need to be a multiplier in $L^1$ or $L^\infty$). This is well-known and the most classical counterexample is just a smoothed Heaviside function $q$ that is $0$ on $(-\infty,1]$, $1$ on $[1,+\infty)$ and whatever you want (just keep it $C^\infty$) in between. One reason it is not a Fourier transform of a measure is that on one hand, such measure should simultaneously have and have not point masses by Wiener's formula <code>$$ \sum_x|\mu(\{x\})|^2=\lim_{|I|\to+\infty}|I|^{-1}\int_I |\widehat \mu(y)|^2\ dy $$</code> (usually it is written for the interval $[-T,T]$, but the truth is that the limit can be taken over any sequence of intervals whose lengths tend to infnity). The existence of this limit is quite a restrictive condition on the absolute value of a bounded function that wants to be a Fourier transform of a finite measure.</p> <p>The full description of Fourier transforms of measures seems beyond reach (not in the sense that it is hard to figure out what happens for each particular function or whether a given condition is necessary or sufficient, but in the sense that there are no easily checkable conditions that would be exactly equivalent to that property).</p>