# fourier transform of radon measure

hi,

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 $q$?

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 $1<p<\infty$. How about the boundary cases $1$ and $\infty$?

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I added backticks around some math to fix the TeX. –  Mark Meckes Sep 10 '10 at 14:44
Fixed your math so that the LHS is not trivially 0. Also for future reference, this type of harmonic analysis $\subset$ ca.classical-analysis in the arXiv classification scheme. –  Willie Wong Sep 10 '10 at 15:02

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 $$\sum_x|\mu(\{x\})|^2=\lim_{|I|\to+\infty}|I|^{-1}\int_I |\widehat \mu(y)|^2\ dy$$ (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.