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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

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up vote 5 down vote accepted

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.

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).

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@fedja: a follow up clarification question is asked mathoverflow.net/questions/48042/… you may be able to provide a reference there? –  Willie Wong Dec 2 '10 at 13:54

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