Let $1 \leq p \leq 2$. A measurable function $m(\xi)$ is called a $L^p(R^n)$ ($L^p$ for convenience) multiplier, if $$\m(D)\varphi\/\\varphi\_{L^p} \leq C , \varphi \in L^p $$ for some constant $C$, and $$ m(D)\varphi(x) = F^{1}(m(\xi)F\varphi(\xi)) $$ where $F, F^{1}$ denotes the Fourier transform and its converse in the tempered distribution sense. The picture of $L^1$ multiplier and $L^2$ multiplier are clear. In fact, $m$ is an $L^2$ multiplier iff $m \in L^\infty$, and $m$ is an $L^1$ multiplier iff $m$ is the fourier transform of a finite Borel measure. For general $p \in (1,2)$, we can apply MihlinHormander's theorem to prove $m$ to be an $L^q$ multiplier. We know $e^{i\xi^2}$ is not an $L^p$ multiplier if $p\neq 2$, but the proof is not trivial and relies on the form of the function. So, does there exists any necessary condition for $L^p$ multiplier in general?
2 Answers
In the case of radial multipliers, Heo, Nazarov, Seeger (Acta Math, 206 (2011) 5592) have recently given a complete characterization of $L^p$ multipliers when the dimension is large compared to $p$. More specifically $d > (2+p)/(2p)$, $1 < p <2 $.
Beals has given necessary conditions for certain subalgebras defined by symbollike estimates to be bounded:
R. Beals, $L^{p}$ and Hölder estimates for pseudodifferential operators: necessary conditions. Proc. Sympos. Pure Math., XXXV (1979), 153157.
http://www.ams.org/mathscinetgetitem?mr=545303
This does not give you a necessary condition for a single operator, though. (Of course, there exists a trivial necessary condition: Any $L^p$multiplier is in particular an $L^2$multiplier, so its symbol has to be bounded.)

$\begingroup$ On a noncompact space, is it trivial that L^p Fourier multiplers are in L^\infty? I know this follows from a more general result in nonabelian harmonic analysis, due to Herz, but is the abelian case trivial as you say? $\endgroup$ May 6, 2012 at 20:59