This question is suggested by that about L^p multipliers (and the answer by Michael Lacey in particular). Let E be a measurable set in the plane and XiE its characteristic function. We say E is an Lp multiplier set iff the translation invariant operator F(Tf) = XiE F(f) (F = Fourier Transform), extends to a bounded operator on L^p (it is a-priori defined on the Schwartz Class or L^2\cap L^p). For example in dimension 1 every interval is an L^p multiplier for every p except 1 and infty. The situation strongly differs in the 2-dimensional case, as highlighted by the solution of the ball multiplier conjecture by C. Fefferman: the ball B is a strict L^2 multiplier (not L^p for p\neq 2) and the same is true for every set with an appropriate curvature. I was wondering whether more generally something can be said about the interval of p for which E is an L^p multiplier, in terms of the "geometry" of E, in dimension 2 or higher.
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asked Oct 31, 2009 at 9:34
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The only result along these lines that I am aware of is due to Vebedev and Olevskii (Idempotents of Fourier multiplier algebra. Geom. Funct. Anal. 4 (1994), no. 5, 539--544. ). Their result states that if a set E \subset R^d does not agree (almost everywhere) with an open set, then it can not be a L^p multiplier for p \neq 2.
