3
$\begingroup$

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.

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.