Comments on: Fefferman’s ball multiplier counterexample

By: Hormander-Minlin | Being simple

Hormander-Minlin | Being simple — Sun, 03 Feb 2013 03:05:32 +0000

[...] for be a Fourier multiplier on (see e.g., [1]). However, Fefferman (see [2],[3]) gave an intricate proof which show us that this condition is not sufficient, that is, he showed that the operator does [...]

By: Thomas

Thomas — Fri, 22 Oct 2010 09:35:03 +0000

Of course, that makes a lot of sense. Thank you very much.

By: Mark Lewko

Mark Lewko — Wed, 20 Oct 2010 22:33:28 +0000

Thomas,

Thanks for reading! You’re right that this follows from self-adjointness. By Parseval’s identity we have

The last equality follows from the definition of where denotes the unit ball. Now the last expression is also equal to (by essentially the same argument just given) .

By: Thomas

Thomas — Wed, 20 Oct 2010 17:36:32 +0000

I really like this post, I’m doing a paper on Kakeya sets and their applications and this (and its prequel) helped me a lot. I hope you have time to answer a (probably simple) question about the very last line of the post: why is

sup sup [S_1 f, g]

equal to

sup sup [f, S_1 g]

It looks like some kind of self-adjointness, do we know that the operator is its own adjoint?

Thanks,
Thomas

By: L^{p} Convergence of Fourier Transforms « Lewko's Blog

L^{p} Convergence of Fourier Transforms « Lewko's Blog — Wed, 05 Aug 2009 01:32:33 +0000

[...] Fourier Multipliers, Marcel Riesz leave a comment « Lattice Points in l^{1} Balls Fefferman’s Ball Multiplier Counterexample [...]