Timeline for Complex structure on $L^2(\mathbb R)$ generalizing the Hilbert transform

10 events

when toggle format	what		by	license	comment
Mar 28, 2019 at 23:37	history	edited	André Henriques	CC BY-SA 4.0	deleted 4 characters in body; edited title
Apr 30, 2011 at 21:58	answer	added	Andrew		timeline score: 0
Apr 30, 2011 at 17:35	history	edited	André Henriques	CC BY-SA 3.0	added 1 characters in body
Apr 30, 2011 at 14:31	comment	added	André Henriques		@Piero: I do not require translation invariance. But I agree with you that it looks like it's not possible to achieve my decay condition.
Apr 30, 2011 at 13:54	comment	added	Piero D'Ancona		Well, if K=K(x-y) then the operator is translation invariant, then it is a constant coefficient pseudodifferential operator with a symbol $a(\xi)$, and by your second condition $a(\xi)$ must take only the values $\pm i$. Thus you have jump singularities in the symbol which should always produce a decay of order $\sim t^{-1}$. I guess.
Apr 30, 2011 at 6:26	answer	added	Alex Gavrilov		timeline score: 2
Apr 30, 2011 at 2:23	comment	added	André Henriques		@:Piero. That's right. And $lim_{t\to\infty}t/(2t-x)$ is not zero. But if you have something like $K=1/(x-y)^\alpha$ with $\alpha>1$, then you get $lim_{t\to\infty}t/(2t-x)^\alpha=0$.
Apr 30, 2011 at 2:18	history	edited	André Henriques	CC BY-SA 3.0	added 320 characters in body
Apr 29, 2011 at 21:53	comment	added	Piero D'Ancona		Could you explain the third condition? E.g. when $K=1/(x-y)$ one gets $t \cdot K(t,x-t) = t/(2t-x) $...
Jan 23, 2011 at 21:19	history	asked	André Henriques	CC BY-SA 2.5