- View SourceStephen Montgomery-Smith wrote:
>

Sorry for clattering people's mailboxes, but I just realised that the

> Terence Tao wrote:

> >

> > Dear Fabrice,

> >

> > For the Schrodinger in 2D, the inhomogeneous estimate blows up with

> > a constant similar to 1/(p-2); this is basically because there

> > is no additional cancellation or other room to improve in the

> > Hardy-Littlewood-Sobolev argument. One can use for instance

> > the counterexample in my paper on radial Strichartz estimates

> > (front.math.ucdavis.edu/math.AP/9811168). But I don't know what happens

> > in the homogeneous estimate; for radial data it is known that the homogeneous

> > estimate is true (either by the paper above, or by a paper of Stefanov)

> > and so there is no blowup in constants. In the non-radial case

> > the counterexample for the homogeneous estimate at the endpoint is due to

> > Montgomery-Smith and it is not clear at all whether this example gives the

> > 1/(p-2)^{1/2} blowup; given that it uses Brownian motion I would expect

> > that the blowup rate it gives is slower.

> >

>

> My feeling is that since a blow up like 1/(p-2)^{1/2} would imply a lack

> of L_2(BMO) estimates, that any counterexample that is only for

> L_2(L_infty) estimates will not provide the counterexamples that are

> needed.

above statement is completely false. This is because when interpolating

against BMO, you do get the dwesired spaces, but the constants will blow

up. Sorry for my mistake. (It is not a case of me rushing with my

statement - I have been under this misapprehension for weeks now - and

just spotted it minutes after I tell the whole world!)

I have looked at the L_2(L_infty) example for the Schrodinger equation

and unless my hurried calculation is wrong, it does look like it will

give the 1/(p-2)^{1/2} blow up.

TOny Carberry and Steve Hoffman also had another example, which I did

included in my paper (with their permission). But also, I now remember

that it was Peter Jones who told me that one could use my approach, and

give an explicit construction rather than using Brownian motion.

Basically the idea was to using the construction of Brownian motion

using the integrals of the Haar functions:

b_t = sum g_n int_0^t h_n(s) ds

where h_n are the normalised Haar functions, and g_n are Gaussian r.v.,

and then replace g_n with something explicit, like g_n=1. This gives a

path with the required properties.

--

Stephen Montgomery-Smith

stephen@...

http://www.math.missouri.edu/~stephen - View SourceOn Wed May 15 2002 at 11:16:31AM -0500, Stephen Montgomery-Smith wrote:
> above statement is completely false. This is because when interpolating

Ok, I had trouble understanding what you said because of this (I

> against BMO, you do get the dwesired spaces, but the constants will blow

initially made the same mistake). Now, would it be possible to turn

the tables and get that sqrt{1/(p-2)} is indeed the right bound, or at

least say that a bound like log{1/(p-2)} is not possible ? I am aware

there are extrapolation theorems about operators bounded on L^p with

bounds like p/(p-1) and one can say something when p -> 1. On the

other hand, the Riesz transforms which are somehow the generic

simplest CZOs have L^p bounds like p when p-> infty, and they are

bounded on BMO... all of this doesn't add up as a proof of anything,

but it just makes me feel like indeed 1/(p-2)^{1/2} is the right thing

wrt to my original question.

> I have looked at the L_2(L_infty) example for the Schrodinger equation

well. Unfortunate for what I had in mind. On the other hand, the real

> and unless my hurried calculation is wrong, it does look like it will

> give the 1/(p-2)^{1/2} blow up.

thing I was after was the bound in the inhomogeneous (retarded)

estimate for the 3D wave. Terry's argument for the failure of the

inhomogeneous endpoint for 2D Schrodinger is convincing that the right

bound there should be 1/(p-2), I don't suppose there should be a

difference with the wave...

Anyway, thanks for all the answers,

F.