Stop wasting his time Johan you’re not going to solve the restriction problem

]]>Why wouldn`t we scale $$||f||_{\infty} = 2$$ and divide my pointwise estimate with $|\widehat{\sigma}|^{q}$ twice. Doesn`t that show that $$2^q|\hat{\sigma}|^{q} < |\widehat{f\sigma}|^{q} ,$$ is impossible?

]]>Yeah, but don`t you lose information that way? If we had slower asymptotic decay to zero in infinity for $|\widehat{f\sigma}|^{q},$ then we would have

$$C|\hat{\sigma}|^{q} \leq |\widehat{f\sigma}|^{q} ,$$ but this kind of contradicts the pointwise estimate. BTW, don`t you divide by zero?

No, since the factors of on both sides cancel each other out and one is only left with the trivial bound .

]]>I guess the Lord tests as all. I meant the following. In any case we have pointwise

so cannot decay slower than . Doesn`t this prove the resctriction conjecture?

]]>In any case we have pointwise

so cannot decay slower than if is integrable.

]]>I meant that the estimate is clearly true for large if we don`t care about universal small constants. More interesting would be to quantify that cancellation as an function of

]]>Yes, there are issues in my latex. I have been moving to an another town, so I make more mistakes than usually and I make mistakes a lot.

I think it better to make a little bit more longer comment, but I cannot prove the whole thing in these comments. I try to run this comment through an editor, because I cannot see my comment otherwise.

Anyway, if is small, it means that there are lot of cancellation. However, that cancellation is *fixed* for each *pointwise* estimate! So the relevant cancellation comes from

The worst cases are when

W.L.O.G let us suppose that

Thus, there exists

so that

In other words

On the other hand we have

for some

What we wan’t to prove is that

The bigger gets for fixed the more probable the estimate gets. IMO this estimate is a priori probable and it is clearly true for large .

]]>There appear to be several LaTeX issues with your comment, but in any event this approach is unlikely to work, because the non-oscillatory integrals are considerably larger than the oscillatory integral . In particular, they do not decay to zero in the limit and so one will not get any non-trivial estimates this way.

]]>This will probably be my last comment on the subject. In general the estimate fails. But I can show that my estimate works for even functions , because then the Fourier transform is real valued.

Now, there exists and such that

and

Thus,

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