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Jan 30, 2015 at 9:34 comment added gsa I think MO is not the best place for such a discussion. I updated my profile so that you can find my email address.
Jan 29, 2015 at 16:58 comment added Sal @gsa Maybe one reason could be that using Strichartz estimates for Klein-Gordon they get a larger range for the values of $\sigma$ of that they would obtain using energy inequality for the wave equation.
Jan 28, 2015 at 18:47 comment added Sal @gsa But the wave equation has also a conservation of energy which can be used. I have to think about it, because it's not very clear to me.
Jan 28, 2015 at 17:31 comment added gsa Maybe it's possible to work with the wave equation as well. But the standard Strichartz estimates for the wave equation are stated in the homogeneous Sobolev spaces $\dot{H}^{s,p}$, which are a bit delicate. My guess is that they prefer to work with the inhomogeneous spaces $H^{s,p}$ and thus use the Strichartz estimates for Klein-Gordon.
Jan 28, 2015 at 8:21 comment added Sal @gsa May I ask you to clarify another doubt I have? Why do you think they use estimates for Klein-Gordon equations and not estimates for the wave equation?
Jan 27, 2015 at 16:39 comment added Sal @gsa Thank you. I stopped before that lemma and I was trying to understand it with previous computations.
Jan 27, 2015 at 16:25 comment added gsa They use the assumption $s\geq 5/3$ in the estimate (6.10). They apply the fractional Leibniz rule to the product in combination with the Sobolev embedding of $H^{s-1,2}$ to $L^q$ where they use $1-1/\tilde{r} = 1/2 +1/q$ and choose $\tilde{r} = 3/(4-2s)$. The Sobolev embedding then only works for $s\geq 5/3$.
S Jan 27, 2015 at 14:34 history suggested user25199 CC BY-SA 3.0
Made title more specific, minor typos.
Jan 27, 2015 at 14:32 review Suggested edits
S Jan 27, 2015 at 14:34
Jan 27, 2015 at 14:31 review First posts
Jan 27, 2015 at 14:33
Jan 27, 2015 at 14:29 history asked Sal CC BY-SA 3.0