# Strichartz estimates over cones

I'm trying to understand Sogge's book Lectures on Non-Linear Wave Equations, the part where he proves global existence for semilinear equations. There is one part he uses the following inequality:

$\|u\|_{L^4 L^{12}(K)} \lesssim \|\nabla u(0)\|_{L^2} + \|\partial_t u(0)\|_{L^2} + \|F\|_{L^1 L^2 (K)},$

where $K$ is the cone $\{(t,x) : |x-x_0| \leq t_0 -t\}$ and $u$ solves $\square u = F$. If $K$ was $[0,T]\times \mathbb{R}^d$, then this would just be Strichartz's inequality for the wave equation. He says that we can put a cone instead of $[0,T]\times \mathbb{R}^d$ by Huygens' principle, but I don't know exactly how he's able to do that.

I tried localizing the initial data to the ball $B(x_0,t_0)$, which by finite speed of propagation will yield the inequality, but I can only localize smoothly, so I end up with a larger cone on the right hand side. How can I use Huygens' principle to prove this?

Thanks

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Concerning F, there is no need to localize smoothly, just take F equal to zero outside the cone and notice that by Huygens the values of u inside K remain unchanged. Concerning the initial data, probably the easiest way is to notice that you can modify them outside the base B of the cone while u remains the same on K; thus, assign the derivatives of the data on B, extend the data outside B as you wish but so that the norm of the extended data on Rn increases at most by a factor C (= the norm of the extension operator; this is ok provided B is not too small) and then apply the global estimate –  Piero D'Ancona Mar 26 '12 at 10:41
I forgot to mention that $F$ depends on $u$, my $F = |u|^4 u$. How can I localize $u$ without braking the smoothness? –  Guillermo Mar 26 '12 at 11:05
By multiplying F by the characteristic function of K? –  Piero D'Ancona Mar 26 '12 at 11:52
It doesn't matter that $F$ depends on $u$. The Strichartz estimate is an a priori estimate. That is, the estimate does not care whether $u$ solves $\Box(\cdot) = (\cdot)^5$ or it solves $\Box (\cdot) = G$ where $G$ just happens to be numerically equal to $(u)^5$. –  Willie Wong Mar 26 '12 at 11:56