Estimates for Klein-Gordon-Equation follow directly from Wave equation Estimates

Question

in this paper http://arxiv.org/pdf/1412.1626.pdf it says that Lemma 3.1/(3.1) follows from Theorem 1.3 in http://arxiv.org/pdf/math/0402192.pdf without extra details. Can somebody please explain that?

I can see, that these two estimates are every similar as $q=2$, $r=\infty$ , $n=3$ are admissable in Theorem 1.3. Furthermore the $\gamma=1$ in Theorem 1.3 fits perfectly with the $\lambda$ in Lemma 3.1.

The only real difference seems to be that Theorem 1.3 concerns the wave equation and Lemma 3.1 the Klein-Gordon-Equation, I know, that the wave equation is just the Klein-Gordon-Equation with $m=0$ but I still don't see why this local Klein-Gordon-Estimate follows directly from the corresponding local Wave-Estimate.

Thanks a lot for your help!

Answer 1

I don't think they mean that you can literally just plug in Theorem 1.3 of Sterbenz-Rodnianski to get Lemma 3.1. I think they mean that the proof is basically the same, with suitable adjustments.

A rough sketch:

The key to Theorem 1.3 of Sterbenz-Rodnianski is the derivation of the $L^\infty$ estimate (14). If you follow the same argument for Klein-Gordon, you will have that in the equation between (13) and (14) you should have, instead of $$ \exp( 2\pi i (t \pm r + k/4)\rho) $$ (which is derived from the wave propagator $e^{it\sqrt{-\triangle}}$) the expression $$ \exp(2\pi i ( c(\rho) t \pm r + k/4)\rho) $$ (which is derived from the Klein-Gordon propagator $e^{it\sqrt{-\triangle + m}}$) where the wave-speed, coming from the dispersion relation of Klein-Gordon, is $$ c(\rho) = \sqrt{1 + \frac{m}{\rho^2}}.$$

Now, since we have inserted a $\rho$ cut-off, we can restrict the $L^\infty$ estimate for $\psi^\pm_k$ to the region where $$ \{(t,r): \exists \rho \in [\lambda/2,2\lambda], [\rho c'(\rho) + c(\rho)] t \pm r + k/4 = 0 \} $$ and its complement. In the complement we can integrate by parts as in the wave case to get basically equation (14) with a wave-speed $C(\lambda)$ inserted in the estimate. Within the region we don't integrate by parts, and bound the $\exp$ term by $1$. This gives us a space-time $L^\infty$ estimate for $\psi^\pm_k$. Plug this in the equation (13) and you can get $L^2$ integrability in time.