By Fourier transform, the flow of the Klein–Gordon wave flow is $e^{it\sqrt{1-\Delta}}$. That is, if we have initial data $\phi(0,x)=\phi_0(x)$, then the solution will be given by $\phi(t,x)=e^{it\sqrt{1-\Delta}}\phi_0(x)$ or $\hat\phi(t,\xi)=e^{it\sqrt{1+|\xi|^2}}\hat\phi_0(\xi)$ on the Fourier side.
It is clear from this representation that the $L^2$ norm is conserved. I want to prove that the $L^\infty$ norm decays like $t^{-d/2}$.
The way to prove for the Schrodinger equation with wave flow $e^{-it\Delta}$ is to take the inverse Fourier transform $S_t(x):=\mathcal F^{-1}[e^{it|\xi|^2}]$ and show that it decays like $t^{-d/2}$, and then to use Young's inequality: $$\|\phi\|_{L^\infty}=\|S_t*\phi_0\|_{L^\infty}\leq\|S_t\|_{L^\infty}\|\phi_0\|_{L^1}.$$ But I can't seem to do this for the Klein–Gordon equation.