Remarks on the geodesically completeness and the smoothing effect on asymptotically Minkowski spacetimes

Abstract

In this note, we study a geometric property of asymptotically Minkowski spacetimes and an analytic property of the wave operator. More precisely, our first main results show that asymptotically Minkowski spacetimes are geodesically complete under a null non-trap** condition. Secondly, we prove that Sobolev index of a real principal type estimate used in a previous work is optimal.

A short proof of smoothing effects on the Minkowski spacetimes

In this appendix, we give a short proof of (1.1) and (1.2) using the explicit formula (3.2).

Lemma A.1

Set

$$\begin{aligned} I(t,y)=\int _{{\mathbb {R}}}e^{-i|t-s|A}f(s,y)ds. \end{aligned}$$

Then, for $\varepsilon >0$ we have

$$\begin{aligned}&\Vert \langle t \rangle ^{-\frac{1}{2}-\varepsilon }I\Vert _{L^2({\mathbb {R}}^{n+1})}\le C\Vert \langle D_y \rangle ^{\frac{1}{2}}f\Vert _{L^2({\mathbb {R}}^{n+1})},\quad \Vert \langle t \rangle ^{-\frac{1}{2}-\varepsilon }I\Vert _{L^2({\mathbb {R}}^{n+1})}\\&\quad \le C\Vert \langle t \rangle ^{\frac{1}{2}+\varepsilon }f\Vert _{L^2({\mathbb {R}}^{n+1})} \end{aligned}$$

Proof

The second inequality immediately follows from Hölder’s inequality. Thus, we shall show the first inequality. We write ${\hat{f}}(t,\eta )={\mathcal {F}}_{y\rightarrow \eta }f(t,\eta )$ and $a_2:=-{\mathrm{Im}\;}a=-{\mathrm{Im}\;}\sqrt{|\eta |^2-i}$, where ${\mathcal {F}}_{y\rightarrow \eta }$ denotes the Fourier transform from the variable y to the variable $\eta$. Fourier transforming in $y\rightarrow \eta$ and using Young’s inequality, we have

$$\begin{aligned} |{\hat{I}}(t,\eta )|&\le |\int _{{\mathbb {R}}} e^{-(t-s)a_2}|{\hat{f}}(s,\eta )|ds|\le \Vert e^{-sa_2}\Vert _{L^2({\mathbb {R}}_s)} \Vert {\hat{f}}(s,\eta )\Vert _{L^2({\mathbb {R}}_s)}\\&\quad \le C\Vert a_2^{-\frac{1}{2}}{\hat{f}}(s,\eta )\Vert _{L^2({\mathbb {R}}_s)}. \end{aligned}$$

This calculation gives $\Vert \langle t \rangle ^{-\frac{1}{2}-\varepsilon }{\hat{I}}(t,\eta )\Vert _{L^2_t}\le \Vert \langle t \rangle ^{-\frac{1}{2}-\varepsilon }\Vert _{L^2_t}\Vert {\hat{I}}(t,\eta )\Vert _{L^{\infty }_t}\le C\Vert a_2^{-\frac{1}{2}}{\hat{f}}(t,\eta )\Vert _{L^2_t}$. Plancherel’s theorem and (3.1) imply

$$\begin{aligned} \Vert \langle t \rangle ^{-\frac{1}{2}-\varepsilon }I\Vert _{L^2({\mathbb {R}}^{n+1}_{t,y})}\le C\Vert a_2^{-\frac{1}{2}}{\hat{f}}\Vert _{L^2({\mathbb {R}}^{n+1}_{t,\eta })}\le C\Vert \langle D_y \rangle ^{\frac{1}{2}}f\Vert _{L^2({\mathbb {R}}^{n+1}_{t,y})}. \end{aligned}$$

$\square$

Now we shall prove (1.1) or a stronger bound: $\Vert \langle t \rangle ^{-\frac{1}{2}-\varepsilon }\langle D_x \rangle ^{\frac{1}{2}}(P-i)^{-1}f\Vert _{L^2({\mathbb {R}}^{n+1})}\le C\Vert f\Vert _{L^2({\mathbb {R}}^{n+1})}$. To do this, it suffices to prove

$$\begin{aligned} \Vert \langle t \rangle ^{-\frac{1}{2}-\varepsilon }\langle D_y \rangle ^{\frac{1}{2}}(P-i)^{-1}f\Vert _{L^2({\mathbb {R}}^{n+1})}\le C\Vert f\Vert _{L^2({\mathbb {R}}^{n+1})}. \end{aligned}$$

(4.1)

where we recall $x=(t,y)\in {\mathbb {R}}\times {\mathbb {R}}^n$. In fact, we take $\varphi (D_x)=\psi (P/(-\Delta _x+1))$, where $\psi \!\in \! C_c^{\infty }({\mathbb {R}};[0,1])$ satisfies $\psi (s)\!=\!1$ on $|s|\!\le \! \frac{1}{4}$ and ${\mathrm{supp}\;}\psi \!\subset \! \{|s|\!\le \! \frac{1}{2}\}$. Moreover, we set $\chi (D_x)=\varphi (D_x)\langle D_x \rangle ^{\frac{1}{2}}\langle D_y \rangle ^{-\frac{1}{2}}$. Since P is elliptic on the essential support of $1-\varphi (D_{x})$, we have $\Vert (1-\varphi (D_x))\langle D_x \rangle ^{\frac{1}{2}}(P-i)^{-1}f\Vert _{L^2({\mathbb {R}}^{n+1})}\le C\Vert f\Vert _{L^2({\mathbb {R}}^{n+1})}$. Moreover, since $c^{-1}|\xi |\le |\eta |\le c|\xi |$ on the essential support of $\varphi (D_x)$ with a constant $c\ge 1$, it turns out that $\chi (D_x)$ is bounded in $L^2({\mathbb {R}}^{n+1})$. Hence,

$$\begin{aligned} \Vert \langle t \rangle ^{-\frac{1}{2}-\varepsilon }\varphi (D_x)\langle D_x \rangle ^{\frac{1}{2}}(P-i)^{-1}f\Vert _{L^2({\mathbb {R}}^{n+1})}\!=&\Vert \langle t \rangle ^{-\frac{1}{2}-\varepsilon }\chi (D_x)\langle D_y \rangle ^{\frac{1}{2}}(P-i)^{-1}f\Vert _{L^2({\mathbb {R}}^{n+1})}\\ \le&C\Vert \chi (D_x)f\Vert _{L^2({\mathbb {R}}^{n+1})}\le C\Vert f\Vert _{L^2({\mathbb {R}}^{n+1})}, \end{aligned}$$

due to (4.1). Combining these inequalities with (4.2), we obtain (1.1). Now we turn to the proof of (4.1). Lemma A.1 with the formula (3.2) immediately implies

$$\begin{aligned} \Vert \langle t \rangle ^{-\frac{1}{2}-\varepsilon }(P-i)^{-1}f\Vert _{L^2({\mathbb {R}}^{n+1})}\le C\Vert A^{-1}\langle D_y \rangle ^{\frac{1}{2}}f\Vert _{L^2({\mathbb {R}}^{n+1})}\le C\Vert \langle D_y \rangle ^{-\frac{1}{2}}f\Vert _{L^2({\mathbb {R}}^{n+1})}, \end{aligned}$$

(4.2)

which shows (4.1).

Finally, we prove (1.2) or a stronger bound: $\Vert \langle t \rangle ^{-\frac{1}{2}-\varepsilon }\langle D_x \rangle (P\!-i)^{-1}\langle t \rangle ^{-\frac{1}{2}-\varepsilon } f \Vert _{L^2({\mathbb {R}}^{n+1})}\le C\Vert f\Vert _{L^2({\mathbb {R}}^{n+1})}$. Lemma A.1 with the formula (3.2) implies

$$\begin{aligned} \Vert \langle t \rangle ^{-\frac{1}{2}-\varepsilon }\langle D_y \rangle (P-i)^{-1}\langle t \rangle ^{-\frac{1}{2}-\varepsilon }f \Vert _{L^2({\mathbb {R}}^{n+1})}\le C\Vert f\Vert _{L^2({\mathbb {R}}^{n+1})}, \end{aligned}$$

where we recall $x=(t,y)\in {\mathbb {R}}\times {\mathbb {R}}^n$. Let $\varphi$ be as in the proof of (4.1) and set $\chi _1(D_x)=\varphi (D_x)\langle D_x \rangle \langle D_y \rangle ^{-1}$. We repeat an argument similar to the proof of (4.1): Since P is elliptic on the essential support of $1-\varphi (D_{x})$, we have $\Vert (1-\varphi (D_x))\langle D_x \rangle (P-i)^{-1}f\Vert _{L^2({\mathbb {R}}^{n+1})}\le C\Vert f\Vert _{L^2({\mathbb {R}}^{n+1})}$. Moreover, since $c^{-1}|\xi | \le |\eta |\le c|\xi |$ on the essential support of $\varphi (D_x)$ with a constant $c\ge 1$, it turns out that $\chi _1(D_x)$ is bounded in $L^2({\mathbb {R}}^{n+1})$. Since $\chi _1(D_x)$ is a pseudodifferential operator of order 0, then $[\langle t \rangle ^{-\frac{1}{2}-\varepsilon },\chi _1(D_x)]$ is a pseudodifferential operator of order $-1$. Hence, $[\langle t \rangle ^{-\frac{1}{2}-\varepsilon },\chi _1(D_x)]\langle D_y \rangle$ is bounded on $L^2({\mathbb {R}}^{n+1})$. Thus, we have

$$\begin{aligned}&\Vert \langle t \rangle ^{-\frac{1}{2}-\varepsilon }\varphi (D_x)\langle D_x \rangle (P-i)^{-1}\langle t \rangle ^{-\frac{1}{2} -\varepsilon }f\Vert _{L^2({\mathbb {R}}^{n+1})}\\&=\Vert \langle t \rangle ^{-\frac{1}{2}-\varepsilon }\chi _1(D_x)\langle D_y \rangle (P-i)^{-1}\langle t \rangle ^{-\frac{1}{2} -\varepsilon }f\Vert _{L^2({\mathbb {R}}^{n+1})}\\&\le \Vert [\langle t \rangle ^{-\frac{1}{2}-\varepsilon },\chi _1(D_x)]\langle D_y \rangle \Vert _{L^2\rightarrow L^2} \Vert (P-i)^{-1}f\Vert _{L^2({\mathbb {R}}^{n+1})}\\&+C \Vert \langle t \rangle ^{-\frac{1}{2}-\varepsilon }\langle D_y \rangle (P-i)^{-1}\langle t \rangle ^{-\frac{1}{2}-\varepsilon }f \Vert _{L^2({\mathbb {R}}^{n+1})}\\&\le C\Vert f\Vert _{L^2({\mathbb {R}}^{n+1})}, \end{aligned}$$

This completes the proof of (1.2). See also the proof of [4, Theorem C.1].