A Remark on the Essential Self-adjointness for Klein–Gordon-Type Operators

Abstract

We discuss a new simplified proof of the essential self-adjointness for formally self-adjoint differential operators of real principal type with the null non-trapping condition, previously proved by Vasy (J Spectr Theory 10: 439–461, 2020) and Nakamura-Taira (Ann Henri Lebesgue 4: 1035–1059, 2021). For simplicity, here we discuss the second-order cases, i.e., Klein–Gordon-type operators only.

Appendices

Appendix A. Proof of Lemma 2.1

Suppose $\psi$ satisfies the conditions of Lemma 2.1. At first we note that if $\varphi \in H^1(\mathbb {R}^n)$ and $P\varphi \in L^2(\mathbb {R}^n)$, then by the definition of the distributional derivative, we learn

$$\begin{aligned} \langle \varphi , P\varphi \rangle&= \sum _{j,k=1}^n \int g^{jk}(x)\overline{D_j\varphi (x)}D_k\varphi (x)\text {d}x \\&\quad +\textrm{Re}\left( \sum _{j=1}^n \int u_j \overline{\varphi (x)}D_j\varphi (x)\text {d}x\right) +\int u_0|\varphi (x)|^2 \text {d}x \in \mathbb {R}. \end{aligned}$$

We choose a smooth function $\chi \in C_0^{\infty }(\mathbb {R}^n;[0,1])$ such that $\chi (x)=1$ for $|x|\le 1$. We set $X_R\varphi (x)=\chi (x/R)\varphi (x)$ for $R>0$ and $\varphi \in L^2(\mathbb {R}^n)$.

Let $\psi \in L^2(\mathbb {R}^n)\cap H^{1,-1}(\mathbb {R}^n)$ such that $(P-z)\psi =0$. Now we apply the above formula to $\varphi =X_R\psi$. Then $X_R\psi \in H^1(\mathbb {R}^n)$ and $P(X_R\psi )\in L^2(\mathbb {R}^n)$ for each $R>0$, and hence we learn

$$\begin{aligned} \textrm{Im}\langle X_R\psi ,(P-z)X_R\psi \rangle = -\textrm{Im}\, z\Vert X_R\psi \Vert ^2. \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \langle X_R\psi ,(P-z)X_R\psi \rangle&=\langle X_R\psi , X_R(P-z)\psi \rangle +\langle X_R\psi ,[P,X_R]\psi \rangle \\&=\langle X_R\psi ,[P,X_R]\psi \rangle . \end{aligned}$$

It is easy to observe that $[P,X_R]$ is a first-order differential operator with the coefficients uniformly bounded by $C\langle x \rangle ^{-1}$, and converges to 0 pointwise as $R\rightarrow \infty$. Thus $[P,X_R]\psi$ is bounded by an $L^2$ function, and then by the dominated convergence theorem, we have $\langle X_R\psi ,[P,X_R]\psi \rangle \rightarrow 0$ as $R\rightarrow \infty$. Now we conclude

$$\begin{aligned} -\textrm{Im}\, z\Vert \psi \Vert ^2 =\lim _{R\rightarrow \infty }\bigl (-\textrm{Im}\, z\Vert X_R\psi \Vert ^2\bigr ) =\lim _{R\rightarrow \infty }\langle X_R\psi ,[P,X_R]\psi \rangle =0, \end{aligned}$$

and thus $\psi =0$. $\square$

Appendix B. Proof of the Basic Commutator Estimate

In this appendix, we prove a basic inequality used in Sect. 2.3. More precisely, we show

$$\begin{aligned} i[B^*B, P] \ge \frac{c}{h}B^* \langle x \rangle ^{-1} B -\tilde{B}^* \langle x \rangle ^{-1}\tilde{B} -E^*E, \end{aligned}$$

(B.1)

implies

$$\begin{aligned}&\frac{c}{2h}\bigl \Vert \langle x \rangle ^{-1/2}B\varphi \bigr \Vert ^2 +2(\textrm{Im}z)\Vert B\varphi \Vert ^2\nonumber \\&\qquad \le \frac{2h}{c}\bigl \Vert \langle x \rangle ^{1/2}B(P-z)\varphi \bigr \Vert ^2 +\Vert \langle x \rangle ^{-1/2}\tilde{B}\varphi \Vert ^2 +\Vert E\varphi \Vert ^2, \end{aligned}$$

(B.2)

where $\Vert \cdot \Vert =\Vert \cdot \Vert _{L^2(\mathbb {R}^n)}$.

At first, we prove (B.2) for $\varphi \in \mathcal {S}(\mathbb {R}^n)$. If $\varphi \in \mathcal {S}(\mathbb {R}^n)$, we have

$$\begin{aligned} i\langle \varphi ,[B^*B,P]\varphi \rangle&= i\langle \varphi , (B^*B P-PB^*B)\varphi \rangle \\&= i(\langle B\varphi ,B(P-z)\varphi \rangle -\langle B(P-\bar{z})\varphi ,B\varphi \rangle )\\&= -2\textrm{Im}(\langle B\varphi ,B(P-z)\varphi \rangle ) -2(\textrm{Im}z)\Vert B\varphi \Vert ^2\\&\le 2\Vert \langle x \rangle ^{-1/2} B\varphi \Vert \,\Vert \langle x \rangle ^{1/2} B(P-z)\varphi \Vert -2(\textrm{Im}z)\Vert B\varphi \Vert ^2. \end{aligned}$$

On the other hand, we have

$$\begin{aligned}&\frac{c}{h}\langle \varphi ,B\langle x \rangle ^{-1}B\varphi \rangle -\langle \varphi ,\tilde{B}^*\langle x \rangle ^{-1}\tilde{B}\varphi \rangle -\langle \varphi ,E^*E\varphi \rangle \\&\qquad = \frac{c}{h}\Vert \langle x \rangle ^{-1/2}B\varphi \Vert ^2-\Vert \langle x \rangle ^{-1/2}\tilde{B}\varphi \Vert ^2-\Vert E\varphi \Vert ^2. \end{aligned}$$

Combining them with our assumption (B.1), we learn

$$\begin{aligned}&\frac{c}{h}\Vert \langle x \rangle ^{-1/2}B\varphi \Vert ^2+2(\textrm{Im}z)\Vert B\varphi \Vert ^2\\&\qquad \le 2\Vert \langle x \rangle ^{-1/2} B\varphi \Vert \,\Vert \langle x \rangle ^{1/2} B(P-z)\varphi \Vert +\Vert \langle x \rangle ^{-1/2}\tilde{B}\varphi \Vert ^2 +\Vert E\varphi \Vert ^2. \end{aligned}$$

Now we use the elementary bound:

$$\begin{aligned} \Vert \langle x \rangle ^{-1/2} B\varphi \Vert \,\Vert \langle x \rangle ^{1/2} B(P-z)\varphi \Vert \le \frac{c}{4h}\Vert \langle x \rangle ^{-1/2} B\varphi \Vert ^2+ \frac{h}{c}\Vert \langle x \rangle ^{1/2} B(P-z)\varphi \Vert ^2 \end{aligned}$$

in the right-hand side, and we obtain (B.2) for $\varphi \in \mathcal {S}(\mathbb {R}^n)$.

In applications, we use (B.2) for $\varphi \in L^2(\mathbb {R}^n)$ such that $(P-z)\varphi \in H^{0,1/2+\gamma }$, and we need to show the inequality extends to such functions. Since $B, \tilde{B}, E\in \bigcap _{m\in \mathbb {R}}\textrm{Op}S^{m,\gamma }$, it is easy to observe that (B.2) is extended to $\varphi \in \bigcap _{\ell \in \mathbb {R}} H^{0,\ell }$.

Now let A be one of the operators $\langle x \rangle ^{-1/2}B$, B, $\langle x \rangle ^{1/2}B(P-z)$, $\langle x \rangle ^{-1/2}\tilde{B}$ and E. Let $X_R$ be the operator used in the last Appendix. Then $[A,X_R]$ is a pseudodifferential operator with the symbol which is bounded in $S^{0,-1/2+\gamma }$ and supported in $\textrm{supp}[\nabla X_R]\subset \{|x|\ge R\}$. These imply

$$\begin{aligned} \Vert [A,X_R] \Vert _{L^2\rightarrow L^2}\le CR^{-1/2+\gamma }\rightarrow 0, \quad \text {as }R\rightarrow \infty \end{aligned}$$

by the $L^2$-boundedness theorem for pseudodifferential operators. Using this, and since $X_R\varphi \in \bigcap _{\ell \in \mathbb {R}} H^{0,\ell }(\mathbb {R}^n)$ if $\varphi \in L^2(\mathbb {R}^n)$, we have

$$\begin{aligned}&\frac{c}{2h}\bigl \Vert \langle x \rangle ^{-1/2}B\varphi \bigr \Vert ^2 +2(\textrm{Im}z)\Vert B\varphi \Vert ^2 \\&\qquad =\lim _{R\rightarrow \infty }\biggl (\frac{c}{2h}\bigl \Vert X_R\langle x \rangle ^{-1/2}B\varphi \bigr \Vert ^2 +2(\textrm{Im}z) \Vert X_R B\varphi \Vert ^2\biggr ) \\&\qquad \le \lim _{R\rightarrow \infty }\biggl (\frac{2h}{c}\bigl \Vert X_R\langle x \rangle ^{1/2}B(P-z)\varphi \bigr \Vert ^2 +\Vert X_R\langle x \rangle ^{-1/2}\tilde{B}\varphi \Vert ^2 +\Vert X_RE\varphi \Vert ^2\biggr )\\&\qquad = \frac{2h}{c}\bigl \Vert \langle x \rangle ^{1/2}B(P-z)\varphi \bigr \Vert ^2 +\Vert \langle x \rangle ^{-1/2}\tilde{B}\varphi \Vert ^2 +\Vert E\varphi \Vert ^2, \end{aligned}$$

provided $\varphi \in L^2(\mathbb {R}^n)$ and $(P-z)\varphi \in H^{0,1/2+\gamma }$. $\square$

Appendix C. Proof of Lemmas 3.1 and 5.1

Proof of Lemma 3.1

It suffices to prove that for each $j=2,3$,

$$\begin{aligned} \{p_2,\zeta _1(x/R,\xi )\} \le 0,\quad \{p_2,\zeta _j(x,\xi )\}\le 0 \end{aligned}$$

on $\textrm{supp}[\zeta _1(\cdot /R,\cdot )\zeta _2\zeta _3]$ for sufficiently large R.

Throughout this proof, we denote $\delta =\sigma -\sigma '>0$ for simplicity. At first, we consider the estimate for $\zeta _1(x/R,\xi )$. We note

$$\begin{aligned} \textrm{supp}[\zeta _1]\subset \bigl \{(x,\xi )\bigm |x_\xi ^\parallel \le -1+\tfrac{1}{2}|x_\xi ^\perp |^2\bigr \} \subset \bigl \{(x,\xi )\bigm ||x|\ge 1\bigr \}, \end{aligned}$$

and

$$\begin{aligned} \textrm{supp}[\partial _{(x,\xi )}\zeta _1]\subset \bigl \{(x,\xi )\bigm |-2+\tfrac{1}{2}|x_\xi ^\perp |^2 \le x_\xi ^\parallel \le -1+\tfrac{1}{2}|x_\xi ^\perp |^2\bigr \}. \end{aligned}$$

Moreover,

$$\begin{aligned} \hat{v}(\xi )\cdot \partial _x \zeta _1(x,\xi ) ={\chi }_1'\bigl (x_\xi ^\parallel -\tfrac{1}{2} |x^\perp _\xi |^2+1\bigr )\le 0. \end{aligned}$$

Since $\partial _\xi p_2=v(\xi )+O(|\xi |\langle x \rangle ^{-\mu })$, we have

$$\begin{aligned} \partial _\xi p_2\cdot \partial _x \zeta _1(x/R,\xi ) = R^{-1} {\chi }_1'\bigl ((x/R)_\xi ^\parallel -\tfrac{1}{2} |(x/R)^\perp _\xi |^2+1\bigr )\cdot \bigl (1+O(R^{-\mu })\bigr )|\xi |. \end{aligned}$$

Similarly, since $\partial _x p_2=O(|\xi |^2\langle x \rangle ^{-1-\mu })$ and $\zeta _1(x,\xi )$ is homogeneous in $\xi$, we learn

$$\begin{aligned} \partial _x p_2\cdot \partial _\xi \zeta _1(x/R,\xi ) = {\chi }_1'\bigl ((x/R)_\xi ^\parallel -\tfrac{1}{2} |(x/R)^\perp _\xi |^2+1\bigr )\cdot O(R^{-1-\mu })|\xi |. \end{aligned}$$

These imply

$$\begin{aligned} \{p_2,\zeta _1(x/R,\xi )\} = R^{-1}{\chi }_1'\bigl ((x/R)_\xi ^\parallel -\tfrac{1}{2} |(x/R)^\perp _\xi |^2+1\bigr )\cdot \bigl (1+O(R^{-\mu })\bigr )|\xi | \le 0 \end{aligned}$$

on $\textrm{supp}[\zeta _1(x/R,\xi )\zeta _2(x,\xi )\zeta _3(x,\xi )]$ for sufficiently large R.

Next, we deal with the estimate for $\zeta _2$. We recall $\zeta _2$ is homogenous in $(x,\xi )$, and we note

$$\begin{aligned} \partial _x\zeta _2(x,\xi )&= (\partial _x\beta (x,\xi )) {\chi }_1'((\beta (x,\xi )-\sigma )/\delta ))/\delta \\&=|x|^{-1}(\hat{v}(\xi )-\hat{x} \beta (x,\xi ))\cdot {\chi }_1'((\beta (x,\xi )-\sigma )/\delta )/\delta , \end{aligned}$$

and in particular

$$\begin{aligned} \hat{v}(\xi )\cdot \partial _x\zeta _2(x,\xi )&=|x|^{-1}(1-\beta (x,\xi )^2) {\chi }_1'((\beta (x,\xi )-\sigma )/\delta )/\delta \\&\le \delta ^{-1}|x|^{-1} (1-\sigma ^2){\chi }_1'((\beta (x,\xi )-\sigma )/\delta ) \le 0. \end{aligned}$$

Similarly to the argument for $\zeta _1$, if $|x|\ge R$ then we have

$$\begin{aligned} \partial _\xi p_2\cdot \partial _x \zeta _2 =\delta ^{-1}|x|^{-1} (1-\sigma ^2){\chi }_1'((\beta (x,\xi )-\sigma )/\delta ) \cdot (1+O(R^{-\mu }))|\xi | \end{aligned}$$

and

$$\begin{aligned} \partial _x p_2\cdot \partial _\xi \zeta _2(x,\xi )= {\chi }_1'((\beta (x,\xi )-\sigma )/\delta )\cdot O(|x|^{-1-\mu }|\xi |). \end{aligned}$$

Combining these, we have

$$\begin{aligned} \{p_2,\zeta _2\} =\delta ^{-1}|x|^{-1} (1-\sigma ^2){\chi }_1'((\beta (x,\xi )-\sigma )/\delta ) \cdot (1+O(R^{-\mu }))|\xi |\le 0 \end{aligned}$$

on $\textrm{supp}[\zeta _1(x/R,\xi )\zeta _2(x,\xi )\zeta _3(x,\xi )]$.

Finally, we consider the estimate for $\zeta _3$. We now note $\tau (x,\xi )$ is the length of the line segment $\bigl \{x+t\hat{v}(\xi )\bigm |t\ge 0\bigr \}$ inside $\bigl \{(x,\xi )\bigm |\beta (x,\xi )\le \sigma _\infty \bigr \}$. We recall

$$\begin{aligned} \tau (x,\xi )=c_0\sqrt{|x|^2-(x\cdot \hat{v}(\xi ))^2}-x\cdot \hat{v}(\xi ) =c_0 |x_\xi ^\perp | -x\cdot \hat{v}(\xi ), \end{aligned}$$

and hence

$$\begin{aligned} \partial _x \tau (x,\xi )= c_0\, \widehat{x_\xi ^\perp } -\hat{v}(\xi ), \end{aligned}$$

and in particular,

$$\begin{aligned} -v(\xi )\cdot \partial _x \tau (x,\xi )=|v(\xi )|. \end{aligned}$$

(C.1)

We also note

$$\begin{aligned} c_1|x|\le \tau (x,\xi )\le C_1|x| \quad \text {for }(x,\xi )\in \textrm{supp}[\zeta _2] \end{aligned}$$

with some $0<c_1<C_1$. We also note

$$\begin{aligned} \partial _\xi \zeta _3(x,\xi ) = {\chi }_2'\bigl (\tfrac{|\xi |^2-1}{\lambda (x,\xi )}\bigr ) \biggl (\frac{2\xi }{\lambda (x,\xi )} -(|\xi |^2-1)\frac{\partial _\xi \lambda (x,\xi )}{\lambda (x,\xi )^2}\biggr ) \end{aligned}$$

where $(\cdots )$ is smooth and uniformly bounded on the support of ${\chi }_2'(\cdots )$. On the other hand,

$$\begin{aligned} \partial _x \zeta _3(x,\xi )&= -{\chi }_2'\bigl (\tfrac{|\xi |^2-1}{\lambda (x,\xi )}\bigr )(|\xi |^2-1) \frac{\partial _x\lambda (x,\xi )}{\lambda (x,\xi )^2}\\&=-{\chi }_2'\bigl (\tfrac{|\xi |^2-1}{\lambda (x,\xi )}\bigr )\frac{|\xi |^2-1}{\lambda (x,\xi )^2}\frac{\nu \delta _0\tau (x,\xi )\partial _x\tau (x,\xi )}{\langle \tau (x,\xi ) \rangle ^{2+\nu }}, \end{aligned}$$

and in particular, by (C.1), we have

$$\begin{aligned} v(\xi )\cdot \partial _x \zeta _3(x,\xi ) = {\chi }_2'\bigl (\tfrac{|\xi |^2-1}{\lambda (x,\xi )}\bigr )\frac{|\xi |^2-1}{\lambda (x,\xi )^2}\frac{\nu \delta _0\tau (x,\xi )|v(\xi )|}{\langle \tau (x,\xi ) \rangle ^{2+\nu }}. \end{aligned}$$

This also implies

$$\begin{aligned} \partial _\xi p_2(x,\xi )\cdot \partial _x\zeta _3(x,\xi )= {\chi }_2'\bigl (\tfrac{|\xi |^2-1}{\lambda (x,\xi )}\bigr )\frac{|\xi |^2-1}{\lambda (x,\xi )^2}\frac{\nu \delta _0\tau (x,\xi )|v(\xi )|}{\langle \tau (x,\xi ) \rangle ^{2+\nu }}(1+O(\langle x \rangle ^{-\mu })). \end{aligned}$$

Noting ${\chi }'_2(t)t\le 0$ for $t\in \mathbb {R}$ and $|t|\ge 1$ on $\textrm{supp}[{\chi }_2'(t)]$, we learn that

$$\begin{aligned} \partial _\xi p_2(x,\xi )\cdot \partial _x\zeta _3(x,\xi ) \le - c_2 \bigl | {\chi }_2'\bigl (\tfrac{|\xi |^2-1}{\lambda (x,\xi )}\bigr ) \bigr |\langle x \rangle ^{-1-\nu } \end{aligned}$$

with some $c_2>0$ on $\textrm{supp}[\zeta _2]$. On the other hand, using $\partial _x p_2(x,\xi )=O(|\xi |^2$$\langle x \rangle ^{-1-\mu })$ again, we have

$$\begin{aligned} \partial _x p_2(x,\xi )\cdot \partial _\xi \zeta _3(x,\xi )= {\chi }_2'\bigl (\tfrac{|\xi |^2-1}{\lambda (x,\xi )}\bigr )\times O(|\xi |^3\langle x \rangle ^{-1-\mu }). \end{aligned}$$

Since $|\xi |$ is bounded on $\textrm{supp}\zeta _3$, these imply

$$\begin{aligned} \{p_2,\zeta _3\} \le -c_2\bigl | {\chi }_2'\bigl (\tfrac{|\xi |^2-1}{\lambda (x,\xi )}\bigr ) \bigr |(\langle x \rangle ^{-1-\nu }-C\langle x \rangle ^{-1-\mu })\le 0 \end{aligned}$$

on $\textrm{supp}[\zeta _1(x/R,\xi )\zeta _2(x,\xi )\zeta _3(x,\xi )]$ with sufficiently large R. $\square$

Proof of Lemma 5.1

At first, we note if we set

$$\begin{aligned} \rho (x,\xi )=\{p_2,\tilde{\zeta }_1(x,\xi )\tilde{\zeta }_2(x,\xi )\}, \end{aligned}$$

then $\rho$ satisfies the properties of the lemma, and it suffices to show $\{p_2,\tilde{\zeta }_3\}$ is non-positive on the support to prove inequality (5.2). The computation is almost identical to the one in the proof of Lemma 3.1 above, but we remark necessary changes. Even though the definition of $\lambda _+(x,\xi )$ is different from $\lambda (x,\xi )$, we have the same derivative formula:

$$\begin{aligned} \partial _x\lambda _+(x,\xi ) =-\nu \delta _0\frac{\tau (x,\xi )\partial _x\tau (x,\xi )}{\langle \tau (x,\xi ) \rangle ^{\nu +2}}, \end{aligned}$$

and we have the same bound eventually:

$$\begin{aligned} \partial _\xi p_2(x,\xi )\cdot \partial _x\tilde{\zeta }_3(x,\xi ) \le - c_2 \bigl | {\chi }_2'\bigl (\tfrac{|\xi |^2-1}{\lambda (x,\xi )}\bigr ) \bigr |\langle x \rangle ^{-1-\nu }. \end{aligned}$$

The rest of the computation is carried out without changes to conclude $\tilde{\zeta }_1\tilde{\zeta }_2\{p_2,$$\tilde{\zeta }_3\}\le 0$ with sufficiently large R. $\square$