Consider the Laplace operator $\Delta:=\sum_{j=1}^{n}\partial_{x_{j}}^{2}$ on $\mathbb{R}^n$. Let $E_{\lambda}$ be the spectral resolution of $\Delta$, and $$ H_{t}[f]:=\cos{(t\sqrt{-\Delta})}f=\int_{0}^{\infty}\cos{t\sqrt{\lambda}}dE_{\lambda}f\qquad \forall f\in L^2(\mathbb{R}^n)$$ be the Huygens operator.
For any $u_{0}\in C_{0}^{\infty}(\mathbb{R}^n)$, the function $$ u(x,t)=(H_{t}[u_{0}])(x)=\int_{0}^{\infty}\cos(t\sqrt{\lambda})dE_{\lambda}u_{0}$$ solves \begin{equation}\label{1-7} \left\{ \begin{array}{ll} \frac{\partial^{2}u}{\partial t^{2}}-\Delta u=0 , & \hbox{on $\mathbb{R}^n \times (0,+\infty)$;} \\ u_{t}(x,0)=0, & \hbox{$\forall x\in \mathbb{R}^n$;} \\ u(x,0)=u_{0}, & \hbox{$\forall x\in \mathbb{R}^n$.} \end{array} \right. \end{equation} The Huygens operator $H_t$ admits a kernel given by \begin{equation} w(x,y,t)=\int_{0}^{\infty}\cos(t\sqrt{\lambda})de(x,y,\lambda) \end{equation} Here, $e(x,y,\lambda)$ is the spectral function on $\mathbb{R}^n$. It follows that, for any fixed $y\in \mathbb{R}^n$, $w(x,y,t)$ satisfies the equation
\begin{equation} \left\{ \begin{array}{ll} (\frac{\partial^{2}}{\partial t^{2}}-\Delta_{x})w(x,y,t) =0 , & \hbox{on $\mathbb{R}^n \times (0,+\infty)$;} \\ w_{t}(x,y,0)=0, & \hbox{$\forall x\in \mathbb{R}^n$;} \\ w(x,y,0)=\delta_{y}(x), & \hbox{$\forall x\in \mathbb{R}^n$.} \end{array} \right. \end{equation}
Let $\Omega$ be an open bounded domain in $\mathbb{R}^n$. Denote by $\Delta_{\Omega}$ be the Dirichlet Laplacian operator on $\Omega$, and $E_{\lambda,\Omega}$ the corresponding spectral resolution. We can also defined the Huygens operator $$ H_{t}^{\Omega}[f]:=\cos{(t\sqrt{-\Delta_{\Omega}})}f=\int_{0}^{\infty}\cos{t\sqrt{\lambda}}dE_{\lambda,\Omega}f\qquad \forall f\in L^2(\Omega)$$ We denote by $w_{\Omega}(x,y,t)$ the kernel of $H_{t}^{\Omega}$: $$ w_{\Omega}(x,y,t)=\int_{0}^{\infty}\cos(t\sqrt{\lambda})de_{\Omega}(x,y,\lambda), $$ where $e_{\Omega}(x,y,\lambda)$ is the spectral function of $-\Delta_{\Omega}$. Similarly, for fixed $y\in \Omega$, $w_{\Omega}(x,y,t)$ satisfies the equations: \begin{equation} \left\{ \begin{array}{ll} (\frac{\partial^{2}}{\partial t^{2}}-\Delta_{x})w_{\Omega}(x,y,t) =0 , & \hbox{on $\Omega \times (0,+\infty)$;} \\ w_{\Omega}(x,y,t)=0, & \hbox{$\forall x\in \partial\Omega$;} \\ (w_{\Omega})_{t}(x,y,0)=0, & \hbox{$\forall x\in \Omega$;} \\ w_{\Omega}(x,y,0)=\delta_{y}(x), & \hbox{$\forall x\in \Omega$.} \end{array} \right. \end{equation}
In (1.6) of the paper Seeley, R., A sharp asymptotic remainder estimate for the eigenvalues of the Laplacian in a domain of (R^3), Adv. Math. 29, 244-269 (1978). ZBL0382.35043. , it claims the following fact that $$ w(x,y,t)=w_{\Omega}(x,y,t) \tag{*}$$ holds for all $|t|\leq d(y,\partial\Omega)$ and all $x\in \Omega$.
My Question: How can we prove the conclusion $(*)$ by means of the finite speed of propagation property of wave equation?
Remark: I review many other papers, but all of them just write "this is due to the "Finite Speed of Propagation Property". I feel very confuse about the details in applying the finite speed of propagation property. Can someone help me with above question? Thank you very much!
Recall that the Finite Speed of Propagation Property states that:
Let $u(x,t)$ solves $\frac{\partial^2 u}{\partial t^2}-\Delta u=0$. If $u(x,t)=u_{t}(x,t)=0$ on the set $B(x_0,t_0)\times\{t=0\}$, then $u(x,t)=0$ on $K(x_0,t_0):=\{(x,t)|0\leq t\leq t_0, |x-x_0|\leq t_0-t\}$.
