of $T_{\rho}$ we note that due to linearity it suffices to show for $u\in L_{2,\rho}({\mathbb{R}};H)$ with ${\operatorname{spt}}u\subseteq{\mathbb{R}}_{\geq a}$ for some $a\in{\mathbb{R}}$ it follows that ${\operatorname{spt}}T_{\rho}u\subseteq{\mathbb{R}}_{\geq a}.$ So, let $u\in L_{2,\rho}({\mathbb{R}};H)$ with ${\operatorname{spt}}u\subseteq{\mathbb{R}}_{\geq a}$ for some $a\in{\mathbb{R}}$. Then $u\in\bigcap_{\mu\geq\rho}L_{2,\mu}({\mathbb{R}};H).$ For $\mu\geq\rho$ we compute $$\begin{aligned}
\intop_{-\infty}^{a}|\left(T_{\rho}u\right)(t)|\text{ d}t & =\intop_{-\infty}^{a}|\left(T_{\mu}u\right)(t)|{\mathrm{e}}^{-\mu t}{\mathrm{e}}^{\mu t}\text{ d}t\\
 & \leq\left|T_{\mu}u\right|_{\mu}\frac{1}{\sqrt{2\mu}}{\mathrm{e}}^{\mu a}\\
 & \le\frac{1}{\sqrt{2\mu}}|T|_{\mathrm{ev},\rho_{0}}|u|_{\mu}{\mathrm{e}}^{\mu a}\\
 & =\frac{1}{\sqrt{2\mu}}|T|_{\mathrm{ev},\rho_{0}}\left(\intop_{a}^{\infty}|u(t)|^{2}{\mathrm{e}}^{-2\mu t}\text{ d}t\right)^{\frac{1}{2}}{\mathrm{e}}^{\mu a}\\
 & =\frac{1}{\sqrt{2\mu}}|T|_{\mathrm{ev},\rho_{0}}\left(\intop_{a}^{\infty}|u(t)|^{2}{\mathrm{e}}^{-2\rho t}{\mathrm{e}}^{2(\rho-\mu)t}\text{ d}t\right)^{\frac{1}{2}}{\mathrm{e}}^{\mu a}\\
 & \leq\frac{1}{\sqrt{2\mu}}|T|_{\mathrm{ev},\rho_{0}}|u|_{\rho}{\mathrm{e}}^{\rho a}.\end{aligned}$$ Letting $\mu\to\infty,$ we infer that $T_{\rho}u=0$ on $]-\infty,a]$ and hence, ${\operatorname{spt}}T_{\rho}u\subseteq\mathbb{R}_{\geq a}.$

\[prop:causality\_aux\]Assume and let $\delta>\|\mathcal{M}'\|_{\mathrm{ev},\rho_{0}}+\|\mathcal{N}\|_{\mathrm{ev},\rho_{0}.}$ Then, for each $\rho\geq\rho_{0}$ the solution operator $\mathcal{S}_{\rho,\mathrm{aux}}$ is causal.

Let $\rho\geq\rho_{0}.$ By it suffices to prove that hold if we replace $\mathcal{N}_{\rho}$ by $\delta-\mathcal{M}_{\rho}'$. Note that only (b) has to be verified for $\mathcal{N}$ replaced by $\delta-\mathcal{M}'.$ So, let $\varphi\in C_{c}^{\infty}({\mathbb{R}};H)$ and $a\in{\mathbb{R}}.$ Note that by we have that $${\operatorname{Re}}\langle(\partial_{t}\mathcal{M}+\mathcal{N})\varphi,\chi_{\mathbb{R}_{\leq a}}\varphi\rangle_{\rho}\geq c|\chi_{\mathbb{R}_{\leq a}}\varphi|_{\rho}^{2}.$$ Moreover, we note that by we have $$\chi_{\mathbb{R}_{\leq a}}(\mathcal{M}'+\mathcal{N})\varphi=\chi_{\mathbb{R}_{\leq a}}(\mathcal{M}'+\mathcal{N})\chi_{\mathbb{R}_{\le a}}\varphi.$$ Hence, we can estimate $$\begin{aligned}
{\operatorname{Re}}\langle(\partial_{t}\mathcal{M}-\mathcal{M}'+\delta)\varphi,\chi_{\mathbb{R}_{\leq a}}\varphi\rangle_{\rho} & ={\operatorname{Re}}\langle(\partial_{t}\mathcal{M}+\mathcal{N})\varphi,\chi_{\mathbb{R}_{\leq a}}\varphi\rangle_{\rho}+{\operatorname{Re}}\langle\delta\varphi-(\mathcal{M}'+\mathcal{N})\varphi,\chi_{{\mathbb{R}}_{\leq a}}\varphi\rangle_{\rho}\\
 & \geq c|\chi_{\mathbb{R}_{\leq a}}\varphi|_{\rho}^{2}+\delta|\chi_{{\mathbb{R}}_{\leq a}}\varphi|_{\rho}^{2}-{\operatorname{Re}}\langle\chi_{{\mathbb{R}}_{\leq a}}(\mathcal{M}'+\mathcal{N})\varphi,\chi_{{\mathbb{R}}_{\leq a}}\varphi\rangle_{\rho}\\
 & =c|\chi_{\mathbb{R}_{\leq a}}\varphi|_{\rho}^{2}+\delta|\chi_{{\mathbb{R}}_{\leq a}}\varphi|_{\rho}^{2}-{\operatorname{Re}}\langle(\mathcal{M}'+\mathcal{N})\chi_{{\mathbb{R}}_{\leq a}}\varphi,\chi_{{\mathbb{R}}_{\leq a}}\varphi\rangle_{\rho}\\
 & \geq\left(c+\delta-(\|\mathcal{M}'\|_{\mathrm{ev},\rho_{0}}+\|\mathcal{N}\|_{\mathrm{ev},\rho_{0}})\right)|\chi_{\mathbb{R}_{\leq a}}\varphi|_{\rho}^{2}\\
 & \ge c|\chi_{\mathbb{R}_{\leq a}}\varphi|_{\rho}^{2}.\tag*{\qedhere}\end{aligned}$$

With this result at hand, we are able to prove the independence on the parameter $\rho$ for the solution 