这是一份andrews圣安德鲁斯大学 MT1001作业代写的成功案例
Proof:
For simplicity let us consider the Lindbladian $\mathcal{L}$ associated with an element $r=\sum_{g \in \mathcal{G}} c_{g} U_{g} \in \mathcal{A}$ such that $|r|{2}:=\sum{g \in \mathcal{G}}\left|c_{g}\right||g|^{2}<\infty$. Here $\mathcal{L}$ takes the form
Denoting these two bounded derivations $\left[r_{k}^{}, .\right]$ and $\left[., r_{k}\right]$ on $\mathcal{A}$ by $\delta_{k}^{\dagger}$ and $\delta_{k}$ respectively, $\mathcal{L}(x)=\frac{1}{2} \sum_{k \in \mathbb{Z}^{d}} \delta_{k}^{\dagger}(x) r_{k}+r_{k}^{} \delta_{k}(x)$.
In order to prove (i), for $x \in \mathcal{C}^{1}(\mathcal{A})$, let us estimate the norm of $\mathcal{L}(x)$ :
$$
|\mathcal{L}(x)| \leq \frac{1}{2} \sum_{k \in \mathbb{Z}^{d}}\left|\delta_{k}^{\dagger}(x) r_{k}+r_{k}^{*} \delta_{k}(x)\right|
$$
MT1001 COURSE NOTES ：
Summing over $\alpha^{\prime}$, it follows that
$$
\left|\left(x_{n}\right)\right|_{1} \leq\left|(\lambda-\Gamma)^{-1}\right||y|_{1}<\infty $$ and hence $x_{n} \in \mathcal{C}^{1}(\mathcal{A})$. Now setting $y_{n}=(\mathcal{L}-\lambda)\left(x_{n}\right)$, we have $$ \left|y_{n}-y\right|=\left|\left(\mathcal{L}-\mathcal{L}^{(n)}\right) x_{n}\right|=\sum_{|k|>n} \mathcal{L}{k}\left(x{n}\right)
$$