arbitrarily $$\lambda:\:\inf\limits_{n\in \mathbb{N}} |if(n)-\lambda|\geq a>0,$$ fix $x=(\mathfrak{f})\sum\limits_{n=1}^{\infty} c_n e_n\in H_1\left(\{e_n\}\right)$ and demonstrate that $A(\lambda) x \in D(A_1).$ For this purpose, taking into consideration (\[Domain\_k\]), it is sufficient to prove that $$\label{da}
\left\{\frac{i f(n) \cdot c_n}{i f(n)-\lambda}\right\}_{n=1}^{\infty}\in \ell_2(\Delta),$$ where $\ell_2(\Delta)=\left\{s=\{\alpha_n\}_{n=1}^{\infty}:\:\Delta s \in \ell_2\right\}$ and $\Delta$ is a difference operator given by $$\Delta=\left( \begin{array}{ccccc} 1 & 0 & 0 & 0 & \dots\\ -1 & 1 & 0 & 0 & \dots\\ 0 & -1 & 1 & 0 & \dots\\ 0 & 0 & -1 & 1 & \dots\\ \vdots & \vdots & \vdots & \vdots & \ddots\end{array} \right).$$

To this end observe that $$\begin{aligned}
\sum\limits_{n=2}^{\infty} &\left| \frac{i f(n)\cdot c_n}{i f(n) -\lambda} - \frac{if(n-1) \cdot c_{n-1}}{if(n-1)-\lambda} \right|^2\\
&=\sum\limits_{n=2}^{\infty} \left| \left(c_n+\frac{\lambda c_n}{if(n) -\lambda}\right) -\left(c_{n-1}+\frac{\lambda c_{n-1}}{if(n-1) -\lambda}\right)\right|^2\\
 & \qquad\leq 2\sum\limits_{n=1}^{\infty} |c_n-c_{n-1}|^2 +  2|\lambda|^2\sum\limits_{n=2}^{\infty} \left|\frac{ c_n}{if(n) -\lambda} -\frac{c_{n-1}}{if(n-1) -\lambda}  \right|^2\\
&\qquad\qquad\leq 2M \|x\|_1^2 + 2|\lambda|^2 \Xi.\end{aligned}$$

For any $n\geq 2$ we have $$\begin{aligned}
\frac{ c_n}{if(n) -\lambda} -\frac{c_{n-1}}{if(n-1) -\lambda} &=  \frac{ c_n}{if(n) -\lambda} -\frac{ c_{n-1}}{if(n) -\lambda} + \frac{ c_{n-1}}{if(n) -\lambda} - \frac{c_{n-1}}{if(n-1) -\lambda}\\
&= \frac{c_n- c_{n-1}}{if(n) -\lambda} + c_{n-1} \left( \frac{i(f(n-1)- f(n))}{(if(n) -\lambda)(if(n-1) -\lambda)}\right).\end{aligned}$$ It follows that $$\begin{aligned}
\Xi &\leq 2 \sum\limits_{n=2}^{\infty} \left|\frac{c_n- c_{n-1}}{if(n) -\lambda}  \right|^2 + 2 \sum\limits_{n=2}^{\infty} \left|c_{n-1} \frac{f(n-1)- f(n)}{(if(n) -\lambda)(if(n-1) -\lambda)} \right|^2\\
