&\leq \frac{2M}{a^2}\|x\|_1^2+ \frac{2}{a^4} \sum\limits_{n=2}^{\infty} n^2\left|\Delta f(n)\right|^2 \frac{\left|c_{n-1}\right|^2}{n^2}\\
&\leq \frac{2M}{a^2} \|x\|_1^2 + \frac{2C^2}{a^4} \sum\limits_{n=2}^{\infty}\frac{\left|c_{n-1}\right|^2}{n^2}\\
&\leq \frac{2M}{a^2} \|x\|_1^2 + \frac{2C^2}{a^4}  \sum\limits_{n=1}^{\infty} \left(\frac{1}{n} \sum\limits_{j=1}^{n}|c_j-c_{j-1}|\right)^2.\end{aligned}$$

By virtue of the Hardy inequality (\[Hardy\]) for $p=2$ we have $$\Xi \leq \frac{2M}{a^2} \|x\|_1^2 + \frac{8 MC^2}{a^4} \|x\|_1^2.$$ Hence (\[da\]) holds. Therefore $A(\lambda) x \in D(A_1)$ and thus, $$\label{18888}
\left(A_1-\lambda I \right)A(\lambda) x = \left(A_1-\lambda I \right)(\mathfrak{f})\sum\limits_{n=1}^{\infty} \frac{1}{\lambda_n-\lambda} c_n e_n =(\mathfrak{f})\sum\limits_{n=1}^{\infty} c_n e_n =x.$$

Now take $z\in D(A_1)$ and consider $x=\left(A_1-\lambda I \right) z.$ Then by (\[18888\]) we have that $$x=\left(A_1-\lambda I \right)A(\lambda) x = \left(A_1-\lambda I \right)A(\lambda)\left(A_1-\lambda I \right) z.$$ Consequently, $$\left(A_1-\lambda I \right)(z - A(\lambda)\left(A_1-\lambda I \right) z)=x-x=0.$$ Since $\lambda \neq if(n),$ $n\in \mathbb{N},$ then for every $z\in D(A_1)$ we have $$z = A(\lambda)\left(A_1-\lambda I \right)z,$$ and, combining this equality with (\[18888\]), we infer that $\lambda \in \rho(A_1)$ and $A(\lambda) = \left(A_1-\lambda I \right)^{-1}$ is the resolvent of $A_1$. Besides, we proved that $$\left\{ \lambda \in \mathbb{C}: \:\lambda \neq if(n), \: n\in \mathbb{N}\right\}\subset \rho(A_1).$$

Finally we observe that since $\lambda_n \in \sigma(A_1),\: n\in\mathbb{N},$ operator $A_1$ is closed as the generator of the $C_0$-group by Theorem \[construction\], the spectrum of closed operator is closed set and the set $\{i f(n)\}_{1}^{\infty}$ contains all its limit points, then $\sigma(A_1)= 