of the spectrum $\sigma(A_k)$ of generators $A_k.$

\[Expl\] Let $k\in\mathbb{N}$ and $A_k$ be the operator from Theorem \[construction\]. Then:

1.  $\sigma(A_k)= \sigma_p(A_k)=\{i f(n)\}_{1}^{\infty}$.

2.  The resolvent of $A_k$ is given by the following formula: $$\label{resolvent not basis fk}
    \left(A_k- \lambda I \right)^{-1} x= (\mathfrak{f})\sum\limits_{n=1}^{\infty} \frac{c_{n}  e_n}{i f(n)-\lambda}, \:\:\lambda \in  \rho(A_k)=\mathbb{C}\setminus \{i f(n)\}_{1}^{\infty},$$ where $x=(\mathfrak{f})\sum\limits_{n=1}^{\infty} c_{n} e_n\in H_k\left(\{e_n\}\right).$

First we prove the Theorem for the case $k=1$.

Let us prove that $\rho(A_1)=\mathbb{C}\setminus \{i f(n)\}_{1}^{\infty}$ is the resolvent set of the operator $A_1$ and the operator $$A(\lambda) x = (\mathfrak{f})\sum\limits_{n=1}^{\infty} \frac{1}{i f(n)-\lambda} c_n e_n,$$ where $\lambda \neq i f(n)$ for all $n\in \mathbb{N}$ and $x=(\mathfrak{f})\sum\limits_{n=1}^{\infty} c_n e_n\in H_1\left(\{e_n\}\right),$ is the resolvent of $A_1$.

To this end denote $\lambda_n=i f(n),$ $n\in \mathbb{N}$. Recall that the norm in Hilbert space $H_1\left(\{e_n\}\right)$ is $$\|x\|_1=\left\|(\mathfrak{f})\sum\limits_{n=1}^{\infty}c_n e_n\right\|_1=\left\|\sum\limits_{n=1}^{\infty}\left(c_{n} - c_{n-1}\right) e_n\right\|,$$ where $\|\cdot \|$ denotes the norm in an initial Hilbert space $H$ and $c_0=0$, see [@Sklyar3]. Observe that $$\begin{aligned}
\left\|A(\lambda) x\right\|_1^2 &= \left\|(\mathfrak{f})\sum\limits_{n=1}^{\infty}\frac{c_n e_n}{\lambda_n-\lambda} \right\|_1^2=\left\|\frac{c_1 e_1}{\lambda_1-\lambda} + \sum\limits_{n=2}^{\infty}\left( \frac{c_n}{\lambda_n-\lambda}-\frac{c_{n-1}}{\lambda_{n-1}-\lambda} \right) e_n \right\|^2\\
&=\left\|\frac{c_1 e_1}{\lambda_1-\lambda}+\sum\limits_{n=2}^{\infty}\left( \frac{c_n}{\lambda_n-\lambda} - \frac{c_{n-1}}{\lambda_n-\lambda} + \frac{c_{n-1}}{\lambda_n-\lambda} -\frac{c_{n-1}}{\lambda_{n-1}-\lambda} \right)  e_n \right\|^2\\
&\leq \left( \left\|\sum\limits_{n=1}^{\infty} \frac{c_n-c_{n-1}}{\lambda_n-\lambda}  e_n \right\| + \left\|\sum\limits_{n=2}^{\infty}\left( \frac{1}{\lambda_n-\lambda}-\frac{1}{\lambda_{n-1}-\lambda} \right) c_{n-1} e_n \right\| \right)^2\\
&= \left(\Sigma_1 +\Sigma_2\right)^2 \leq 2 \Sigma_1^2 + 2\Sigma_2^2.\end{aligned}$$

Now consider 