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

If $\{e_n\}_{n=1}^{\infty}$ is a symmetric basis of $\ell_p$, then there exist two constants $\widetilde{M}\geq \widetilde{m}>0$ such that for every $\widetilde{y}=\sum\limits_{n=1}^{\infty} \alpha_n e_n\in \ell_p$ we have $$\widetilde{m} \|\widetilde{y}\|^p \leq \sum\limits_{n=1}^{\infty} |\alpha_n|^p \leq \widetilde{M} \|\widetilde{y}\|^p,$$ see [@Sklyar3] for details. Thus the proof repeats ideas and lines of the proof of Theorem \[Expl\].

Asymptotic behaviour of the resolvent
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For any closed linear operator $A$ on a Hilbert space $H$ the following bound for the norm of the resolvent is true: $$\label{resolvent bound}
\left\| \left(A-\lambda I\right)^{-1}\right\| \geq \frac{1}{\varrho\left(\lambda, \sigma(A)\right)},$$ provided that $\lambda\in \rho(A)$. Here $\varrho\left(\lambda, \sigma(A)\right)$ is the standard Euclidean distance between the point $\lambda$ and the spectrum $\sigma(A)$. If $A$ is normal operator on $H$, then by the spectral theorem for normal operators we immediately obtain that $$\label{resolvent b}
\left\| \left(A-\lambda I\right)^{-1}\right\| = \frac{1}{\varrho\left(\lambda, \sigma(A)\right)},$$ i.e. the inequality (\[resolvent bound\]) turns into an equality. However, equality (\[resolvent b\]) is not satisfied even for $2\times2$ nonselfadjoint matrix $$B=\left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right),$$ since $\sigma(B)=\{1\}$ and 