$\lambda:\:\inf\limits_{n\in \mathbb{N}} |\lambda_n-\lambda|\geq a>0$. Since $\{e_n\}_{n=1}^{\infty}$ is a Riesz basis of a Hilbert space $H$ (see Theorem \[construction\]), there exist two constants $M\geq m>0$ such that for every $y=\sum\limits_{n=1}^{\infty} \alpha_n e_n\in H$ we have $$\label{Riesz}
  m \|y\|^2 \leq \sum\limits_{n=1}^{\infty} |\alpha_n|^2 \leq M \|y\|^2.$$ Applying (\[Riesz\]) we obtain that $$\label{est Sigma 1}
\Sigma_1^2 \leq \frac{1}{m} \sum\limits_{n=1}^{\infty} \frac{|c_n-c_{n-1}|^2}{|\lambda_n-\lambda|^2}\leq \frac{1}{m a^2} \sum\limits_{n=1}^{\infty}|c_n-c_{n-1}|^2 \leq \frac{M}{m a^2} \|x\|_1^2.$$

Since $$\frac{1}{\lambda_n-\lambda}-\frac{1}{\lambda_{n-1}-\lambda} = \frac{\lambda_{n-1}-\lambda_n}{(\lambda_n-\lambda)(\lambda_{n-1}-\lambda)}$$ for $n\geq 2$, by virtue of (\[Riesz\]) we conclude that $$\Sigma_2^2 \leq \frac{1}{m} \sum\limits_{n=2}^{\infty}  \frac{|\lambda_{n-1}-\lambda_n|^2 |c_{n-1}|^2}{|\lambda_n-\lambda|^2 |\lambda_{n-1}-\lambda|^2}\leq \frac{1}{m a^4} \sum\limits_{n=2}^{\infty} \frac{|c_{n-1}|^2}{n^2} n^2 |\Delta f(n)|^2.$$ Note that $\left\{f(n)\right\}_{n=1}^{\infty}\in\mathcal{S}_1$, hence $n|\Delta f(n)|\in \ell_{\infty}$ by the definition of the class $\mathcal{S}_1$, see Definition \[Class Sk\]. Denote $$C=\sup\limits_{n\in\mathbb{N}} n|\Delta f(n)|.$$ Then, since for $n\geq 2$ $$c_{n-1}= \sum\limits_{j=1}^{n-1} \left(c_j-c_{j-1}\right),$$ we obtain that $$\begin{aligned}
\Sigma_2^2 &\leq \frac{C^2}{m a^4} \sum\limits_{n=2}^{\infty} \frac{|c_{n-1}|^2}{n^2}  =  \frac{C^2}{m a^4} \sum\limits_{n=2}^{\infty}  \left(\frac{1}{n} \left|\sum\limits_{j=1}^{n-1}(c_j-c_{j-1})\right|\right)^2 \\
&\leq \frac{C^2}{m a^4} \sum\limits_{n=1}^{\infty} \left(\frac{1}{n} \sum\limits_{j=1}^{n}|c_j-c_{j-1}|\right)^2.\end{aligned}$$

By virtue of the Hardy inequality (\[Hardy\]) for $p=2$ and (\[Riesz\]) we obtain $$\Sigma_2^2  \leq \frac{4C^2}{m a^4}  \sum\limits_{n=1}^{\infty} |c_n-c_{n-1}|^2 \leq  \frac{4 MC^2}{m a^4} \|x\|_1^2.$$ Combining this with (\[est Sigma 1\]) we finally arrive at the estimate $$\label{alambda}
\left\|A(\lambda) x\right\|_1^2 \leq \left(\frac{2}{a^2}+\frac{8C^2}{a^4} \right) \frac{M}{m} \|x\|_1^2$$ and $A(\lambda)\in[H_1\left(\{e_n\}\right)],$ i.e. $A(\lambda)$ is a linear bounded operator.

Further we choose 