verify conditions 1-3 for a concrete system (\[Cauchy problem\]), while the important Riesz basis property of eigenvectors ($A$-invariant subspaces) is difficult to prove, the XYZ Theorem provides us a powerful machinery for the analysis of various applications.

From the other hand, it was totally unclear: What if the eigenvalues of $A$ lie in a strip, parallel to imaginary axis, and do not satisfy the condition of separation (\[1\])? In particular:

\[qq\] Is it possible to construct the unbounded generator of the $C_0$-group with eigenvalues $\{\lambda_n\}_{n=1}^{\infty}\subset i\mathbb{R}$ not satisfying (\[1\]) and dense family of eigenvectors, which does not form a Riesz basis?

In [@Sklyar3] the authors obtained an affirmative answer to the Question \[qq\] and presented the class of infinitesimal operators with such eigenvalues $\{\lambda_n\}_{n=1}^{\infty}$ and complete minimal family of eigenvectors, which, however, does not form even a Schauder basis. To formulate the corresponding result we need to consider the following classes of sequences.

\[Class Sk\]([@Sklyar3]) Let $k\in\mathbb{N}$ and $\Delta$ stands for the difference operator. Then we define $$\mathcal{S}_k=\Bigl\{\left\{f(n)\right\}_{n=1}^{\infty}\subset\mathbb{R}:\: \lim\limits_{n\rightarrow \infty} f(n)=+\infty; \left\{n^j \Delta^j f(n) \right\}_{n=1}^{\infty}\in \ell_{\infty}\:\: \text{for} \:\: 1\leq j\leq k\Bigr\}.$$

Clearly $\left\{\ln n\right\}_{n=1}^{\infty}\in \mathcal{S}_k$ for all $k$.

\[construction\]([@Sklyar3]) Assume that $\{e_n\}_{n=1}^{\infty}$ is a Riesz basis of a Hilbert space 