the eigenvalues $\{\lambda_n\}_{n=1}^{\infty}$ of $\mathcal{L}$ obey the condition $$\lim\limits_{n\rightarrow \infty} |\lambda_{n+1}-\lambda_n|=0$$ and, hence, the set $\{\lambda_n\}_{n=1}^{\infty}$ cannot be decomposed into a finite number of sets $\Lambda_k$ satisfying (\[1\]).

The eigenfunctions of $\mathcal{L}$ are $$\tilde{\psi}_n=Ai\left(e^{\frac{\pi i}{6}} x +\mu_n\right)\in H_{0}^{2} \left(\mathbb{R_{+}},\mathbb{C}\right),\: n\in\mathbb{N}.$$ Normalized eigenfunctions $\psi_n=\frac{\tilde{\psi}_n}{\left\| \tilde{\psi}_n\right\|},$ $n\in\mathbb{N},$ are dense in $L_2\left(\mathbb{R},\mathbb{C}\right)$, as it is proved in [@Almog], but do not form a Schauder basis of $L_2 \left(\mathbb{R},\mathbb{C}\right)$, since $\mathcal{L}$ is spectrally wild [@Davies3].

The sharpness of the XYZ Theorem
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We will use the notation from [@Sklyar3], see also Theorem \[construction\]. By Proposition 3 of [@Sklyar3] we have that for any $k\in\mathbb{N}$ the sequence $\{e_n\}_{n=1}^{\infty}$ is dense and minimal in $H_k\left(\{e_n\}\right)$, but it is not uniformly minimal in $H_k\left(\{e_n\}\right)$. It means that for each $n\in\mathbb{N}$ $$\varrho\left(e_n, \overline{Lin}\{e_j\}_{j\neq n} \right)>0,$$ but $$\inf\limits_{n\in\mathbb{N}}\varrho\left(e_n, \overline{Lin}\{e_j\}_{j\neq n} \right)=0,$$ where $\varrho\left(x,Y\right)$ denotes a standard distance from the point $x$ to a set $Y$, defined by $$\varrho\left(x,Y\right)=\inf\limits_{y\in Y} \|x-y\|.$$

Let $\{\phi_n\}_{n=1}^{\infty}$ be dense and minimal sequence in a Hilbert space $H$, but is not uniformly minimal in $H$. Then it can happen that there exists a splitting of $\{\phi_n\}_{n=1}^{\infty}$ into infinite number of disjoint groups with at most $K<\infty$ elements in each of them, i.e. $$\{\phi_n\}_{n=1}^{\infty}=\left\{\{\phi_j\}_{j\in A_n}\right\}_{n=1}^{\infty},$$ where 