j\in A_n,\\
0,\quad j\notin A_n.\\
\end{array}
\right .$$

Thus $\left\|P_n x^{\ast} \right\|_k \nrightarrow 0$ as $n\to\infty$, which means that $x^{\ast}$ can not be represented in a convergent series $$\sum\limits_{n=1}^{\infty} x_n^{\ast}=\sum\limits_{n=1}^{\infty} P_n x^{\ast},$$ where $x_n^{\ast}\in \mathfrak{M}_n$ for each $n\in\mathbb{N}$. So we arrived at a contradiction with the definition of the Schauder decomposition.

Theorem \[sharp\] leads to the following.

\[cor sharp\] The XYZ Theorem is sharp. None of its conditions can be weakened.

Indeed, condition $3$ of the XYZ Theorem obviously can not be weakened. If one weakens condition $2$ but conditions $1$ and $3$ are fulfilled, then, by virtue of Theorem \[sharp\], the class of counterexamples are given by Theorem \[construction\].

Let us weaken condition $1$. Suppose that conditions $2$ and $3$ are satisfied, operator $A$ does not generate the $C_0$-group on $H$ but $A$ generates the $C_0$-semigroup on $H$. Then the counterexample is given as follows.

Let $\{\phi_n\}_{n=1}^{\infty}$ be a bounded non-Riesz basis of $H$, i.e. bounded conditional basis. It means that $\{\phi_n\}_{n=1}^{\infty}$ constitutes a Schauder basis of $H$, but does not form a Riesz basis of $H$, and we have $$0<\inf\limits_{n}\|\phi_n\|,\quad \sup\limits_{n}\|\phi_n\|<\infty.$$ Since $\{\phi_n\}_{n=1}^{\infty}$ is a Schauder basis of $H$, every $x\in H$ has a unique norm-convergent expansion $$x=\sum\limits_{n=1}^{\infty} 