$\bar{T}_{\psi,\bm{e}}$ has a densely defined inverse.

\(iii) $T_{\psi,\bm{e}}^{\ast}(=T_{\bm{e},\psi})$ has a densely defined inverse.\
If this holds, then $T_{\bm{e},\psi}^{-1}= ( \bar{T}_{\psi,\bm{e}}^{-1})^{\ast}$.\
\

[*Proof.*]{} This is shown similarly to Lemma 2.4.\
\

Semi-regular biorthogonal pairs and generalized Riesz bases
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In Ref. [@h-t], the author has defined the notion of generalized Riesz bases. First we redefine the notion of generalized Riesz bases.\

[*Definition 3.1.*]{} [*If there exists a densely defined closed operator $T$ in ${\cal H}$ with a densely defined inverse and there exists an ONB $\bm{e}= \{ e_{n} \}$ in ${\cal H}$ such that $$\begin{aligned}
\{ e_{n} \} \subset D(T) \cap D \left( (T^{-1})^{\ast} \right) \;\;\; {\rm and} \;\;\; Te_{n} = \phi_{n}, \;\;\; n=0,1, \cdots , \nonumber\end{aligned}$$ then a sequence $\{ \phi_{n} \}$ in ${\cal H}$ is called a generalized Riesz basis with a constructing pair $( \bm{e} , T )$.*]{}\
\
Here, we delete the conditions of Definition 2.1, (ii) and (iii) in Ref. [@h-t], that is, $D_{\phi}$ and $D_{\psi}$ are not necessarily dense in ${\cal H}$. Then we have the following\

[*Lemma 3.2.*]{}

*Let $\{ \phi_{n} \}$ be a generalized Riesz basis. Then, we have the following statements.*

\(1) $T^{\ast}$ has a densely defined inverse and $(T^{\ast})^{-1}= (T^{-1})^{\ast}$.

\(2) 