. \nonumber\end{aligned}$$ In Ref. [@hiroshi1], the author have defined an operator $T_{\bm{e}}$ on $D_{\bm{e}} \equiv Span \{ e_{n} \}$ by $$\begin{aligned}
T_{\bm{e}} \left( \sum_{k=0}^{n} \alpha_{k} e_{k} \right)
= \sum_{k=0}^{n} \alpha_{k} \phi_{k}. \nonumber\end{aligned}$$ For the operators $T_{\phi,\bm{e}}$, $T_{\bm{e},\phi}$ and $T_{\bm{e}}$ we have the following\

[*Lemma 2.1.*]{}

*The following statements hold.*

\(1) $T_{\phi, \bm{e}}$ is a densely defined linear operator in ${\cal H}$ such that $$\begin{aligned}
T_{\phi,\bm{e}} \supset T_{\bm{e}} \;\;\; {\rm and} \;\;\; T_{\phi, \bm{e}} e_{n} =\phi_{n}, \;\;\; n=0,1, \cdots . \nonumber\end{aligned}$$

\(2) $$\begin{aligned}
D( T_{\bm{e},\phi} ) = D(\phi) \equiv \left\{ x \in {\cal H} ; \sum_{k=0}^{\infty} | (x | \phi_{k})|^{2} < \infty \right\} 
\;\;\; {\rm and} \;\;\; T_{\bm{e}}^{\ast}=T_{\phi, \bm{e}}^{\ast} =T_{\bm{e}, \phi}. \nonumber\end{aligned}$$

[*Proof.*]{} The statements (1) and (2) are easily proved by the definitions of $T_{\phi, \bm{e}}$, $T_{\bm{e}, \phi}$ and $T_{\bm{e}}$.\
\
By Lemma 2.1, (2), $T_{\bm{e},\phi}$ is closed. However $D(T_{\phi,\bm{e}}^{\ast})$ is not necessarily dense in ${\cal H}$, equivalently, $T_{\bm{e}}$ and $T_{\phi,\bm{e}}$ are not necessarily closable. Thus we investigate the conditions under what $T_{\phi,\bm{e}}$ is closable.\

[*Lemma 2.2.*]{}

*The following statements are equivalent:*

\(i) $T_{\bm{e}}$ is closable.

\(ii) $T_{\phi,\bm{e}}$ is closable.

\(iii) $D(\phi)$ is dense in ${\cal H}$.\
If this holds, then $$\begin{aligned}
\bar{T}_{\bm{e}}=\bar{T}_{\phi,\bm{e}}=(T_{\bm{e},\phi})^{\ast}. \nonumber\end{aligned}$$

[*Proof.*]{} This follows from Lemma 2.1, 