$D(\phi)$ is dense in ${\cal H}$, by Lemma 2.2 and Lemma 2.3 we have $T_{\phi,\bm{e}}$ is closable and $\bar{T}_{\phi,\bm{e}}$ has an inverse. Furthermore, since $D(\bar{T}_{\phi,\bm{e}}^{-1})=\bar{T}_{\phi,\bm{e}}D(\bar{T}_{\phi,\bm{e}}) \supset D_{\phi}$ and $D_{\phi}$ is dense in ${\cal H}$, $\bar{T}_{\phi,\bm{e}}^{-1}$ is densely defined.\
(ii)$\Rightarrow$(iii) By Ref. [@hiroshi1], Lemma 2.2 and Lemma 2.3, we have $$\begin{aligned}
(\bar{T}_{\phi,\bm{e}}^{-1})^{\ast}
= (\bar{T}_{\phi,\bm{e}}^{\ast})^{-1}
=( T_{\bm{e},\phi})^{-1}. \nonumber\end{aligned}$$ Hence we have $$\begin{aligned}
D \left( ( \bar{T}_{\phi,\bm{e}}^{\ast})^{-1} \right)
= T_{\bm{e},\phi} D(T_{\bm{e},\phi}) \supset T_{\bm{e},\phi} D_{\psi} =D_{\bm{e}}. \nonumber\end{aligned}$$ Thus we have $( T_{\phi,\bm{e}}^{\ast} )^{-1}$ is densely defined.\
(iii)$\Rightarrow$(i) Take an arbitrary $x \in D_{\phi}^{\perp}$. Then, $$\begin{aligned}
0=(\phi_{n}|x) =( T_{\phi,\bm{e}} e_{n}|x)=(T_{\bm{e}} e_{n}|x), \;\;\; n=0,1, \cdots . \nonumber\end{aligned}$$ Hence, by Lemma 2.1, (2) we have $$\begin{aligned}
x \in D(T_{\bm{e}}^{\ast})=D\left( T_{\phi,\bm{e}}^{\ast} \right) \;\;\; {\rm and} \;\;\; T_{\bm{e}}^{\ast}x=T_{\phi,\bm{e}}^{\ast}x=0. \nonumber\end{aligned}$$ By (iii), it follows that $$\begin{aligned}
x= \left( T_{\phi,\bm{e}}^{\ast} \right)^{-1} T_{\phi,\bm{e}}^{\ast} x =0. \nonumber\end{aligned}$$ Thus, $D_{\phi}$ is dense in ${\cal H}$. This completes the proof.\
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Similarly we have the following statements.\

[*Lemma 2.5.*]{}

*Suppose $( \{ \phi_{n} \} , \{ \psi_{n} \} )$ is a biorthogonal pair such that $D(\psi)$ is dense in ${\cal H}$. Then the following statements are equivalent:*

\(i) $D_{\psi} \equiv Span \{ \psi_{n} \}$ is dense in ${\cal H}$.

\(ii) $T_{\psi,\bm{e}}$ is closable and 