follows that (a) and (b) are equivalent.

Conditions (c) and (d) are obviously equivalent. To finish the proof we shall show that (b) and (c) are equivalent. Clearly, (b) implies (c). Since $Y^{o}(v,x)w\in W((x^{-1}))$, from (c) we get $$\begin{aligned}
x^{l}(x-z)^{k}\<\alpha,Y^{o}(v,x)w\>\in {\C}[[x]]\cap {\C}((x^{-1}))
={\C}[x].\end{aligned}$$ This proves that (c) implies (b).$\;\;\;\;\Box$

The following result gives the closeness of ${\cal{D}}_{P(z)}(W)$ under the actions of the component operators $v^{*}_{n}$ of each vertex operator $Y^{*}(v,x)$.

For $u,v\in V, \;\alpha\in {\cal{D}}_{P(z)}(W)$, there exists $k\in {\N}$ such that [*for all $n\in {\Z}$*]{}, $$\begin{aligned}
(x-z)^{k}Y^{*}(u,x)v^{*}_{n}\alpha\in W^{*}((x)).\end{aligned}$$ Furthermore, for $v\in V, \;n\in {\Z},\;\alpha\in {\cal{D}}_{P(z)}(W)$, $$\begin{aligned}
v^{*}_{n}\alpha\in {\cal{D}}_{P(z)}(W).\end{aligned}$$ That is, the space ${\cal{D}}_{P(z)}(W)$ is a $g(V)$-submodule of $W^{*}$.

Let $u,v\in V,\;n\in {\Z}$. From the commutator formula we have $$\begin{aligned}
\label{e3.7}
Y^{*}(u,x)v^{*}_{n}\alpha
=v^{*}_{n}Y^{*}(u,x)\alpha-\sum_{i\in {\N}}{n\choose i}x^{n-i}
Y^{*}(v_{i}u,x)\alpha.\end{aligned}$$ Since $v_{i}u=0$ for all but finitely many $i\in {\N}$, there exists $k\in {\N}$ (independent of $n$) such that $$(x-z)^{k}Y^{*}(u,x)\alpha,\;\;
(x-z)^{k}Y^{*}(v_{i}u,x)\alpha\in W^{*}((x))$$ for all $i\in {\N}$. Then using (\[e3.7\]) we have $$(x-z)^{k}Y^{*}(u,x)v^{*}_{n}\alpha\in W^{*}((x)).$$ It follows immediately from Lemma \[lequivconds\] that $v_{n}^{*}\alpha\in {\cal{D}}_{P(z)}(W)$. $\;\;\;\;\Box$

Let ${\C}(x)$ be the algebra of rational functions in $x$. Define linear maps $\iota_{x;\infty}$ and $\iota_{x;0}$ (cf. \[FLM\], \[FHL\]) from ${\C}(x)$ to ${\C}[x,x^{-1}]]$ and ${\C}((x))$, respectively, such that for $f(x)\in {\C}(x)$, $\iota_{x;\infty}f(x)$ and 