the following result:

For $v\in V,\; \alpha\in {\cal{D}}_{P(z)}(W)$, $$\begin{aligned}
\label{e3.11}
(x-z)^{k}Y^{R}(v,x)\alpha=(x-z)^{k}Y^{*}(v,x)\alpha\end{aligned}$$ whenever $(x-z)^{k}Y^{*}(v,x)\alpha\in W^{*}((x))$ for $k\in {\N}$. Furthermore $$\begin{aligned}
\label{e3.12}
(x-z)^{k}\<Y^{R}(v,x)\alpha,w\>
=(x-z)^{k}\<\alpha,Y^{o}(v,x)w\>\end{aligned}$$ for all $w\in W$. $\;\;\;\;\Box$

Set $$\begin{aligned}
Y^{R}(v,x)=\sum_{n\in {\Z}}v^{R}_{n}x^{-n-1}.\end{aligned}$$ For $v=\omega$ (the Virasoro element), following the tradition we set $$\begin{aligned}
Y^{R}(\omega,x)=\sum_{n\in {\Z}}L^{R}(n)x^{-n-2}.\end{aligned}$$

For now, $v^{R}_{n}$ are only linear maps from ${\cal{D}}_{P(z)}(W)$ to $W^{*}$. Next, we shall show that $v^{R}_{n}$ are in fact linear endomorphisms of ${\cal{D}}_{P(z)}(W)$. We shall need the following result:

Let $U$ be a vector space and let $$\begin{aligned}
f(x)=\sum_{n\in {\Z}}f_{n}x^{-n-1},\;\;g(x)=\sum_{n\in {\Z}}g_{n}x^{-n-1}
\in U[[x,x^{-1}]].\end{aligned}$$ Suppose that either $f(x)\in U((x))$ or $f(x)\in U((x^{-1}))$ and that there exist $k\in {\N}$ and $z\in {\C}^{\times}$ such that $$\begin{aligned}
\label{ebasic52}
(x-z)^{k}f(x)=(x-z)^{k}g(x).\end{aligned}$$ Then for $n\in {\Z}$, $$\begin{aligned}
f_{n}\in \mbox{\rm linear span }\{g_{m}\;|\;m\ge n\}\end{aligned}$$ if $f(x)\in U((x))$ and $$\begin{aligned}
f_{n}\in \mbox{\rm linear span }\{g_{m}\;|\;m\le n+k\}\end{aligned}$$ if $f(x)\in U((x^{-1}))$.

Assume $f(x)\in U((x))$. For $n\in {\Z}$, let $s$ be a positive integer such that $x^{n+s}f(x)\in U[[x]]$. Then $\Res_{x}x^{n+j}f(x)=0$ for $j\ge s$ and furthermore, $$\begin{aligned}
\Res_{x}x^{n+i}(x-z)^{k}f(x)=0\end{aligned}$$ for $i\ge s$. Using this and (\[ebasic52\]) we get $$\begin{aligned}
\label{e3.30f}
f_{n}&=&\Res_{x}x^{n}f(x)\nonumber\\
&=&\Res_{x}x^{n}(-z+x)^{-k}(x-z)^{k}f(x)\nonumber\\
&=&\Res_{x}\sum_{i\ge 0}{-k\choose i}(-z)^{-k-i}x^{n+i}(x-z)^{k}f(x)\nonumber\\
&=&\Res_{x}\sum_{i= 0}^{s-1}{-k\choose i}(-z)^{-k-i}x^{n+i}(x-z)^{k}f(x)
\nonumber\\
&=&\Res_{x}\sum_{i= 0}^{s-1}{-k\choose i}(-z)^{-k-i}x^{n+i}(x-z)^{k}g(x)
\nonumber\\
&\in& \mbox{linear span }\{g_{n}, g_{n+1},\cdots, g_{n+k+s-1}\}.\end{aligned}$$

If $f(x)\in U((x^{-1}))$, let $s'$ be a positive integer such 