Invertibility of Partial Differential Operators Consider the strong dual $\mathcal{S}_{b}^{'}$ of the Schwartz space $\mathcal{S}(\mathbb{R}^n)$, and let $P$ be a complex polynomial in $n$ variables $z=(z_1,\dots,z_n)$:
\begin{equation}
P(z)=\sum_{|\alpha| \leq N} c_{\alpha} z^{\alpha},
\end{equation}
where as usual for every $\alpha=(\alpha_1,\dots,\alpha_n) \in \mathbb{N}^{n}$ we set $|\alpha|=\alpha_1+\dots+\alpha_n$, and $z^{\alpha}=z_1^{\alpha_1}\dots z_n^{\alpha_n}$. Consider the differential operator $P(D): \mathcal{S}_{b}^{'} \rightarrow \mathcal{S}_{b}^{'}$ defined by
\begin{equation}
P(D)=P \left( -i \frac{\partial}{\partial x_1},\dots, - i \frac{\partial}{\partial x_n} \right).
\end{equation}
Clearly,  $P(D): \mathcal{S}_{b}^{'} \rightarrow \mathcal{S}_{b}^{'}$ is a continuous linear map. 
Define also the multiplication operator $M:\mathcal{S}(\mathbb{R}^n) \rightarrow \mathcal{S}(\mathbb{R}^n)$ defined as
\begin{equation}
M(\phi)=P \phi \quad (\phi \in \mathcal{S}(\mathbb{R}^n)).
\end{equation}
I am trying to prove the equivalence of the following two statements:
(i) there exists a continuous linear map $R: \mathcal{S}_{b}^{'} \rightarrow \mathcal{S}_{b}^{'}$ such that $P(D) \circ R = \mathbb{1}_{\mathcal{S}_{b}^{'}}$,
(ii) there exists a continuous linear map $L:\mathcal{S}(\mathbb{R}^n) \rightarrow \mathcal{S}(\mathbb{R}^n)$ such that $L \circ M = \mathbb{1}_{\mathcal{S}(\mathbb{R}^n)}$.
It is easy to prove that (ii) implies (i). Indeed, by denoting with $\hat{\phi}$ and $\check{\phi}$ respectively the Fourier and anti-Fourier transform of $\phi \in \mathcal{S}(\mathbb{R}^n)$, if we set for each $T \in \mathcal{S}_{b}^{'}$
\begin{equation}
R(T)(\phi)=T \left[ \widehat{L ( \check{\phi}) } \right] \quad (\phi \in \mathcal{S}(\mathbb{R}^n)),
\end{equation}
it is immediate to see, by using the fact that the Fourier transform on $\mathcal{S}(\mathbb{R}^n)$ is a continuous linear bijection with continuous inverse, that $R(T) \in \mathcal{S}_{b}^{'}$, that $R: \mathcal{S}_{b}^{'} \rightarrow \mathcal{S}_{b}^{'}$ is a continuous linear map. By using the properties of the Fourier transform on $\mathcal{S}_{b}^{'}$ it is also immediate to check that $P(D) \circ R = \mathbb{1}_{\mathcal{S}_{b}^{'}}$.
As for the converse, I am stuck. I don't know how to prove that (i) implies (ii), so any help is welcome.
Thank you very much in advance for your attention.
NOTE. The equivalence of (i) and (ii) is stated in Langenbruch, Real roots of polynomials and right inverses for partial differential operators in the space of tempered distributions, Proc. Roy. Soc. Edinburgh 114A (1990) 169-179. Langenbruch says that the equivalence is trivial and maybe it really is, but I could not prove it.
 A: Finally, I found the answer to my question, and it is quite trivial, but only if you know the theory of duality for topological vector spaces.
Forgive me, if I give references also for elementary results, but I hope this will help the readers of this post who are not expert in functional analysis. 
First of all, some notation. Let $F:\mathcal{S}(\mathbb{R}^n) \rightarrow \mathcal{S}(\mathbb{R}^n)$ be the Fourier transform in $\mathcal{S}(\mathbb{R}^n)$ and $\mathscr{F}: \mathcal{S}_{b}^{'} \rightarrow \mathcal{S}_{b}^{'}$ the Fourier transform in $\mathcal{S}_{b}^{'}$. Both these maps are isomorphisms of topological vector spaces. 
If (ii) holds, then set $G= F \circ L \circ F^{-1}$ and define
\begin{equation}
R(T) = T \circ G \quad (T \in \mathcal{S}_{b}^{'}).
\end{equation}
We have $R(T) \in \mathcal{S}_{b}^{'}$ for each $T \in \mathcal{S}_{b}^{'}$, since $G$ is a continuous linear map. Moreover, since a continuous linear map is bounded (Rudin, Functional Analysis, Second Edition, Theorem(1.32)), we easily deduce from the definition of strong dual that $R:\mathcal{S}_{b}^{'} \rightarrow \mathcal{S}_{b}^{'}$ is continuous. We also get from the definition of $R$:
\begin{equation}
\widehat{R(T)}(P\phi)=\hat{T}(\phi) \quad (\phi \in \mathcal{S}(\mathbb{R}^n)),
\end{equation}
so that from the properties of Fourier transforms we get $P(D) \circ R= \mathbb{1}_{\mathcal{S}_{b}^{'}}$.
To prove that (i) implies (ii), we need the following fact. Let $\mathcal{S}''$ be the strong dual of $\mathcal{S}_{b}^{'}$. Associate to each $\phi \in \mathcal{S}(\mathbb{R}^n)$ the continuous linear functional $E(x)$ on $\mathcal{S}_{b}^{'}$ defined by
\begin{equation}
E(x)(T)=T(x) \quad (T \in \mathcal{S}_{b}^{'}).
\end{equation}
Since $\mathcal{S}(\mathbb{R}^n)$ is a Montel Space (see e.g. Treves, Topological Vector Spaces, Distributions and Kernels, p. 357), and since a Montel space is reflexive (see Treves, cit., Proposition (36.10), p. 376), we know that the evaluation map $E:\mathcal{S}(\mathbb{R}^n) \rightarrow \mathcal{S}''$ is an isomorphism of topological vector spaces. Define $\mathscr{G}=\mathscr{F} \circ R \circ \mathscr{F}^{-1}$ and
\begin{equation}
\mathscr{L}(\Lambda)=\Lambda \circ \mathscr{G} \quad (\Lambda \in \mathcal{S}'' ).
\end{equation}
Since $\mathscr{G}$ is a continuous linear map,as above we see that $\mathscr{L}(\Lambda) \in \mathcal{S}''$ for each $\Lambda \in \mathcal{S}'' $, and that the map $\mathscr{L}:\mathcal{S}'' \rightarrow \mathcal{S}''$ is linear and continuous. Finally, define
\begin{equation}
L=E^{-1} \circ \mathscr{L} \circ E.
\end{equation}
Since we have
\begin{equation}
\widehat{R(T)}(P\phi)=\hat{T}(\phi) \quad (\phi \in \mathcal{S}(\mathbb{R}^n)),
\end{equation}
we easily check that $L(P\phi)=\phi$ for each $\phi \in \mathcal{S}(\mathbb{R}^n)$, so that $L \circ M=\mathbb{1}_{\mathcal{S}(\mathbb{R}^n)}$.
QED
