Let $V$ be a finite dimensional real / complex vector space and consider the space $L(V,V)$ of linear operators on $V$.
Fix $n \in \mathbb{N}$ and let $\mathcal{M}$ be the real / complex vector space of smooth functions $F : \mathbb{R}^n \to L(V,V)$.
Then, $\mathcal{M}$ may be given the Frechet topology of uniform convergence of all derivatives on any compact subset of $\mathbb{R}^n$ in analogy to $C^\infty(\mathbb{R}^n)$. Here, I am not specifying the norm to be used on $V$ or $L(V,V)$ etc, because on a finite dimensional vector space, all norms are equivalent.
Fix $k \in \mathbb{N}$ and let $F_1, \cdots, F_k \in \mathcal{M}$. Then we may define a scalar-valued smooth mapping on $\mathbb{R}^n$ by $$ x \in \mathbb{R}^n \to \mathrm{Tr}\Bigl( F_1(x) \cdots F_k(x)\Bigr) $$ We denote such mapping simply as $\mathrm{Tr}\Bigl( F_1 \cdots F_k\Bigr)$.
Next, consider the $k$-fold tensor product space $[\mathcal{M}]^{\otimes k}$ and the mapping $$ F_1 \otimes \cdots \otimes F_k \in [\mathcal{M}]^{\otimes k} \to \mathrm{Tr}\Bigl( F_1 \cdots F_k\Bigr) \in C^\infty(\mathbb{R}^n) $$
Here, my question is as follows:
In analogy to linear extension in finite dimensions, I suspect that the above mapping can be uniquely extended to a continuous linear map from $[\mathcal{M}]^{\otimes k}$ to $C^\infty(\mathbb{R}^n)$. However, I do not see how to find or even define a "basis" for $\mathcal{M}$ to prove this.
Could anyone please help me?
Add) As pointed out in the comment below, $\mathcal{M}$ is a nuclear Frechet space and therefore we can have some sensible topology on $[\mathcal{M}]^{\otimes k}$ according to the Wikipedia article here.
