Let $\mathcal{S}(\mathbb{R}^4)$ denote the Schwartz space on $\mathbb{R}^4$, $\mathcal{S}^{\otimes n}(\mathbb{R}^4)$ the tensor product of $n$ of these spaces and $\mathcal{S}(\mathbb{R}^{4n})$ the Schwartz space on $\mathbb{R}^{4n}$. Let $W$ be a tempered distribution $$ W : \mathcal{S}^{\otimes n}(\mathbb{R}^4) \rightarrow \mathbb{C}.$$
My questions are:
Is it correct to say that, from the kernel theorem, there exists a distribution $$ W' : \mathcal{S}(\mathbb{R}^{4n}) \rightarrow \mathbb{C}$$ such that for $f \in \mathcal{S}^{\otimes n}(\mathbb{R}^4) \subset \mathcal{S}(\mathbb{R}^{4n})$, $W'(f) = W(f)$, i.e. $W'$ restricted to the tensor product just yields $W$?
Suppose then that $W$ is translation invariant $W(f) = W(f_a)$, where $f_a(x) = f(x-a)$. Is $W'$ then also translation invariant?
And finally, I have a statement that says if $W'$ is translation invariant, there exists a distribution $W''$ "that only depends on the differences $x_1-x_2,\dots,x_{n-1}-x_n$" and then they write $$W'(x_1,\dots,x_n) = W''(x_1-x_2,\dots,x_{n-1}-x_n).$$ But this would mean $$W'' : \mathcal{S}(\mathbb{R}^{4n-4}) \rightarrow \mathbb{C},$$ and I have no idea how I would show this without using the above notation. The book references the original work by Schwartz, but my French isn't good enough to understand what he writes. So if anyone could help me show that there exists such a distribution $W''$ that would be much appreciated.