After my [previous post](https://math.stackexchange.com/questions/3616506/derivative-as-an-integral-kernel?noredirect=1#comment7433473_3616506) I got curious about the following very simple question (which I don't seem to find the answer). Given a tempered distribution $K \in \mathcal{S}'(\mathbb{R}^{n\_{1}+\cdots+n\_{N}})$, does if follow that:
\begin{eqnarray}
K(\varphi\_{1}\otimes\cdots\otimes\varphi\_{N}) = \int k(x\_{1},...,x\_{N})\varphi\_1(x\_{1})\cdots\varphi\_{N}(x\_{N})dx\_{1}\cdots dx\_{N}\tag{1}\label{1}
\end{eqnarray}
for some integral kernel $k$ and $\varphi\_{j}\in \mathcal{S}(\mathbb{R}^{n\_{j}})$? Here 
$$
(\varphi\_{1}\otimes \cdots \otimes \varphi\_{N})(x\_{1},...,x\_{N}) := \varphi\_1(x\_{1})\cdots\varphi\_{N}(x\_{n}).
$$



Integral representation of tempered distributions
Let $\mathcal L$ be a continuous linear mapping from $\mathscr S(\mathbb R^n)$ into
$\mathscr S'(\mathbb R^n)$. The Laurent Schwartz kernel theorem asserts that there exists $K\in \mathscr S'( \mathbb R^n\times \mathbb R^n)$ such that for all $\phi, \psi\in
\mathscr S(\mathbb R^n)$
$$
\langle\mathcal L\phi,\psi\rangle\_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}
=\langle K,\phi\otimes\psi\rangle\_{\mathscr S'(\mathbb R^{2n}), \mathscr S(\mathbb R^{2n})}.
$$
If $K$ happens to be locally integrable you get indeed that 
$$
\langle K,\phi\otimes\psi\rangle\_{\mathscr S'(\mathbb R^{2n}), \mathscr S(\mathbb R^{2n})}
=\iint K(x,y) \phi(y) \psi(x) dx dy,
$$
but most of the time $K$ will be a tempered distribution: take $\mathcal L=Id$, you will get $K=\delta\_0(x-y)$, take $\mathcal L=\partial/\partial x\_1$, you get
$$
K=\delta'\_0(x\_1-y\_1)\otimes \delta\_0(x\_2-y\_2)\otimes\dots\otimes \delta\_0(x\_n-y\_n).
$$
So somehow your question is tautological: some authors prefer to use the integral sign instead of the brackets of duality, but that notational device does not make all distributions locally integrable, you get two examples above. 


One point among others is remarkable in the Schwartz KT: the kernel $K$ is a tempered distribution and this is a similarity with finite dimensional linear mappings, a sharp contrast for instance with the bounded linear mappings from $L^2(\mathbb R^n)$ into itself, which have also kernels, but in general much more singular than $L^2(x,y)$ (these are compact operators, Hilbert-Schmidt type), think for instance about the Hilbert transform whose kernel is 
$$
\text{pv}\frac{1}{x-y}
$$
and is bounded on $L^2(\mathbb R)$.


