After my previous post 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 $$ (\varphi_{1}\otimes \cdots \otimes \varphi_{N})(x_{1},...,x_{N}) := \varphi_1(x_{1})\cdots\varphi_{N}(x_{n}). $$
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 $K=\delta_0(xy)$, take $\mathcal L=\partial/\partial x_1$, you get $$ K=\delta'_0(x_1y_1)\otimes \delta_0(x_2y_2)\otimes\dots\otimes \delta_0(x_ny_n). $$ 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, HilbertSchmidt type), think for instance about the Hilbert transform whose kernel is $$ \text{pv}\frac{1}{xy} $$ and is bounded on $L^2(\mathbb R)$.

1$\begingroup$ This is precisely what concerned me: a lot of authors use the integral representation and this'd confuse me. You explanation is perfect and it was my first guess. Thank you! $\endgroup$ – IamWill Apr 9 at 13:51
This is a comment since your question has been answered but it will be too long. Often, a first version of a result can be wrong but it can be tweaked to give a correct formulation. In your case, this holds, namely, the existence of a kernel $K$ which is $O(x^\alpha)$ for some positive $\alpha$ in the classical sense such that your formula is valid but with each $\phi_i(x_i)$ replaced by $D_i^{r_i}\phi_i(x_i)$ for some index $r_i$. If you are interested, I can provide references.

1$\begingroup$ +1, indeed this follows from the combo: Kernel Theorem plus the structure theorem about temperate distributions being distributional derivatives of continuous functions. $\endgroup$ – Abdelmalek Abdesselam Apr 9 at 14:00
No, and a simple example goes as follows: $$ \begin{split} K(\varphi_{1}\otimes\cdots\otimes\varphi_{N}) & =\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\varphi_1(0)\cdot\ldots\cdot\varphi_{N}(0)\\ & \triangleq\left(\prod_{i=1}^{N}\prod_{k=1}^{n_i}\frac{\partial}{\partial x_k}\right)\delta(x_1,\ldots,x_n)\quad x_k\in\Bbb R^{n_k}, k= 1,\ldots, N \end{split} $$ where $\delta$ is the customary Dirac distribution in $\mathcal{S}'(\mathbb{R}^{n_{1}+\cdots+n_{N}})$. The distribution $K$ is then obviously a Schwartz distribution but it is not a measure, in a similar way as the example $\text{(NIF)}$ in this answer is not.
The standard kernel theorem ([1] chapter 1, §1.3 pp. 1120 and §3.5 pp. 7379) guarantees "only" that there is a distribution which does the job, but does not guarantee that there exists an integral representation, though this can be true for general $n$linear functionals on a particular function space, or for particular functionals on general function spaces: also, the integral representation may not have the form \eqref{1}. For example, in the case of linear functionals, an integral representation as a contour integral $$ f(\varphi)=\oint_\gamma v(z)\varphi(z)\mathrm{d}z\quad \varphi\in \mathscr{H\!\!o\!l}(D) $$ where
 $v(z)$ is the Fantappiè indicatrix of the functional, i.e. the (in general meromorphic) function $$ v(z)=f\left(\frac{1}{\zetaz}\right)\quad(\zeta\text{ is the "inner" variable}) $$
 $\gamma\subset D\subseteq \Bbb C$ is a curve enclosing all singularities of $v$.
holds for every linear analytic functional $f\in\mathscr{O}^\prime\!(\Bbb C)$. And there are also more general "integral" representations (if you allow for the kernel $K$ to be a Radon Measure) which include large classes of linear functionals and more general generalized functions: for example, see the paper [2].
References
[1] Gel’fand, I. M.; Vilenkin, N. Ya., Generalized functions. Vol. 4: Applications of harmonic analysis, Translated from the Russian by Amiel Feinstein. (English) New York and London: Academic Press. XIV, 384 p. (1964), MR0173945, Zbl 0136.11201.
[2] Kaneko, Akira, "Representation of hyperfunctions by measures and some of its applications", (English) Journal of the Faculty of Science, Section I A 19, 321352 (1972), MR0336328, Zbl 0247.35007.

1$\begingroup$ An even simpler counterexample: In dimension 1 (with $N=1$ and $n_1=1$) let $K$ be the Dirac mass distribution at the origin $K(\phi):=\phi(0)$. Then obviously $K$ is tempered, and it is a classical exercise to show that it cannot be written as an integral w.r.t. the Lebesgue measure $dx$ on the real line. No need for derivatives or tensor products... $\endgroup$ – leo monsaingeon Apr 9 at 6:31

4$\begingroup$ @leomonsaingeon: I agree with you, but $\delta$ it is a Radon measure, so it can be stated that $$ K(\varphi)=\int_{\Bbb R} \varphi \mathrm{d}\mu_{\delta_0}=\varphi(0)$$ at least by (though customary) abuse of notation: I wanted to give an example where the distribution, thought being slowly increasing, in not even a measure. $\endgroup$ – Daniele Tampieri Apr 9 at 6:38

$\begingroup$ @DanieleTampieri thank you for the answer! You mentioned the answer to my previous question on the functional derivative. In this case (where $K\in \mathcal{S}'(\mathbb{R}^{d})$ is a derivative of some function $f: \mathcal{S}(\mathbb{R}^{d}) \to \mathbb{C}$, can we garantee the integral representation? This was my first post and it seems we can, but i not sure yet. $\endgroup$ – IamWill Apr 9 at 13:57

$\begingroup$ @IamWill surely yes, since $f\in\mathcal{S}(\Bbb R^d)$ implies it is $C^\infty$smooth and rapidly decreasing, thus \eqref{1} has a precise, standard meaning as a $n$linear functional. I am adding a few remarks on the topic of integral representations of functionals to my answer, by the way. $\endgroup$ – Daniele Tampieri Apr 9 at 14:04