Representation of translation invariant distribution Suppose $F \in \mathcal{D}'(\mathbb{R}^d \times \mathbb{R}^d)$ is a distribution which is translation invariant in the sense that for $h \in \mathbb{R}^d$ and $f,g \in \mathcal{D}(\mathbb{R}^d)$ we have that $F(f \otimes g) = F(\tau_h f \otimes \tau_h g)$ where $\tau_h f(x) = f(x-h)$.

I would like to prove that there exists a distribution $\mathcal{R}[F] \in \mathcal{D}'(\mathbb{R}^d)$ such that $F(f \otimes g) = \mathcal{R}[F](f \ast \bar{g})$ where $\bar{g}(x) = g(-x)$ and $\ast$ denotes convolution.


I am aware of a couple of results that say translation invariant maps on spaces of test functions or distributions are convolutions. For example, $G: \mathcal{D}(\mathbb{R}^d) \to \mathcal{D}(\mathbb{R}^d)$ is translation invariant (in the sense that $\tau_h \circ G = G \circ \tau_h$ for $h \in \mathbb{R}^d$) if and only if it is given by convolution with a compactly supported distribution. If $G$ is translation invariant, one can check that the distribution is given by $\phi \mapsto G\phi(0)$. 
From this result and the reflexivity of $\mathcal{D}(\mathbb{R}^d)$ I can also show that if $G: \mathcal{D}'(\mathbb{R}^d) \to \mathcal{D}'(\mathbb{R}^d)$ is translation invariant then it is again given by convolution with a compactly supported distribution by exploiting the fact that for $\phi \in \mathcal{D}(\mathbb{R}^d)$, $u \mapsto Gu(\phi)$ is a continuous linear functional on $\mathcal{D}'(\mathbb{R}^d)$ and hence is given by testing against some $\phi_G$. One then checks that $\phi \mapsto \phi_G$ is continuous and translation invariant and exploits the above result.
You can recast the problem in this question in a similar form as the above by considering $F$ as a map $\mathcal{D}(\mathbb{R}^d) \to \mathcal{D}'(\mathbb{R}^d)$ given by $F(f)(g) := F(f \otimes g)$ (allowing an abuse of notation) which is then translation invariant in the sense that $\tau_h \circ F = F \circ \tau_h$. Unfortunately, at this point I fail to make progress since the crux of the above argument was having a good candidate distribution by at some point using pointwise evaluations or by trying to reduce to that case by testing against fixed test functions. The second technique doesn't seem to go anywhere here since $F(\cdot)(g) \in \mathcal{D}'(\mathbb{R}^d)$ and so we can't apply previous results.
 A: This answer is based on the comment above and Lemma 2.9 of the thesis of Ajay, as mentioned in that comment. Up to minor adaptation, the result proved in the Lemma can be stated as follows

Lemma: For $z \in \mathbb{R}^n$, let $T_z: \mathcal{D}(\mathbb{R}^m \times \mathbb{R}^n) \to \mathcal{D}(\mathbb{R}^m \times \mathbb{R}^n)$ be defined by $T_zf(x,y) = f(x, y+z)$ where $x \in \mathbb{R}^m, y \in \mathbb{R}^n$. Suppose that $\phi \in \mathcal{D}'(\mathbb{R}^m \times \mathbb{R}^n)$ has the property that $T_z^*\phi = \phi$ for each $z \in \mathbb{R}^n$. Then there is a $\psi \in \mathcal{D}'(\mathbb{R}^m)$ such that
  $$\phi(f) = \psi \bigg( \int_{\mathbb{R}^n} f(\cdot, y) dy \bigg).$$

To prove this, first note that it is sufficient to prove the case $n = 1$ since iterating this argument will give the result for general $n$. The key is then to note that $\phi$ vanishes on test functions which are of the form $\partial_{x_{m+1}}g$ for some $g$, by translation invariance. This means that if we pick $h \in \mathcal{D}(\mathbb{R})$ with $\int h = 1$ and set $\psi(g) = \phi(g \otimes h)$ then we have
$$\phi(f) - \psi\bigg( \int_{\mathbb{R}} f(\cdot, y) dy \bigg) = \phi(f -  \tilde{f})$$
where $\tilde{f}(x,y) = h(y) \int f(x,z) dz$. Then since $\int f(\cdot, z) - \tilde{f}(\cdot,z) dz = 0$, it is a standard fact that there is a $g$ such that $\partial_{x_{m+1}}g = f - \tilde{f}$ which implies that $\phi(f-\tilde{f}) = 0$ so that
$$\phi(f) = \psi\bigg(\int f(\cdot,y) dy \bigg)$$
as desired.

To conclude the result of the question, we just need to make a change of variables. Define $\tau(x,y) = (x-y,-y)$ and let $\tilde{F} = F \circ \tau^*$. Then $\tilde{F}(T_zf) = F((T_zf) \circ \tau)$. If $\tau_z f(x,y) = f(x-z,y-z)$ then we have that $(T_zf) \circ \tau = \tau_z (f \circ \tau)$ so by translation invariance of $F$, $\tilde{F}(T_zf) = \tilde{F}(f)$. Hence, by the Lemma there is an $\mathcal{R}[F]$ such that 
$$\tilde{F}(f) = \mathcal{R}[F]\bigg( \int_{\mathbb{R}^n} f(\cdot, y) dy \bigg).$$
Then, given $f$, $F(f) = \tilde{F}(\tau^*f)$ since $\tau^2 = \operatorname{Id}$. Hence 
$$F(f) = \mathcal{R}[F]\bigg(\int_{\mathbb{R}^n} \tau^*f(\cdot, y) dy \bigg)$$
which is exactly the desired result when $f$ is a tensor product.
