Let $f, g\in \mathcal{S}(\mathbb R)$ (Schwartz class function), $\delta_0$ (dirac delta distribution).
Consider distribution as follows: $$G(x, y)= f(x)g(x)\delta_0(y)-f(y)g(y)\delta_0(x), \ (x, y\in \mathbb R)$$
Let $h(x,y)= e^{-(x^2+y^2)}.$
My Question is:
Can we expect that $G\ast h \in L^{1}(\mathbb R^2)$?
where $\ast$ denotes the convolution.
You can just do the explicit computation: $$ (G\ast h)(x,y)=\int\int G(u,v)h(x-u,y-v)dudv $$ $$ =\int\int f(u)g(u)\delta(v)h(x-u,y-v)dudv- \int\int f(v)g(v)\delta(u)h(x-u,y-v)dudv $$ $$ =\int f(u)g(u)h(x-u,y)du-\int f(v)g(v)h(x,y-v)dv=\gamma(y)((fg)\ast\gamma)(x)- \gamma(x)((fg)\ast\gamma)(y) $$ where $\gamma(z)=e^{-z^2}$ which is in $\mathcal{S}(\mathbb{R})$. The latter is stable by product and convolution. Moreover, the product of a Schwartz function in $x$ and a Schwartz function in $y$ is in $\mathcal{S}(\mathbb{R}^2)$. So the result is not only in $L^1(\mathbb{R}^2)$ but in fact in $\mathcal{S}(\mathbb{R}^2)$.