# Convolution with Schwartz class function

Let $$f, g\in \mathcal{S}(\mathbb R)$$ (Schwartz class function), $$\delta_0$$ (dirac delta distribution).

Consider distribution as follows: $$H(x, y)= f(x)g(x)\delta_0(y)-f(y)g(y)\delta_0(x), \ (x, y\in \mathbb R)$$

Let $$h(x,y)= e^{-(x^2+y^2)}.$$

My Question is:

Can we expect that $$H\ast h \in L^{1}(\mathbb R^2)$$?

where $$\ast$$ denotes the convolution.

The answer is yes. Indeed, for any Schwartz function $$\phi$$ we have $$(h*G)(\phi)=G(h^-*\phi)=G(h*\phi)$$, where $$h^-(x,y):=h(-x,-y)=h(x,y)=e^{-x^2-y^2}$$. So, $$\begin{multline} (h*G)(\phi)=G(h*\phi)=\int dx\,f(x)g(x)\iint du\,dv\,\phi(u,v)e^{-(x-u)^2-v^2} \\ -\int dy\,f(y)g(y)\iint du\,dv\,\phi(u,v)e^{-u^2-(y-v)^2} \\ =\int dx\,f(x)g(x)\iint du\,dv\,\phi(u,v)[e^{-(x-u)^2-v^2}-e^{-u^2-(x-v)^2}] \\ =\iint du\,dv\,\phi(u,v)p(u,v), \end{multline}$$ where $$$$p(u,v):=\int dx\,f(x)g(x)[e^{-(x-u)^2-v^2}-e^{-u^2-(x-v)^2}].$$$$ We have $$\begin{multline} \iint du\,dv\,|p(u,v)|\le2\int dx\,|f(x)g(x)|\iint du\,dv\,e^{-(x-u)^2-v^2} \\ =2\int dx\,|f(x)g(x)|\,c^2<\infty, \end{multline}$$ where $$c:=\int dv\,e^{-v^2}<\infty$$. So, $$$$(h*G)(\phi)=\iint p\phi,$$$$ and $$p\in L^1$$.