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 \begin{equation} p(u,v):=\int dx\,f(x)g(x)[e^{-(x-u)^2-v^2}-e^{-u^2-(x-v)^2}]. \end{equation} 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, \begin{equation} (h*G)(\phi)=\iint p\phi, \end{equation} and $p\in L^1$.

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