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.

  • $\begingroup$ Do you mean $G\ast h$? $\endgroup$ – Marco Nov 19 '18 at 16:51
  • $\begingroup$ @Marco: Yes. Thanks. I'll correct the typo. $\endgroup$ – Math Learner Nov 19 '18 at 16:55
  • $\begingroup$ Also, could you please be more precise about how the distribution $G$ acts on Schwarz functions? For example for a writing like $f(x)\delta_0(y)$ you mean that $\left\langle f(x)\delta_0(y), \phi(x,y)\right\rangle=?$ $\endgroup$ – Marco Nov 19 '18 at 17:07
  • $\begingroup$ BTW it is a bit strange to formulate the question as you did instead of replacing $fg$ by a single Schwartz function. This is because every Schwartz function can be written as $fg$ for some Schwartz functions $f,g$. $\endgroup$ – Abdelmalek Abdesselam Nov 19 '18 at 17:12
  • $\begingroup$ @AbdelmalekAbdesselam I'd say if $T \in S', \varphi \in S$ then $f \mapsto f \ast (\varphi T)$ is a continuous map $S \to S$. Not sure by which argument. When replacing $S$ by $C^\infty_c$ it is obvious. $\endgroup$ – reuns Nov 19 '18 at 20:03

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 $\mathcal{S}(\mathbb{R}^2)$.


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