Let $K:\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ be a so-called "integral-kernel": we certainly require $K(x,.)$ and $K(.,y)$ to be Lebesgue measurable for almost all $x,y \in \mathbb{R}^n$. An integral kernel may fail to be translation-invariant, meaning that the equation $K(x,y)=K(x+\Delta,y+\Delta)$ fails on a set of positive Lebesgue measure. A natural way to define the self-convolution $K*K$ of $K$ (with itself) is $$K*K: \mathbb{R}^n\times \mathbb{R}^n \to \mathbb{R}\cup\{+\infty\}:\\(x_1,x_2) \mapsto \int_{\mathbb{R}^n} dy\,K(x_1,y)K(y,x_2)\text{ if well-defined, otherwise }+\infty.$$ Is it not true that this type of self-convolution inherits some of the smoothing properties of ordinary convolution? 'Rephrasing' the question towards a concrete application: can you not prove the Weyl lemma/hypo-ellipticity by merely analysing the repercussions of the Chapman-Kolmogorov equation (associated to a given time-homogeneous Markov process).

Extra notes/context:

• If $K(x,.)\geq 0$ and $\|K(x,.)\|_1=1$ for all $x \in \mathbb{R}^n$, then $(K*K)(x,y)$ is obviously finite for a.e. $y\in \mathbb{R}^n$, $(K*K)(x,.)>0$ and $\|K(x,.)\|_1=1$

• If $K$ is translation-invariant, then my question essentially boils down to asking for smoothing-properties of ordinary convolution, which are well-documented and well-known.

• As for the application in Markov-processes, self-convolution occurs when writing the following instance of the Chapman-Kolmogorov equation of a time-homogeneous Markov-process: $$p(x_1|x_2,2t)=\int_{\mathbb{R}^n}dy\,p(x_1|y,t)p(y|x_2,t)$$

• One could generalize the question to distributional $K$ and/or $K$ with co-domain $\mathbb{R}^{d\times d}$ (and defining $(K * K)_{ij}(x_1,x_2)=\int dy \,\sum_{m=1}^dK_{im}(x_1,y)K_{m j}(x_1,y)$). In this extended setting, I can think of obvious counterexamples (i.e. examples where $K*K$ is not "smoother" than $K$).

• This question has a counterpart on MSE, but as so often these days the indifference on that platform is biting. On the other hand, I hope that my question is suffiently research-level to be acceptable for this venue.

Let $K:\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ be a so-called "integral-kernel": we certainly require $K(x,.)$ and $K(.,y)$ to be Lebesgue measurable for almost all $x,y \in \mathbb{R}^n$. An integral kernel may fail to be translation-invariant, meaning that the equation $K(x,y)=K(x+\Delta,y+\Delta)$ fails on a set of positive Lebesgue measure. A natural way to define the self-convolution $K*K$ of $K$ (with itself) is $$K*K: \mathbb{R}^n\times \mathbb{R}^n \to \mathbb{R}\cup\{+\infty\}:\\(x_1,x_2) \mapsto \int_{\mathbb{R}^n} dy\,K(x_1,y)K(y,x_2)\text{ if well-defined, otherwise }+\infty.$$ Is it not true that this type of self-convolution inherits some of the smoothing properties of ordinary convolution? 'Rephrasing' the question towards a concrete application: can you not prove the Weyl lemma/hypo-ellipticity by merely analysing the repercussions of the Chapman-Kolmogorov equation (associated to a given time-homogeneous Markov process).

Extra notes/context:

• Intuitively, there's still the picture that this extended form of convolution is about blurring and since blurring and smoothing are nearly synonymous, one would a priori expect affirmative answers to my question.

• If $K(x,.)\geq 0$ and $\|K(x,.)\|_1=1$ for all $x \in \mathbb{R}^n$, then $(K*K)(x,y)$ is obviously finite for a.e. $y\in \mathbb{R}^n$, $(K*K)(x,.)>0$ and $\|K(x,.)\|_1=1$

• If $K$ is translation-invariant, then my question essentially boils down to asking for smoothing-properties of ordinary convolution, which are well-documented and well-known.

• As for the application in Markov-processes, self-convolution occurs when writing the following instance of the Chapman-Kolmogorov equation of a time-homogeneous Markov-process: $$p(x_1|x_2,2t)=\int_{\mathbb{R}^n}dy\,p(x_1|y,t)p(y|x_2,t)$$

• One could generalize the question to distributional $K$ and/or $K$ with co-domain $\mathbb{R}^{d\times d}$ (and defining $(K * K)_{ij}(x_1,x_2)=\int dy \,\sum_{m=1}^dK_{im}(x_1,y)K_{m j}(x_1,y)$). In this extended setting, I can think of obvious counterexamples (i.e. examples where $K*K$ is not "smoother" than $K$).

• This question has a counterpart on MSE, but as so often these days the indifference on that platform is biting. On the other hand, I hope that my question is suffiently research-level to be acceptable for this venue.

Let $K:\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ be a so-called "integral-kernel": we certainly require $K(x,.)$ and $K(.,y)$ to be Lebesgue measurable for almost all $x,y \in \mathbb{R}^n$. An integral kernel may fail to be translation-invariant, meaning that the equation $K(x,y)=K(x+\Delta,y+\Delta)$ fails on a set of positive Lebesgue measure. A natural way to define the self-convolution $K*K$ of $K$ (with itself) is $$K*K: \mathbb{R}^n\times \mathbb{R}^n \to \mathbb{R}\cup\{+\infty\}:\\(x_1,x_2) \mapsto \int_{\mathbb{R}^n} dy\,K(x_1,y)K(y,x_2)\text{ if well-defined, otherwise }+\infty.$$ Is it not true that this type of self-convolution inherits some of the smoothing properties of ordinary convolution? 'Rephrasing' the question towards a concrete application: can you not prove the Weyl lemma/hypo-ellipticity by merely analysing the repercussions of the Chapman-Kolmogorov equation (associated to a given time-homogeneous Markov process).

Extra notes/context:

• If $K(x,.)\geq 0$ and $\|K(x,.)\|_1=1$ for all $x \in \mathbb{R}^n$, then $(K*K)(x,y)$ is obviously finite for a.e. $y\in \mathbb{R}^n$, $(K*K)(x,.)>0$ and $\|K(x,.)\|_1=1$

• If $K$ is translation-invariant, then my question essentially boils down to asking for smoothing-properties of ordinary convolution, which are well-documented and well-known.

• As for the application in Markov-processes, self-convolution occurs when writing the following instance of the Chapman-Kolmogorov equation of a time-homogeneous Markov-process: $$p(x_1,x_2,2t)=\int_{\mathbb{R}^n}dy\,p(x_1,y,t)p(y,x_2,t)$$

• One could generalize the question to distributional $K$ and/or $K$ with co-domain $\mathbb{R}^{d\times d}$ (and defining $(K * K)_{ij}(x_1,x_2)=\int dy \,\sum_{m=1}^dK_{im}(x_1,y)K_{m j}(x_1,y)$). In this extended setting, I can think of obvious counterexamples (i.e. examples where $K*K$ is not "smoother" than $K$).

• This question has a counterpart on MSE, but as so often these days the indifference on that platform is biting. On the other hand, I hope that my question is suffiently research-level to be acceptable for this venue.

Let $K:\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ be a so-called "integral-kernel": we certainly require $K(x,.)$ and $K(.,y)$ to be Lebesgue measurable for almost all $x,y \in \mathbb{R}^n$. An integral kernel may fail to be translation-invariant, meaning that the equation $K(x,y)=K(x+\Delta,y+\Delta)$ fails on a set of positive Lebesgue measure. A natural way to define the self-convolution $K*K$ of $K$ (with itself) is $$K*K: \mathbb{R}^n\times \mathbb{R}^n \to \mathbb{R}\cup\{+\infty\}:\\(x_1,x_2) \mapsto \int_{\mathbb{R}^n} dy\,K(x_1,y)K(y,x_2)\text{ if well-defined, otherwise }+\infty.$$ Is it not true that this type of self-convolution inherits some of the smoothing properties of ordinary convolution? 'Rephrasing' the question towards a concrete application: can you not prove the Weyl lemma/hypo-ellipticity by merely analysing the repercussions of the Chapman-Kolmogorov equation (associated to a given time-homogeneous Markov process).

Extra notes/context:

• If $K(x,.)\geq 0$ and $\|K(x,.)\|_1=1$ for all $x \in \mathbb{R}^n$, then $(K*K)(x,y)$ is obviously finite for a.e. $y\in \mathbb{R}^n$, $(K*K)(x,.)>0$ and $\|K(x,.)\|_1=1$

• If $K$ is translation-invariant, then my question essentially boils down to asking for smoothing-properties of ordinary convolution, which are well-documented and well-known.

• As for the application in Markov-processes, self-convolution occurs when writing the following instance of the Chapman-Kolmogorov equation of a time-homogeneous Markov-process: $$p(x_1|x_2,2t)=\int_{\mathbb{R}^n}dy\,p(x_1|y,t)p(y|x_2,t)$$

• One could generalize the question to distributional $K$ and/or $K$ with co-domain $\mathbb{R}^{d\times d}$ (and defining $(K * K)_{ij}(x_1,x_2)=\int dy \,\sum_{m=1}^dK_{im}(x_1,y)K_{m j}(x_1,y)$). In this extended setting, I can think of obvious counterexamples (i.e. examples where $K*K$ is not "smoother" than $K$).

• This question has a counterpart on MSE, but as so often these days the indifference on that platform is biting. On the other hand, I hope that my question is suffiently research-level to be acceptable for this venue.