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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.

5th decile
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    This is a delicate issue: such kernels can well decrease regularity. For example, if $K(x, y) = k_1(x) k_2(y)$, then $K f(x) = \int K(x,y) f(y) dy$ is exactly as smooth as $k_1$, no matter how smooth $f$ is. Do you have a specific class of kernels $K$ in mind? – Mateusz Kwaśnicki Feb 13 '20 at 14:15
  • @Mateusz: fair point... On first sight it strikes me that your counterexample violates the property $\forall x \in \mathbb{R}^n:|K(x,.)|_1=1$. Could such a constraint make it harder for you to come up with counterexamples? – 5th decile Feb 13 '20 at 14:19
  • Not really: just set $K(x,y) = k_1(x)k_2(y) + (1-k_1(x))k_3(y)$, where $0<k_1(x)<1$ and $k_2, k_3$ are arbitrary probability density functions. – Mateusz Kwaśnicki Feb 13 '20 at 14:21
  • Okay, your example is also a positive kernel... Well, you mentioned that it's a delicate issue. So, you're aware of already-existing discussions, debates, treatments in the literature? Note that I tagged my question with the "reference-request"-label: you could write an answer outlining these counterexamples and giving literature references? – 5th decile Feb 13 '20 at 14:25
  • I am aware of a large number of papers where people prove some regularity (say: Hölder regularity) of heat kernels for various operators, which are likely not smooth. I do not know if anyone was interested in proving rigorously how irregular these heat kernels are, though. Transition probabilities of any non-Feller (but still Markov) process should also work as a counter-example; man such processes are known (for example the usual Brownian motion reflected at $0$, but running in all of $\mathbb{R}$). It is difficult to give references without knowing what exactly you are looking for. – Mateusz Kwaśnicki Feb 13 '20 at 14:39
  • It doesn't surprise me that (in contrast to ordinary convolution) you can't go beyond proving (Hölder-)continuity via this way. On the other hand, I must have been deluded in thinking that my Kolmogorov-Chapman-convolution would be universally smoothing. That is: I didn't think of a more specific set of assumptions. What if you assume a uniform-in-$x$ bound on the second moment $\int dy,K(x,y)(x-y)^2$? Does that accomplish something? – 5th decile Feb 13 '20 at 14:47
  • Or what about imposing continuity-in-$x$ for $\int dy,K(x,y)\chi_{y\in O}$ for all open $O$? – 5th decile Feb 13 '20 at 14:53
  • Again: I do not really know what kind of kernels you have in mind, but there are at least two standard notions. A kernel is Feller if it maps $C_0(\mathbb{R}^n)$ into itself (so in some sense at least it does not break continuity). A kernel is strong Feller if it maps bounded Borel functions into continuous ones (so it provides some minimal smoothing). Both are well-studied, and both (particularly the former one) have various variants. None of them imply any further regularity of iterated compositions of $K$ without (severe) further restrictions. – Mateusz Kwaśnicki Feb 13 '20 at 21:42

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