Timeline for “Chapman-Kolmogorov”-convolution vs. smoothness
Current License: CC BY-SA 4.0
11 events
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| Feb 13, 2020 at 21:42 | comment | added | Mateusz Kwaśnicki | 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. | |
| Feb 13, 2020 at 14:53 | comment | added | 5th decile | Or what about imposing continuity-in-$x$ for $\int dy\,K(x,y)\chi_{y\in O}$ for all open $O$? | |
| Feb 13, 2020 at 14:47 | comment | added | 5th decile | 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? | |
| Feb 13, 2020 at 14:39 | comment | added | Mateusz Kwaśnicki | 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. | |
| Feb 13, 2020 at 14:25 | comment | added | 5th decile | 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? | |
| Feb 13, 2020 at 14:21 | comment | added | Mateusz Kwaśnicki | 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. | |
| Feb 13, 2020 at 14:19 | comment | added | 5th decile | @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? | |
| Feb 13, 2020 at 14:17 | history | edited | 5th decile | CC BY-SA 4.0 |
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| Feb 13, 2020 at 14:15 | comment | added | Mateusz Kwaśnicki | 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? | |
| Feb 13, 2020 at 12:26 | history | edited | 5th decile | CC BY-SA 4.0 |
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| Feb 13, 2020 at 12:20 | history | asked | 5th decile | CC BY-SA 4.0 |