Timeline for Measurability of the heat semigroup in $L^\infty$
Current License: CC BY-SA 4.0
10 events
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| Mar 17, 2019 at 11:11 | comment | added | Rabat | This suppose that the heat semigroup can be expressed by means of the Gauss–Weierstrass kernel. I have made some research on this, but It seems that such an expression holds when the domain $\Omega$ is the whole space $R^n$. | |
| Mar 16, 2019 at 23:28 | comment | added | Mateusz Kwaśnicki | Joint continuity of $S(t) x$ is a consequence of the dominated convergence theorem and joint continuity of heat kernel. The latter property is classical, and I suppose it can be found in most textbooks on heat equation or parabolic PDEs (or Brownian motion); but I do not have a reference off the top of my head. | |
| Mar 16, 2019 at 22:23 | comment | added | Rabat | Thank you for these useful information. Could you please indicate me any reference for that, especially for the continuity. | |
| Mar 16, 2019 at 21:28 | comment | added | Mateusz Kwaśnicki | I leave the above as a comment, as I believe this question is more suitable for Math.SE. | |
| Mar 16, 2019 at 21:27 | comment | added | Mateusz Kwaśnicki | The integral is indeed well-defined. Firste, $|S(t) x|$ is bounded (pointwise) by the convolution of the Gauss–Weierstrass kernel $k_t$ and $|x|$, and by Cauchy–Schwarz, this convolution is bounded by $\|k_t\|_2 \|x\|_2 = c t^{-1/2} \|x\|_2$. Second, $S(t) x$ is jointly continuous as a function on $(0, \infty) \times \Omega$, and thus $\|S(t) x\|_\infty$ is continuous on $(0, \infty)$. | |
| Mar 16, 2019 at 20:52 | comment | added | Rabat | Many thanks to Prof. Alex for the editing. | |
| S Mar 16, 2019 at 18:06 | history | edited | Alex M. | CC BY-SA 4.0 |
added 29 characters in body; edited title
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| Mar 16, 2019 at 17:46 | review | Suggested edits | |||
| S Mar 16, 2019 at 18:06 | |||||
| Mar 16, 2019 at 17:45 | review | First posts | |||
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| Mar 16, 2019 at 17:44 | history | asked | Rabat | CC BY-SA 4.0 |