Timeline for How to get $\int_{\mathbb R^d} |\partial_i\partial_j(1-\Delta)^{-\frac{\delta}{2}}p_t(\cdot-y)(x)| \, \mathrm d x \lesssim t^{\frac{\delta}{2}-1}$?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Oct 8, 2023 at 20:58 | history | edited | Terry Tao | CC BY-SA 4.0 |
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| Oct 8, 2023 at 14:44 | comment | added | Akira | Thank you so much for your explanation! I will check it out. Have a nice weekend! | |
| Oct 8, 2023 at 14:39 | comment | added | Terry Tao | One can use the integral kernel of $(1-\Delta)^{-\delta/2}$ (which is even, as is $p_t$) to move them over to $g$, and move the derivatives over by integration by parts. (one can also restrict attention to smooth $g$ if desired to make it easier to justify the calculations). Alternatively, you can express everything in Fourier space where the self adjointness of the relevant operators becomes transparent. | |
| Oct 8, 2023 at 13:10 | comment | added | Akira | Dear professor Tao, thank you so much for your help! Unfortunately, I could not see how to use integration by parts to get $$ \begin{align*} &\int_{\mathbb R^d} \partial_i \partial_j (1- \Delta)^{- \frac{\delta}{2}} p_t ( \cdot - y) (x) g(x) \, \mathrm d x \\ = &\partial_i \partial_j (1- \Delta)^{- \frac{\delta}{2}} \left ( \int_{\mathbb R^d} p_t ( x - \cdot) (x) g(x) \, \mathrm d x \right ) (y). \end{align*} $$ Could you elaborate more on this point? | |
| Oct 8, 2023 at 13:09 | vote | accept | Akira | ||
| Oct 7, 2023 at 23:30 | history | answered | Terry Tao | CC BY-SA 4.0 |