Timeline for How much "room" in inequality $\displaystyle \int_a^b \varphi' ov ~\mathrm{d}x \leq 0$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 23, 2023 at 8:49 | vote | accept | Hyperbolic PDE friend | ||
| Mar 20, 2023 at 13:40 | comment | added | Hyperbolic PDE friend | Thank you, the first argument I certainly understand. The other one I will take a look at! | |
| Mar 20, 2023 at 13:38 | comment | added | Willie Wong | In the rough case, using the mollifier argument I gave above, what you find is that the smoothed function $\Psi(x) := \int_{-\epsilon}^{\epsilon} \chi(y) v(x + y) o(x+y) ~dy$ must be increasing on $[a+\epsilon,b-\epsilon]$. A similar argument as the previous comment shows that $\Psi(b-\epsilon) - \Psi(a-\epsilon) = O(\epsilon)$ (using that near $a$ you can write $|o(x) - o(a)| \leq 2 |o'(a)| (x-a)$. | |
| Mar 20, 2023 at 13:33 | comment | added | Willie Wong | @Meowdog: suppose $ov$ were not constant, then $\int_a^b \varphi (ov)' \geq 0$, so $ov$ must be non-decreasing (otherwise $(ov)' < 0$ on an open interval, and you can take $\varphi$ to be supported in that interval). Since $ov$ is non-decreasing, you have $o(a)v(a) \leq o(b)v(b)$. But your hypothesis also give $o(b) \leq v(a) o(a)$ and $v(b) \leq 1$, which together implies $o(b)v(b) \leq o(a) v(a)$. For a non-decreasing function $ov$ to have equal values at the end points, the function must be constant. | |
| Mar 20, 2023 at 8:16 | comment | added | Hyperbolic PDE friend | Thank you as well. Can you maybe explain how from integration by parts, we can see that $ov$ is constant? Of course, after ibp we have $\int_a^b \varphi(-ov'-v'o) \leq 0$, but it is unclear for me how we can conclude that $-ov'-v'o=0$... | |
| Mar 17, 2023 at 17:33 | history | answered | Willie Wong | CC BY-SA 4.0 |