Timeline for Gronwall-type bound for a mix-effect inequality?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 23, 2020 at 15:25 | comment | added | Nikolayevich | Hmm I think you are right. Maybe I worried too much, there are some upper-continuity property due to our settings. | |
| Dec 23, 2020 at 12:37 | comment | added | fedja | $Z$ is explicit, so its continuity is not a problem. As to $Y$, OK, if you care about bad functions, note that it has to be assumed at least locally integrable (otherwise the statement is just false), so by the inequality it is bounded from above on compacts. Fix $T>1$ and consider the least $C$ such that $Y\le Z$ on $[1,T]$ (it exists). If $C$ is too large, then plugging the estimate into the integral results in a continuous upper bound strictly less than $Z$ on $[0,T]$, so $C$ can be dropped - a contradiction. The bound you obtain for $C$ does not depend on $T$. | |
| Dec 23, 2020 at 12:27 | comment | added | Nikolayevich | Sure that seems a nice argument! In that case, the continuity of Y and Z is required right? It is not clear to me how to justify this ... | |
| Dec 22, 2020 at 4:24 | comment | added | fedja | Just consider the smallest $T$ for which $Y(T)=Z(T)$ (if it exists). Then $T>1$ and $Y(t)<Z(t)$ for $1\le t<T$. Then integrating from $1$ to $T$, you derive the strict inequality $Y(T)<Z(T)$ (because $Z$ is a strict supersolution, or because the inequality is strict up to $T$, whichever reason you like more). This is a clear contradiction. | |
| Dec 20, 2020 at 17:23 | comment | added | Nikolayevich | I grasped a hint of your proposal -- yet there is a recursive argument going on here: you need to show Y<=Z by assuming Y<=Z -- why is this a valid operation? I was thinking of comparison theorems in ODE but not quite sure how to use it. | |
| Dec 19, 2020 at 19:44 | comment | added | fedja | If $Y$ is non-negative (which, I presume, it is), I do not see what problem you have with it: Just show that $Z(t)=CY(1)\frac{\alpha^2+e^{-t}}{t^2}$ satisfies the reverse inequality when $C>0$ is large enough, so you'll not be able to break through $Z$ (consider the first moment when $Y=Z$, etc.). Your $100$ in the denominator is, certainly, large enough for that. | |
| Dec 16, 2020 at 21:33 | history | asked | Nikolayevich | CC BY-SA 4.0 |