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This popped up in my research: we have the following mix-effect inequality that $\forall T \geq 1$ \begin{equation}\tag{*} Y(T) - \frac{1}{100T^2}\int_1^T[\alpha^2 + e^{-(T - t)}]Y(t)dt \lesssim \frac{1}{T^2}[\alpha^2 + e^{-T}] Y(1) \end{equation} where $\alpha\in [0,1]$ is a small constant. We denote by $a\lesssim b$ when $a\le Cb$ for some absolute constant $C>0$.

Note we can have a coarsened bound ($\alpha^2 + e^{-T} \lesssim 1$)) which gives $$ Y(T) - \frac{1}{100T^2}\int_1^T Y(t) dt \lesssim \frac{1}{T^2}Y(1) $$ Then the classical Gronwall-type inequality would give (or simply integrating on both sides) $$ Y(T) \lesssim \frac{1}{T^2}Y(1) $$ But this bound is not tight in the dependency on $\alpha$: when $\alpha$ is set as 0 $$ e^T Y(T) - \frac{1}{100T^2} \int_1^T e^t Y(t) dt \lesssim \frac{1}{T^2} Y(1) $$ Integrating on both sides we get $$ e^T Y(T) \lesssim \frac{1}{T^2}Y(1) $$ so there is an exponential decaying prefactor.

Now back to (*), we would conjecture the following fine-grained bound $$ Y(T) \lesssim \frac{\alpha^2 + e^{-T}}{T^2}Y(1) $$ But how do we achieve this (or something similar)?

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  • $\begingroup$ 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. $\endgroup$
    – fedja
    Commented Dec 19, 2020 at 19:44
  • $\begingroup$ 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. $\endgroup$
    – Nikolayevich
    Commented Dec 20, 2020 at 17:23
  • $\begingroup$ 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. $\endgroup$
    – fedja
    Commented Dec 22, 2020 at 4:24
  • $\begingroup$ 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 ... $\endgroup$
    – Nikolayevich
    Commented Dec 23, 2020 at 12:27
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    $\begingroup$ $Z$ 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 $T$. $\endgroup$
    – fedja
    Commented Dec 23, 2020 at 12:37

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