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Quick question: Are we able to show a Gronwall-type inequality assuming that $$u(t)\le\alpha(t)+\int_0^t\max(u(s),\beta(s))\:{\rm d}s,$$ where $\alpha$ is nondecreasing (or constant) and $\beta$ is bounded and measurable (or nondecreasing, if this is helpful)?

I'm willing to assume that $\alpha, u$ and $\beta$ are nonnegative.

We can then clearly apply the usual Gronwall inequality after using $\max(u(s),\beta(s))\le u(s)+\beta(s)$, but I'm looking for a sharper inequality.

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  • $\begingroup$ Upon the suggestion of a friend of mine, if you add $\beta(t)$ on each side and use the inequality, you can then apply gronwall on $\psi(t) = \max(u(t), \beta(t))$. $\endgroup$
    – Theleb
    Commented Jan 28, 2022 at 12:54
  • $\begingroup$ @Theleb Thank you for the suggestion. I guess you're intending to assume that $u$ and $\beta$ are nonnegative, since then $\psi(t)\le u(t)+\beta(t)\le\alpha(t)+\beta(t)+\int_0^t\psi(s)\:{\rm d}s$. Applying Gronwall yields $\psi(t)\le(\alpha(t)+\beta(t))e^t$. But is this result really better than using $u(t)\le\alpha(t)+\int_0^t\beta(s)\:{\rm d}s+\int_0^tu(s)\:{\rm d}s$, which yields $u(t)\le\left(\alpha(t)+\int_0^t\beta(s)\:{\rm d}s\right)e^t$? $\endgroup$
    – 0xbadf00d
    Commented Jan 30, 2022 at 17:12
  • $\begingroup$ Yes it is better : the second bound is worse than the first by a factor $t$ for $t$ large. $\endgroup$
    – username
    Commented Jan 30, 2022 at 19:56
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    $\begingroup$ One can save a factor of $2$ or so by taking max with $\beta(t)$ rather than summation, thus using $\max(u(t),\beta(t)) \leq \max(\alpha(t),\beta(t)) + \int_0^t \max(u(s), \beta(s))\ ds$ as the starting point (assuming everyone is non-negative). $\endgroup$
    – Terry Tao
    Commented Jan 30, 2022 at 20:23
  • $\begingroup$ @TerryTao Thank you for your comment, but doesn't taking max with $\beta(t)$ yield $\max(u(t),\beta(t))\le\max(\alpha(t),\beta(t))+\max\left(\int_0^t\max(u(s),\beta(s))\:{\rm d}s,\beta(t)\right)$ instead? $\endgroup$
    – 0xbadf00d
    Commented Jan 31, 2022 at 7:29

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