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Let's suppose to have the following inequality $$f(t)\leq g(t)+\int_s^t\int_s^{t'}f(r) dr dt'\,.$$ Is there a Gronwall's type inequality to bound $f(t)$?

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  • $\begingroup$ Is $s$ just some fixed value? $\endgroup$ – Umberto P. Jan 24 '17 at 18:48
  • $\begingroup$ @UmbertoP. Yes it is ($s<t$) $\endgroup$ – user404629 Jan 24 '17 at 18:56
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    $\begingroup$ I don't know if it helps, but the double integral is equal to $\int_s^t (t-r)f(r)dr$. $\endgroup$ – Martin R Jan 24 '17 at 20:18
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Use the hint from Martin R plus the generalized Gronwall inequality from Lemma 2.7 in G. Teschl, Ordinary Differential Equations and Dynamical Systems.

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  • $\begingroup$ How exactly would the Lemma 2.7 apply in this case? As the the comment of Martin R, the integrant contains a dependency in $t$ (the integral bound) in the term $(t-r)$. Such a dependency does not appear in your Lemma. $\endgroup$ – user3371583 Nov 18 '17 at 12:16
  • $\begingroup$ You are right, a comment about this is in order! However, note that you can replace the $t-r$ by the maximum over the interval under consideration, say $[0,T]$ (I am using $s=0$ for simplicity of notation). Then you get $f(t) \le g(t) + \int_0^t g(s)(T-s) \exp(\int_s^t (T-r) dr) ds$ for $t\le T$. Now finally note that you can choose $T=t$ which gives $f(t) \le g(t) + \int_0^t g(s)(t-s) \exp(\int_s^t (t-r) dr) ds$. $\endgroup$ – gerald Nov 19 '17 at 9:22

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