# Generalized Gronwall inequality

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 $f(t)$?

• Is $s$ just some fixed value? – Umberto P. Jan 24 '17 at 18:48
• @UmbertoP. Yes it is ($s<t$) – user404629 Jan 24 '17 at 18:56
• I don't know if it helps, but the double integral is equal to $\int_s^t (t-r)f(r)dr$. – Martin R Jan 24 '17 at 20:18

• 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. – user3371583 Nov 18 '17 at 12:16
• 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 $f(t) \le g(t) + \int_0^t g(s)(t-s) \exp(\int_s^t (t-r) dr) ds$. – gerald Nov 19 '17 at 9:22