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 $f(t)$?
Generalized Gronwall inequality
1
$\begingroup$
real-analysis
functional-analysis
analysis
inequality
-
0Is $s$ just some fixed value? – 2017-01-24
-
0@UmbertoP. Yes it is ($s
-
1I don't know if it helps, but the double integral is equal to $\int_s^t (t-r)f(r)dr$. – 2017-01-24
1 Answers
0
Use the hint from Martin R plus the generalized Gronwall inequality from Lemma 2.7 in G. Teschl, Ordinary Differential Equations and Dynamical Systems.
-
0How 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. – 2017-11-18
-
0You 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 $f(t) \le g(t) + \int_0^t g(s)(t-s) \exp(\int_s^t (t-r) dr) ds$. – 2017-11-19