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I know the following statement for gronwall inequality: Given $f$ non negative and absolutely continuous on $[0;T]$ and $\phi \in L^1(0;T)$if we have, $f' \leq \phi f$ and $f(0)=0$ then $f=0$

Now is it still true if we have $f'(t) \leq \phi(t) sup_{s \in [0;t]}f(s)$ ?

In fact I have an inequality that looks like this: $f(t) \leq \int_0^t \phi(s)sup_{x \in [0;s]}f(x) ds$ and I was wondering if i could still deduce that f is 0

Thanks for any response if anyone did already encounter this !

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    $\begingroup$ Define $F(t)=\max_{0\le s\le t}f(s)$. Then it's "clear" that $F'(t)=0$ or $=f'(t)$, so $F'\le \phi F$, and thus $F\equiv 0$. The only (hopefully small) technical problem here is to establish that $F\in AC$ also, but that sounds right. $\endgroup$
    – Christian Remling
    Jul 4, 2014 at 19:18
  • $\begingroup$ $F\in AC$ is in fact obvious as $F(b)-F(a)$ is estimated by the total variation of $f$ on this interval, which is an AC measure by assumption. $\endgroup$
    – Christian Remling
    Jul 4, 2014 at 20:00
  • $\begingroup$ That seems correct, I will have to work out all the details but everything seems good. Thank you for your answer ! $\endgroup$
    – incas
    Jul 5, 2014 at 0:08

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