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 !