I´ve got these lines from an article:
( where $b:\mathbb{R}_+\to \mathbb{R}_+$ is non-decreasing and $(X_t)$ is an $\mathbb{R}_+$-valued process - it doesn't matter very much, I guess, anyway-.)
$$ X_t \leqslant X_0 + \int_0^t {[-X_s+ E(V) E(b(X_S))\ ] ds} $$ Using Gronwall Lemma we have: $$ X_t \leqslant X_0\ e^{-t} + E(V) \int_0^t {e^{(t-s)} E(b(X_S))\ ds} $$
Now, from what I know, the cited lemma can be stated as Follows(not the most general form): Let $\alpha$, $u$ and $\beta$ be real-valued functions on $\mathbb{R}_+$, $u$ and $\beta$ continuous; and let $\alpha \in \mathbb{R}$. If $$ u(t) \leqslant \alpha(t) + \int_0^t{\beta(s) u(s)\ ds}\ \ \ \ \ \ \text{and}\ \\ \beta \geqslant 0 $$ Then $$ u(t) \leqslant \alpha (t) + \int_0^t { \alpha (s) \beta (s) \exp{ \left( \int_s^t {\beta(r) \ dr} \right)} \ ds} $$
Moreover, if $\alpha$ is non-decreasing, then $$ u(t) \leqslant \alpha(t) \exp{\left(\int_0^t \beta(s) \ ds \right)} $$
So, first of all, I have the feeling that there's a mistake in the article, and where it says $ e^{(t-s)} $ there should be a $ e^{(s-t)} $.
Anyway, I can't identify in the procedure of the article who would be $\alpha$, $u$ and $\beta$
This is my Progress so far: I'm almost sure there's a mistake , as I said , and where it says $ e^{(t-s)} $ there should be a $ e^{(s-t)} $; because later in the article the inequality is used with the "corrected" version; and because , I think the term $$X_0\ e^{-t} $$ that appears when applied the lemma is there because in the original inequality ( the one before applying the lemma) The author maybe multiplied by $e^t$ in order to apply the lemma (otherwise I don't know why does $e^{-t}$ appear )
I think that if what I said is correct, I could work backwards and take $e^{-t} $ as a common factor and multiply by $ e^t$ , so the result would be:
$$ X_t e^t \leqslant X_0\ + E(V) \int_0^t {e^{s} E(b(X_s))\ ds} $$
Still, I can't seem to find the elements to construct the form of the G-Lemma
EDIT:
If I could say that Gronwall Lemma holds even when $\beta$ is not necessarily positive, then I could take $$ u(t) = X_t \ \ \ \ \ \ \ \alpha(t) = X_0 + \int_0^t{E(V)E(b(X_s))\ ds} $$ and $$ \beta(s)=-1 $$
But I'm not sure this could be the case, since everywhere I saw, $\beta(t) \geq 0 $ was explicitly recquired...
EDIT 2: Although I've not proven that Gronwall doesn't hold for $\beta$ possibly negative, I did have checked that the condition $\beta(t)\geq 0$ is necessary for any demonstration like the one that apeears in wikipedia . There's an inequality that gets inverted when $\beta(t)<0$....