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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 $\beta(t)<0$....

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  • $\begingroup$ not enough information to look into it. Even the equation given about is not in the form to apply Gronwall lemma, $\endgroup$ – runaround Feb 18 '16 at 22:48
  • $\begingroup$ @runaround I already know that!!!!! That's exactly why I'm posting it! If it was exactly in the form to apply Gronwall Lemma, there would be nothing to ask!! Unfortunately, that's the only information I have available... $\endgroup$ – Max Feb 19 '16 at 14:28
  • $\begingroup$ If $X$ and $b$ are chosen to be constant and the integrand $[-X_s+ E(V) E(b(X_S)]$ could be negative in this case, this would show that the first inequation is wrong. $\endgroup$ – aventurin Feb 23 '16 at 21:43
  • $\begingroup$ @aventurin Thank you, but it's not the case. I've not put all the context in which this equation appears in order not to distract from the question. $\endgroup$ – Max Feb 23 '16 at 22:42

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