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My question comes from a proof in Daniel Stroock's book 'An introduction to the Analysis of Paths on a Riemannian Manifold' (lemma 3.60, page 86). He proves that a function F satisfies the integral inequality

$F(t) \le U + \int_0^t g(\tau)d\tau + \alpha \int_0^t F(\tau)d\tau$

where $F,g$ are are positive continuous functions, $\alpha, U >0$ are constants and $t>0$. He then writes 'an easy application of Gronwall's inequality' yields

$e^{-\alpha t}F(t) \le U + \int_0^t e^{-\alpha \tau}g(\tau)d\tau$.

If I apply Gronwall's inequality (for example the integral version on wikipedia) I only get the weaker estimate

$e^{-\alpha t}F(t) \le U + \int_0^t g(\tau)d\tau$

Are there stronger versions of Gronwall that he could have used? A more clever way of applying it? Is his conclusion as stated here even true in general?

My weaker estimate is not sufficient for his applications later in the book, so I don't think it is just a typo, but I don't see where else in the proof he could have gotten the extra $e^{-\alpha \tau}$ term. Thanks.

quarague
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  • "If I apply Gronwall's inequality (for example the integral version on wikipedia)" Please show the details, this is probably where the mistake lies. – Did Aug 14 '14 at 13:04
  • Gronwall's inequality, as in the wikipedia version reads: if $u(t) \le \alpha(t) + \int_0^t \beta(s)u(s)ds$ and $\alpha(t)$ is non decreasing, then $u(t) \le \alpha(t) exp(\int_0^t\beta(s)ds)$. In my example $\alpha(t)=U+\int_0^tg(s)ds$ and $\beta(s)=\alpha$, so the conclusion is $F(t) \le (U + \int_0^tg(s)ds)e^{\alpha t}$. – quarague Aug 14 '14 at 16:15
  • Try version (a) here. – Did Aug 14 '14 at 17:14
  • This does work, after half a page of algebra and an integration by parts you get the claim as stated by Stroock. I guess I was thrown off by thinking that additional assumptions should give you better estimates, which at least in this example is not the case. Thanks. – quarague Aug 15 '14 at 08:11
  • Actually I find the WP page rather misleading since it suggests that version (b), using stronger hypothesis, would yield a stronger conclusion... – Did Aug 15 '14 at 10:02

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