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I am having a problem with the following exercise.

Let $f$, $g$ be continuous non-negative functions on $[a,b]$, and let $C$ a positive constant.

Suppose that: $f(x) \leq C+ \int_{a}^x f(t)g(t)dt$,

for all $x \in [a,b]. $ Show that:

$f(x) \leq C\exp\left(\int_{a}^x g(t)dt\right).$

Thank you in advance

2 Answers 2

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Let $h(x):=\int_a^xf(t)g(t)dt$. Then $g(x)(C+h(x))\geq h'(x)$. So $h'(x)-g(x)h(x)\leq Cg(x)$. Now multiplying by $\exp\left(-\int_0^xg(t)dt\right)$, we get $\frac d{dx}\left(h(x)\exp\left(-\int_0^xg(t)dt\right)\right)\leq Cg(x)\exp\left(-\int_0^xg(t)dt\right),$ and integrating $h(x)\exp\left(-\int_0^xg(t)dt\right)\leq C\int_0^xg(u)\exp\left(-\int_0^ug(t)dt\right)du.$ The RHS is $C-C\exp\left(-\int_0^xg(t)dt\right)$, so $h(x)\leq C\exp\left(\int_0^xg(t)dt\right)-C.$ As $h(x)\geq f(x)-C$, we get the wanted result.

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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6288/discussion-between-user43418-and-davide-giraudo)2012-10-31
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you can use the idea of wansik inequality in ODE

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    That being said, what is the "wansik inequality"?2012-10-31