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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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    Can you detail a bit more. I do not see how to use this..2012-10-31
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    $h'(x)=f(x)g(x)$ and $f(x)\le C+h(x)$.2012-10-31
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    Why do I multiply by $\exp\left(-\int_0^xg(t)dt\right)$ ?2012-10-31
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    The LHS will be the derivative to a product.2012-10-31
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    what is does LHS stand for?2012-10-31
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    Left Hand Side.2012-10-31
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    If I understood correctly, we will obtain an inequality related to f'(x) and then I can conclude the inequality of f(x). However, I don't find this expression. Please help with this last step2012-10-31
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    I've added details.2012-10-31
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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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    I can't use that2012-10-31
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    I can only use properties regarding differentiation and integrals2012-10-31
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    If you need to answer a question in a certain way, you should say so when you ask the question, by providing context and a description of what attempts you have made. It is frustrating to answer a question only to have the goalposts move.2012-10-31
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    ok sorry about that. I am still new to the website2012-10-31
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    That being said, what is the "wansik inequality"?2012-10-31