# Inequality with integrals

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:

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

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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.. –  user43418 Oct 31 '12 at 11:05
$h'(x)=f(x)g(x)$ and $f(x)\le C+h(x)$. –  Berci Oct 31 '12 at 11:08
Why do I multiply by $\exp\left(-\int_0^xg(t)dt\right)$ ? –  user43418 Oct 31 '12 at 14:02
The LHS will be the derivative to a product. –  Davide Giraudo Oct 31 '12 at 14:02
what is does LHS stand for? –  user43418 Oct 31 '12 at 14:05