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I have on domain $[0,\infty)$ a known and positive function $f(x)$ and two unknown functions $g(x), h(x)$ that start positive when $x=0$.

I also know that if $h(x)$ is positive, then $g(x)$ is also positive.

I define the induction

$$h(x)=\int_0^x f(t)g(t)h(t) dt$$

I imagine that is possible to say that $h(x)$ (and so $g(x)$) remains positive, but I could not find a formal argument.

Thank you in advance.

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    $\begingroup$ By taking the derivative w.r.t. $x$, you get a differential equation for $h$ that should help you. $\endgroup$
    – gmvh
    Commented Feb 15, 2021 at 16:34
  • $\begingroup$ Please, why the down vote? Thank you $\endgroup$
    – Quiet_waters
    Commented Feb 15, 2021 at 17:19
  • $\begingroup$ @gmvh Thank you! $\endgroup$
    – Quiet_waters
    Commented Feb 15, 2021 at 17:20
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    $\begingroup$ I didn't downvote, but some might feel that the question isn't really research-level. $\endgroup$
    – gmvh
    Commented Feb 15, 2021 at 17:41
  • $\begingroup$ @gmvh thank you so much. I migrated a little bit from math exchange because here the answers have been faster, but I will try to police myself better on the level. Thank you for clue! $\endgroup$
    – Quiet_waters
    Commented Feb 15, 2021 at 18:00

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Taking the derivative of your "induction" w.r.t. $x$, one gets $$ h'(x) = f(x)g(x)h(x) $$ and thus $h'(x)>0$ if $h(x)>0$, according to your conditions. Since $h(0)>0$, one concludes that $h(x)>0$ for $x\in[0;\infty)$.

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  • $\begingroup$ Note, though, that your "induction" implies $h(0)=0$. $\endgroup$
    – gmvh
    Commented Feb 16, 2021 at 9:02

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