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Let $y,\,z:[0,1]\rightarrow \mathbb{R}$ be integrable functions.

$y\left(p\right)$ is weakly increasing and positive ($0 < y\left(p\right)\leq y\left(p'\right)$ for $0 \leq p\leq p'$).

Assuming that for any $0<p\leq1$, $$ \int_0^p z\left(p'\right) dp'\geq 0 $$

is it true that for any $0<p\leq1$

$$ \int_0^p \frac{z\left(p'\right)}{y\left(p'\right)} dp'\geq 0 $$

?

I'm pretty sure it is true, but the proof is sneaky.

Thanks!

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If $y(p')$ is right-continuous I think you can do this readily via Lebesgue-Stieltjes integration, if you are familiar with that. Let $F(p) = \int_0^p z(p')dp'$. Then if $dh(p')$ denotes the Lebesgue-Stieltjes measure associated with the monotone decreasing function $h(p') = {1 \over y(p')}$ you can integrate by parts to say that $$\int_0^p {z(p') \over y(p')}dp' = {F(p) \over y(p)} - \int_0^p F(p') dh(p')$$ Since $y(p')$ is increasing, ${1 \over y(p')}$ is decreasing, so the integral of the nonnegative function $F(p')$ with respect to $dh(p')$ will give a nonpositive result. Thus the right hand side here is the sum of two nonnegative terms and gives a nonnegative result.

Even if $y(p')$ is not right-continuous, you can change the function at its jumps to make it right-continuous, and then the above should once again apply.

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  • $\begingroup$ Thanks. What about the other direction? Let's assume $y$ is continuous, $y,\,z:[0,1]\rightarrow \mathbb{R}$ be integrable functions. $y\left(p\right)$ is weakly increasing and positive ($0 < y\left(p\right)\leq y\left(p'\right)$ for $0 \leq p\leq p'$). Assuming that for any $0<p\leq1$, $$ \int_0^p \frac{z\left(p'\right)}{y\left(p'\right)} dp'\geq 0 $$ What are the conditions on $y$ so that also $$ \int_0^p z\left(p'\right) dp'\geq 0 ? $$ It is not generally true, but curious about the conditions for which it might work. $\endgroup$ – Serb Dec 6 '19 at 13:53

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