# Generalized Jensen's equality on averages

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 $$0, $$\int_0^p z\left(p'\right) dp'\geq 0$$

is it true that for any $$0

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

?

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

Thanks!

## 1 Answer

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.

• 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. – Serb Dec 6 '19 at 13:53