The continuous version of the Chebyshev Sum Inequality say that if $f,g:[0,1] \to \mathbb{R}$ are either both decreasing or both increasing, then $$\int_0^1 fg \;dx \geq \int_0^1 f \;dx \int_0^1 g \;dx$$ If $f, g$ are differentiable this can be generalized to the situation where $f'$ and $g'$ have the same sign everywhere since then the integrals can be split up into underlying sets where $f$ and $g$ are either both increasing or decreasing. I was wondering the following:
Is there any way to generalize this to functions $f,g: B \to \mathbb{R}$ where $B$ is the closed unit ball in $\mathbb{R}^n$?
Can anything be said about the case $g = f'$? Ideally, I would like it to be true that $$\int_0^1 ff' \;dx \geq \int_0^1 f \;dx \int_0^1 f' \;dx$$ But this seems too much to ask. It would suffice for me that $$\int_0^1 f \;dx \int_0^1 f' \;dx \leq 0$$ if $$\int_0^1 ff' \;dx \leq 0$$ Are there any conditions on $f,f'$ under which this implication would be true?
I would also be interested in any similar results.
Thanks in advance.