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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:

  1. Is there any way to generalize this to functions $f,g: B \to \mathbb{R}$ where $B$ is the closed unit ball in $\mathbb{R}^n$?

  2. 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.

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up vote 1 down vote accepted

1) It can be generalized easely to cubes, but not in a nice way:

$$\int_a^b \int_c^d fg dy dx \geq \int_a^b \left( \frac{1}{d-c}\int_c^d f dy \int_c^d g dy \right) dx \geq \geq \frac{1}{(d-c)(b-a)}\left( \int_a^b \int_c^d f dy dx \right)\left( \int_a^b \int_c^d g dy dx \right) $$

And then by induction:

$$ \int_{\prod [a_i,b_i]} fg \geq \frac{1}{\prod(b_1-a_i)} (\int_{\prod [a_i,b_i]} f)( \int_{\prod [a_i,b_i]} g) \,.$$

The condition is that at every step $f,g$ must be both increasing or decreasing in the direction of $e_i=( 0, 0, 0, .., 1, 0,..0)$.

2) Since $\int_a^b ff'dx = \frac{f^2(x)}{2}|_a^b$, your question is:

Under what conditions

$$f^2(1)-f^2(0) \leq 0 \Rightarrow (f(1)-f(0)) \int_0^1 f(x) dx \leq 0$$

If $f$ is positive and decreasing, but this is too much to ask for.

You basically have 3 situations: $f(0)=f(1)$, $f(0)<f(1)$ and $f(0) >f(1)$, it is easy to decide what happens in each of the three situations...

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