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Let $X\subseteq \mathbb{R}$. Let $q\colon X\to \mathbb{R}$ and $f \colon X\times X\to\mathbb{R}$ be integrable functions.

My question: Is it true that \begin{align} \int_{X} \int_{X}q(x)^{2} f(x,y)\, \mathrm{d} x\, \mathrm{d} y\geq \int_{X} \int_{X} q(x)q(y) f(x,y)\, \mathrm{d} x\,\mathrm{d} y ?\quad (\star) \end{align}

Some comments. I sketch my attempt to prove it in the spoiler box below.

We can rewrite the LHS of the to-be-proven inequality as \begin{align} \int_{X} \int_{X}q(x)^{2} f(x,y)\, \mathrm{d} x\,\mathrm{d} y&\overset{(\#)}{=} \int_{X} \int_{X}q(y)^{2} f(y,x)\, \mathrm{d} y\,\mathrm{d} x \notag\\ & = \int_{X} \int_{X}\left(\frac{q(x)^{2}+q(y)^{2}}{2}\right) f(x,y)\, \mathrm{d} x\,\mathrm{d} y, \end{align} and since $q(x)^{2}+q(y)^{2}\geq 2q(x)q(y)$ for all $x,y\in X$, we are done.

For $(\#)$ to hold $f(\cdot,\cdot)$ must be symmetric. Is this a necessary condition for $(\star)$ to be true?

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  • $\begingroup$ In # you are just renaming variables. The next equality doesn't really follow from # alone, though. $\endgroup$ – axioman Sep 23 '16 at 20:23
  • $\begingroup$ Doesn't the last equality require $f(x,y)=f(y,x)$ for it to be true? $\endgroup$ – πr8 Sep 23 '16 at 20:23
  • $\begingroup$ Yes, in my attempt symmetry of $f$ is required. Is this also a necessary condition for the inequality in my question to hold true? $\endgroup$ – Ludwig Sep 23 '16 at 20:31
  • $\begingroup$ You definitely need to assume that $f$ is non-negative, also. $\endgroup$ – Willie Wong Sep 23 '16 at 20:47
  • $\begingroup$ i'm not entirely sure wheter wolfram just put out crap, but try $q(x)=e^x and f(x,y)=(1/x)*e^y$ as a counterexample. (X=(0,2)) $\endgroup$ – axioman Sep 23 '16 at 20:47
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First, obviously you need $f(x,y) \geq 0$. Otherwise there are easy counter examples.

Next, let me give a counterexample when $f(x,y)$ is not symmetric in $x,y$.

Let $X = (0,2)$ and $f(x,y)$ be defined to be $0$ when $y\in (0,1)$ and $1$ when $y\in [1,2)$.

Let $q(x)$ be defined to be $\frac12$ when $x\in (0,1)$ and $1$ when $x\in [1,2)$.

$$ \iint_{X\times X} q(x)^2 f(x,y) ~\mathrm{d}x\mathrm{d}y = \frac14 + 1 = \frac54 $$

while

$$ \iint_{X\times X} q(x) q(y) f(x,y) ~\mathrm{d}x \mathrm{d}y = \frac12 + 1 = \frac32 $$

(Notice that

$$ \iint_{X\times X} q(y)^2 f(x,y) ~\mathrm{d}x \mathrm{d}y = 2 $$

and $2 + \frac54 \geq 3 = 2 \times \frac32$.)

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  • $\begingroup$ As an aside: typically integral inequalities of these forms do not depend on the precise form of the measure used, which means that often times it suffices to first think about the case of sums. For sums examples are much easier to compute. $\endgroup$ – Willie Wong Sep 23 '16 at 21:03

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