2
$\begingroup$

Does anyone know a simple proof of the following Prékopa-Leindler style inequality:

If we have $f_1,f_2,g_1,g_2$ strictly positive functions on $\mathbb{R}$ such that, for any $x_1,x_2 \in \mathbb{R}$, one has $$f_1(x_1)^2 f_2(x_2)^3 \leq g_1(x_1)^2g_2(x_2)^3 $$ then $$ \left(\int_\mathbb{R} f_1\right)^2\left(\int_\mathbb{R}f_2\right)^3 \leq \left(\int_\mathbb{R}g_1\right)^2\left(\int_\mathbb{R} g_2\right)^3.$$

$\endgroup$

1 Answer 1

5
$\begingroup$

Is not it obvious (unlike Prékopa-Leindler)?

We are given that for all $x_1,x_2$ we have $(f_1/g_1)^2 (x_1)\leqslant (g_2/f_2)^3(x_2)$, thus there exists $c>0$ such that $(f_1/g_1)^2 (x_1)\leqslant c^6\leqslant (g_2/f_2)^3(x_2)$, i.e. $f_1(x)\leqslant c^3 g_1(x)$, $c^2 f_2(x)\leqslant g_2(x)$ for all $x$, we integrate these pointwise inequalities, take appropriate powers of integrals and multiply.

$\endgroup$
2
  • $\begingroup$ Hi ! Thx for your answer ! Then if you mix the right-hand side of the pointwise inequality as $g_1(x_1 + x_2)^2 g_2(x_1 - x_2)^3$, i guess it is not trivial anymore.. $\endgroup$
    – Anthony
    Apr 18 at 8:25
  • 4
    $\begingroup$ At least, not so trivial. But if you want to ask this, please make a new question with all details. $\endgroup$ Apr 18 at 8:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.