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For $x \geq 0$,let $H_1(x) > H_2(x) > 0 $ and $f_1(x) > f_2(x) > 0 $ be real valued, smooth functions on $\mathbb{R}$. Assume the products $H_1f_1,H_1f_2,H_2f_1$ and $H_2 f_2$ are all integrable on $\mathbb{R}_+$. Can one say that

$$ \frac{\int_0^\infty H_1(x)f_1(x) \mathrm{d}x }{\int_0^\infty H_1(x)f_2(x) \mathrm{d}x} \geq \frac{\int_0^\infty H_2(x)f_1(x) \mathrm{d}x }{\int_0^\infty H_2(x)f_2(x) \mathrm{d}x} $$

?

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Write $H_3 = H_1 - H_2 > 0$ and $f_3 = f_1 - f_2 > 0$. Also denote

$$ I_{ij} = \int_{0}^{\infty} H_i(x)f_j(x) \, dx, \qquad i, j \in \{2, 3\}. $$

Then the inequality is equivalent to

$$ (I_{22} + I_{23} + I_{32} + I_{33})I_{22} \geq (I_{22} + I_{23})(I_{22} + I_{32}), $$

which is then equivalent to the following question

Question. Assume that $H_2, H_3, f_2, f_3$ are positive and smooth functions on $\mathbb{R}$ such that $H_i f_j$ are integrable on $\mathbb{R}_+$ for all $i, j\in\{2,3\}$. Then does it true that

$$ I_{22}I_{33} \geq I_{23}I_{32} ? $$

But the answer is obviously false, since we can always interchange the role of $f_2$ and $f_3$ to reverse the inequality. As a concrete example, consider

$$ H_2(x) = e^{-2x}, \quad H_3(x) = e^{-x}, \quad f_2(x) = e^{-x^2}, \quad f_3(x) = e^{-2x^2}. $$

Then $I_{22}I_{33} \approx 0.166043$ but $I_{23}I_{32} \approx 0.178883$, and hence with $H_1 = H_2 + H_3$ and $f_1 = f_2 + f_3$ the original inequality is also invalidated:

$$ \frac{\int_{0}^{\infty} H_1(x)f_1(x) \, dx}{\int_{0}^{\infty} H_1(x)f_2(x) \, dx} \approx 1.82851, \qquad \frac{\int_{0}^{\infty} H_2(x)f_1(x) \, dx}{\int_{0}^{\infty} H_2(x)f_2(x) \, dx} \approx 1.86516.$$

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  • $\begingroup$ wow, great! thanks. $\endgroup$ – smorbrod Apr 9 '18 at 15:28

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