1
$\begingroup$

For fixed $ 0 < \alpha < \beta $, is there a positive constant $C_0$, depending only on $\alpha$ and $\beta$, such that for any bounded measurable function $ \varphi : \mathbb{R}^+\rightarrow [0 ,1] $ the inequality $$ \int_0^\infty \varphi(x) x^2 e^{-\alpha x^2} d x \leq C_0 \int_0^\infty \varphi(x) x^2 e^{-\beta x^2} d x $$ is valid? If yes, please give me some hints, and otherwise show me an example. Thank you very much!

$\endgroup$
  • $\begingroup$ Any bounded function? No, you need at least measurability in addition. $\endgroup$ – zhw. May 17 '16 at 16:34
  • $\begingroup$ If the measurability is given, does the statement hold? $\endgroup$ – Loring May 17 '16 at 16:47
  • $\begingroup$ See my answer.$\,$ $\endgroup$ – zhw. May 17 '16 at 16:55
0
$\begingroup$

Well, as $\alpha < \beta$ you have $-\alpha = -\beta + \underbrace{\beta - \alpha}_{>0}$, hence: $$\phi(x) x^2 e^{-\alpha x^2} = \phi (x) x^2 e^{-\beta x^2}\cdot \underbrace{e^{(\beta -\alpha)x^2}}_{\geq 1} \geq \phi (x) x^2 e^{-\beta x^2}\; .$$ Therefore your inequality with a universal constant $C_0=C_0(\alpha ,\beta)$ is hardly true for every bounded measurable function.

To prove this, let $n\in \mathbb{N}$ and: $$\phi(x) = \phi_n(x)= \frac{1}{x}\cdot \chi_{[n,2n[}(x) = \begin{cases} \frac{1}{x} &\text{, if } n\leq x < 2n \\ 0 &\text{, otherwise}\end{cases}$$ which is measurable and bounded (because $0\leq \phi_n(x)\leq 1$); using such a test function we can explicitly compute: $$\begin{split}\int_0^\infty \phi_n(x) x^2\ e^{-\alpha x^2}\ \text{d} x &= \int_n^{2n} x\ e^{-\alpha x^2}\ \text{d} x\\ &=\frac{1}{2\alpha}\ e^{-\alpha n^2}\ (1-e^{-3\alpha n^2})\\ \int_0^\infty \phi_n(x) x^2\ e^{-\beta x^2}\ \text{d} x &= \frac{1}{2\beta}\ e^{-\beta n^2}\ (1-e^{-3\beta n^2})\; . \end{split}$$ Therefore the ratio: $$\frac{\int_0^\infty \phi_n(x) x^2\ e^{-\alpha x^2}\ \text{d} x}{\int_0^\infty \phi_n(x) x^2\ e^{-\beta x^2}\ \text{d} x} = \frac{\beta}{\alpha}\ e^{(\beta - \alpha)n^2} \frac{1-e^{-3\alpha n^2}}{1-e^{-3\beta n^2}}$$ approaches $+\infty$ when $n\to +\infty$ (because $\beta-\alpha >0$); this fact implies that no universal constant $C_0$ can make your inequality work for every bounded measurable function $\phi$.

$\endgroup$
  • $\begingroup$ That should be $e^{(\beta - \alpha)}x^2$ in the middle term. $\endgroup$ – zhw. May 17 '16 at 16:36
  • $\begingroup$ I think the first equality in the second line has something wrong. Thank you for your answer. $\endgroup$ – Loring May 17 '16 at 16:38
  • $\begingroup$ Yep, somehow I reversed the exponents... Sorry! :-( I'm going to edit that. $\endgroup$ – Pacciu May 17 '16 at 23:00
0
$\begingroup$

Hint: Fix any $y>0.$ Let $\phi_n= n\chi_{[y,y+1/n]}.$ What is

$$\lim_{n\to \infty} \int_0^\infty x^2 e^{-\alpha x^2}\phi_n(x)\, dx?$$

$\endgroup$
  • $\begingroup$ Exuse me! When trying the functions $\phi(x) = \chi_{[n,n+1]}$, one need to consider the sequences $ \int_n^{n+1} x^2 e^{-\alpha x^2} d x $ and $ \int_n^{n+1} x^2 e^{-\beta x^2} d x $, which remains having something trouble for me. So can you give me more details ? $\endgroup$ – Loring May 17 '16 at 16:58
  • $\begingroup$ I changed my hint, and hopefully improved it. $\endgroup$ – zhw. May 17 '16 at 17:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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