# A real analysis problem on the integral inequality.

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!

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

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 $\phi$.

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

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

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

• 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 ? – Loring May 17 '16 at 16:58
• I changed my hint, and hopefully improved it. – zhw. May 17 '16 at 17:20