Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $g:\mathbb{R}\rightarrow\mathbb{R}$ be bounded with derivative $g'$. I have shown that the following inequality holds for all $w\in\mathbb{R}$, \begin{equation}\bigg|w\int_0^1\int_{-\infty}^{\infty}tg'(tw+x\sqrt{1-t^2})e^{-x^2/2}dxdt\bigg|\leq 8\sqrt{\pi}\|g\|_{\infty}, \end{equation} where $\|g\|_{\infty}=\sup_{x\in\mathbb{R}}|g(x)|$.

$\mathbf{Question:}$ I would like to prove that the following statement, which extends the above result:

Let $g:\mathbb{R}\rightarrow\mathbb{R}$ be bounded with derivative $g'$ and let $n$ be a postive integer. Then, for all $w\in\mathbb{R}$, \begin{equation}\bigg|w\int_0^1\int_{-\infty}^{\infty}t^ng'(tw+x\sqrt{1-t^2})e^{-x^2/2}dxdt\bigg|\leq 8\sqrt{\pi}\|g\|_{\infty}. \end{equation} I used an indirect approach to prove the first inequality, which is not applicable to the second inequality. However, I think that it should be possible to deduce the second inequality from the first inequality.

$\mathbf{PS:}$ Note that for the case that $g'$ has no sign changes (suppose, without lose of generality, that $g'$ is non-negative on $\mathbb{R}$), we can used the first inequality to prove the second: \begin{align*}&\bigg|w\int_0^1\int_{-\infty}^{\infty}t^ng'(tw+x\sqrt{1-t^2})e^{-x^2/2}dxdt\bigg| \\ &\leq |w|\int_0^1\int_{-\infty}^{\infty}t^n|g'(tw+x\sqrt{1-t^2})|e^{-x^2/2}dxdt \\ &\leq |w|\int_0^1\int_{-\infty}^{\infty}tg'(tw+x\sqrt{1-t^2})e^{-x^2/2}dxdt \\ &\leq 8\sqrt{\pi}\|g\|_{\infty}, \end{align*} as required.

share|improve this question
1  
I think you can easily reduce to the case of a monotone $g$ if you replace it with a primitive function of either $g'_+$ or $g'_-$. The double integral increases and $\|g\|_\infty$ does not change. –  Pietro Majer Nov 3 '13 at 21:12
    
Many thanks for your answer! That answers my question. –  red271 Nov 9 '13 at 11:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.