Hello,(adsbygoogle = window.adsbygoogle || []).push({});

I'm interested in the following problem. We are given a probability density p on R, that is a continuous positive function which integrates to 1. What are the weakest possible conditions on p such that there exists a K>0 satisfying

[tex]

\int_{\mathbb{R}}{|p(x+y)-p(x)|dx}<K|y|,\quad\forall y\in\mathbb{R}

[/tex]

or equivalently

[tex]

\sup_{y\in\mathbb{R}}\frac{\int_{\mathbb{R}}{|p(x+y)-p(x)|dx}}{|y|}<\infty.

[/tex]

I assume there must be conditions like bounded variation, continuous first derivative or something similar.

If we denote

[tex]

\Delta(y)=\int_{\mathbb{R}}{|p(x+y)-p(x)|dx}

[/tex]

it seems clear that only the behavior around y=0 matters; because [itex]\Delta[/itex] is continuous it must take on its maximum on the compact set [itex]\{y:\varepsilon\leq |y|\leq R\}[/itex], R>2, which we denote by M. This means that

[tex]

\sup_{\{y:\varepsilon\leq |y|\leq R\}}\frac{\Delta(y)}{|y|}\leq\frac{M}{\varepsilon}.

[/tex]

The easy estimate [itex]\Delta(u)\leq 2[/itex] implies

[tex]

\sup_{\{y:|y|\geq R\}}\frac{\Delta(y)}{|y|}\leq 1

[/tex]

so one only needs to worry about [itex]\{y:|y|\leq \varepsilon\}[/itex] and I think that it is here that the derivative of p enters the game.

I would appreciate very much any thoughts on the problem,

regards,

Pere

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Uniform boundedness of integral of differences

Loading...

Similar Threads - Uniform boundedness integral | Date |
---|---|

I Understanding uniform continuity.... | May 9, 2016 |

I Uniform current in cylinder and straight wire: same field? | Apr 28, 2016 |

Difference between continuity and uniform continuity | May 7, 2015 |

Uniform Convergence & Boundedness | Apr 12, 2010 |

Uniform continuity and boundedness | Dec 9, 2007 |

**Physics Forums - The Fusion of Science and Community**