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Uniform boundedness of integral of differences

  1. Oct 30, 2009 #1
    Hello,

    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
     
  2. jcsd
  3. Nov 20, 2009 #2
    Sorry to bring this thread up, but it seems to me that bounded variation is indeed one way to go.

    At least for univariate step functions f it seems clear that

    [tex]
    \lim_{u\to 0}\frac{1}{|u|}\int_{\mathbb{R}}{dv|f(v-u)-fv|}=||f||_{\text{Variation}}
    [/tex]

    Does anyone know if a similar result holds for continuous functions? Or maybe even in a multivariate setting.

    I'd very much appreciate any thoughts,

    Thanks,

    Pere
     
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