Uniform boundedness of integral of differences

1. Oct 30, 2009

Pere Callahan

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
$$\int_{\mathbb{R}}{|p(x+y)-p(x)|dx}<K|y|,\quad\forall y\in\mathbb{R}$$
or equivalently
$$\sup_{y\in\mathbb{R}}\frac{\int_{\mathbb{R}}{|p(x+y)-p(x)|dx}}{|y|}<\infty.$$

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

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

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

The easy estimate $\Delta(u)\leq 2$ implies
$$\sup_{\{y:|y|\geq R\}}\frac{\Delta(y)}{|y|}\leq 1$$
so one only needs to worry about $\{y:|y|\leq \varepsilon\}$ 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. Nov 20, 2009

Pere Callahan

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

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

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