Uniform boundedness of integral of differences

[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

 

[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