I have the following question concerning an estimate of the total variation norm. Let $\mu$ be a bounded Borel measure on $\mathbb{R}$ and denote by $\mu_t$ the measure defined by $\mu_t(\Omega):=\mu(\Omega+t)$ for $\Omega$ a Borel set and $t$ a real number. I want to show that: $$ \sup_{t\in(0,1]}\frac{1}{t}\|\mu_t-\mu\|<\infty $$ By the definition of the variation norm one obtains: $$ \frac{1}{t}\|\mu_t-\mu\|=\frac{1}{t}\sup_{\pi}\sum_{\Omega\in\pi}{|\mu(\Omega+t)-\mu(\Omega)}| $$ where $\pi$ runs over partitions of $\mathbb{R}$ into countable disjoint measurable subsets. Under what kind of conditions is this uniformly bounded on $(0,1]$? Some continuity property maybe? Another idea is to ask $\mu$ to be differentiable but then I still do not see why the expression should be finite.

Does someone have an idea? Thank you very much in advance.

  • $\begingroup$ @ChristianRemling : Thank you very much for your comment? What do you mean with maximal function? Does continuity of measures would help? $\endgroup$ – Mabel Caldwell Jan 14 '19 at 7:08
  • $\begingroup$ I was sloppy; while your quantity is perhaps somewhat reminiscent of some kind of maximal function, it's certainly not very similar to the standard maximal function (of a locally integrable function, say). $\endgroup$ – Christian Remling Jan 14 '19 at 19:10

This is not true in general. The easiest counterexamples come from discrete $\mu$, for which $(\mu_t)_{t \ge 0}$ are singular. For example, if $\mu=\delta_0$ is a Dirac at zero, then $\mu_t=\delta_{-t}$, and $\|\mu_t-\mu\| = \sup_{A}|\mu_t(A)-\mu(A)|=1$ for each $t > 0$.

On the other hand, if $\mu(dx)=f(x)dx$ has a density, then $\mu_t(dx) = f(x+t)dx$, and we have $\frac{1}{t}\|\mu_t-\mu\| = \frac{1}{2t}\int_{\mathbb{R}} |f(x+t)-f(x)|dx$. If $f$ has a sufficiently nice derivative, then this stays bounded as $t \downarrow 0$.

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