# Total variation norm estimate

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.

• @ChristianRemling : Thank you very much for your comment? What do you mean with maximal function? Does continuity of measures would help? – Mabel Caldwell Jan 14 at 7:08
• 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). – Christian Remling Jan 14 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 $$t \downarrow 0$$.