Prove $\int_\mathbb{R}|F(x+h)-F(x)|dx\leq A |h|$ for bounded variation function F(x)

Question

A bounded function $F$ is said to be of bounded variation on $\mathbb{R}$ if $F$ is of bounded variation on any finite sub-interval $[a, b]$, and $\sup_{a,b}T_F (a,b)<\infty$ $~~~~$(The $sup_{a,b}T_F(a,b)$ means the supreme taken over total variation of F on all finite interval $[a,b]$ )

And I want to prove that:

$\int_\mathbb{R}|F(x+h)-F(x)|dx\leq A |h|$

There is a Hint: write $F = F_1 − F_2$, where $F_1$ and $F_2$ are monotonic and bounded.

But I don't konw how to get the decomposite and it will be much easier for me if I get it.

Answer 1

If $F$ is of bounded variation on $\mathbb{R}$, then there is a finite signed measure $\mu_F$ whose variation $|\mu_F|$ has finite total mass, i.e. $|\mu_F|(\mathbb{R})<\infty$, and such that $\mu_F((a,b])=F(b)-F(a)$. Suppose $h>0$. Then by Fubini's theorem \begin{align} \int_\mathbb{R}|F(x+h)-F(x)|\,dx&=\int_\mathbb{R}\Big|\int_{(x,x+h]}\,d\mu_F(dy)\Big|\\ &\leq\int_\mathbb{R}\int_{(x,x+h]}|\mu_F|(dy)dx\\ &=\int_\mathbb{R}\Big(\int_{[y-h,y]}dx\Big)|\mu_F|(dy)\\ &=|\mu_F|(\mathbb{R})h \end{align}

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The justification of the statements made here concerning existence and uniqueness of the finite signed measure $\mu_F$ such that $\mu_F((a,b])=F(b)-F(b)$ is based on standard results cover in many integration books and is all under the umbrella of Riemann-Stieltjes theorem.

Starting with:

Proposition 1: bounded right-continuous monotone nondecreasing function on $\mathbb{R}$ $F$, there is a unique measure $\mu_F$ such that $\mu_F((a,b])=F(b)-F(b).

There are many ways to prove this Theorem. In many textbooks (Folland, G. Real Analysis: Modern Techniques and Applications, Wiley, Theorem 1.16 pp. 35) they use the Caratheodory extension theorem.

For any function $F:\mathbb{R}\rightarrow\mathbb{R}$, the variation of a function $F$ at $x$ is defined as: $$T_F(x) = \sup\{\sum^n_{k=1}|F(x_j)-F(x_{j-1})|:n\in\mathbb{N}, -\infty<x_0<\ldots<x_n=x\}$$ $F$ is of bounded variation if $T_F(x)$ there is $M>0$ such that $T_F(x)\leq M$ for all $x\in\mathbb{R}$.

The following result covered in many textbooks:

Proposition 2: If $F$ is of bounded variation, then $F$, $F_1:x\mapsto F_T(x)+F(x)$ and $F_2:x\mapsto F_T(x)-F(x)$ are both bounded, and $F_1$ and $F_2$ are monotone nondecreasing functions. Clearly $F(x)=\frac12(F_1(x)-F_2(x))$

(See Folland, op. cit, pp. 102-103)

If $F$ is ringhtcontinuous, then an application of Proposiiotn 2 to $F_1$ and $F_2$ yields a signed measure $\mu_F=\mu_{F_1}-\mu_{F_2}$ on $\mathbb{R}$ such that $$\mu_F((a,b])=F(b)-F(a)$$ In fact, (see Folland, op. cit. pp. 104-105)

Theorem: If $\mu$ is a finite signed measure on $\mathbb{R}$ and $F(x)=\mu((-\infty,x])$, then $F$ is of bounded variation with $\lim_{x\rightarrow-\infty}F(x)=0$. Conversely, if $F$ bounded and of bounded variation, then by replacing $F$ with $F-\lim_{x\rightarrow-\infty}F(x)$ if necessary), there is a unique measure signed measure $\mu_F$ such that $\mu_F((-\infty,x])=F(x)$ and the variation function $|\mu_F|$ if $\mu_F$ is the measure $\mu_{T_F}$ associated to the nondecreasing right-continuous function $T_F$.