Function coincides with a function of bounded variation almost everywhere

Question

Problem

Suppose $f\in L^1(\mathbb R)$ satisfies that there exists $A\ge0$ such that $$\int_{\mathbb R}\lvert f(x+h)-f(x)\rvert dx\le A\lvert h\rvert$$ for all $h\in\mathbb R$. We need to show that $f$ coincides with some function of bounded variation on $\mathbb R$, i.e., there exists $g\colon\mathbb R\to\mathbb C$, such that $g$ is of bounded variation on $\mathbb R$, and $f=g$ a.e.

Discussion

Conversely, if $f$ is of bounded variation, then $f$ could be written as the difference of two bounded increasing functions, and then the absolute value could be expanded and the approximation is somewhat easier. However, I need to prove the converse proposition, which seems tough to me.

Any idea? Thanks!

Answer 1

Given a measurable function $f:\Bbb R\to \Bbb C$, denote $$T:\Bbb R\to [0,+\infty]:\quad h\mapsto \int_{\Bbb R}|f(x+h)-f(x)|dx. \tag{1}$$ Your assumption on $f$ is simply $$T(h)\le A|h|,\quad \forall~h\in\Bbb R. \tag{2}$$ When $f$ is locally integrable, i.e. integrable on every compact set, by Lebesgue differentiation theorem,

$$F:\Bbb R\to \Bbb C,\quad x\mapsto \int_0^x f(t)dt \tag{3}$$ is well defined, $$\quad E:=\{x\in\Bbb R: F'(x)\text { exists}\} \tag{4}$$ is of full measure and $F'=f$ a.e. . Moreover, denote the total variation of $F'$ on $E$ by ${\rm V}$, i.e. $$V:=\sup\big\{\sum_{i=1}^n|F'(x_i)-F'(x_{i-1})|: x_0,x_1,\cdots, x_n\in E,\ x_0<x_1<\cdots <x_n\big\}. \tag{5}$$ We claim that for every local integrable function $f:\Bbb R\to \Bbb C$, $$V\le \liminf_{h\to 0}\frac{T(h)}{|h|}. \tag{6}$$ Once $(6)$ is proved, from $(2)$ we know that $F'$ is of bounded variation on $E$. To complete the proof, without of loss of generality, we may suppose that $f$ is real valued(otherwise, consider real part and imaginary part separately). Then $F'$ is real valued and of bounded variation on $E$. By imitation the argument for functions of bounded variation on an interval, we can see that there are bounded increasing functions $f_\pm: E\to \Bbb R$, such that $F'=f_+-f_-$ on $E$. Extend $f_\pm$ to bounded increasing functions on $\Bbb R$, still denoted by $f_\pm$. As a result, for $g:=f_+-f_-$, $g$ is of bounded variation on $\Bbb R$. Since $E$ is of full measure in $\Bbb R$, $g=F'=f$ a.e. . Therefore, it suffices to prove $(6)$.

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Proof of $(6)$:

Given $h\ne 0$ and points $x_0<\cdots < x_n$ in $E$, \begin{eqnarray} && \sum_{i=1}^n\left|\frac{F(x_i+h)-F(x_i)}{h}-\frac{F(x_{i-1}+h)-F(x_{i-1})}{h}\right| \\ &=& \frac{1}{|h|}\sum_{i=1}^n\left|\int_{x_{i-1}}^{x_i}\left(f(x+h)-f(x)\right)dx \right| \\ &\le& \frac{1}{|h|}\sum_{i=1}^n\int_{x_{i-1}}^{x_i}|f(x+h)-f(x)|dx \\ &\le& \frac{T(h)}{|h|} . \end{eqnarray} Letting $h\to 0$ first and then taking supremum for $x_0<\cdots < x_n$ in $E$, $(6)$ follows. $\quad\square$

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Remark: With some additional work, it can be shown that:

1. Let $f:\Bbb R\to \Bbb C$ be measurable. If $(2)$ holds for some $A>0$, then $f$ is locally integrable.
2. Now suppose $f$ is locally integrable, so $(3)$, $(4)$ and $(5)$ are well defined. Define $$\mathscr F:=\{g:\Bbb R\to \Bbb C \mid g=f \text{ a.e.} \},\tag{7}$$ and for every $g:\Bbb R\to \Bbb C$, denote its total variation on $\Bbb R$ by ${\rm Var}(g)$. Then we have: $$V=\sup_{h\ne 0}\frac{T(h)}{|h|}=\lim\limits_{h\to 0}\frac{T(h)}{|h|}=\inf_{g\in \mathscr F} {\rm Var}(g).\tag{8}$$
3. If $V<+\infty$, then there exists a unique right(or left) continuous function $g\in \mathscr F$; moreover, for this particular $g$, ${\rm Var}(g)=V$.