Prove that there exist $M$ such that when $|h|$ is sufficiently small, $ \frac{1}{h}\int_a^b|f(x+h)-f(x)|\mathrm{d}x\leq M. $

Question

Suppose $f(x)\in$ BV$([a,b])$(bounded variation). Prove that there exist $M$ such that when $|h|$ is sufficiently small, $$ \dfrac{1}{h}\int_a^b|f(x+h)-f(x)|\mathrm{d}x\leq M. $$

By Jordan decomposition, we only need to prove the $f$ increasing case. I know that in this condition, $f$ is differentiable almost every where, and $$ \int_a^bf'(x)\mathrm{d}x\leq f(b)-f(a). $$ But I don't know how to proceed.

Appreciate any help.

Answer 1

As you said, it suffices to consider the case that $f$ is a non-decreasing function.

For $0 < h < b-a$ is $$ \int_a^b |f(x+h)-f(x)| \, dx = \int_a^b (f(x+h)-f(x)) \, dx \\ = \int_{a+h}^{b+h} f(x) \, dx - \int_a^b f(x) \, dx = h f(b) - \int_a^{a+h} f(x) \, dx \\ \le h f(b) - \int_a^{a+h} f(a) \, dx = h(f(b) - f(a)) \, . $$ Similarly, $$ \int_a^b |f(x+h)-f(x)| \, dx \le (-h)(f(b)-f(a)) $$ $-(b-a)< h < 0$, so that $$ \frac{1}{|h|}\int_a^b|f(x+h)-f(x)| \, dx \leq f(b)-f(a) $$ for $0 < |h| < b-a$.

Answer 2

Case 1: $f$ is increasing. Note that $f(x)=f(a)$ if $x<a$ and $f(x)>f(b)$ if $x>b$. Under this extension, $f$ is increasing over $\mathbb{R}$. Consider the case that $h>0$. Recall that monotone function is integrable over compact set $[a,b]$. We have that \begin{eqnarray*} & & \int_{a}^{b}|f(x+h)-f(x)|dx\\ & = & \int_{a}^{b}\left[f(x+h)-f(x)\right]dx\\ & = & \int_{a}^{b}f(x+h)dx-\int_{a}^{b}f(x)dx\\ & = & \int_{a+h}^{b+h}f(x)dx-\int_{a}^{b}f(x)dx\\ & = & \int_{b}^{b+h}f(x)dx-\int_{a}^{a+h}f(x)dx\\ & = & hf(b)-\int_{a}^{a+h}f(x)dx. \end{eqnarray*} Therefore, \begin{eqnarray*} \frac{1}{h}\int_{a}^{b}|f(x+h)-f(x)|dx & = & f(b)-\frac{1}{h}\int_{a}^{a+h}f(x)dx\\ & \leq & f(b)-\frac{1}{h}\int_{a}^{a+h}f(a)dx\\ & = & f(b)-f(a). \end{eqnarray*}

Then, we consider the case that $h<0$. We have that \begin{eqnarray*} & & \int_{a}^{b}|f(x+h)-f(x)|dx\\ & = & \int_{a}^{b}\left[f(x)-f(x+h)\right]dx\\ & = & \int_{a}^{a+h}f(x)dx-\int_{b}^{b+h}f(x)dx\\ & = & f(a)h+\int_{b+h}^{b}f(x)dx\\ & \leq & f(a)h+\int_{b+h}^{b}f(b)dx\\ & = & f(a)h-f(b)h. \end{eqnarray*} Hence, $\frac{1}{|h|}\int_{a}^{b}|f(x+h)-f(x)|dx\leq f(b)-f(a)$.

In short, if we take $M=f(b)-f(a)\geq0$ at the very beginning, then for any $h\neq0$, $\frac{1}{|h|}\int_{a}^{b}|f(x+h)-f(x)|dx\leq M$.

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Case 2: $f$ is of bounded variation over $[a,b]$ and that $f(x)=f(a)$ if $x<a$, $f(x)=f(b)$ if $x>b$. Choose increasing functions $g_{1},g_{2}$ such that $f=g_{1}-g_{2}$ and $g_{i}(x)=g_{i}(a)$ if $x<a$ and $g_{i}(x)=g_{i}(b)$ if $x>b$, for $i=1,2$. By Case 1, there exists $M_{1}>0$ and $M_{2}>0$ such that $\frac{1}{|h|}\int_{a}^{b}|g_{i}(x+h)-g_{i}(x)|dx\leq M_{i}$ whenever $h\neq0$, for $i=1,2.$ We have that \begin{eqnarray*} & & \frac{1}{|h|}\int_{a}^{b}|f(x+h)-f(x)|dx\\ & = & \frac{1}{|h|}\int_{a}^{b}\left|[g_{1}(x+h)-g_{1}(x)]-[g_{2}(x+h)-g_{2}(x)]\right|dx\\ & \leq & \frac{1}{|h|}\int_{a}^{b}\left|g_{1}(x+h)-g_{1}(x)\right|dx+\frac{1}{|h|}\int_{a}^{b}\left|g_{2}(x+h)-g_{2}(x)\right|dx\\ & \leq & M_{1}+M_{2}. \end{eqnarray*}

Answer 3

Extend $f(x)=f(b)$ for $x\gt b$; then $$\newcommand{\Var}{\operatorname*{Var}} \begin{align} h\,\Var_{[a,b]}(f) &=\int_a^{a+h}\color{#C00}{\Var_{[a,b]}(f)}\,\mathrm{d}x\tag{1a}\\ &\ge\int_a^{a+h}\color{#C00}{\sum_{k=0}^{\left\lfloor\frac{b-a}h\right\rfloor}|f(x+(k+1)h)-f(x+kh)|}\,\mathrm{d}x\tag{1b}\\ &=\int_a^{a+\left\lfloor\frac{b-a+h}h\right\rfloor h}|f(x+h)-f(x)|\,\mathrm{d}x\tag{1c}\\ &\ge\int_a^b|f(x+h)-f(x)|\,\mathrm{d}x\tag{1d} \end{align} $$ Explanation:
$\text{(1a)}$: integral of a constant over an interval of width $h$
$\text{(1b)}$: the variation is the supremum of such sums
$\text{(1c)}$: collect the essentially disjoint intervals into one
$\text{(1d)}$: $a+\left\lfloor\frac{b-a+h}h\right\rfloor h\ge b$

Therefore, $$ \frac1h\int_a^b|f(x+h)-f(x)|\,\mathrm{d}x\le\Var_{[a,b]}(f)\tag2 $$