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Consider $F:\mathbb{R}\rightarrow\mathbb{R}$ such that $\sup_{a,b}T_F (a,b)<\infty$ where $T_F (a,b)$ is the total variation of $F$ on the interval $[a,b]$. Then we have

i) $\int_\mathbb{R}|F(x+h)-F(x)|dx\leq A |h| $, for some constant $A$ and $\forall h\in \mathbb{R}$;

ii)$\int_\mathbb{R}F(x)\phi'(x)dx\leq A $ whenever $\phi \in C^1$ with compact support and $|\phi|_\infty\leq 1$.

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  • We can write for $h\neq 0$, using Fubini's theorem for non-negative functions that \begin{align} \int_{\mathbb R}|f(x+h)-f(x)|dx&=\sum_{j\in\mathbb Z}\int_{j|h|}^{(j+1)|h|}|f(x+h)-f(x)|dx\\\ &=\sum_{j\in\mathbb Z}\int_0^{|h|}|f(x+(j+1)h)-f(x+jh)|dx\\\ &=\int_0^{|h|}\sum_{j\in\mathbb Z}|f(x+(j+1)h)-f(x+jh)|dx. \end{align} For integers $M$ and $N$ and $x\in\Bbb R$, we have $$\sum_{j=-N}^M |f(x+(j+1)h)-f(x+jh)|\leqslant T_F(x-Nh,x+(M+1)h)\leqslant \sup_{a,b\in\mathbb R}T_F(a,b),$$ because we took the subdivision $x+jh,-N\leqslant j\leqslant M$ of $[x-Mh,x+(M+1)h]$. We conclude that $\int_{\mathbb R}|f(x+h)-f(x)|dx\leqslant |h|\sup_{a,b\in\mathbb R}T_F(a,b)$.
  • Put $f_n(x)=F(x)n(\phi(x+n^{-1})-\phi(x))$ and check that we have the hypothesis to apply the dominated convergence theorem. We get \begin{align*} \int_{\mathbb R}F(x)\phi'(x)dx&=\lim_{n\to \infty}\int_{\mathbb R}nF(x)(\phi(x+n^{-1})-\phi(x))dx\\\ &=\lim_{n\to \infty}n\int_{\mathbb R}(F(x-n^{-1})-F(x))\phi(x)dx\\\ &\leqslant \limsup_{n\to\infty}n\int_{\mathbb R}|F(x-n^{-1})-F(x)|\cdot |\phi(x)|dx\\\ &\leqslant \sup_{a,b\in\mathbb R}T_F(a,b) \mbox{ by the first step}. \end{align*}
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  • $\begingroup$ I couldn't get the gap in the last equality. I could not see also how to dominate that sequence. $\endgroup$ – checkmath Mar 5 '12 at 3:22
  • $\begingroup$ I think I got it you use the Intermediate Value theorem on $\phi$ right? $\endgroup$ – checkmath Mar 5 '12 at 3:52
  • $\begingroup$ The domination is clear the problem is the last equality since $F\phi$ is not absolutely continuous. $\endgroup$ – checkmath Mar 5 '12 at 4:07
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    $\begingroup$ If a sequence of non-negative real numbers $(a_n)$ is such that there exists a constant $C$ for which for all $N, M$, $\sum_{k=M}^Na_k\leqslant C$, then $\sum_{k\in\mathbb Z}a_k\leqslant C$. $\endgroup$ – Davide Giraudo Jun 24 at 12:12
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    $\begingroup$ $f_n(x)$ can be bounded by $\lvert F(x)\rVert \sup_{t\in\mathbb R}\lvert \phi'(x)\rVert $ and the last supremum is finite since $\phi$ has a compact support. For you second question, I splitted the integral into two parts, and do the substitution $t=x+n^{-1}$. $\endgroup$ – Davide Giraudo Jun 24 at 14:53

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