# Functions of bounded variation on all $\mathbb{R}$

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*}
I think I got it you use the Intermediate Value theorem on $\phi$ right? –  checkmath Mar 5 '12 at 3:52
The domination is clear the problem is the last equality since $F\phi$ is not absolutely continuous. –  checkmath Mar 5 '12 at 4:07
@DavideGiraudo, Could you please elaborate this one, I am having a hard time to follow this "Since for integers $M$ and $N$ we have $$\sum_{j=-N}^M |f(x+(j+1)h)-f(x+jh)|\leq T_F(-Mh,(N+2)h)\leq \sup_{a,b\in\mathbb R}T_F(a,b)$$" –  Deepak Jan 2 '13 at 1:00
@Deepak My previous estimate didn't take account of $x$. Now I've corrected. –  Davide Giraudo Jan 2 '13 at 11:50