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I'm finding some trouble in showing this fact... It seems to me that I should go trough riemann approximations of the integral, but i cannot formalize this process. That's why i'm asking for some help, at least for some hints...

Let $f:[0,1]\to\mathbb R$ be a function with bounded variation, and let $TV_0^1(f)$ be its total variation. Then show that for any $h\in (0,1)$, the following holds: $$\int_0^{1-h}|f(x+h)-f(x)|\mathrm dx\leq hTV_0^1(f).$$ Thanks for your attention

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up vote 8 down vote accepted

Hint: Let $f(x)=f(1)$ for every $x\geqslant1$ and introduce, for every $x$ in $(0,h)$, $$ g(x)=\sum\limits_{n=0}^{+\infty}|f(x+nh)-f(x+(n+1)h)|, $$ that is, $$ g(x)=|f(x)-f(x+h)|+|f(x+h)-f(x+2h)|+|f(x+2h)-f(x+3h)|+\ldots. $$ Then the integral on the LHS is bounded by the integral of $g$ on $(0,h)$. You can show this by splitting the integral from $0$ to $1-h$ into a sum of integrals on the intervals $(nh,(n+1)h)$ for $n\geqslant0$.

You are done if you know that $g(x)\leqslant \mathrm{TV}_0^1(f)$ uniformly over $x$ in $(0,h)$. And this follows from the definition of $\mathrm{TV}$.

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Oh, yes. Missed your comment on making f zero after 1. – Ilya Sep 29 '11 at 12:21

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