Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

1 Answer 1

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 $\mathrm{TV}$.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.