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Show the Lemma following:

Let the function $f$ be of bounded variation on the closed, bounded interval $[a,b]$. Then $f$ has the following explicit expression as the difference of two increasing functions on $[a,b]$:

$f(x)=[f(x)+TV(f_{[a,x]})]-TV(f_{[a,x]})$ for all $x\in [a,b].$

Hello! I don't clear how show this... If $f$ is a function of bounded variation then $TV(f_{[a,x]})<\infty$. I think that can showed that $f(x)+TV(f_{[a,x]})$ and $TV(f_{[a,x]})$ are increasing functions, yes?

Julio_fmat
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  • Yes. It's obvious that $TVf$ is increasing. To show that $f+TVf$ is increasing, suppose $x<y$. You need to show that $TVf(y)-TVf(x)\ge f(x)-f(y)$. Which is also close to obvious, from the definition of $TVf$. – David C. Ullrich Jun 03 '16 at 15:29
  • Thanks, then can expressed the function as $f(x)=TV(f_{[a,x]})+[f(x)-TV(f_{[a,x]})]$ yes? Not problem? But I think that this step is done to show the Jordan's Theorem? (The direct implication...) – Julio_fmat Jun 03 '16 at 15:44
  • I have no idea what theorem you're referring to. You asked about the lemma... – David C. Ullrich Jun 03 '16 at 15:46
  • Hey David. My proof for $f(x)+TV(f_{[a,x]})$ increasing is this... Let $x<y$ we can show that $(f+TV(f_{[a,x]}))(x)\le (f+TV(f_{[a,x]}))(y).$ Indeed, note that $f$ is a function of bounded variation, hence that $f$ is increasing. Therefore, $f(x)\le f(y).$ Furthermore, at $TV(f)$ is increasing, we have that $TV(f)(x)\le TV(f)(y).$ Thus, we conclude that $f(x)+TV(f_{[a,x]})(x)\le f(y)+TV(f_{[a,x]})(y)$. Finally, $ f(x)+TV(f_{[a,x]})$ is increasing. ¿Yes? :) – Julio_fmat Jun 03 '16 at 16:20
  • No. Saying $f$ has bounded variation does not say that $f$ is increasing. I gave you the inequality you need to prove in my first comment. That inequality follows from the definition of $TVf$. – David C. Ullrich Jun 03 '16 at 16:22
  • Aaahm, but all function monotone is of bounded variation... – Julio_fmat Jun 03 '16 at 16:30
  • Think before you type. Of course every monotone function (on a compact interval) has bounded variation. You said that every function of bounded variation is monotone. That's not the same thing. And that's obviously false - if $f$ has bounded variation then $-f$ also has bounded variation. – David C. Ullrich Jun 03 '16 at 16:43
  • Another little hint regarding that inequality: If $a\le x<y$ then $TV_{[a,y]}f-TV_{[a,x]}f=TV_{[x,y]}f$. – David C. Ullrich Jun 03 '16 at 16:45

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