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Let $f$ be of bounded variation on $[a,b]$ and define $v(x) = TV(f_{[a,x]}$ for all $x \in [a,b]$.

Show that $\int_a^b |f'| \leq TV(f)$, and show that this is equality $iff$ $f$ is absolutely continuous.

$proof$:

proof: Let $P$ be the trivial partition of $[a,x]$ for all $x \in (a,b]$. Then we have that:

$v(x+h) \geq V(f,P) = |f(x+h)-f(a)|$

$v(x) \geq V(f,P) = |f(x)-f(a)|$

And thus we have:

$v'(x) = \lim_{h \rightarrow 0^+} \frac{v(x+h)-v(x)}{h}$

$\geq \lim_{h \rightarrow 0^+} \frac{|f(x+h)-f(a) - f(x) + f(a)|}{h}$

$\geq \lim_{h \rightarrow 0^+} \frac{|f(x+h)-f(x)|}{h}$

$=f'(x)$

Furthermore, $f$ can be written as a sum of monotonic function, and $v$ is a monotonic function, thus both are differential almost everywhere, and so the previous inequality holds almost everywhere.

Furthermore, by the monotonicty of the integral we have:

$\int_a^b |f'| \leq \int_a^b v' = TV(f,[a,b]) - TV(f,[a,a]) = TV(f,[a,b])$ $\\$


Is this correct? Furthermore, how do I prove that equality holds if $f$ is absolutely continuous? Thanks!!!

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    $\begingroup$ You have an error when you seemingly write $-v(x) \geq -|f(x)-f(a)|$ in the sequence of inequalities starting with $v'(x)$. Why don't you just use the same argument to show $$v(x+h)-v(x) = \text{ variation on } [x,x+h] \geq |f(x+h)-f(x)|$$ directly? $\endgroup$ Nov 29, 2019 at 20:44
  • $\begingroup$ edited. I think i fixed that error. $\endgroup$
    – user637978
    Nov 29, 2019 at 20:46
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    $\begingroup$ Are you assuming $f$ is differentiable everywhere in $[a,b]?$ $\endgroup$
    – zhw.
    Nov 29, 2019 at 20:56
  • $\begingroup$ I don't think so. I address in one of the paragraphs that both $f$ and $v$ are diff almost everywhere (by Lebesgues theorem) $\endgroup$
    – user637978
    Nov 29, 2019 at 20:57
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    $\begingroup$ do you know the result: "to every monotonic function there corresponds a measure $\mu$ such that ..." $\endgroup$
    – zhw.
    Nov 29, 2019 at 21:40

1 Answer 1

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For any partition $a=t_{0}<t_{1}<\cdots<t_{n}=b$, we have \begin{align*} \sum_{i=1}^{n}|f(t_{i})-f(t_{i-1})|&=\sum_{i=1}^{n}\left|\int_{t_{i-1}}^{t_{i}}f'(x)dx\right|\\ &\leq\sum_{i=1}^{n}\int_{t_{i-1}}^{t_{i}}|f'(x)|dx\\ &=\int_{a}^{b}|f'(x)|dx, \end{align*} so $TV(f)\leq\displaystyle\int_{a}^{b}|f'(x)|dx$ is established.

Next we denote $\text{sgn}f'(x)=1$ for $f'(x)\geq 0$ and $\text{sgn}f'(x)=0$ for $f'(x)<0$, then $\text{sgn}f'\in L^{1}[a,b]$ and hence can be approximated by step functions $\varphi$ on $[a,b]$ in $L^{1}$ sense.

Note that for \begin{align*} \varphi(x)=\sum_{i=1}^{n}a_{i}\chi_{(t_{i-1},t_{i})}(x), \end{align*} we have \begin{align*} \max(\min(\varphi(x),1),-1)=\sum_{i=1}^{n}\max(\min(a_{i},1),-1)\chi_{(t_{i-1},t_{i})}(x), \end{align*} and \begin{align*} |\max(\min(\varphi(x),1),-1)-\text{sgn}f'(x)|\leq|\varphi(x)-\text{sgn}f'(x)|, \end{align*} so we can assume that all $a_{i}$ are such that $|a_{i}|\leq 1$.

As a consequence, \begin{align*} \int_{a}^{b}|f'(x)|dx&=\int_{a}^{b}f'(x)\text{sgn}f'(x)dx\\ &=\int_{a}^{b}f'(x)(\text{sgn}f'(x)-\varphi(x))dx+\int_{a}^{b}f'(x)\varphi(x)dx. \end{align*} We can use Lebesgue Dominated Convergence Theorem to make the term \begin{align*} \int_{a}^{b}f'(x)(\text{sgn}f'(x)-\varphi(x))dx \end{align*} to be arbitrarily small.

Now we estimate that \begin{align*} \left|\int_{a}^{b}f'(x)\varphi(x)dx\right|&\leq\sum_{i=1}^{n}|a_{i}|\left|\int_{a}^{b}f'(x)\chi_{(t_{i-1},t_{i})}(x)dx\right|\\ &\leq\sum_{i=1}^{n}\left|\int_{t_{i-1}}^{t_{i}}f'(x)dx\right|\\ &=\sum_{i=1}^{n}\left|f(t_{i})-f(t_{i-1})\right|\\ &\leq TV(f), \end{align*} we are done.

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  • $\begingroup$ This is a pretty technical proof, know of any other methods?? Thanks btw!! $\endgroup$
    – user637978
    Dec 1, 2019 at 1:05
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    $\begingroup$ I believe that Lebesgue Stieltjes measure will do the job, as @zhw. noted in the comment. $\endgroup$
    – user284331
    Dec 1, 2019 at 1:06
  • $\begingroup$ How come we need to be able to assume that $|a_i| \leq 1$?? Is it so we know $f'(x)a_i(x)$ is integrable??? $\endgroup$
    – user637978
    Dec 1, 2019 at 15:30
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    $\begingroup$ Do you see the $\max\min$ stuff? Actually I let $b_{m}=\max\min...$, and these $b_{n}$ is less than $1$, and I use this version of $\varphi$. $\endgroup$
    – user284331
    Dec 1, 2019 at 16:29
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    $\begingroup$ Yes, I want to bound it so that I can use Lebesgue. $\endgroup$
    – user284331
    Dec 1, 2019 at 16:44

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