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

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!!!

• 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? Nov 29, 2019 at 20:44
• edited. I think i fixed that error.
– user637978
Nov 29, 2019 at 20:46
• Are you assuming $f$ is differentiable everywhere in $[a,b]?$
– zhw.
Nov 29, 2019 at 20:56
• I don't think so. I address in one of the paragraphs that both $f$ and $v$ are diff almost everywhere (by Lebesgues theorem)
– user637978
Nov 29, 2019 at 20:57
• do you know the result: "to every monotonic function there corresponds a measure $\mu$ such that ..."
– zhw.
Nov 29, 2019 at 21:40

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

• This is a pretty technical proof, know of any other methods?? Thanks btw!!
– user637978
Dec 1, 2019 at 1:05
• I believe that Lebesgue Stieltjes measure will do the job, as @zhw. noted in the comment. Dec 1, 2019 at 1:06
• 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???
– user637978
Dec 1, 2019 at 15:30
• 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 $\varphi$. Dec 1, 2019 at 16:29
• Yes, I want to bound it so that I can use Lebesgue. Dec 1, 2019 at 16:44