About a function of bounded variation

Question

If $f:[a,b]\to\Bbb{R}$ is of bounded variation. Let be $f^+(x)=\max\{f(x),0\}$ and $f^-(x)=\max\{-f(x),0\}$, then $f^+,f^-$ are functions of bounded variation. It is true? I know the inverse is true.

Answer 1

The statement is true: let's prove it, by first recalling some elementary properties of positive and negative parts $f^+$ and $f^-$ of a given function $f:[a,b]\to\Bbb{R}$. $$ \begin{split} f&=f^+-f^-\\ |f|&=f^++f^- \end{split} \iff \begin{split} f^+&=\frac{1}{2}\big(f-|f|\big)\\ f^-&=\frac{1}{2}\big(|f|-f\big) \end{split}\tag{1}\label{1} $$

From \eqref{1}, recalling that $BV([a,b])$ is a (topological) vector (as a matter of fact, Banach) space (see for example Natanson (1983), §8.3, pp.216-217, theorem 3), we can prove that $f\in BV([a,b])$ implies $f^+,f^-\in BV([a,b])$ by proving that the former property implies $|f|\in BV([a,b])$: that's precisely what we'll do below.

Definition 1. A function $f:[a,b]\to\Bbb{R}$ is said to be of bounded variation, symbolized as $f\in BV([a,b])$, if $$ V_a^b(f)=\sup_{P \in \mathscr{P}} \sum_{i=0}^{n_{P}-1} | f(x_{i+1})-f(x_i) |<\infty $$ where

• $P$ is a partition of $[a,b]$ and
• $\mathscr{P}$ is the set of all partitions of the same interval.

Lemma. If $f\in BV([a,b])$ then $|f|\in BV([a,b])$.

Proof. $f\in BV([a,b])$ implies that $V_a^b(f)<\infty$ $$ \begin{split} V_a^b(f)&=\sup_{P \in \mathscr{P}} \sum_{i=0}^{n_{P}-1} | f(x_{i+1})-f(x_i) |\\ &\geq \sup_{P \in \mathscr{P}} \sum_{i=0}^{n_{P}-1} \big| |f(x_{i+1})|-|f(x_i)| \big|\\ &=V_a^b(|f|) \end{split} $$ therefore $|f|\in BV([a,b])$. $\blacksquare$

Then, by the vector space properties of $BV([a,b])$, $$ f\in BV([a,b])\:\Longrightarrow \begin{split} f^+&=\frac{1}{2}\big(f-|f|\big)\in BV([a,b])\\ f^-&=\frac{1}{2}\big(|f|-f\big)\in BV([a,b]) \end{split} $$ Notes

• It is not necessary to have a look at one of the classical monographs in real analysis in order to prove that $BV([a,b])$ is a vector space: the property follow easily from the definition of bounded variation ad from the triangle inequality, by proceeding in a similar way to what is done in the above lemma.
• Again the above lemma, jointly with other standard propositions, implies that $$ f\in BV([a,b])\iff|f|\in BV([a,b])\iff f^+,f^-\in BV([a,b]) $$

[1] Natanson, I. P. (1983)[1961] "Theory of Functions of a Real Variable", revised ed., New York: Ungar, pp. 277, ISBN 0-8044-7012, MR 0067952, Zbl 0064.29102.

Answer 2

Here is a self contained argument:

If $h:=f+g$ then one has for all $x$ and $y$ the inequalities $$|h(x)-h(y)|\leq|f(x)-f(y)|+|g(x)-g(y)|,\quad \bigl|\,|f(x)|-|f(y)|\,\bigr|\leq|f(x)-f(y)|\ .$$ From this we can immediately conclude that when both $f$ and $g$ are of bounded variation on an interval $[a,b]$ then so are $f+g$ and $|f|$.

It remains to note that $$f^+={1\over2}\bigl(|f|+f\bigr),\qquad f^-={1\over2}\bigl(|f|-f\bigr)\ .$$