Bounded variation, difference of two increasing functions
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Prove that
if $f$ is of bounded variation in $[a,b]$, it is the difference of two positive, monotonic increasing functions; and
the difference of two bounded monotonic increasing functions is a function of bounded variation.
real-analysisbounded-variation
asked 2012-05-05
user id:27106
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1 Answers
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Let $f$ a function of bounded variation. Let $F(x):=\sup \sum_{j=1}^{n-1}|f(x_{j+1})-f(x_j)|=:\operatorname{Var}[a,x]$, where the supremum is taken over the $x_1,\ldots,x_n$ which satisfy $a=x_1. Since $f$ is of bounded variation, $F$ is bounded, and by definition increasing. Let $G:=F-f$. We have to show that $G$ is bounded and increasing. Boundedness follows from this property for $f$ and $F$, now fix $a\leq x_1. We have $G(x_2)-G(x_1)=F(x_2)-f(x_2)-F(x_1)+f(x_1)\geq 0$ because $\operatorname{Var}[a,x_1]+f(x_2)-f(x_1)\leq \operatorname{Var}[a,x_1]+|f(x_2)-f(x_1)|\leq \operatorname{Var}[a,x_2]$.
If $f$ and $g$ are of bounded variation so is $f-g$. If $f$ is increasing then we have, if $a=x_0 that $\sum_{j=1}^{n-1}|f(x_{j+1})-f(x_j)|=|f(b)-f(a)|$, so $f$ is of bounded variation. So the difference of two bounded monotonic increasing functions is of bounded variation.