Let $f: [a,b] \rightarrow \mathbb{R}$ be a bounded variation function.
Let $f_a(x) = f(a)+ \int_a^x f' dm$ be the indefinite Lebesgue integral of $f$.
Let $g= f – f_a$.
Show if $f$ is monotone, then $g$ is also monotone.
Attempt:
Consider $f_a(x) = f(a)+ \int_a^x f' dm$.
We know since $f \in BV[a,b]$, $f'$ is integrable and $\int_a^x f' dm$ is monotone increasing.
So $f_a$ is also monotone.
But we know that the sum of a monotone increasing and monotone decreasing function is not necessarily monotone.
And that's where I got stuck.
I don't have any clues how to use the monotonicity other than this approach.
Any help will be appreciated!
Thank you.
Best Answer
Let's assume $f$ is increasing. Let $y < x$. Then we have $$ g(x) - g(y) = f(x) - \bigg[f(a) + \int_{a}^{x} f'(s) \mathrm{d} s\bigg] - \bigg[f(y) - \bigg[f(a) + \int_{a}^{y} f'(s) \mathrm{d} s\bigg]\bigg]. $$ Distributing all of those minus signs and cancelling, we obtain $$ g(x) - g(y) = f(x) - f(y) - \int_{y}^{x} f'(s) \mathrm{d} s. $$ Now we state a lemma that will finish the proof.
Lemma: Let $I \subseteq \mathbb{R}$ be an interval and let $u \colon I \rightarrow \mathbb{R}$ be a monotone function. Then $u'$ is a Lebesgue measurable function and for every $[a, b] \subseteq I$, $$ \int_{a}^{b} |u'(x)| \mathrm{d} x \leq |u(b) - u(a)|. $$ By this lemma and the fact that $f$ is increasing, $$ g(x) - g(y) \geq f(x) - f(y) - \int_{y}^{x} |f'(s)| \mathrm{d} s \geq 0. $$ For reference, this lemma is Corollary 1.25 in Giovanni Leoni's A First Course in Sobolev Spaces (2nd edition), or Corollary 1.37 in the 1st edition.