I have a question about the following problem:
Let $f:[a,b] \to \mathbb{R}$ be a function of bounded variation with $f(a) = 0$ and $f_1$, $f_2$ be two increasing function such that $f_1(a) = 0$ and $f_2(a)=0$. and $f=f_1-f_2$. Define positive and negative variation $$TV^{+}(f)_a^x = \frac{TV(f)_a^x + f(x)}{2},TV^{-}(f)_a^x = \frac{TV(f)_a^x - f(x)}{2} $$ Then by Jordan decomposition we know $$f = TV^+(f)_a^x - TV^-(f)_a^x$$ where two terms are increasing. Show this is a decomposition of $f$ as the difference of two increasing functions in the lowest terms ($TV^+(f) \leq f_1$ and $TV^-(f) \leq f_2$)
I first of all don't get what the question is telling us to show. So prove that for all increasing function positive and negative variation is the smallest?
Thank you!
The question wants you to show that ${TV}^{+}(f)\leq f_1$ and ${TV}^{-}(f)\leq f_2$ as I think you've surmised :)
What that means in English is that any function of bounded variation can be decomposed into the difference of two increasing functions and that the smallest${}^{1}$ pair of such functions are the positive and negative variations of $f$ itself. This is useful in that it prescribes a way to find a decomposition of $f$ into the difference of increasing functions so that each time doesn't require a special effort (and proof).
${}^{1}$Normally for 'smallest' we'd say $|{TV}^{±}(f)|\leq f_i$ where $i=1,2$ respectively, but since we're talking about strictly non-negative functions we can drop the absolute values.
EDIT: To show that ${TV}^+(f)\leq f_1$ as well:
First of all note that it's enough to show that ${TV}^+(f)\leq f_1$, as ${TV}^-(f)\leq f_2$ is an immediate consequence (just compare the two decompositions of $f$). Next we need a technical lemma to help us:
Lemma: if there exists an increasing function $f_1:[a,b]\rightarrow \mathbb{R}$ and a function $f$ such that $|f(y)-f(x)|\leq f_1(y)-f_1(x)$ for all $[x,y] \subseteq [a,b]$ then $f$ is of bounded variation and, in particular, ${TV}(f)^y_x \leq f_1(y)-f_1(x)$.
Proof: ${TV}(f)^b_a =: \sup_{P\in \cal{P}} \{ \sum_{i=1}^{N}\left|f(x_i)-f(x_{i-1})\right| \} \leq \sup_{P\in \cal{P}} \{ \sum_{i=1}^{N}\left|f_1(x_i)-f_1(x_{i-1})\right| \} \leq f_1(b)-f_1(a)$ since the middle sum telescopes because $f_1$ is a non-negative increasing function.
Finally then: let $f=f_1 - f_2$ be the decomposition into non-negative increasing functions and let $a\leq x < y \leq b$. Then: $$\left|f(y)-f(x)\right| = \left|(f_1(y)-f_1(x)) - (f_2(y)-f_2(x)) \right| $$ $(f_2(y)-f_2(x)) \geq 0$ because $f_2$ is non-negative and increasing so it can only make the RHS smaller. Hence $$\left|f(y)-f(x)\right| \leq f_1(y)-f_1(x) $$ where we've dropped the absolute value because this RHS is also non-negative. We know that ${TV}(f)^y_x = {TV}^+(f)^y_x + {TV}^-(f)^y_x$ and that the RHS here is also two positive quantities. Putting this altogether: $$ {TV}^+(f)^y_x \leq {TV}^+(f)^y_x + {TV}^-(f)^y_x = {TV}(f)^y_x \leq f_1(y)-f_1(x) \leq f_1(y) $$ where the second-to-last inequality uses the technical lemma we proved.