User:Addemf/sandbox/Lebesgue Differentiation (version 2.0)/Functions of Bounded Variation

Bounded Variation
This lesson tries to understand many of the major consequences, if a function's variation is bounded on a compact interval.

Decomposable
We will now prove that a function has a decomposition into increasing and decreasing parts, if and only if it is has bounded variation.

Suppose that $$f:[a,b]\to\Bbb R$$ has two monotonically increasing functions, $$f^+,f^-$$ such that $$f=f^+-f^-$$ and prove that f has bounded variation.

Hint: Consider any partition, P, and use the triangle inequality to show that $$V_f(P) \le (f^+(b)-f^+(a))+(f^-(b)-f^-(a))$$.

Suppose that f is of bounded variation. We have already remarked that we will try to accomplish the decomposition $$f^+=T(x)$$ as defined above, and that this is monotonically increasingly. We will also identify $$f^-(x) = T(x)-f(x)$$ which we know exists -- all that remains is to show that this is monotonically increasing.

Let $$a\le x<y\le b$$ be any points in the domain and show that $$T(x)-f(x)\le T(y)-f(y)$$. It makes sense to, in some sense, "group like terms", so that you will prove the equivalent inequality


 * $$f(y)-f(x)\le T(y)-T(x)$$

It makes some intuitive sense that $$T(y)-T(x)=\mathcal V_f([x,y])$$ and you should prove that this is true. It may help you to organize your work, to state a simpler lemma, and derive this equation as a corollary.

It should also be a one- or two-sentence argument, to show that $$f(y)-f(x) \le \mathcal V_f([x,y])$$.

Prove that if a function has bounded variation on a compact interval, then it is differentiable a.e.