I read this sentence THEOREM 5.62. A function $f : [a,b] → R $ is of bounded variation iff f can be written as the difference of two increasing functions , that is, f = g−h on [a,b] for two increasing functions g,h. PROOF (⇐=) This part is clear, by Example 5.59 (ii)...
Example5.59(ii) is (ii) Every monotonic function on [a,b] is of bounded variation. For example, suppose that f is a decreasing function over [a,b]. Then for any ... (Example on decreasing function)
Hence every decreasing function is of bounded variation. Increasing function is also of bounded variation.
And i have question that difference of 2 increasing functions can be not of bounded variation? let. $h=f-g$ and $f,g$ are increasing functions. if $f(x)$ is sinusoidal like $ y=x+sin(\frac{1}{{x}^n}) $ vibrating function near y=x, then $f$ and let its inverse function ${f}^{-1}=g$, $h=f-g$ can becomes difference of two increasing functions, but not of bounded variation.
If it is true and i was wrong, i also think how to disprove it.
- If a function is vibrating near $y=x$ greater than y=x on any point, and smaller than $y=x$ at next point,... infinitely, then a function$f$ can not be inecrasing. 2)Even if a function$f$ is vibrating near $y=x$ infinitely, its total variation converges to a finite number If $f$ is increasing.
Please explain to me with example, and also I'm not fluent on english, thank you for any comment or correction on any fault in english.