Bounded Variation - Difference of Functions

[joypav]

Define $f(x)=sinx$ on $[0, 2\pi]$. Find two increasing functions h and g for which f = h−g
on $[0, 2\pi]$.

I know that if f is of bounded variation in $[a,b]$, it is the difference of two positive, monotonic increasing functions. However, we didn't do any examples of this in class. Is there a way to choose these functions?

 

[Greg]

At a casual glance and given that I know virtually nothing of Real Analysis, it seems the following wiki might help. My apologies if I am incorrect.

 

[I like Serena]

How about h(x)=sin x + x?

 

[S.G. Janssens]

Define $f(x)=sinx$ on $[0, 2\pi]$. Find two increasing functions h and g for which f = h−g
on $[0, 2\pi]$.

I know that if f is of bounded variation in $[a,b]$, it is the difference of two positive, monotonic increasing functions. However, we didn't do any examples of this in class. Is there a way to choose these functions?

For this particular case, you find your answer above.

In general, if you let $V_f : [a,b] \to \mathbb{R}$ denote the variation function associated with $f$ (i.e. $V_f(t)$ is the variation of $f$ over $[0,t]$) then you can use $V_f$ to construct such a decomposition of $f$. Can you see how $h$ and $g$ may easily be expressed in terms of $V_f$?

P.S. It may be slightly clearer to say "non-decreasing" instead of "increasing".

 

[joypav]

For this particular case, you find your answer above.

In general, if you let $V_f : [a,b] \to \mathbb{R}$ denote the variation function associated with $f$ (i.e. $V_f(t)$ is the variation of $f$ over $[0,t]$) then you can use $V_f$ to construct such a decomposition of $f$. Can you see how $h$ and $g$ may easily be expressed in terms of $V_f$?

P.S. It may be slightly clearer to say "non-decreasing" instead of "increasing".

In class we did define two functions to obtain the desire result. We defined them as..
$h(x) = V(f, [a,x])$
$g(x) = -f(x) + h(x)$

I guess I don't understand how to apply this. With the given function we'd have...
$h(x) = V(sinx, [0,x])$
$g(x) = -sinx + V(sinx, [0,x])$

I mean, obviously $f(x) = h(x) - g(x)$, but I didn't think this was all that needed done.

 

[joypav]

How about h(x)=sin x + x?

Yes, thank you.

sinx = (sinx+x)-x