Bounded variation: Jordan's theorem on all of R?
By Sarah Smith
I know that a function of bounded variation on a closed interval can be written as a difference of increasing functions. The same is true is the function is of bounded variation on all of R, but I can't come up with a proof.
Any hints?
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$\begingroup$You can use the positive and negative variation functions as usual. The only difference is that you don't have a left endpoint to use. Assuming BV on all of $\mathbb{R}$, a finite limit exists for each of these functions as you approach $-\infty$, and you could use this in place of the left endpoint. More generally, it is enough to assume that the function is BV on all bounded intervals, but for this you will want to fix a finite base point.
Suppose $f$ is BV on all bounded intervals. We want to find nondecreasing functions $g$ and $h$ such that $f=g-h$. For nonnegative $x$, define $g(x)$ to be $f(0)$ plus the positive variation of $f$ on $[0,x]$, and define $h(x)$ to be the negative variation of $f$ on $[0,x]$. For negative $x$, define $g(x)$ to be $f(0)$ minus the positive variation of $f$ on $[x,0]$, and define $h(x)$ to be minus the negative variation of $f$ on $[x,0]$.
$\endgroup$ 1 $\begingroup$I imagined a demonstration to case $f:\mathbb{R}\to\mathbb{R}$. We know that $\overline{ \mathbb{R}}$ are the homeomorphic $ (a, b) $ by a non-decreasing $\sigma: (a,b)\to\mathbb{R}$ homomorphism. For exemple $$ \sigma(t)= \begin{cases} %-\infty & \mbox{ if } t=a \\ \tan\big( \frac{t-a}{b}\cdot\frac{\pi}{2} \big)& \mbox{if } t\leq 0, \\ \tan\big( \frac{t-a}{b}\cdot\frac{\pi}{2} \big)& \mbox{if } t\geq 0, \\ %+\infty & \mbox{ if } t=b. \end{cases} $$
The theorem is valid to $(f\circ\sigma) : (a,b)\to\mathbb{R}$, i.e. $(f\circ\sigma)=(f\circ\sigma)^+-(f\circ\sigma)^-$. As a monotonic function of composition is still a monotonic function follows the theorem: $$ f: (f\circ\sigma)^-\circ\sigma^{-1}-(f\circ\sigma)^-\circ\sigma^{-1} $$ were $(f\circ\sigma)^-\circ\sigma^{-1}$ and $(f\circ\sigma)^+\circ\sigma^{-1}$ are no decresing.
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