# positive and negative variation of a function

Let $${F: {\mathbb R} \rightarrow {\mathbb R}}$$ be of bounded variation. Define the positive variation $${F^+: {\mathbb R} \rightarrow {\mathbb R}}$$ of $${F}$$ by the formula

$$\displaystyle F^+(x) := \sup_{x_0 < \ldots < x_n \leq x} \sum_{i=1}^n \max(F(x_{i}) - F(x_{i-1}),0)$$

and the negative variation $$F^-$$ by

$$\displaystyle F^-(x) := \sup_{x_0 < \ldots < x_n \leq x} \sum_{i=1}^n \max(F(x_{i-1}) - F(x_{i}),0)$$

Establish the identity

$$\displaystyle F(x) = F(-\infty) + F^+(x) - F^-(x)$$, where $${F(-\infty) := \lim_{x \rightarrow -\infty} F(x)}$$.(Hint: The main difficulty comes from the fact that a partition $${x_0 < \ldots < x_n \leq x}$$ that is good for $${F^+}$$ need not be good for $${F^-}$$, and vice versa. However, this can be fixed by taking a good partition for $${F^+}$$ and a good partition for $${F^-}$$ and combining them together into a common refinement.)

From the hint,I assume we want good partitions $$x_0 < \ldots < x_n \leq x$$ for $$F^-$$ and $$y_0 < \ldots < y_m \leq x$$ for $$F^+$$ respectively, such that $$F(x) < F(- \infty) + F^+(x) - \sum_{i=1}^n \max(F(x_{i-1}) - F(x_i), 0) < F(x) + \varepsilon_1$$ and $$F(x) - \varepsilon_2 < F(- \infty) + \sum_{i=1}^m \max(F(y_{i}) - F(y_{i-1}, 0) - F^-(x) < F(x)$$. We then form the common refinement and take the sup so that $$\varepsilon_1, \varepsilon_2 \rightarrow 0$$. Yet I’m not sure how to explicitly construct these partitions. Am I confusing what the hint is suggesting at?

Let's fix $$x\in\mathbb{R}$$ for the rest of the proof. By analogy with Riemann sums, I'm going to call a partition any finite sequence $$X=(x_i)_{i=0,\ldots,n}$$ such that $$x_0<\ldots. We'll also say that a partition $$X'$$ is a refinement of $$X$$ if every element of $$X$$ is also in $$X'$$.

For a partition $$X=(x_i)_{i=0,\ldots,n}$$, let's define the sums: $$S^+(X) = \sum_{i=1}^n \max\left(F(x_i) - F(x_{i-1}),0\right) ,\ \ \ \ \ \ \ S^-(X) = \sum_{i=1}^n \max\left(F(x_{i-1}) - F(x_i),0\right).$$ Those are just the sums present in definitions of $$F^+$$ and $$F^-$$, this will make notation easier. We can see that we have $$F^+(x)=\sup_X S^+(X)$$ and $$F^-(x)=\sup_X S^-(X)$$. By definition of the supremum, we also have for any partition $$X$$, $$S^+(X) \leq F^+(x)$$ and $$S^-(X) \leq F^-(x)$$.

Now here's an important property that might not be obvious at first: let $$X'$$ be a refinement of another partition $$X$$. Then $$S^+(X)\leq S^+(X')$$ and $$S^-(X)\leq S^-(X')$$.

To see that this is true, consider a partition $$X=(x_i)_{i=0,\ldots,n}$$, and for some $$i$$, insert an element $$x_{i-1/2}$$ between $$x_{i-1}$$ and $$x_i$$. Then: \begin{align*} \max\left(F(x_i) - F(x_{i-1}),0\right) &= \max\left(F(x_i) - F(x_{i-1/2}) + F(x_{i-1/2}) - F(x_{i-1}),0\right) \\ &\leq \max\left(F(x_i) - F(x_{i-1/2}),0\right) + \max\left(F(x_{i-1/2}) - F(x_{i-1}),0\right). \end{align*} Similarly, inserting an element less than $$x_0$$ or greater than $$x_n$$ just adds a non-negative term to the sum, so by recursion we get $$S^+(X)\leq S^+(X')$$ for any refinement $$X'$$ of $$X$$ (we can use the same arguments for $$S^-$$).



Now, on to the main proof. Let's fix some $$\varepsilon>0$$. By definition of the supremum, we know there exist partitions $$X^+$$ and $$X^-$$ such that: $$0 \leq F^+(x) - S^+(X^+) < \varepsilon, \ \ \ \ \ \ \ \ 0 \leq F^-(x) - S^-(X^-) < \varepsilon.$$ We can also choose some $$\bar{x}\in\mathbb{R}$$ far enough to the left so that $$|F(-\infty)-F(\bar{x})|<\varepsilon$$.

Now let's define a new partition $$X=(x_i)_{i=0,\ldots,n}$$ that's made up of every element of $$X^+$$, every element of $$X^-$$, $$\bar{x}$$, and $$x$$. $$X$$ is a refinement of $$X^+$$ and $$X^-$$, so based on what we've shown before: $$S^+(X^+) \leq S^+(X) \leq F^+(x), \ \ \ \ \ \ \ \ S^-(X^-) \leq S^-(X) \leq F^-(x),$$ which gives us: $$0 \leq F^+(x) - S^+(X) < \varepsilon, \ \ \ \ \ \ \ \ 0 \leq F^-(x) - S^-(X) < \varepsilon.$$ Since every element of $$X$$ is less than $$x$$ and we've included $$x$$, we have $$x_n=x$$. Moreover, we can assume to have chosen $$\bar{x}$$ far enough to the left so that $$x_0=\bar{x}$$.

Now that we've set everything up, we can write: \begin{align*} |F(x) - F(-\infty) - F^+(x) + F^-(x)| &\leq |F(x)-F(-\infty) - S^+(X) + S^-(X)| + |S^+(X)-F^+(x)| + |S^-(X)-F^-(x)| \\ &\leq |F(x)-F(-\infty) - S^-(X) + S^-(X)| + 2\varepsilon. \end{align*}

We can see that we have: $$S^+(X) - S^-(X) = \sum_{i=1}^n F(x_i)-F(x_{i-1}) = F(x_n) - F(x_0) = F(x) - F(\bar{x}),$$ and so: $$|F(x)-F(-\infty) - S^+(X) + S^-(X)| = |F(-\infty)-F(\bar{x})| < \varepsilon,$$ finally giving us: $$|F(x) - F(-\infty) - F^+(x) + F^-(x)| < 3\varepsilon.$$ This is true for every $$\varepsilon>0$$ and for every $$x\in\mathbb{R}$$, and so we have $$F(x) = F(-\infty) + F^+(x) - F^-(x)$$.

• The fact that a common refinement increases both $S^+$ and $S^-$ is obvious, by simply noting that points added between any interval on which $F$ is monotone increasing or decreasing don't change the value of $S^+$ and $S^-$ on that interval, and points which fall outside of such intervals contribute new segment of rise and fall, which contribute to the total sum. Commented Jul 14, 2023 at 1:04