关于函数正变差的定义及具体实例含义的技术问询

Great question—let's break down the positive variation definition and your example to clear up the confusion, using Folland's notation as a guide.

First, let's recap the core ideas behind Jordan's decomposition for functions of bounded variation, which is what Folland builds on:

• The total variation $T_F(x)$ on $[0,x]$ measures the total "amount of change" (both upward and downward) that $F$ undergoes from the start of the interval to $x$.
• The positive variation $P_F(x)$ is the cumulative total of all upward changes (increases) that $F$ experiences over $[0,x]$.
• The negative variation $N_F(x)$ is the cumulative total of all downward changes (decreases) over $[0,x]$.

These three quantities satisfy two key identities:

1. $T_F(x) = P_F(x) + N_F(x)$ (total variation is the sum of all upward and downward changes)
2. $F(x) - F(0) = P_F(x) - N_F(x)$ (the net change in $F$ from $0$ to $x$ is upward changes minus downward changes)

If we solve these equations for $P_F(x)$, we get the standard formula from Folland's text:
$$P_F(x) = \frac{1}{2}\left(T_F(x) + F(x) - F(0)\right)$$

Now let's apply this correctly to your example: $F(x) = 1 - x$ on $[0,1]$. Here's what we know:

• $F(0) = 1$ (the starting value of the function)
• $T_F(x) = x$ (since $F$ decreases linearly from $1$ to $1-x$, the total amount of change over $[0,x]$ is exactly $x$)
• $F(x) - F(0) = (1 - x) - 1 = -x$ (the net change is a decrease of $x$)

Plugging into the correct formula:
$$P_F(x) = \frac{1}{2}\left(x + (-x)\right) = 0$$

This makes perfect intuitive sense: $F$ is strictly decreasing on $[0,1]$—it never increases at any point. So its positive variation (the total of all upward changes) should be $0$ everywhere on the interval.

Now, why did you get $\frac{1}{2}$ using $\frac{1}{2}(T_F + F)$? It looks like you might have misremembered the full formula—you're missing the $-F(0)$ term that anchors the variation to the function's starting value. The expression you calculated:
$$\frac{1}{2}(T_F(x) + F(x)) = \frac{1}{2}(x + 1 - x) = \frac{1}{2}$$
is just a constant, which doesn't align with the actual meaning of positive variation. That constant result is an artifact of using an incomplete version of the formula.

To sum up: positive variation tracks only the increasing portions of a function. For your strictly decreasing $F$, that's zero. The $\frac{1}{2}$ value you found doesn't represent the positive variation—it's a result of an incomplete formula.

备注：内容来源于stack exchange，提问作者kam