Monotone Functions And Exact Variation

Why Monotonicity Determines Variation

Total variation measures the total size of a function’s changes along an interval, counting both increases and decreases. In general, the endpoint values alone do not record all those changes: a function may rise and then fall while returning to its starting value. For a monotone function, however, all changes have the same direction. The increments along any partition therefore have a common sign, and their absolute values add by telescoping. This gives an exact formula, even when the function is discontinuous.

Throughout, let \(a,b\in\mathbb{R}\) with \(a\le b\), and let \(f:[a,b]\to\mathbb{R}\) be real-valued. No measure space or integral is involved in this topic. A partition of \([a,b]\), when \(a<b\), is a finite sequence \(P=(x_0,\ldots,x_n)\) with \(a=x_0<x_1<\cdots<x_n=b\). Its variation sum is the sum of the absolute increments of \(f\) across the partition.

The total variation of \(f\) on \([a,b]\) is the supremum of these sums over all partitions. If \(a=b\), we define the variation to be zero; the sum for the partition consisting of the single endpoint is empty and equals zero.

The function \(f\) is nondecreasing if \(f(x)\le f(y)\) whenever \(x\le y\). It is nonincreasing if \(f(x)\ge f(y)\) whenever \(x\le y\). “Monotone” will mean either nondecreasing or nonincreasing; strict inequalities are not required. Thus constant portions are allowed. Since \(f\) is real-valued, its endpoint values are finite. Monotonicity also gives \(f(a)\le f(x)\le f(b)\) for every \(x\in[a,b]\) in the nondecreasing case, with the inequalities reversed in the nonincreasing case.

The Exact Variation Formula

Proof. First suppose \(a<b\) and \(f\) is nondecreasing. For any partition \(P=(x_0,\ldots,x_n)\), the ordering of the partition points implies \(f(x_{i-1})\le f(x_i)\) for every \(i\). Therefore \(\left|f(x_i)-f(x_{i-1})\right|=f(x_i)-f(x_{i-1})\). Summing these increments gives $$ S_P(f)=\sum_{i=1}^{n}\bigl(f(x_i)-f(x_{i-1})\bigr) =f(x_n)-f(x_0)=f(b)-f(a). $$ The middle equality follows because each interior value \(f(x_i)\) appears once with a plus sign and once with a minus sign. This holds for every partition, so the supremum of the variation sums is \(f(b)-f(a)=|f(b)-f(a)|\).

Now suppose \(f\) is nonincreasing. Then \(f(x_{i-1})\ge f(x_i)\) at every partition step, so \(\left|f(x_i)-f(x_{i-1})\right|=f(x_{i-1})-f(x_i)\). Thus $$ S_P(f)=\sum_{i=1}^{n}\bigl(f(x_{i-1})-f(x_i)\bigr) =f(a)-f(b)=|f(b)-f(a)|. $$ Again this value is the same for every partition, and hence is its supremum. If \(a=b\), both the defined variation and \(|f(b)-f(a)|\) equal zero. This proves the formula in all cases. \(\square\)

The proof gives more than finiteness: refining a partition cannot change the sum for a monotone function, because the refined increments still have the same sign and still telescope. No continuity, differentiability, or strict monotonicity has been assumed. In particular, jumps and flat stretches do not obstruct the formula.

Proof. The restriction of a nondecreasing function to \([c,d]\) is nondecreasing, and the restriction of a nonincreasing function is nonincreasing. Applying the theorem on \([c,d]\) gives the claimed formula when \(c<d\). When \(c=d\), both sides are zero by the definition of variation and because \(f(d)-f(c)=0\). \(\square\)

Worked Computations

What Monotonicity Does—and Does Not—Give

For a monotone function, every partition step moves in one direction or stays level. Absolute values therefore remove the same sign from every increment, and the resulting sum telescopes. This is the whole reason the endpoint formula works. It also shows that a monotone real-valued function on a closed bounded interval has bounded variation: its variation is the finite number \(|f(b)-f(a)|\).

A common mistake is to use the endpoint formula for any function. For example, define \(q:[0,2]\to\mathbb{R}\) by \(q(0)=0\), \(q(1)=1\), and \(q(2)=0\), with linear interpolation on \([0,1]\) and \([1,2]\). This function first increases and then decreases, so it is not monotone. Its endpoint difference is \(|q(2)-q(0)|=|0-0|=0\), but the partition \((0,1,2)\) gives $$ |q(1)-q(0)|+|q(2)-q(1)|=|1-0|+|0-1|=1+1=2. $$ Consequently \(V_0^2(q)\ge2\), and the endpoint difference cannot be its variation. The corrected calculation counts both changes; in this example every partition sum is at most \(2\), since the total increase on the first half is \(1\) and the total decrease on the second half is \(1\). Hence \(V_0^2(q)=2\).

Another potential confusion is to think that “monotone” must mean strictly increasing or strictly decreasing. The theorem requires only nondecreasing or nonincreasing behavior. A constant function is both: if \(f(x)=c\) for every \(x\in[a,b]\), then every increment is \(c-c=0\), and the formula gives \(|c-c|=0\). Likewise, a nondecreasing function may have intervals on which it is constant, as the jump example illustrates.