Jordan Decomposition Of A Bounded-Variation Function

From Total Variation to Monotone Components

A function can move both upward and downward while still having finite total variation. The variation function records the accumulated size of those movements. Its increase over any subinterval is at least the absolute change in the original function, which is exactly the inequality needed to build two nondecreasing functions whose difference recovers that change.

Let \(a\le b\) be real numbers, and let \(f:[a,b]\to\mathbb{R}\) be real-valued. For \(a\le u<v\le b\), define the variation of \(f\) on \([u,v]\) by \(V_u^v(f)=\sup\{\sum_{i=1}^n|f(x_i)-f(x_{i-1})|:u=x_0<x_1<\cdots<x_n=v\}\). Set \(V_u^u(f)=0\). The function \(f\) has bounded variation on \([a,b]\), written \(f\in BV[a,b]\), if \(V_a^b(f)<\infty\). Its variation function, based at \(a\), is \(v(x)=V_a^x(f)\) for \(x\in[a,b]\).

The construction rests on how variation behaves when an interval is split. We record that fact and its justification, including the case of intervals with coincident endpoints.

Proof. If \(u=w\) or \(w=z\), the assertion follows from \(V_r^r(f)=0\). Otherwise, take any partition of \([u,z]\) and insert \(w\) if it is not already a partition point. The triangle inequality shows that inserting a point cannot decrease the partition sum. The refined sum is the sum of a partition sum on \([u,w]\) and one on \([w,z]\), so it is at most \(V_u^w(f)+V_w^z(f)\). Taking the supremum over the original partitions gives \(V_u^z(f)\le V_u^w(f)+V_w^z(f)\). For the reverse inequality, let \(\varepsilon>0\). By the definition of the supremum, choose partitions of \([u,w]\) and \([w,z]\) whose sums exceed \(V_u^w(f)-\varepsilon/2\) and \(V_w^z(f)-\varepsilon/2\), respectively. Joining these partitions at \(w\) gives a partition of \([u,z]\), so \(V_u^z(f)>V_u^w(f)+V_w^z(f)-\varepsilon\). Since this holds for every \(\varepsilon>0\), the reverse inequality follows. \(\square\)

The Jordan Decomposition on a Compact Interval

Proof. For \(a\le x\le y\le b\), additivity of variation gives \(v(y)-v(x)=V_x^y(f)\). The one-interval partition of \([x,y]\) shows \(V_x^y(f)\ge |f(y)-f(x)|\); when \(x=y\), both sides are zero. Write \(\Delta v=v(y)-v(x)\) and \(\Delta f=f(y)-f(x)\). Then \(\Delta v\ge|\Delta f|\), so both \(\Delta v+\Delta f\ge0\) and \(\Delta v-\Delta f\ge0\). The definitions give \(P(y)-P(x)=(\Delta v+\Delta f)/2\ge0\) and \(N(y)-N(x)=(\Delta v-\Delta f)/2\ge0\). Thus \(P\) and \(N\) are nondecreasing. Since \(v(a)=0\), their values at \(a\) are both zero. Subtracting their definitions gives \(P(x)-N(x)=f(x)-f(a)\), which proves the stated identity at every \(x\), including both endpoints. The functions are real-valued: \(v(x)\le V_a^b(f)<\infty\), and \(f(x)\) is finite by assumption. If \(a=b\), then \(v(a)=P(a)=N(a)=0\), and the same identity reduces to \(f(a)=f(a)\). \(\square\)

The theorem gives an explicit existence construction. It does not require continuity: jumps are included in the variation measured up to each point. The normalization also makes the role of the initial value precise. The components record all changes after \(a\), while the constant \(f(a)\) supplies the starting value.

What the Construction Tracks—and a Common Mistake

For any \(x\), \(v(x)\) is the total amount of movement up to \(x\), while \(f(x)-f(a)\) is the net change. The formulas split those two quantities into upward and downward movement: \(P(x)+N(x)=v(x)\) and \(P(x)-N(x)=f(x)-f(a)\). These identities follow by adding and subtracting the defining formulas. In particular, the components are not obtained by simply taking the positive and negative parts of the value \(f(x)\); they account for the path of changes from the initial point.

Common mistake: forgetting the initial value. For the constant function \(f(x)=3\) on \([0,1]\), the variation function is \(v(x)=0\) for all \(x\). The normalized formulas give \(P(x)=N(x)=0\). Thus \(P-N=0\), not \(f=3\). The corrected identity includes the starting value: \(f(x)=f(0)+P(x)-N(x)=3+0-0=3\). Equivalently, if the goal is an unshifted difference \(f=F-G\), take \(F=f(a)+P\) and \(G=N\); both remain nondecreasing, and \(F-G=f\) at every point.

This decomposition is useful because questions about a function that changes direction can be expressed through two functions that never decrease. The construction works for every real-valued function of bounded variation on a compact interval, whether or not it is continuous. Its central verification is local: on each \([x,y]\), the variation increase dominates the absolute function increment, forcing both component increments to be nonnegative.