Variation Across a Joining Point
Total variation measures the accumulated size of a function’s changes, not merely the difference between its final and initial values. When an interval is split at a point, the changes before and after that point can be considered separately. The key fact is exact additivity: no change is lost and none is counted twice, provided the two subintervals share the joining point and use the function’s actual value there.
Throughout, let \(a\le b\le c\) be real numbers and let \(f:[a,c]\to\mathbb{R}\) be real-valued. All variations below are pointwise quantities, with no almost-everywhere qualification; no measure space is involved. A partition of a nondegenerate closed interval \([u,v]\) is a finite list \(u=x_0<x_1<\cdots<x_n=v\). Its variation sum is \(\sum_{i=1}^n|f(x_i)-f(x_{i-1})|\), and \(V_u^v(f)\) is the supremum of these sums over all partitions. We set \(V_u^u(f)=0\). The supremum may be \(+\infty\); \(f\) has bounded variation on the interval precisely when this supremum is finite.
Why Refining a Partition Helps
A partition is refined by inserting additional points while keeping its original points. Refinement cannot decrease the variation sum: if a single step from \(x\) to \(z\) is split at \(y\), then the triangle inequality gives \(|f(z)-f(x)|\le |f(y)-f(x)|+|f(z)-f(y)|\). This elementary observation is the mechanism behind additivity.
Proof. It is enough to check what happens when one point is inserted, since a finite refinement can be obtained by finitely many insertions. Suppose \(y\) is inserted between consecutive points \(x_i<x_{i+1}\). The old contribution \(|f(x_{i+1})-f(x_i)|\) is replaced by \(|f(y)-f(x_i)|+|f(x_{i+1})-f(y)|\). The triangle inequality shows that the replacement is no smaller. All other terms are unchanged. Repeating this argument for each inserted point proves the claim. \(\square\)
Additivity on Adjacent Closed Intervals
Proof. First suppose \(a<b<c\). Take any partition of \([a,c]\) and insert \(b\) if it is not already present. By the refinement proposition, this does not decrease the sum. The refined partition consists of a partition of \([a,b]\) followed by a partition of \([b,c]\). Its sum is at most \(V_a^b(f)+V_b^c(f)\), by the definition of each supremum. Therefore every original partition sum is at most this quantity, and taking the supremum gives \(V_a^c(f)\le V_a^b(f)+V_b^c(f)\).
For the reverse inequality, let \(P_1\) be any partition of \([a,b]\) and \(P_2\) any partition of \([b,c]\). Their concatenation, listing \(b\) only once, is a partition of \([a,c]\). Its sum is the sum for \(P_1\) plus the sum for \(P_2\), so \(V_a^c(f)\) is at least that combined sum. If either subinterval variation is infinite, its partition sums can be made larger than any prescribed finite bound; extending such a partition to \([a,c]\) shows \(V_a^c(f)=\infty\). If both subinterval variations are finite, then for any \(\varepsilon>0\), the definition of supremum gives partitions whose sums exceed \(V_a^b(f)-\varepsilon/2\) and \(V_b^c(f)-\varepsilon/2\), respectively. Concatenating them yields \(V_a^c(f)\ge V_a^b(f)+V_b^c(f)-\varepsilon\). Since this holds for every \(\varepsilon>0\), \(V_a^c(f)\ge V_a^b(f)+V_b^c(f)\). If \(b=a\) or \(b=c\), the identity follows from the convention \(V_u^u(f)=0\). The two inequalities prove the theorem in all cases. \(\square\)
By applying the theorem repeatedly, variation on an interval split at finitely many ordered points is the sum of the variations on the resulting closed subintervals. In particular, if \(a=x_0\le x_1\le\cdots\le x_k=c\), then \(V_a^c(f)=\sum_{j=1}^k V_{x_{j-1}}^{x_j}(f)\). This follows by induction on \(k\): split first at \(x_{k-1}\), then apply the identity to the first resulting interval. The shared endpoints matter: each restriction uses the same value \(f(x_j)\) at a join.
Worked Piecewise Calculations
Worked Example: A Piecewise-Linear Tent
Define \(f:[0,1]\to\mathbb{R}\) by \(f(x)=x\) for \(0\le x\le \tfrac12\), and \(f(x)=1-x\) for \(\tfrac12<x\le1\). The formulas agree at the joining point: the first gives \(f(\tfrac12)=\tfrac12\), and the second formula would give the same value there.
On \([0,\tfrac12]\), every partition \(0=x_0<\cdots<x_n=\tfrac12\) has sum \(\sum_i|x_i-x_{i-1}|=\sum_i(x_i-x_{i-1})=\tfrac12\). Thus \(V_0^{1/2}(f)=\tfrac12\). On \([\tfrac12,1]\), each increment has absolute value \(|(1-x_i)-(1-x_{i-1})|=x_i-x_{i-1}\), so every partition sum is \(\sum_i(x_i-x_{i-1})=\tfrac12\). Hence \(V_{1/2}^1(f)=\tfrac12\). Additivity gives \(V_0^1(f)=\tfrac12+\tfrac12=1\).
Check: the partition \(0,\tfrac12,1\) has sum \(|\tfrac12-0|+|0-\tfrac12|=\tfrac12+\tfrac12=1\), matching the computed variation. The partition argument on each half also shows that no finer partition can give a larger sum.
Worked Example: A Jump at the Joining Point
Define \(g:[0,1]\to\mathbb{R}\) by \(g(x)=0\) for \(0\le x<\tfrac12\), and \(g(x)=2\) for \(\tfrac12\le x\le1\). The value at the join is \(g(\tfrac12)=2\).
On \([0,\tfrac12]\), any partition has all values \(0\) before its final point, while the final value is \(2\). Its sum is therefore \(0+\cdots+0+|2-0|=2\), and \(V_0^{1/2}(g)=2\). On \([\tfrac12,1]\), the function is constantly \(2\), so every partition sum is \(0\) and \(V_{1/2}^1(g)=0\). Consequently, \(V_0^1(g)=2+0=2\).
Check: the partition \(0,\tfrac12,1\) gives \(|2-0|+|2-2|=2+0=2\). More generally, a partition that omits \(\tfrac12\) has exactly one step crossing the join; that step contributes \(2\), while every other step contributes \(0\). Thus the supremum is exactly \(2\).
Worked Example: Three Pieces with Two Jumps
Define \(h:[0,3]\to\mathbb{R}\) by \(h(x)=0\) for \(0\le x<1\), \(h(x)=3\) for \(1\le x<2\), and \(h(x)=1\) for \(2\le x\le3\). The assigned values are \(h(1)=3\) and \(h(2)=1\).
On \([0,1]\), the function is \(0\) up to but not including \(1\), where it equals \(3\). As in the preceding example, every partition sum is \(3\), so \(V_0^1(h)=3\). On \([1,2]\), it equals \(3\) up to but not including \(2\), where it equals \(1\); each partition sum is \(0+\cdots+0+|1-3|=2\), so \(V_1^2(h)=2\). On \([2,3]\) it is constantly \(1\), giving \(V_2^3(h)=0\). Therefore the finite-splitting identity gives \(V_0^3(h)=3+2+0=5\).
Check: the partition \(0,1,2,3\) gives \(|3-0|+|1-3|+|1-1|=3+2+0=5\). Any partition can have at most one transition from the value \(0\) to \(3\), contributing at most \(3\), and at most one transition from \(3\) to \(1\), contributing at most \(2\); all other increments are zero. Thus no partition sum exceeds \(5\).
Endpoint Values and a Common Pitfall
The additivity theorem concerns restrictions to adjacent closed intervals. It does not say that variation on the open pieces can be added while ignoring the boundary values. At a join, the function has one actual value, and increments leading to or from that value are part of the variation. A jump need not be represented by a discontinuity inside either open piece; it can be recorded by a partition step that reaches the endpoint.
Common mistake. For the function \(g\) in the second example, one might say “\(g\) is constant on each side of \(\tfrac12\), so its variation is \(0+0=0\).” This calculation leaves out the value at the join. On the closed left interval, the endpoint changes from \(0\) to \(g(\tfrac12)=2\), so \(V_0^{1/2}(g)=2\); the right interval contributes \(0\). The corrected total is \(2\), as the partition \(0,\tfrac12,1\) verifies. Splitting into closed intervals with the actual shared endpoint avoids this error.
Additivity is useful whenever a function is described by different formulas on successive intervals. It reduces a global variation calculation to separate calculations on those intervals, while preserving the contribution of any endpoint changes. The result is exact even when a subinterval has infinite variation: the full interval then has infinite variation as well.
Check Your Understanding
Use the definition of variation and the adjacent-interval theorem to justify each answer.
- Let \(f(x)=2x\) on \([0,1]\) and \(f(x)=2-2x\) on \((1,2]\). Compute \(V_0^1(f)\), \(V_1^2(f)\), and \(V_0^2(f)\).
- For \(q(x)=0\) on \([0,2)\) and \(q(2)=5\), compute \(V_0^2(q)\). Which partition displays the change?
- Explain why inserting the splitting point into a partition cannot decrease its variation sum.
- If \(V_a^b(f)=\infty\), what must \(V_a^c(f)\) be when \(a\le b\le c\)? Justify using additivity.
- Give an example where adding the variations on two open pieces while ignoring the value at their common boundary gives the wrong total.