Partition Refinement And Variation Bounds

Partitions as a Tool for Controlling Variation

For a real-valued function \(f:[a,b]\to\mathbb{R}\), a partition records successive points at which we measure changes in \(f\). The variation sum adds the absolute sizes of those changes. Refinement is useful because it lets us compare measurements made with different partitions: adding points cannot make the sum smaller, and a common refinement gives a shared comparison for any two partitions.

Here \(a\le b\) are finite real numbers, and no measure space is involved. A partition of a nondegenerate interval \([a,b]\) is a finite ordered set \(P=\{x_0,\ldots,x_n\}\) with \(a=x_0<x_1<\cdots<x_n=b\). Its variation sum is

If \(a=b\), the partition consists of the single point \(a\), and its sum is \(0\). A partition \(Q\) refines \(P\) if it contains every point of \(P\). The total variation on \([a,b]\) is the supremum of all partition sums, with \(+\infty\) allowed:

The function has bounded variation on \([a,b]\), written \(f\in BV[a,b]\), when this supremum is finite. A supremum need not be attained by any one partition, so arguments about \(V_a^b(f)\) must control all partitions or take an appropriate supremum.

Refinement Cannot Decrease a Variation Sum

Proof. Write the points of \(P\) in increasing order. Between any two consecutive points \(x_{i-1}\) and \(x_i\) of \(P\), the refinement \(Q\) inserts a finite list of points. Write that whole block as \(x_{i-1}=y_0<y_1<\cdots<y_k=x_i\), where \(k\ge1\). Repeated application of the triangle inequality gives

Sum this inequality over every interval of \(P\). The right-hand sums together include exactly the successive increments of \(Q\), so their total is \(S(f;Q)\). The left-hand total is \(S(f;P)\). Thus \(S(f;P)\le S(f;Q)\). If \(a=b\), both sums are \(0\), so the conclusion holds in that case as well. \(\square\)

Common Refinements Compare Different Partitions

Two partitions need not refine one another. Their union, with its points placed in increasing order, is a partition that refines both. This is called a common refinement. The refinement theorem therefore gives an upper comparison for both original sums, even when neither original sum is the larger one.

Proof. Form the finite set \(P\cup Q\), remove any repeated points, and list the remaining points in increasing order. Since both partitions contain \(a\) and \(b\), this list is a partition \(R\) of \([a,b]\). It contains every point of \(P\) and every point of \(Q\), so it refines each. The refinement theorem applied first to \(P,R\) and then to \(Q,R\) yields the two inequalities. If the interval is degenerate, all three partitions consist of \(a=b\), and all sums are \(0\). \(\square\)

Bounds for All Partitions and Subintervals

Proof. If \(V_a^b(f)<\infty\), every partition sum is at most its supremum, so \(C=V_a^b(f)\) is a finite uniform bound. Conversely, if every partition sum is at most some finite \(C\), their supremum is at most \(C\), hence is finite. For the subinterval assertion, take any partition of \([u,v]\). Add \(a\) and \(b\), and include \(u,v\), omitting repeated points; the resulting ordered set is a partition of \([a,b]\) refining the chosen partition of \([u,v]\). By refinement, the original sum is at most the sum on the extended partition, which is at most \(V_a^b(f)\). Taking the supremum over partitions of \([u,v]\) proves \(V_u^v(f)\le V_a^b(f)\). If \(u=v\), the only sum on that interval is \(0\), and the assertion still holds. \(\square\)

A useful way to obtain the uniform bound is a Lipschitz estimate. A function is Lipschitz with constant \(L\ge0\) on \([a,b]\) if \(|f(y)-f(x)|\le L|y-x|\) for every \(x,y\in[a,b]\). For any partition \(P=\{x_0,\ldots,x_n\}\), applying this bound to each consecutive pair gives

The equality in the final sum follows by cancellation of the intermediate partition points. This estimate controls every partition at once and therefore proves \(V_a^b(f)\le L(b-a)\).

A Common Refinement Error

A frequent mistake is to reverse the triangle inequality when a point is inserted. For the quadratic example, one might incorrectly write \(|f(1)-f(-1)|\ge |f(0)-f(-1)|+|f(1)-f(0)|\), concluding that refinement can reduce the sum. Substitution gives \(0\ge1+1=2\), which is false. The correct inequality is

Thus the refined sum is at least the original sum. The signed increments can cancel, as they do here, but taking absolute values before adding prevents that cancellation from lowering the refined sum. Refinement provides an upper comparison between selected partition sums; a common refinement does not imply that the two original sums are equal or directly ordered.