The Variation Function

Accumulating Variation Along an Interval

A function can move up and down while having the same value at two endpoints. The endpoint difference records only the net change; variation records the total movement allowed by all finite samples of the function along the interval. Accumulating that variation from a fixed starting point produces a nondecreasing function that gives a useful bound on every increment of the original function.

Throughout, $f:[a,b]\to\mathbb{R}$ is real-valued, and all statements about $f$ and its variation are pointwise. No measure space or almost-everywhere qualification is involved. A partition of $[s,t]$, where $a\le s<t\le b$, is a finite sequence $s=t_0<t_1<\cdots<t_n=t$. Its variation sum is $\sum_{i=1}^n|f(t_i)-f(t_{i-1})|$. For a degenerate interval $[s,s]$, define its variation to be zero.

Definition: For $a\le s\le t\le b$, the variation of $f$ on $[s,t]$ is $$ V_s^t(f)=\sup\left\{\sum_{i=1}^n|f(t_i)-f(t_{i-1})|:\ s=t_0<t_1<\cdots<t_n=t\right\} $$ when $s<t$, and $V_s^s(f)=0$. The function $f$ is of bounded variation on $[a,b]$ if $V_a^b(f)<\infty$. For such an $f$, its variation function, accumulated from $a$, is $V:[a,b]\to[0,\infty)$ defined by $V(x)=V_a^x(f)$.

The supremum in this definition ranges over all finite partitions, not just a chosen partition. The value $V(x)$ is finite: any partition of $[a,x]$ can be extended to a partition of $[a,b]$ by appending $b$ when $x<b$, and the added term is nonnegative. Thus $V_a^x(f)\le V_a^b(f)$. In particular, $V(a)=0$. The following additivity fact will let us compare the accumulated variation at two points.

Proposition (Additivity on adjacent intervals): If $f:[a,b]\to\mathbb{R}$ has bounded variation and $a\le s\le c\le t\le b$, then $V_s^t(f)=V_s^c(f)+V_c^t(f)$.

Proof. If $c=s$ or $c=t$, the identity follows from the definition $V_s^s(f)=0$. Suppose $s<c<t$. Given any partition of $[s,t]$, insert $c$ if it is not already present. The triangle inequality shows that inserting a point cannot decrease the variation sum. The refined partition splits into partitions of $[s,c]$ and $[c,t]$; their sums are at most $V_s^c(f)$ and $V_c^t(f)$. Taking the supremum over the original partitions gives $V_s^t(f)\le V_s^c(f)+V_c^t(f)$.

For the reverse inequality, let $\varepsilon>0$. By the definition of supremum, choose a partition of $[s,c]$ with sum greater than $V_s^c(f)-\varepsilon/2$, and a partition of $[c,t]$ with sum greater than $V_c^t(f)-\varepsilon/2$. Concatenating them gives a partition of $[s,t]$, so $V_s^t(f)>V_s^c(f)+V_c^t(f)-\varepsilon$. Since this holds for every $\varepsilon>0$, $V_s^t(f)\ge V_s^c(f)+V_c^t(f)$. The two inequalities prove the identity. $\square$

Monotonicity and the Increment Bound

Theorem: If $f:[a,b]\to\mathbb{R}$ has bounded variation and $V(x)=V_a^x(f)$, then $V$ is nondecreasing. Moreover, for every $a\le x\le y\le b$, $$ |f(y)-f(x)|\le V(y)-V(x)=V_x^y(f). $$

Proof. Additivity gives $V(y)=V(x)+V_x^y(f)$. Since variation is a supremum of nonnegative sums, $V_x^y(f)\ge0$, so $V(x)\le V(y)$. If $x<y$, the partition consisting of the two endpoints $x<y$ is one of the partitions used to define $V_x^y(f)$. Its sum is $|f(y)-f(x)|$, so $|f(y)-f(x)|\le V_x^y(f)=V(y)-V(x)$. If $x=y$, both sides of the inequality are zero. This proves both claims, including the degenerate case. $\square$

The theorem says that the increase in $V$ over an interval pays for at least the net change of $f$ across that interval. It does not assert that the two quantities are equal: when $f$ changes direction, variation counts movement that cancels in the endpoint difference.

Worked Example: A Piecewise Linear Function

Let $f:[0,3]\to\mathbb{R}$ be linear on each interval between the nodes $(0,0)$, $(1,2)$, $(2,-1)$, and $(3,1)$. Compute $V(x)=V_0^x(f)$.

On $[0,1]$, the slope is $2$; on $[1,2]$, it is $-3$; and on $[2,3]$, it is $2$. On any partition lying within one of these linear pieces, each absolute increment is the absolute value of the slope times the length of that subinterval. Summing over the partition therefore gives $2$ times the length on the first piece, $3$ times the length on the second, and $2$ times the length on the third. For an arbitrary partition, insert any nodes it crosses. The triangle inequality shows that this cannot decrease its sum, and the refined sum is bounded by the corresponding quantities just computed. Taking the partition through the nodes up to $x$ attains that bound. Thus

V(x)= \begin{cases} 2x, & 0\le x\le1,\\ 2+3(x-1), & 1\le x\le2,\\ 5+2(x-2), & 2\le x\le3. \end{cases}

For example, at $x=1.5$, $f(1.5)=2-3(0.5)=0.5$ and $V(1.5)=2+3(0.5)=3.5$. At $x=2.5$, $f(2.5)=-1+2(0.5)=0$ and $V(2.5)=5+2(0.5)=6$. The partition $0<1<2<2.5$ has sum $2+3+1=6$, checking the latter value. Also $V(3)=2+3+2=7$, exactly the sum along the full node partition. As an increment check, $x=1.5$ and $y=2.5$ give $|f(y)-f(x)|=|0-0.5|=0.5$, while $V(y)-V(x)=6-3.5=2.5$, as required.

Worked Example: A Step Function and Its Right-Side Jumps

On $[0,3]$, define $f(0)=0$, $f(x)=2$ for $0<x\le1$, $f(x)=-1$ for $1<x\le2$, and $f(x)=3$ for $2<x\le3$. Compute $V(x)$, paying attention to the value at each threshold.

For $0<x\le1$, every partition begins at value $0$, and all its later values are $2$. Its sum is therefore $2$, attained by $0<x$. At $x=0$, the variation is $0$. For $1<x\le2$, the partition $0<1<x$ has sum $|2-0|+|-1-2|=2+3=5$. No partition can have a larger sum: its values can move through the levels $0,2,-1$ only in that order, so the total of its successive changes is at most $2+3$. For $2<x\le3$, the partition $0<1<2<x$ has sum $2+3+4=9$. Again, no partition can have a larger sum, because its values pass through $0,2,-1,3$ in that order; omitting an intermediate level cannot increase the sum, by the triangle inequality. Hence

V(x)= \begin{cases} 0, & x=0,\\ 2, & 0<x\le1,\\ 5, & 1<x\le2,\\ 9, & 2<x\le3. \end{cases}

The endpoint of each partition matters: when $1<x\le2$, $0<1<x$ ends at $x$; when $2<x\le3$, $0<1<2<x$ ends at $x$. As a check on the increment bound, take $x=2$ and $y=2.5$. Then $f(2)=-1$, $f(2.5)=3$, so $|f(y)-f(x)|=4$; also $V(2.5)-V(2)=9-5=4$. Equality holds here.

Worked Example: An Affine Function

Let $f:[-1,2]\to\mathbb{R}$ be $f(x)=-2x+1$. Compute its accumulated variation and check the increment inequality.

For any partition $-1=t_0<\cdots<t_n=x$, each summand is \[ |f(t_i)-f(t_{i-1})|=|-2(t_i-t_{i-1})|=2(t_i-t_{i-1}). \] Adding these terms gives $2\sum_{i=1}^n(t_i-t_{i-1})=2(x+1)$, independently of the partition. Therefore $V(x)=2(x+1)$ for $-1\le x\le2$. In particular, $V(1.25)=2(2.25)=4.5$. Since $f(-1)=3$ and $f(1.25)=-2(1.25)+1=-1.5$, their absolute difference is $|-1.5-3|=4.5$. The variation increase from $-1$ to $1.25$ is $V(1.25)-V(-1)=4.5-0=4.5$, which independently checks equality in the increment bound.

What the Variation Function Does—and Does Not—Measure

The variation function gives a running budget for changes in $f$: for any $x\le y$, the theorem bounds the increment by $V(y)-V(x)$. It is useful precisely because it remains nondecreasing even when $f$ oscillates. The step-function example also shows that $V$ can jump: the value $V(1)=2$ does not include the change immediately to the right of $1$, while $V(y)=5$ for every $y>1$ up to $2$. The definition uses intervals ending at the stated point, so endpoint values cannot be silently replaced by nearby values.

Common mistake. It is incorrect to calculate accumulated variation as $V(y)-V(x)=|f(y)-f(x)|$ in general. In the piecewise linear example, $f(0)=0$ and $f(2)= -1$, so the endpoint difference is $|-1-0|=1$. But the partition $0<1<2$ has sum $|2-0|+|-1-2|=2+3=5$, and the computed variation increase is $V(2)-V(0)=5$, not $1$. The correction is to use the endpoint difference as a lower bound for interval variation, not as its definition; the supremum over all partition sums captures intermediate movement.

Check Your Understanding

Use the definition through partitions and the increment bound to answer these questions.

Why does $V(a)=0$, and which convention handles the interval $[a,a]$?
For the piecewise linear example, compute $V(0.75)$ and verify it using the slope on the first segment.
For the step function, why is $V(1)=2$, while $V(y)=5$ for $1<y\le2$? Give a partition ending at each relevant point.
State the increment bound for $x<y$, and explain why it also holds when $x=y$.
Give an example above where the endpoint difference is strictly smaller than the variation increase, and calculate both quantities.

What You'll Learn