Function Jumps and Variation Jumps
A bounded-variation function can jump, and its variation function records the size of that jump. The relevant size is absolute: an upward jump and a downward jump both increase variation. At an interior point, the value of the function at the point matters as well. It can differ from both one-sided limits, producing a jump in the variation function on each side of the point.
Fix a compact interval \([a,b]\) and a real-valued function \(f:[a,b]\to\mathbb{R}\) of bounded variation. For \(a\le u\le v\le b\), write \(V_u^v(f)\) for the total variation of \(f\) on \([u,v]\): it is the supremum, over all finite partitions \(u=t_0<\cdots<t_n=v\), of \(\sum_{i=1}^n|f(t_i)-f(t_{i-1})|\). Set \(V_u^u(f)=0\). The variation function based at \(a\) is \(V(x)=V_a^x(f)\). It is nondecreasing and finite. We consider interior points \(c\in(a,b)\), where both one-sided limits \(f(c-)\) and \(f(c+)\) exist for a bounded-variation function.
The formulas below match left with left and right with right. They do not assert that the two function jumps have the same sign, or that the value \(f(c)\) equals either one-sided limit.
The Local Variation Behind a Jump
Proof. First, variation is additive across a point: if \(u\le v\le w\), then \(V_u^w(f)=V_u^v(f)+V_v^w(f)\). To see the upper bound, insert \(v\) into any partition of \([u,w]\); the triangle inequality says insertion cannot decrease its sum, and the resulting sum is at most \(V_u^v(f)+V_v^w(f)\). For the reverse bound, choose partitions on \([u,v]\) and \([v,w]\) whose sums are within any prescribed positive error of their respective suprema, and join them at \(v\). Taking the error to zero gives the reverse inequality. This also covers degenerate intervals, whose variation is zero.
We prove the left formula. For \(x<c\), define \(R(x)\) as the supremum of partition sums on \([x,c)\), where a partition starts at \(x\) and its final point is strictly less than \(c\). Additivity gives $$ R(x)=\sup_{x\le y<c}V_x^y(f)=\sup_{x\le y<c}\bigl(V(y)-V(x)\bigr)=V(c-)-V(x). $$ Here the last equality follows from \(V(c-)=\lim_{y\uparrow c}V(y)\) and monotonicity of \(V\). In particular, \(R(x)\to0\) as \(x\uparrow c\).
For any partition of \([x,c]\), let \(y<c\) be its last point before \(c\) (if the partition has only its endpoints, take \(y=x\)). The sum before the final increment is at most \(R(x)\), and that final increment is \(|f(c)-f(y)|\). Consequently, $$ |f(c)-f(x)|\le V_x^c(f)\le R(x)+\sup_{x\le z<c}|f(c)-f(z)|. $$ The lower bound uses the partition consisting of \(x,c\). Since \(f(z)\to f(c-)\) as \(z\uparrow c\), the supremum on the right tends to \(|f(c)-f(c-)|\): for every \(\varepsilon>0\), all sufficiently late \(z\) satisfy \(\bigl||f(c)-f(z)|-|f(c)-f(c-)|\bigr|<\varepsilon\). The lower bound tends to the same value, and \(R(x)\to0\). Thus \(V_x^c(f)\to|f(c)-f(c-)|\). By additivity, \(V_x^c(f)=V(c)-V(x)\); taking \(x\uparrow c\) proves \(V(c)-V(c-)=|f(c)-f(c-)|\).
For the right formula, apply the left formula just proved to the reversed function \(g(s)=f(-s)\) on \([-b,-a]\), at the point \(-c\). Its jump from the left is \(g(-c)-g((-c)-)=f(c)-f(c+)\), whose absolute value is \(|f(c+)-f(c)|\). Its accumulated variation at \(-c\) is \(V_c^b(f)\); the left limit of that accumulated variation is \(\lim_{x\downarrow c}V_x^b(f)\). By additivity, their difference is \(\lim_{x\downarrow c}V_c^x(f)=V(c+)-V(c)\). The left formula for \(g\) therefore gives \(V(c+)-V(c)=|f(c+)-f(c)|\). This proves both identities. \(\square\)
The proof explains why the formula concerns local variation, not merely the net change between two distant points. As the interval shrinks to \(c\) from one side, the variation that remains is exactly the magnitude of the endpoint jump; the variation accumulated strictly before that endpoint tends to zero.
Worked Example: A Step Function with Four Jumps
On \([0,3]\), define \(f\) to equal \(1\) on \([0,1)\), \(4\) on \([1,\tfrac32)\), \(-2\) on \([\tfrac32,2)\), \(0\) on \([2,\tfrac52)\), and \(3\) on \([\tfrac52,3]\). The function takes its new value at each listed interior step point. Compute its total variation and its variation function.
Every change occurs at one of \(1,\tfrac32,2,\tfrac52\). The jump magnitudes are, in order, \(|4-1|=3\), \(|-2-4|=6\), \(|0-(-2)|=2\), and \(|3-0|=3\). A partition containing points from each constant piece gives the full sum explicitly:
This is the total variation, not just a lower bound. For any partition, an increment across several constant pieces is at most the sum of the magnitudes of the jumps it crosses, by the triangle inequality. Each step point can be crossed by at most one consecutive pair of partition points. Thus every partition sum is at most \(3+6+2+3=14\), and the displayed partition attains that bound. The variation function is \(0\) on \([0,1)\), \(3\) on \([1,\tfrac32)\), \(9\) on \([\tfrac32,2)\), \(11\) on \([2,\tfrac52)\), and \(14\) on \([\tfrac52,3]\). Its jumps are \(3,6,2,3\), matching the absolute function jumps. Check: the final value is \(V(3)=14\), equal to the total variation just computed.
Worked Example: A Point Value Different from Both Limits
Fix \(c\in(0,1)\) and define \(f(x)=0\) for \(x\ne c\), while \(f(c)=2\), on \([0,1]\). Its left and right limits at \(c\) are both zero. The left function jump is \(2-0=2\), and the right function jump is \(0-2=-2\). The theorem predicts a variation-function jump of \(2\) on each side.
Indeed, if \(x<c\), the function is constant on \([0,x]\), so \(V(x)=0\). At \(c\), the only change on \([0,c]\) is from zero to two, so \(V(c)=2\). For \(x>c\), a partition that includes \(c\) and \(x\) has sum \(|2-0|+|0-2|=4\). No partition can exceed \(4\), since all values are either \(0\) or \(2\) and the value \(2\) occurs at just one point; hence \(V(x)=4\). Therefore \(V(c-)=0\), \(V(c+)=4\), and the two variation jumps are \(2-0=2\) and \(4-2=2\), as asserted. Check: the total variation on \([0,1]\) is \(4\), attained by any partition containing \(c\) and at least one point on each side.
Worked Example: Separate Left and Right Jump Sizes
On \([0,2]\), let \(f(x)=1\) for \(0\le x<1\), let \(f(1)=-1\), and let \(f(x)=2\) for \(1<x\le2\). At \(c=1\), the left limit is \(1\) and the right limit is \(2\). The left function jump is \(-1-1=-2\), and the right function jump is \(2-(-1)=3\). Thus the variation function should jump by \(2\) on the left and by \(3\) on the right.
Before \(1\), the function is constant, so \(V(1-)=0\). The change from a point to the left of \(1\) to \(f(1)=-1\) has magnitude \(|-1-1|=2\), so \(V(1)=2\). Immediately to the right, the change from \(-1\) to \(2\) has magnitude \(|2-(-1)|=3\), giving \(V(1+)=5\). Directly, the partition \(0,1,\tfrac32,2\) has function values \(1,-1,2,2\) and sum \(2+3+0=5\); the triangle inequality bounds every partition sum by these same two jump magnitudes. Thus the total variation is \(5\), confirming the calculation. The signed jumps \(-2\) and \(3\) are not the variation jumps; their absolute values are.
A Common Pitfall: Counting Net Change Instead of Variation
A frequent error is to use the signed change in \(f\) where the formula requires its absolute value, or to omit a step point from a partition and thereby miss a jump. In the preceding example, the net change from \(f(0)=1\) to \(f(2)=2\) is \(2-1=1\). That is not the total variation: the function first falls by \(2\) and then rises by \(3\), so its variation is \(2+3=5\).
There is a similar issue when a function has an exceptional value at one point. In the point-value example, looking only at the limits on either side gives zero on both sides, but the value \(f(c)=2\) creates two changes, each of magnitude \(2\). The variation function registers both. The theorem is useful precisely because it keeps these local contributions visible: for a step function, each one-sided jump contributes its absolute size to the corresponding one-sided jump of \(V\), regardless of whether the original function moves up or down.
Check Your Understanding
For each question, distinguish the signed function jump from the nonnegative jump of the variation function.
- If \(f(c-)=3\) and \(f(c)= -2\), what is \(V(c)-V(c-)\)?
- If \(f(c)=4\) and \(f(c+)=-1\), what is \(V(c+)-V(c)\)?
- For a function equal to zero except that \(f(c)=5\), compute the left and right jumps of the variation function at \(c\).
- For a right-continuous step from \(-3\) to \(2\), what is the variation-function jump at the step, and why is it not the signed jump?
- Explain why a partition containing every step point verifies the total variation of a finite step function, and why the triangle inequality gives the matching upper bound.