Worked Problems In Variation And Jordan Decomposition

A Reliable Workflow for Variation Problems

For a function $f:[a,b]\to\mathbb{R}$, total variation records the accumulated size of its changes, not just the difference between its endpoint values. On a smooth monotone piece, the variation is the absolute change across that piece; at a jump, the contribution depends on the actual value of the function as well as its one-sided limits. These distinctions matter when locating discontinuities and constructing Jordan components.

Definition: For a partition $P=\{a=x_0<x_1<\cdots<x_n=b\}$, its variation sum is $\sum_{j=1}^{n}|f(x_j)-f(x_{j-1})|$. The total variation is $V_a^b(f)=\sup_P\sum_{j=1}^{n}|f(x_j)-f(x_{j-1})|$. The function $f$ is of bounded variation, or $f\in BV[a,b]$, if $V_a^b(f)<\infty$. For $x\in[a,b]$, write $V_a^x(f)$ for the variation on $[a,x]$, with $V_a^a(f)=0$.

A function is monotone increasing if $f(x)\le f(y)$ whenever $x<y$; monotone decreasing is defined by reversing that inequality. On a monotone interval, every partition sum telescopes after taking absolute values, so variation equals the absolute endpoint change. For a function with finitely many monotone pieces, one can therefore compute variation piece by piece, including each transition between pieces. We will use the additivity of variation on adjacent intervals.

Proposition: If $a\le c\le b$ and $f:[a,b]\to\mathbb{R}$, then $V_a^b(f)=V_a^c(f)+V_c^b(f)$, allowing infinite values.

Proof. Suppose first that $a<c<b$. Given partitions of $[a,c]$ and $[c,b]$, joining them gives a partition of $[a,b]$, so the supremum on $[a,b]$ is at least the sum of the two suprema: this follows by choosing each partition sum within any prescribed positive $\varepsilon$ of its supremum when the suprema are finite, and by choosing arbitrarily large sums when either is infinite. Conversely, take any partition of $[a,b]$ and insert $c$ if it is not already present. The triangle inequality shows that inserting a point cannot decrease the partition sum. The resulting sum is a sum on $[a,c]$ and a sum on $[c,b]$, hence is at most $V_a^c(f)+V_c^b(f)$. Taking the supremum over the original partitions proves the reverse inequality. If $c=a$ or $c=b$, one interval is degenerate and has variation zero, so the equality follows directly. $\square$

From Variation to Normalized Jordan Components

The normalized Jordan components describe the increase and decrease of a function separately. Their increments are nonnegative, and their starting values are both zero. The following formulas are useful for computations as well as verification.

Theorem: Let $f\in BV[a,b]$, and define $V(x)=V_a^x(f)$ for $x\in[a,b]$. Set $$ P(x)=\frac{V(x)+f(x)-f(a)}{2}, \qquad N(x)=\frac{V(x)-f(x)+f(a)}{2}. $$ Then $P$ and $N$ are nondecreasing, $P(a)=N(a)=0$, and $f(x)=f(a)+P(x)-N(x)$ for every $x\in[a,b]$. Also $V_a^x(f)=P(x)+N(x)$. These are the normalized Jordan components of the increment $f-f(a)$.

Proof. For $a\le x\le y\le b$, additivity gives $V(y)-V(x)=V_x^y(f)$. The partition consisting only of the endpoints $x,y$ shows that $V_x^y(f)\ge |f(y)-f(x)|$. Therefore $$ P(y)-P(x)=\frac{V_x^y(f)+f(y)-f(x)}{2}\ge0, \qquad N(y)-N(x)=\frac{V_x^y(f)-f(y)+f(x)}{2}\ge0. $$ Thus both functions are nondecreasing. Since $V(a)=0$, the definitions give $P(a)=N(a)=0$. Subtracting the formulas gives $P(x)-N(x)=f(x)-f(a)$, while adding them gives $P(x)+N(x)=V(x)$. These identities verify the decomposition and the variation formula at every point, including the endpoints. $\square$

Worked Example: A Smooth Function That Changes Direction

Find the variation and normalized Jordan components of $f(x)=x^2$ on $[-1,2]$, and determine where $f$ is differentiable.

The function decreases from $f(-1)=1$ to $f(0)=0$, then increases from $f(0)=0$ to $f(2)=4$. Its variation on $[-1,0]$ is $1-0=1$, and on $[0,2]$ it is $4-0=4$. Additivity therefore gives $V_{-1}^2(f)=1+4=5$. More generally, the variation function is $$ V_{-1}^x(f)= \begin{cases} 1-x^2,&-1\le x\le0,\\ 1+x^2,&0\le x\le2. \end{cases} $$

Here $f(-1)=1$. For $-1\le x\le0$, the formulas give $P(x)=((1-x^2)+x^2-1)/2=0$ and $N(x)=((1-x^2)-x^2+1)/2=1-x^2$. For $0\le x\le2$, they give $P(x)=((1+x^2)+x^2-1)/2=x^2$ and $N(x)=((1+x^2)-x^2+1)/2=1$. Thus $P$ is nondecreasing and $N$ is nondecreasing. The decomposition check is $1+P-N=x^2$ on both pieces: on the first, $1+0-(1-x^2)=x^2$; on the second, $1+x^2-1=x^2$. At $x=2$, $P(2)+N(2)=4+1=5$, agreeing with the computed total variation. Since $f'(x)=2x$ for every $x\in(-1,2)$, it is differentiable throughout the interior, including $x=0$.

Jumps, Actual Values, and Discontinuities

For a point $c$ in the interior of an interval, the left and right limits of a function are denoted $f(c-)$ and $f(c+)$, when they exist. Continuity at $c$ requires both limits to equal the actual value $f(c)$. A mismatch on either side is a discontinuity. For a function with a jump, do not replace the function’s actual value by a one-sided limit when computing variation: partition sums use $f(c)$.

Worked Example: A Jump and Its Jordan Components

On $[-2,2]$, define $$ f(x)= \begin{cases} x+2,&-2\le x<0,\\ 5-x,&0\le x\le2. \end{cases} $$ Compute its variation, discontinuities, normalized Jordan components, and differentiability set in the interior.

On the left of zero, $f$ increases from $f(-2)=0$ toward the left limit $f(0-)=2$, contributing variation $2-0=2$. The actual value is $f(0)=5$, so the transition from values approaching $2$ to the value at zero contributes $5-2=3$. On the right, $f$ decreases from $5$ to $f(2)=3$, contributing $5-3=2$. Hence $$ V_{-2}^2(f)=2+3+2=7. $$

The one-sided limits at zero are $f(0-)=2$ and $f(0+)=5$, while $f(0)=5$. Thus $f$ is right-continuous at zero but discontinuous there because the left limit differs from the value. On each open interval $(-2,0)$ and $(0,2)$, the function is linear, with derivative $1$ and $-1$, respectively. It is not differentiable at zero: differentiability implies continuity, and continuity fails there.

The variation function is $V_{-2}^x(f)=x+2$ for $-2\le x<0$, $V_{-2}^0(f)=5$, and $V_{-2}^x(f)=5+x$ for $0<x\le2$. Since $f(-2)=0$, the normalized components are $$ P(x)= \begin{cases} x+2,&-2\le x<0,\\ 5,&0\le x\le2, \end{cases} \qquad N(x)= \begin{cases} 0,&-2\le x\le0,\\ x,&0<x\le2. \end{cases} $$

Both are nondecreasing, including across zero: $P$ rises from values below $2$ to $5$, and $N$ stays zero at zero before increasing. For $x<0$, $P-N=x+2=f(x)$; at zero, $P-N=5=f(0)$; and for $x>0$, $P-N=5-x=f(x)$. At $x=2$, $P(2)+N(2)=5+2=7$, checking the total variation.

Monotone Examples and Differentiability

Monotone functions and functions of bounded variation are differentiable almost everywhere with respect to Lebesgue measure on an interval. This is the Lebesgue differentiability theorem for monotone functions and the almost-everywhere differentiability theorem for BV functions established in the earlier material. Those theorems assert differentiability outside a set of Lebesgue measure zero; they do not say that a function is differentiable at every point, nor that its derivative determines all of its pointwise behavior. The examples below locate differentiability directly rather than infer it only from the almost-everywhere statement.

Worked Example: A Monotone Step Function

On $[0,1]$, let $g(x)=0$ for $0\le x<1/2$ and $g(x)=1$ for $1/2\le x\le1$. Find its variation, normalized Jordan components, and interior differentiability set.

The function is nondecreasing, with $g(0)=0$ and $g(1)=1$. For any partition, every increment $g(x_j)-g(x_{j-1})$ is nonnegative, and their sum telescopes to $g(1)-g(0)=1$. Thus every partition has variation sum $1$, and $V_0^1(g)=1$. More generally, $V_0^x(g)=g(x)$: it is zero for $x<1/2$ and one for $x\ge1/2$. The normalized components are consequently $P(x)=g(x)$ and $N(x)=0$. The check $g(0)+P(x)-N(x)=g(x)$ holds because $g(0)=0$.

At every interior point $x<1/2$, $g$ is locally constant and has derivative zero; at every interior point $x>1/2$, it is also locally constant and has derivative zero. At $x=1/2$, the left limit is zero and the value and right limit are one, so $g$ is discontinuous and cannot be differentiable. Thus the interior differentiability set is $(0,1/2)\cup(1/2,1)$. The function is nonconstant even though its derivative is zero almost everywhere. This does not contradict the differentiability theorem: a derivative existing almost everywhere does not, by itself, imply that a BV function is recovered from integrating its derivative; that issue requires additional hypotheses treated in the theory of absolute continuity.

Common Pitfall: Omitting the Jump

A frequent incorrect calculation for the jump example is to add only the smooth variation on either side of zero and report $2+2=4$. This misses the change from the left-hand limiting value $2$ to the actual value $f(0)=5$. The corrected calculation includes that contribution: $2+|5-2|+2=2+3+2=7$. The partition definition confirms why: partitions may include zero, and points immediately to its left have function values approaching $2$, so the corresponding partition increments approach $5-2=3$.

A related error is to use an unnormalized decomposition without checking its starting values. The formulas for $P,N$ above are normalized by $P(a)=N(a)=0$, while the original function is recovered as $f(a)+P-N$, not necessarily as $P-N$. In the $x^2$ example, $f(-1)=1$, and omitting that constant would give the wrong function. Always verify the normalization, the identity $f=f(a)+P-N$, and the identity $V_a^x(f)=P(x)+N(x)$.

Check Your Understanding

For each problem, track actual function values at transition points and check any proposed components against the defining identities.

For $h(x)=|x|$ on $[-2,1]$, compute $V_{-2}^1(h)$ and the normalized Jordan components.
For the jump function on $[-2,2]$, what is the right limit at zero, and why does it not remove the discontinuity?
For the step function on $[0,1]$, verify directly from an arbitrary partition that its variation sum is one.
Give the derivative of $x^2$ at zero and explain why the change in monotonicity there does not prevent differentiability.
In the jump example, identify the specific variation contribution omitted by the incorrect answer $4$.