Uniqueness Of Normalized Jordan Components

Normalization and the Jordan Decomposition

A real-valued function of bounded variation can be written as the difference of two nondecreasing functions. That representation is not unique, even after both increasing functions are set to zero at the left endpoint: the same increasing function can be added to both components without changing their difference. The canonical Jordan components are distinguished by a further condition: their sum at each point equals the variation accumulated up to that point. With this condition, the components are uniquely determined.

Throughout, let $a<b$, and let $f:[a,b]\to\mathbb{R}$ have bounded variation. For $x\in[a,b]$, define the variation from $a$ to $x$ by $V_a^x(f)$, the supremum of $\sum_{i=1}^n|f(x_i)-f(x_{i-1})|$ over all partitions $a=x_0<x_1<\cdots<x_n=x$. Set $V_a^a(f)=0$. Write $V(x)=V_a^x(f)$ and $V=V_a^b(f)$. Bounded variation means $V<\infty$.

We use the additivity of variation on adjacent intervals, established in the earlier material on total variation: for $a\le s\le t\le b$, $V_a^t(f)=V_a^s(f)+V_s^t(f)$. Also, $V_s^t(f)\ge |f(t)-f(s)|$, since the partition consisting of $s,t$ is one of the partitions used in the supremum. These facts are the key ingredients in establishing the canonical components.

Definition: The normalized Jordan components of $f$ are the functions $P,Q:[a,b]\to[0,\infty)$ satisfying, for every $x\in[a,b]$, $P(a)=Q(a)=0$, $f(x)=f(a)+P(x)-Q(x)$, and $P(x)+Q(x)=V_a^x(f)$. The term “normalized” refers to the values at $a$; the sum condition specifies the canonical components.

Existence and Uniqueness

Theorem: For every real-valued $f:[a,b]\to\mathbb{R}$ of bounded variation, its normalized Jordan components exist and are unique. They are given by $$ P(x)=\frac{V_a^x(f)+f(x)-f(a)}{2}, \qquad Q(x)=\frac{V_a^x(f)-f(x)+f(a)}{2}. $$ Both $P$ and $Q$ are nondecreasing, $P(a)=Q(a)=0$, and the defining difference and sum identities hold everywhere on $[a,b]$.

Proof. Define $P,Q$ by the displayed formulas. Since $V_a^x(f)$ and $f(x)$ are finite real numbers, these formulas define finite real-valued functions. For $a\le s\le t\le b$, additivity of variation gives $V_a^t(f)-V_a^s(f)=V_s^t(f)$. Therefore $$ P(t)-P(s)=\frac{V_s^t(f)+f(t)-f(s)}{2}\ge0, \qquad Q(t)-Q(s)=\frac{V_s^t(f)-f(t)+f(s)}{2}\ge0, $$ because $V_s^t(f)\ge |f(t)-f(s)|$. Thus both functions are nondecreasing. At $a$, $V_a^a(f)=0$, so $P(a)=Q(a)=0$. Adding their formulas gives $P(x)+Q(x)=V_a^x(f)$; subtracting gives $P(x)-Q(x)=f(x)-f(a)$. The sum is nonnegative, and the difference identities together with monotonicity and the zero initial values also show $P(x),Q(x)\ge0$.

For uniqueness, suppose $\widetilde P,\widetilde Q$ satisfy the defining conditions. At each $x$, their sum is $V_a^x(f)$, and their difference is $f(x)-f(a)$. Adding these equations gives $2\widetilde P(x)=V_a^x(f)+f(x)-f(a)$; subtracting the difference equation from the sum equation gives $2\widetilde Q(x)=V_a^x(f)-f(x)+f(a)$. These are exactly the displayed formulas, so $\widetilde P=P$ and $\widetilde Q=Q$ at every $x$. $\square$

The proof separates two roles of the conditions. The difference identity represents $f$, and the sum identity fixes how much total increase is assigned to the two components. Once both are specified, the components are determined by solving two linear equations at each point.

Calculating the Components

Worked Example: A Step Function with Jumps of Both Signs

On $[0,3]$, let $f(x)=2$ for $0\le x<1$, $f(x)=-1$ for $1\le x<2$, and $f(x)=1$ for $2\le x\le3$. The values at the breakpoints are included in the intervals beginning there, so the jump at $1$ is downward and the jump at $2$ is upward. Find the normalized components.

Here $f(0)=2$. Before $1$, the function is constant, so $V_0^x(f)=0$ for $0\le x<1$. At $1$, its change is $2$ to $-1$, of magnitude $3$; there is no further change before $2$. Thus $V_0^x(f)=3$ for $1\le x<2$. At $2$, the change from $-1$ to $1$ has magnitude $2$, so $V_0^x(f)=5$ for $2\le x\le3$. Substitution into the formulas gives $$ (P(x),Q(x))= \begin{cases} (0,0),&0\le x<1,\\ (0,3),&1\le x<2,\\ (2,3),&2\le x\le3. \end{cases} $$

For example, on $1\le x<2$, $P=(3-1)/2=1$ would be incorrect because it omits the baseline $f(0)$ in the formula for $P$. The correct substitution is $P=(3+(-1)-2)/2=0$, while $Q=(3-(-1)+2)/2=3$. On $2\le x\le3$, $P=(5+1-2)/2=2$ and $Q=(5-1+2)/2=3$. Check: on this last interval, $P-Q=2-3=-1=f(x)-f(0)=1-2$, and $P+Q=2+3=5=V_0^x(f)$. Before $1$, both components are zero and their difference agrees with $f(x)-f(0)=0$.

Worked Example: A Piecewise Linear Function with Two Changes of Direction

Define $f:[0,3]\to\mathbb{R}$ by $$ f(x)= \begin{cases} 1+2x,&0\le x\le1,\\ 5-2x,&1\le x\le2,\\ 3-x,&2\le x\le3. \end{cases} $$ The formulas agree at the joins: they give $3$ at $1$ and $1$ at $2$. Calculate its normalized components.

The function rises by $2$ on $[0,1]$, falls by $2$ on $[1,2]$, and falls by $1$ on $[2,3]$. On each linear segment, the variation from the left endpoint to $x$ is the absolute change in function value. Adding the segment variations gives $$ V_0^x(f)= \begin{cases} 2x,&0\le x\le1,\\ 2+2(x-1)=2x,&1\le x\le2,\\ 4+(x-2)=x+2,&2\le x\le3. \end{cases} $$

Since $f(0)=1$, on $[0,1]$ the formulas yield $P=(2x+1+2x-1)/2=2x$ and $Q=(2x-(1+2x)+1)/2=0$. On $[1,2]$, they yield $P=(2x+5-2x-1)/2=2$ and $Q=(2x-(5-2x)+1)/2=2x-2$. On $[2,3]$, they yield $P=(x+2+3-x-1)/2=2$ and $Q=(x+2-(3-x)+1)/2=x$. Hence $$ P(x)= \begin{cases} 2x,&0\le x\le1,\\ 2,&1\le x\le3, \end{cases} \qquad Q(x)= \begin{cases} 0,&0\le x\le1,\\ 2x-2,&1\le x\le2,\\ x,&2\le x\le3. \end{cases} $$

Check: at $x=1$, $(P,Q)=(2,0)$, so $P-Q=2=f(1)-f(0)$ and $P+Q=2=V_0^1(f)$. At $x=2$, $(P,Q)=(2,2)$, so the difference is $0=f(2)-f(0)$ and the sum is $4=V_0^2(f)$. At $x=3$, $(P,Q)=(2,3)$, so the difference is $-1=f(3)-f(0)$ and the sum is $5=V_0^3(f)$. The computed components are nondecreasing: $P$ rises from $0$ to $2$ and then stays fixed; $Q$ stays at $0$, rises to $2$, and then rises to $3$.

Worked Example: The V-Shaped Function

Let $f(x)=|x-1|$ on $[0,2]$. This function decreases from $1$ to $0$ and then increases from $0$ to $1$. Its accumulated variation is $V_0^x(f)=x$: for $0\le x\le1$, the change from $f(0)=1$ is $x$; for $1\le x\le2$, the variation is $1+(x-1)=x$. Since $f(0)=1$, for $0\le x\le1$, $$ P(x)=\frac{x+(1-x)-1}{2}=0, \qquad Q(x)=\frac{x-(1-x)+1}{2}=x. $$ For $1\le x\le2$, $$ P(x)=\frac{x+(x-1)-1}{2}=x-1, \qquad Q(x)=\frac{x-(x-1)+1}{2}=1. $$ Thus $P(x)=0$ up to $1$, then increases as $x-1$; $Q(x)$ increases as $x$ up to $1$, then remains $1$. Check: at $x=2$, $P-Q=1-1=0=f(2)-f(0)$, and $P+Q=1+1=2=V_0^2(f)$. At the turning point $x=1$, the components are $0$ and $1$, whose difference is $-1=f(1)-f(0)$ and whose sum is $1=V_0^1(f)$.

Why Endpoint Normalization Alone Is Not Enough

The endpoint conditions $P(a)=Q(a)=0$, monotonicity, and $f=f(a)+P-Q$ do not by themselves imply uniqueness. For example, let $f(x)=x$ on $[0,1]$. One increasing decomposition is $P(x)=x$, $Q(x)=0$. Another is $\widetilde P(x)=2x$, $\widetilde Q(x)=x$. All four functions are nondecreasing, both pairs vanish at $0$, and $\widetilde P-\widetilde Q=x=P-Q$. But the second pair is not the normalized Jordan pair: $\widetilde P(x)+\widetilde Q(x)=3x$, whereas $V_0^x(f)=x$. The variation-sum condition excludes this extra common increase.

A common mistake is to assert uniqueness from normalization alone, or to compute the two components by splitting the values of $f$ into positive and negative parts. The Jordan components track changes in $f$, not whether the value $f(x)$ itself is positive or negative. For $f(x)=|x-1|$, for instance, the function values are nonnegative, but $Q(x)=x$ on $[0,1]$ because the function decreases there. The correct calculation uses both the accumulated variation and the change from $f(a)$.

In practical calculations, the formulas give a systematic check: first find $V_a^x(f)$, then combine it with $f(x)-f(a)$. The increasing component $P$ records the upward variation, while $Q$ records the downward variation. Their difference recovers the net change, and their sum recovers all variation, including changes that cancel in the net change.

Check Your Understanding

For each question, keep the endpoint normalization and the variation-sum condition distinct.

For $f(x)=3-x$ on $[0,2]$, calculate $V_0^x(f)$, $P(x)$, and $Q(x)$, and verify their sum and difference identities.
Suppose $f$ is constant on $[a,b]$. What are its normalized Jordan components, and why?
For $f(x)=x$ on $[0,1]$, explain why $P(x)=2x,\ Q(x)=x$ is an increasing decomposition but not the normalized Jordan decomposition.
Use the formulas for $P$ and $Q$ to show how the inequality $V_s^t(f)\ge |f(t)-f(s)|$ ensures that both components are nondecreasing.
In the step-function example, identify which jump increases $P$ and which jump increases $Q$, and verify the sizes of those increases.

Uniqueness Of Normalized Jordan Components

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