Jordan Decomposition

Assume $f$ is of bounded variation on $[a, b]$ . Prove that with $g(x) = V_a^x(f)$ and $h(x) = g(x) - f(x)$ , both $g$ and $h$ are increasing, so $f = g - h$ (one direction of the Jordan decomposition).

V_a^{x_2}(f) = V_a^{x_1}(f) + V_{x_1}^{x_2}(f) \qquad (a \le x_1 < x_2 \le b)

h(x_2) - h(x_1) = V_{x_1}^{x_2}(f) - \bigl(f(x_2) - f(x_1)\bigr) \ge 0

At

x_0 = 0.7

: the plot shows

f = 0.4851

,

g = 0.6776

,

h = 0.1926

— both

g

and

h

climb as you move right, and their difference is exactly

f

.

Show proof

First prove additivity of variation: for $a \le x_1 < x_2 \le b$ , $V_a^{x_2}(f) = V_a^{x_1}(f) + V_{x_1}^{x_2}(f)$ . The inequality $\le$ goes as follows: given any partition $P$ of $[a, x_2]$ , if $x_1$ is absent we may insert it, since the sum can only grow ( $\lvert f(v) - f(u)\rvert \le \lvert f(x_1) - f(u)\rvert + \lvert f(v) - f(x_1)\rvert$ ). A partition containing $x_1$ has its sum split into a sum over $[a, x_1]$ plus a sum over $[x_1, x_2]$ , each bounded by the variation on that interval; taking the supremum over $P$ gives $\le$ .
The reverse $\ge$ : concatenating a partition $P_1$ of $[a, x_1]$ with a partition $P_2$ of $[x_1, x_2]$ gives a partition $P_1 \cup P_2$ of $[a, x_2]$ whose sum is the sum of the two sums. Taking suprema on each interval gives $\ge$ . All suprema are finite since $f$ is BV.
Now $g(x) = V_a^x(f)$ is increasing: for $x_1 < x_2$ , additivity gives $g(x_2) = g(x_1) + V_{x_1}^{x_2}(f) \ge g(x_1)$ (variation is nonnegative). And $h(x_2) - h(x_1) = (g(x_2) - g(x_1)) - (f(x_2) - f(x_1)) = V_{x_1}^{x_2}(f) - (f(x_2) - f(x_1))$ ; but $V_{x_1}^{x_2}(f) \ge \lvert f(x_2) - f(x_1)\rvert \ge f(x_2) - f(x_1)$ from the single partition $\{x_1, x_2\}$ , so $h(x_2) - h(x_1) \ge 0$ — $h$ is increasing too. The identity $f = g - h$ is the definition of $h$ .

Prove the converse: if $g, h$ are increasing on $[a, b]$ and $f = g - h$ , then $f$ is of bounded variation with $V_a^b(f) \le (g(b) - g(a)) + (h(b) - h(a))$ .

\lvert f(x_i) - f(x_{i-1})\rvert \le (g(x_i) - g(x_{i-1})) + (h(x_i) - h(x_{i-1}))

V_a^b(f) \le (g(b) - g(a)) + (h(b) - h(a))

At

x_0 = 0.7

:

g

climbs from

0

to

0.6776

and

h

from

0

to

0.1926

— their sum

0.8702

bounds the variation of

f

on

[0, 0.7]

,

0.6776

, exactly as the inequality below says.

Show proof

Take any partition $a = x_0 < x_1 < \dots < x_n = b$ of $[a, b]$ . On each piece, $\lvert f(x_i) - f(x_{i-1})\rvert = \lvert (g(x_i) - g(x_{i-1})) - (h(x_i) - h(x_{i-1}))\rvert \le \lvert g(x_i) - g(x_{i-1})\rvert + \lvert h(x_i) - h(x_{i-1})\rvert$ (triangle inequality). Since $g$ and $h$ are increasing, each difference is nonnegative, so the absolute values drop away.
Summing over $i = 1, \dots, n$ , $\sum_{i=1}^{n} \lvert f(x_i) - f(x_{i-1})\rvert \le \sum_{i=1}^{n}(g(x_i) - g(x_{i-1})) + \sum_{i=1}^{n}(h(x_i) - h(x_{i-1}))$ , and the two sums telescope to $(g(b) - g(a)) + (h(b) - h(a))$ . This bound is partition-independent, so taking the supremum over all partitions gives $V_a^b(f) \le (g(b) - g(a)) + (h(b) - h(a)) < \infty$ — $f$ is of bounded variation.

Prove the Jordan decomposition is never unique: if $f = g - h$ with $g, h$ increasing, then for any increasing $u$ , $f = (g + u) - (h + u)$ is another one.

f = g - h = (g + u) - (h + u), \qquad u\ \text{increasing}

At

x_0 = 0.7

: adding the same increasing

u

(say

u(x) = x

) to both the

g

and

h

curves shifts them up in lockstep, leaving the difference

f

untouched — there are infinitely many decompositions.

Show proof

A sum of increasing functions is increasing: if $x_1 < x_2$ then $g(x_1) \le g(x_2)$ and $u(x_1) \le u(x_2)$ , so $(g+u)(x_1) \le (g+u)(x_2)$ — and the same for $h + u$ .
Then $(g + u) - (h + u) = g - h = f$ — the $u$ cancels. Since there are infinitely many increasing functions $u$ (e.g. $u(x) = x$ , $u(x) = x^2$ , $u(x) = 2$ constant), there are infinitely many Jordan decompositions. Uniqueness would need an extra normalization (e.g. $g(a) = 0$ ).

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