Decomp. of Functions of B.V. as the Diff. of Two Inc. Functs.

# Decomposition of Functions of Bounded Variation as the Difference of Two Increasing Functions

Lemma 1: Let $f$ be a function of bounded variation on the interval $[a, b]$ and define $V : [a, b] \to \mathbb{R}$ by $V(x) = V_f(a, x)$. Then $V$ is an increasing function on $[a, b]$. |

**Proof:**Let $f$ be a function of bounded variation on the interval $[a, b]$ and define $V : [a, b] \to \mathbb{R}$ as above. Let $x, y \in [a, b]$ where $x < y$. Then:

\begin{align} \quad V(y) = V_f(a, y) = V_f(a, x) + V_f(x, y) = V(x) + V_f(x, y) \end{align}

- Since the total variation of $f$ on $[x, y]$ is nonnegative, we have that $V(y) \geq V(x)$ so $V$ is increasing on $[a, b]$. $\blacksquare$

Lemma 2: Let $f$ be a function of bounded variation on the interval $[a, b]$ and defined $V : [a, b] \to \mathbb{R}$ by $V(x) = V_f(a, x)$. Then $V - f$ is an increasing function on $[a, b]$. |

**Proof:**Let $f$ be a function of bounded variation on the interval $[a, b]$ and define $V : [a, b] \to \mathbb{R}$ as above. Let $x, y \in [a, b]$ where $x < y$. We want to show that $(V - f)(x) \leq (V - f)(y)$, i.e., $V(x) - f(x) \leq V(y) - f(y)$, or equivalently, $f(y) - f(x) \leq V(y) - V(x)$. We have that:

\begin{align} \quad V(y) - V(x) = V_f(a, y) - V_f(a, x) = V_f(x, y) \end{align}

- Consider the partition $P = \{ x, y \} \in \mathscr{P}[x, y]$. This is the smallest such partition on $[x, y]$ and as we proved on the Additivity of the Total Variation of a Function page, any refinement $\hat{P}$ of a partition $P$ is such that $V_f(\hat{P}) \geq V_f(P)$. Since $V_f(\{x, y\})$ is the smallest such partition of $[x, y]$ we have that:

\begin{align} \quad V(y) - V(x) = V_f(x, y) \geq V_f(\{x, y \}) = \mid f(y) - f(x) \mid \geq f(y) - f(x) \end{align}

- Therefore $f(y) - f(x) \leq V(y) - V(x)$ so $(V - f)(x) \leq (V - f)(y)$ so $V - f$ is an increasing function on $[a, b]$. $\blacksquare$

Theorem 1: Let $f$ be a function of bounded variation on the interval $[a, b]$ and define $V : [a, b] \to \mathbb{R}$ by $V(x) = V_f(a, x)$. Then $f$ can be decomposed as the difference of two increasing functions. |

**Proof:**Let $f$ be a function of bounded variation on the interval $[a, b]$ and define $V : [a, b] \to \mathbb{R}$ as above. Then:

\begin{align} \quad f(x) = V(x) - [V(x) - f(x)] \end{align}

- From Lemmas 1 and 2 we have that $V$ and $V - f$ are both increasing functions. Therefore $f$ can be decomposed into the difference of two increasing functions. $\blacksquare$

From Theorem 1, we see that every function $f$ of bounded variation on $[a, b]$ can be written as the difference of two increasing functions, namely $V$ and $V - f$, however, if $g$ is any increasing function then:

(5)\begin{align} \quad f(x) = [V(x) + g(x)] - [[V(x) - f(x)] + g(x)] \end{align}

So the representation of $f$ as the difference of two increasing functions is not unique.

We will see the importance of Theorem 1 in the next section on Riemann-Stieltjes integrals.