Monotonic Functions and Functions of Bounded Variation Review

# Monotonic Functions and Functions of Bounded Variation Review

We will now review some of the recent pages regarding monotonic functions and functions of bounded variation.

- On the
**Partitions of a Closed Interval**page we said that if $I = [a, b]$ is a closed interval then a**Partition**is a finite set $P = \{ a = x_0, x_1, ..., x_n = b \}$ that satisfies the following inequality:

\begin{align} \quad a = x_0 < x_1 < ... < x_n = b \end{align}

- We denoted the set of all partitions on $[a, b]$ by $\mathscr{P}[a, b]$. We saw that the partition $P$ divides the interval $[a, b]$ into $n$ subintervals $[x_{k-1}, x_k]$, $k \in \{1, 2, ..., n \}$ whose length is denoted $\Delta x_k = x_k - x_{k-1}$. We also noted that the sum of the lengths of these subintervals is equal to the length of the interval $[a, b]$, i.e.,:

\begin{align} \quad \sum_{k=1}^{n} \Delta x_k = b - a \end{align}

- On the
**Monotonic Functions**page we said that a function $f$ is**Monotonic**on $[a, b]$ if it is either increasing or decreasing. We said that a function is**Increasing**on $[a, b]$ if for all $x, y \in [a, b]$ with $x \leq y$ we have that $f(x) \leq f(y)$, and similarly, $f$ is**Decreasing**on $[a, b]$ if for all $x, y \in [a, b]$ with $x \leq y$ we have that $f(x) \geq f(y)$.

- If $f$ is an increasing function, $x < y < z$, and $f(y^-) = \lim_{a \to y^-} f(a)$ (the lefthand limit of $f$ at $c$) and $f(y^+) = \lim_{a \to y^+}$ (the righthand limit of $f$ at $c$) then the following inequality always holds:

\begin{align} \quad f(x) \leq f(y^-) \leq f(y) \leq f(y^+) \leq f(z) \end{align}

- Similarly, if $f$ is a decreasing function and $x < y < z$ then:

\begin{align} \quad f(x) \geq f(y^-) \geq f(y) \geq f(y^+) \geq f(z) \end{align}

- On the
**Countable Discontinuities of Monotonic Functions**we looked at an important lemma which says that if $f$ is an increasing function on $[a, b]$ and $P = \{ a = x_0, x_1, ..., x_n = b \}\in \mathscr{P}[a, b]$ then the sum of the "jumps" of $f$ at $x_1, x_2, ..., x_{n-1}$ is less than the sum of the jump of $f$ at $a$ to $b$, i.e.,:

\begin{align} \quad \sum_{k=1}^{n-1} [f(x^+) - f(x^-)] \leq f(b) - f(a) \end{align}

- We then went on to prove that if $f$ is monotonic on $[a, b]$ then $f$ has at most countably infinite many discontinuities on $[a, b]$.

- Then on the
**Functions of Bounded Variation**we said that a function $f$ is of**Bounded Variation**on the interval $[a, b]$ if there exists an $M \in \mathbb{R}$, $M > 0$ such that for every partition $P \in \mathscr{P}[a, b]$ we have that:

\begin{align} \quad V_f(P) = \sum_{k=1}^{n} \mid f(x_k) - f(x_{k-1}) \mid \leq M \end{align}

- The value $V_f(P)$ is called the
**Variation**of $f$ associated to $P$.

- We also saw a nice result that showed that if $f$ (not necessarily continuous) is of bounded variation on $[a, b]$ then $f$ is also bounded on $[a, b]$.

- On
**The Sum and Difference of Functions of Bounded Variation**page we saw that if $f$ and $g$ were of bounded variation on $[a, b]$ then the sum $f + g$ and difference $f - g$ are both of bounded variation on $[a, b]$.

- Similarly, from the
**Multiples and Products of Functions of Bounded Variation**page we saw that if $f$ and $g$ were of bounded variation and $t \in \mathbb{R}$ then the multiple $tf$ and product $fg$ are of bounded variation on $[a, b]$.

- We then looked at some more specific examples of functions of bounded variation. On the
**Monotonic Functions as Functions of Bounded Variation**we learned that all monotonic functions are of bounded variation.

- Furthermore, from the
**Continuous Differentiable-Bounded Functions as Functions of Bounded Variation**page we saw that if $f$ is a continuous function whose derivative $f'$ exists and is bounded is also a function of bounded variation. On the**Polynomial Functions as Functions of Bounded Variation**we noted that all polynomials are therefore functions of bounded variation on any interval $[a, b]$ from meeting the criterion of being continuous and having an derivative that exists and is bounded.

- On the
**Total Variation of a Function**page we saw that if $f$ is a function of bounded variation that then the**Total Variation**of $f$ on $[a, b]$ denoted $V_f (a, b)$ was said to be equal to the supremum of the variation of $f$ associated will all partitions $P \in \mathscr{P}[a, b]$, that is:

\begin{align} \quad V_f (a, b) = \sup \{ V_f (P) : P \in \mathscr{P}[a, b] \} \end{align}

- We then looked at some properties of the total variation of a function on the
**Additivity of the Total Variation of a Function**page. We first looked at a property regarding the variation of a function $f$. We saw that if $P = \{ a = x_0, x_1, ..., x_n = b \}$ and $\hat{P} = \{ a = x_0, x_1, ..., x_{k-1}, c, x_k, ..., x_n = b \}$, i.e., $\hat{P}$ is $P$ with an additional point $c$ added, then the variation of $f$ associated to $P$ is less than the variation of $f$ associated to $\hat{P}$:

\begin{align} \quad V_f (P) \leq V_f (\hat{P}) \end{align}

- We used this to prove the additivity of the total variation of a function of bounded variation. If $f$ is of bounded variation on $[a, b]$ and $c \in (a, b)$ then $[a, b] = [a, c] \cup [c, b]$ and the total variation of $f$ on $[a, b]$ equals the total variation of $f$ on $[a, c]$ plus the total variation of $f$ on $[c, b]$:

\begin{align} \quad V_f (a, b) = V_f(a, c) + V_f(c, b) \end{align}

- On the
**Decomposition of Functions of Bounded Variation as the Difference of Two Increasing Functions**page, we finally came to the meat-and-potatoes of this section. For a function $f$ of bounded variation, we considered the function $V : [a, b] \to \mathbb{R}$ defined by:

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

- We proved in a lemma that $V$ is an increasing function. We also proved that the function $V - f$ is an increasing function. Hence, we were able to show that every function $f$ of bounded variation could be written as the difference of two monotonically increasing functions:

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