Continuous Differentiable-Bounded Functions as Functions of B. V.

# Continuous Differentiable-Bounded Functions as Functions of Bounded Variation

Recall from the Functions of Bounded Variation page that if $f$ is a function on the interval $[a, b]$ and $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ then the variation of $f$ associated with $P$ is defined to be:

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

Furthermore, $f$ is said to be of bounded variation on $[a, b]$ if there exists a positive real number $M > 0$ such that for all partitions $P \in \mathscr{P}[a, b]$ we have that:

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

We will now show that if $f$ is continuous on $[a, b]$, $f'$ exists, and $f'$ is bounded on $(a, b)$ then $f$ is of bounded variation on $[a, b]$.

Theorem 1: If $f$ is a continuous function on the interval $[a, b]$, $f'$ exists, and $f'$ is bounded for all $x \in (a, b)$ then $f$ is a function of bounded variation on $[a, b]$. |

**Proof:**Let $f$ be continuous on $[a, b]$, let $f'$ exist, and let $f'$ be bounded for all $x \in (a, b)$. Let $P \in \mathscr{P} [a, b]$ be any partition $P = \{ x_0, x_1, ..., x_n \}$. Then the variation of $f$ associated to the partition $P$ is:

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

- Since $f$ is continuous for all $x \in [a, b]$ and $f'$ exists for all $x \in (a, b)$ we have by the Mean Value Theorem that there for each $k \in \{ 1, 2, ..., n \}$ there exists $\xi_k \in (x_k, x_{k-1})$ such that $f(x_k) - f(x_{k-1}) = f'(\xi_k)(x_k - x_{k-1})$ and so:

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

- Since $x_{k-1} < x_k$ for all $k \in \{ 1, 2, ..., n \}$ we must have that $x_k - x_{k-1} > 0$ so $\mid x_k - x_{k-1} \mid = x_k - x_{k-1}$. Furthermore, since $f'$ is bounded for all $x \in (a, b)$ we have that there exists a positive real number $A > 0$ such that for all $x \in (a, b)$ we have $\mid f'(x) \mid \leq A$. Since $\xi_k \in (a, b)$ we must have that $\mid f(\xi_k) \mid \leq A$ for all $k$ so:

\begin{align} \quad V_f (P) = \sum_{k=1}^{n} A (x_k - x_{k-1}) = A \sum_{k=1}^{n}(x_k - x_{k-1}) = A(b - a) \end{align}

- Let $M = A(b - a) > 0$. Then for all $P \in \mathscr{P} [a, b]$ there exists an $M > 0$ such that $V_f (P) \leq M$ so $f$ is a function of bounded variation on $[a, b]$.