Continuous Functions of Bounded Variation

2.3. Continuous Functions of Bounded Variation#

Previously, we have shown that a function $f$ of bounded variation can be written as the difference of two increasing functions, $f = g - h$. Now, suppose $f$ is continuous. We will show in this section that the two increasing functions $g$ and $h$ can also be made continuous as well.

Theorem 2.6

Suppose $f$ is of bounded variation on $[a, b]$. Then $f$ is continuous at $x_0 \in [a, b]$ if and only if $V_a^x(f)$ is continuous at $x_0$. In other words, every point of continuity of $f$ is also a point of continuity of $V_a^x(f)$ and vice versa.

Proof. We first suppose that $V_a^x(f)$ is continuous at $x_0$ and show that $f$ is also continuous at $x_0$, which is the easier direction to prove.

We will only show that $f$ is continuous from the right at $x_0$($x_0 \neq b$), and the continuity from the left is similarly proved (including $x_0 = b$). Let $\varepsilon > 0$ be arbitrary. Because $V_a^x(f)$ is continuous at $x_0$, there exists $\delta > 0$ such that

\[\abs{x - x_0} < \delta\implies\abs{V_a^x(f) - V_a^x(f)} < \varepsilon\]

For all $h$ satisfying $0 < h < \delta$, we have

\[\begin{split}\abs{f(x_0 + h) - f(x_0)}& = v(P, f) \quad\text{where $P = \{x_0, x_0+h\}$ is a partition of $[x_0, x_0+h]$}\\& \leq V_{x_0}^{x_0+h}(f) \\& = V_a^{x_0+h}(f) - V_a^{x_0}(f) \\& < \varepsilon\end{split}\]

This shows that $f$ is continuous at $x_0$ from the right. Applying a similar argument, one may show that it is also continuous at $x_0$ from the left by considering the interval $[x_0-h, x_0]$.

We now prove the reverse direction. Suppose $f$ is continuous at $x_0$. Again, we will only prove that $V_a^x(f)$ is continuous at $x_0$($x_0 \neq b$) from the right. Let $\varepsilon > 0$ be arbitrary. Since $f$ is continuous at $x_0$, there exists $\delta > 0$ such that

\[\abs{x-x_0} < \delta\implies\abs{f(x) - f(x_0)} < \varepsilon/2 \]

Consider the total variation $V_{x_0}^b(f)$. For any $h$ satisfying $0 < h < \delta$, There exists a partition $P$ such that

\[x_1 - x_0 \leq\delta\]

where $x_1 = x_0 + h$ and

(2.13)#\[v(P, f) > V_{x_0}^b(f) - \varepsilon/2\]

Note

If one is confusing about how finding such $P$ is possible, we can think of finding it with the following process. First, find a partition $P$ of $[x_0, b]$ such that

\[v(P, f) > V_{x_0}^b(f) - \varepsilon/2 \]

and then refine $P$ to $P^\prime$ by adding a point $c$ in between $x_0$ and $x_1$ such that $c - x_0 < \delta$. Note that $v(P^\prime, f) \geq v(P, f)$(Proposition 2.4). Therefore,

\[v(P^\prime, f) > V_{x_0}^b(f) - \varepsilon/2 \]

is satisfied. Finally, rename $P^\prime$ to $P$.

We can express $v(P, f)$ as

(2.14)#\[\begin{split}v(P, f) & = \abs{\Delta f_1} + \underbrace{ \abs{\Delta f_2} + \cdots + \abs{\Delta f_n} }_{ \text{$= v(P^\prime, f)$ where $P^\prime$ is a partition of $[x_1, b]$}}\nonumber\\& = \abs{f(x_1) - f(x_0)} + v(P^\prime, f) \nonumber\\& \leq\abs{f(x_1) - f(x_0)} + V_{x_1}^b(f) \nonumber\\& \text{Recall that $x_1 - x_0 < \delta$ and $f$ is continuous at $x_0$}\nonumber\\& < \varepsilon/2 + V_{x_1}^b(f)\end{split}\]

Combining (2.13) and (2.14), we obtain

\[\varepsilon/2 + V_{x_1}^b(f) > v(P, f) > V_{x_0}^b(f) - \varepsilon/2 \]

Rearranging the terms yields

\[\abs{V_a^{x_0+h}(f) - V_a^{x_0}(f)} = V_{x_0}^{x_0 + h}(f) = V_{x_0}^{x_1}(f) = V_{x_0}^b(f) - V_{x_1}^b(f) < \varepsilon\]

(Specially, it also holds for $x_0 = 1$.) This shows $V_a^x(f)$ is continuous at $x_0$ from the right. And considering $V_a^{x_0}(f)$ and $V_a^{x_0-h}(f)$ and applying a similar argument, one can also show that $V_a^{x}(f)$ is continuous at $x_0$ from the left.