The Sum and Difference of Functions of Bounded Variation

# The Sum and Difference of Functions of Bounded Variation

Recall from the Functions of Bounded Variation page that $f$ is of bounded variation on the interval $[a, b]$ if there exists a positive real number $M > 0$ such that for all partitions $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P} [a, b]$ we have that:

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

We will now look at some nice theorems which tell us that if $f$ and $g$ are both of bounded variation on the interval $[a, b]$ then so is the sum $f + g$ and the difference $f - g$.

 Theorem 1: If $f$ and $g$ are of bounded variation on the interval $[a, b]$ then $f + g$ is of bounded variation on the interval $[a, b]$.
• Proof: Let $f$ and $g$ be of bounded variation on the interval $[a, b]$. Then there exists positive real numbers $M_1, M_2 > 0$ such that for all partitions $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P} [a, b]$ we have that:
(2)
\begin{align} \quad V_f (P) = \sum_{k=1}^{n} \mid f(x_k) - f(x_{k-1}) \mid < M_1 \quad \mathrm{and} \quad V_g (P) = \sum_{k=1}^{n} \mid g(x_k) - g(x_{k-1}) \mid < M_2 \end{align}
• Let $h = f + g$ and consider the variation of $h$ for any partition $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$:
(3)
\begin{align} \quad V_h (P) = \sum_{k=1}^{n} \mid h(x_k) - h(x_{k-1}) \mid = \sum_{k=1}^{n} \mid (f + g)(x_k) - (f + g)(x_{k-1}) \mid = \sum_{k=1}^{n} \mid [f(x_k) - f(x_{k-1})] + [g(x_k) - g(x_{k-1})] \mid \end{align}
• By the triangle inequality we see that:
(4)
\begin{align} \quad \mid [f(x_k) - f(x_{k-1})] + [g(x_k) - g(x_{k-1})] \mid \leq \mid f(x_k) - f(x_{k-1}) \mid + \mid g(x_k) - g(x_{k-1}) \mid \end{align}
• Therefore we have that:
(5)
\begin{align} \quad V_h (P) \leq \sum_{k=1}^{n} \left [ \mid f(x_k) - f(x_{k-1}) \mid + \mid g(x_k) - g(x_{k-1}) \mid \right ] = \sum_{k=1}^{n} \mid f(x_k) - f(x_{k-1}) \mid + \sum_{k=1}^{n} \mid g(x_k) - g(x_{k-1}) \mid \leq M_1 + M_2 \end{align}
• Let $M = M_1 + M_2 > 0$. Then for all $P \in \mathscr{P}[a, b]$ there exists a positive real number $M > 0$ such that $V_h(P) = V_{f+g}(P) \leq M$ so $h = f + g$ is of bounded variation on the interval $[a, b]$. $\blacksquare$
 Theorem 2: If $f$ and $g$ are of bounded variation on the interval $[a, b]$ then $f - g$ is of bounded variation on the interval $[a, b]$.
• Proof: Let $f$ and $g$ be of bounded variation on the interval $[a, b]$. Then there exists positive real numbers $M_1, M_2 > 0$ such that for all partitions $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P} [a, b]$ we have that:
(6)
\begin{align} \quad V_f (P) = \sum_{k=1}^{n} \mid f(x_k) - f(x_{k-1}) \mid < M_1 \quad \mathrm{and} \quad V_g (P) = \sum_{k=1}^{n} \mid g(x_k) - g(x_{k-1}) \mid < M_2 \end{align}
• Let $h = f - g$ and consider the variation of $h$ for any partition $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$:
(7)
\begin{align} \quad V_h (P) = \sum_{k=1}^{n} \mid h(x_k) - h(x_{k-1}) \mid = \sum_{k=1}^{n} \mid (f - g)(x_k) - (f - g)(x_{k-1}) \mid = \sum_{k=1}^{n} \mid [f(x_k) - f(x_{k-1})] + [g(x_{k-1}) - g(x_{k})] \mid \end{align}
• By the triangle inequality we see that:
(8)
\begin{align} \quad \mid [f(x_k) - f(x_{k-1})] + [g(x_{k-1}) - g(x_{k})] \mid \leq \mid f(x_k) - f(x_{k-1}) \mid + \mid g(x_{k-1}) - g(x_{k}) \mid = \mid f(x_k) - f(x_{k-1}) \mid + \mid g(x_k) - g(x_{k-1}) \mid \end{align}
• Therefore we have that:
(9)
\begin{align} \quad V_h (P) \leq \sum_{k=1}^{n} \left [ \mid f(x_k) - f(x_{k-1}) \mid + \mid g(x_k) - g(x_{k-1}) \mid \right ] = \sum_{k=1}^{n} \mid f(x_k) - f(x_{k-1}) \mid + \sum_{k=1}^{n} \mid g(x_k) - g(x_{k-1}) \mid \leq M_1 + M_2 \end{align}
• Let $M = M_1 + M_2 > 0$. Then for all $P \in \mathscr{P}[a, b]$ there exists a positive real number $M > 0$ such that $V_h(P) = V_{f-g}(P) \leq M$ so $h = f - g$ is of bounded variation on the interval $[a, b]$. $\blacksquare$