Let \(\{a_n\}_{n=1}^\infty \subseteq \mathbb{R}\) and a strictly increasing sequence \(\{x_n\}_{n=1}^\infty \subseteq (0,1)\) be given. Assume that \(\sum_{n=1}^\infty a_n\) is absolutely convergent, and define \(\alpha \colon [0,1] \to \mathbb{R}\) by \[\alpha(x):= \begin{cases} a_n & x=x_n \\ 0 & \text{otherwise} \end{cases}.\] Prove or disprove: \(\alpha\) has bounded variation on \([0,1]\).
Solution
1. \(\alpha\) is bounded: \(|\alpha(x)| \le \sum_n |a_n| < \infty\) for all \(x\).
\(\alpha(x) = a_n\) if \(x = x_n\) and \(0\) otherwise, so \(|\alpha(x)| \le \sup_n |a_n| \le \sum_n |a_n|\), and \(\sum_n |a_n|\) converges absolutely.
2. Estimate the variation against a partition.
let \(0 = t_0 < t_1 < \cdots < t_m = 1\) be any partition.
3. \(\alpha\) has bounded variation.
by 2, every partition has variation \(\le 2\sum_n |a_n| < \infty\); taking the supremum, \(V(\alpha) \le 2\sum_n|a_n| < \infty\).
4. Q.E.D.