Define \(f \colon [0,1] \to [-1,1]\) by \[f(x):= \begin{cases} x\sin\big({\frac{1}{x}}\big) & 0 < x \leq 1 \\ 0 & x = 0 \end{cases}\] Determine, with justification, whether \(f\) is if bounded variation on the interval \([0,1]\).
Solution
1. \(f\) is NOT of bounded variation on \([0,1]\).
show the total variation is infinite by evaluating at a sequence of points with large accumulated oscillation.
2. Consider the points \(t_k = \frac{2}{(2k+1)\pi} \in (0, 1]\), \(k = 0, 1, 2, \ldots\) (the critical points where \(\sin(1/t_k) = \pm 1\)).
\(1/t_k = (2k+1)\pi/2\), so \(\sin(1/t_k) = \pm 1\) alternating with \(k\); and \(t_k \downarrow 0\).
3. At these points \(|f(t_k)| = t_k\) (since \(|\sin(1/t_k)| = 1\)), and \(f\) alternates sign: \(f(t_k) = (-1)^k t_k\).
\(f(t_k) = t_k \sin((2k+1)\pi/2) = t_k \cdot (-1)^k\).
4. For the partition \(0 = t_{K} < t_{K-1} < \cdots < t_1 < t_0 = 2/\pi\) (with \(f(0) = 0\)), the variation sum is \(\sum_{k=0}^{K-1}|f(t_{k+1}) - f(t_k)| = \sum_{k=0}^{K-1}(t_k + t_{k+1})\).
\(f(t_k) = (-1)^k t_k\) and \(f(t_{k+1}) = (-1)^{k+1} t_{k+1}\) have opposite signs (and \(f(0) = 0\)), so \(|f(t_k) - f(t_{k+1})| = t_k + t_{k+1}\) exactly; write \(t_K = 0\).
5. \(V(f) \ge \sum_{k=0}^{K-1}(t_k + t_{k+1}) = t_0 + 2\sum_{k=1}^{K-1} t_k + t_K \ge 2\sum_{k=1}^{K-1} t_k\); and \(\sum_{k=1}^{K-1} t_k = \frac{2}{\pi}\sum_{k=1}^{K-1}\frac{1}{2k+1} \to \infty\).
the harmonic tail \(\sum \frac{1}{2k+1}\) diverges.
6. Q.E.D.: \(V_{[0,1]}(f) = \infty\), so \(f \notin \text{BV}[0,1]\).
4 and 5 give arbitrarily large finite-variation lower bounds; \(f\) is continuous (including at \(0\)) but has unbounded variation.