Let \(f\) be a function of bounded variation on \([a,b]\). Furthermore, assume that for some \(c>0\), \(|f(x)| \geq c\) on \([a,b]\). Show that \(g(x) = 1/f(x)\) is of bounded variation on \([a,b]\).
Solution
1. For \(x, y \in [a,b]\): \(\left|\frac{1}{f(x)} - \frac{1}{f(y)}\right| = \frac{|f(x) - f(y)|}{|f(x)f(y)|} \le \frac{1}{c^2}|f(x) - f(y)|\).
\(\frac{1}{f(x)} - \frac{1}{f(y)} = \frac{f(y) - f(x)}{f(x)f(y)}\); and \(|f(x)f(y)| \ge c^2\) by the hypothesis \(|f| \ge c > 0\).
2. For any partition \(a = x_0 < \cdots < x_n = b\): \(\sum_{i=1}^n \left|\frac{1}{f(x_i)} - \frac{1}{f(x_{i-1})}\right| \le \frac{1}{c^2}\sum_{i=1}^n |f(x_i) - f(x_{i-1})| \le \frac{1}{c^2}V_{[a,b]}(f)\).
1 applied term by term; the last inequality is the definition of total variation.
3. Q.E.D.: \(g = 1/f\) is of bounded variation on \([a,b]\), with \(V(g) \le V(f)/c^2\).
2 bounds every partition sum of \(g\) uniformly, so the total variation of \(g\) is finite.