Let \(f \colon [a,b] \to \mathbb{R}\). Suppose \(f \in \text{BV}[a,b]\). Prove \(f\) is the difference of two increasing functions.
Solution
1. Define the total variation \(V(x) = \sup \sum_{i=1}^n |f(x_i) - f(x_{i-1})|\) over all partitions \(a = x_0 < \cdots < x_n = x\) of \([a, x]\), for \(a \le x \le b\).
\(V\) is well-defined and finite because \(f \in \text{BV}[a,b]\) (so \(V(b) < \infty\)), and \(V(a) = 0\).
2. \(V\) is increasing on \([a,b]\): for \(a \le x < y \le b\), \(V(y) \ge V(x)\).
partitions of \([a,x]\) extend to partitions of \([a,y]\) by adding the point \(y\), so the sup over the larger interval is \(\ge\).
3. \(V + f\) and \(V - f\) are increasing on \([a,b]\).
3.1. For \(a \le x < y \le b\): \(V(y) - V(x) \ge |f(y) - f(x)|\).
\(V(y) \ge V(x) + |f(y) - f(x)|\): a partition of \([a,x]\) achieving within \(\eps\) of \(V(x)\), extended by \(y\), gives a partition of \([a,y]\) with variation \(\ge V(x) + |f(y)-f(x)| - \eps\); let \(\eps \to 0\).
3.2. \((V + f)(y) - (V + f)(x) = (V(y) - V(x)) + (f(y) - f(x)) \ge |f(y)-f(x)| + (f(y)-f(x)) \ge 0\).
1 and \(|u| + u \ge 0\).
3.3. \((V - f)(y) - (V - f)(x) = (V(y) - V(x)) - (f(y) - f(x)) \ge |f(y)-f(x)| - (f(y)-f(x)) \ge 0\).
1 and \(|u| - u \ge 0\).
4. Q.E.D.: \(f = \frac12(V + f) - \frac12(V - f)\) is a difference of two increasing functions.
2 and 3 show both \(V + f\) and \(V - f\) are increasing (multiplying by \(1/2\) preserves monotonicity), and the identity \(f = \frac12(V+f) - \frac12(V-f)\) is a direct algebraic rearrangement.