([Rud76, Exercise 5.1]) If \(f:\mathbb R\to\mathbb R\) satisfies \(|f(x)-f(y)|\le(x-y)^2\) for all \(x,y\), then \(f\) is constant.
Solution
1. Show \(f\) is differentiable everywhere with derivative \(0\).Proof: fix \(y \in \mathbb{R}\).For \(x \ne y\), \[\left|\frac{f(x) - f(y)}{x - y}\right| \le \frac{(x-y)^2}{|x-y|} = |x - y| \to 0 \quad \text{as } x \to y.\] Hence \(f'(y) = \lim_{x\to y}\frac{f(x)-f(y)}{x-y}\) exists and equals \(0\), for every \(y\).
2. Apply the zero-derivative theorem.Proof: by the mean value theorem (D5-W1), a function differentiable on \(\mathbb{R}\) with \(f' \equiv 0\) is constant.
3. Q.E.D.