Let \(f : \mathbb{R} \to \mathbb{R}\) be a measurable function. For \(M>0\), prove that \(|f(x)-f(y)|\leq M|x-y|\) for every \(x,y\in\mathbb R\) if and only if \(f\) is absolutely continuous and \(|f'|\leq M\) almost everywhere.
Solution
1. The Lipschitz bound implies absolute continuity and the derivative bound.
Assume the stated Lipschitz inequality. For every finite collection of pairwise disjoint intervals \((a_j,b_j)\), \[ \sum_j|f(b_j)-f(a_j)|\leq M\sum_j(b_j-a_j). \] Given \(\varepsilon>0\), taking \(\delta=\varepsilon/M\) proves absolute continuity, with a single \(\delta\) valid on the whole line. In particular \(f\) is absolutely continuous on each \([-n,n]\), hence differentiable almost everywhere there [@Fol13]. The countable union of the exceptional sets is null, so \(f\) is differentiable almost everywhere on \(\mathbb R\).
At any differentiability point \(x\), every nonzero \(h\) satisfies \(|(f(x+h)-f(x))/h|\leq M\). Passing to the limit gives \(|f'(x)|\leq M\).
2. Absolute continuity and the almost-everywhere bound imply the Lipschitz inequality.
Fix \(x<y\). The fundamental theorem for absolutely continuous functions on \([x,y]\) gives [@Fol13] \[ f(y)-f(x)=\int_x^y f'(t)\,dt. \] Consequently \[ |f(y)-f(x)|\leq\int_x^y|f'(t)|\,dt\leq M(y-x). \] Interchanging \(x\) and \(y\) covers \(x>y\), and the equality case is immediate. This proves the global Lipschitz bound. This step uses absolute continuity only on each compact interval. With step 1, the equivalence therefore also holds when absolute continuity on \(\mathbb R\) means absolute continuity on every compact interval.