TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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$C^1$ implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation

Statement

Let $f:[a,b]\to\mathbb R$ .

If $f$ is continuous on $[a,b]$ , differentiable on $(a,b)$ , and $f'$ extends continuously to $[a,b]$ , then $f$ is Lipschitz.
Every Lipschitz $f$ is absolutely continuous.
Every absolutely continuous $f$ is continuous and has bounded variation.

Thus, with $C^1$ understood in the endpoint-extension sense of claim 1, $C^1\subseteq\mathrm{Lipschitz}\subseteq AC\subseteq C\cap BV$ on a compact interval.

Facts & Assumptions

Given: A compact interval $[a,b]$ and a function $f:[a,b]\to\mathbb R$ .

[L1]

Absolute continuity is the finite disjoint-interval condition of Absolute continuity on a compact interval.

[L2]

Derivatives and continuity use The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ of $f : A \to \mathbb{R}$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set and Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point.

[L3]

A continuous real function on $[a,b]$ is bounded (A continuous real function on a compact subset of $\mathbb{R}$ is bounded).

[L4]

A continuous function with bounded derivative on an interval is Lipschitz (If $f$ is continuous on an interval $I$ and $|f'| \le M$ at every interior point, then $|f(x) - f(y)| \le M|x-y|$ for all $x,y \in I$ , so $f$ is Lipschitz with constant $M$ and uniformly continuous on $I$ ).

[L5]

The Lipschitz condition is $|f(x)-f(y)|\le L|x-y|$ for one $L\ge0$ (Lipschitz map, $\alpha$ -Hölder map for rational $0 < \alpha \le 1$ , and contraction, Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$ , continuity and uniform continuity of $f : A \to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace).

[L6]

Total variation is the supremum of partition sums (Bounded variation and total variation on an interval, Partition of $[a,b]$ as a finite strictly increasing list $a = t_0 < t_1 < \dots < t_n = b$ , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions).

[L7]

Finite sums split and telescope (Laws of finite sums and finite products).

[L8]

The canonical naturals are cofinal in $\mathbb R$ (Every complete ordered field is Archimedean).

Proof

technique · direct

1.1

Under claim 1, the continuous extension of $f'$ is bounded by some $M\ge0$ on $[a,b]$ by [L3]. The bounded-derivative theorem [L4] then makes $f$ Lipschitz with constant $M$ .

L2L3L4

1.2

If $f$ is Lipschitz with constant $L$ , then for every finite disjoint family, $\sum_{j<m}|f(v_j)-f(u_j)|\le L\sum_{j<m}(v_j-u_j)$ . For $L=0$ any positive $\delta$ works; for $L>0$ choose $\delta=\varepsilon/L$ . This proves absolute continuity, including the empty family.

L1L5L7

1.3

If $f$ is absolutely continuous, apply [L1] to the single interval with endpoints $x,y$ to obtain the usual $\varepsilon$ - $\delta$ continuity condition, so $f$ is continuous.

L1L2

1.4

For bounded variation, take $\delta>0$ from absolute continuity with $\varepsilon=1$ . By [L8] choose a natural $N\ge1$ with $(b-a)/N<\delta$ . Insert the points of the uniform $N$ -partition into an arbitrary partition $P$ . Inside each uniform block, the refined subintervals have disjoint interiors and total length at most $(b-a)/N<\delta$ , so their endpoint oscillations sum to less than $1$ . Summing over the $N$ blocks gives $V(f,P)\le N$ , independent of $P$ . Thus $f$ is BV. If $a=b$ , its variation is $0$ .

L1L6L7L8

2.1

Steps 1.1 through 1.4 prove all three inclusions and the asserted hierarchy.

step 1.1step 1.2step 1.3step 1.4∎

Depends on

Used by

|x| is Lipschitz and absolutely continuous but not C¹ on [-1,1] Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 115 results over 24 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Christopher Heil, Absolute Continuity and the Banach-Zaretsky Theorem (standard reference, not scraped)
William F. Trench, Introduction to Real Analysis, Ch. 3 (standard reference, not scraped)

C1C^1C1 implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation