TheoremStatement: AI-adaptedProof: AI-generatedprecheck 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 $f : [a, b] \to R$ .

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, $C^{1} \subseteq L i p s c h i t z \subseteq A C \subseteq C \cap B V$ on a compact interval.

Facts & Assumptions

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

[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 $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set and Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ 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 $R$ is bounded).

[L4]

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

[L5]

The Lipschitz condition is $∣ f (x) - f (y) ∣ \leq L ∣ x - y ∣$ for one $L \geq 0$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction, Dictionary: for $A \subseteq R$ with the metric $d (x, y) = ∣ x - y ∣$ , continuity and uniform continuity of $f : A \to R$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $R$ is compact in the open-cover sense of $R$ 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 $R$ (Every complete ordered field is Archimedean).

Proof

technique · direct

1.1L2L3L4

Under claim 1, the continuous extension of $f^{'}$ is bounded by some $M \geq 0$ on $[a, b]$ by [L3]. The bounded-derivative theorem [L4] then makes $f$ Lipschitz with constant $M$ .

1.2L1L5L7

If $f$ is Lipschitz with constant $L$ , then for every finite disjoint family, $\sum_{j < m} ∣ f (v_{j}) - f (u_{j}) ∣ \leq L \sum_{j < m} (v_{j} - u_{j})$ . For $L = 0$ any positive $δ$ works; for $L > 0$ choose $δ = ε / L$ . This proves absolute continuity, including the empty family.

1.3L1L2

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

1.4L1L6L7L8

For bounded variation, take $δ > 0$ from absolute continuity with $ε = 1$ . By [L8] choose a natural $N \geq 1$ with $(b - a) / N < δ$ . 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 $(b - a) / N < δ$ , so their endpoint oscillations sum to less than $1$ . Summing over the $N$ blocks gives $V (f, P) \leq N$ , independent of $P$ . Thus $f$ is BV. If $a = b$ , its variation is $0$ .

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

C1C1 implies Lipschitz, Lipschitz implies absolutely continuous, and absolutely continuous implies continuous and bounded variation

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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