LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Banach mean value estimate on a convex set

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $U$ be an open convex subset of a real Banach space $X$ , let $f : U \to Y$ be Fréchet differentiable on $U$ with values in a real Banach space $Y$ , let $x, y \in U$ and let $M \geq 0$ be a real number with

$∥ D f (z) ∥ \leq M for every z \in [x, y] : = {x + t (y - x) : 0 \leq t \leq 1} \subseteq U .$

Then

$∥ f (y) - f (x) ∥ \leq M ∥ y - x ∥ .$

Facts & Assumptions

Given: An assumed AC, an open convex $U$ in a real Banach space $X$ , a differentiable $f : U \to Y$ into a real Banach space $Y$ , points $x, y \in U$ and a real $M \geq 0$ bounding $∥ D f ∥$ on the segment $[x, y]$ .

[L1]

Differentiability of $f$ at each $z \in U$ means $f (z + h) - f (z) - D f (z) h = o (∥ h ∥)$ for $h \to 0$ in the sense of the $ε$ - $δ$ remainder estimate (Fréchet derivative between Banach spaces).

[L2]

Since $U$ is convex and $x, y \in U$ , the point $γ (t) : = x + t (y - x)$ lies in $U$ for every real $t \in [0, 1]$ ; a convex set contains all convex combinations of its points with real coefficients in $[0, 1]$ (Convex sets and continuous real-hyperplane separation in a normed space).

[L3]

Under AC, every nonzero $w \in Y$ admits $φ \in Y^{*}$ with $∥ φ ∥ = 1$ and $φ (w) = ∥ w ∥$ (Every nonzero vector has a norming functional); here $Y^{*}$ is the dual space of bounded linear functionals (The dual space X^* of a normed space and its dual norm) and $∣ φ (v) ∣ \leq ∥ φ ∥ ∥ v ∥$ .

[L4]

The chain rule applies to maps between open domains when the image of the first map lies in the domain of the second (Chain sum product and composition rules for Banach derivatives). A bounded linear map is Fréchet differentiable everywhere with derivative equal to itself, and Fréchet differentiability implies continuity (Fréchet derivative between Banach spaces, Remarks).

[L5]

The mean value theorem: a real function continuous on $[0, 1]$ and differentiable on $(0, 1)$ has some $c \in (0, 1)$ with $g (1) - g (0) = g^{'} (c)$ (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

[L6]

The operator norm satisfies $∥ T u ∥ \leq ∥ T ∥ ∥ u ∥$ for all $u$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[L7]

Norm axioms: triangle inequality and absolute homogeneity (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms); the absolute value of a real number is its norm, so $∣ g^{'} (c) ∣ \leq M ∥ y - x ∥$ reads as $g^{'} (c) \geq - M ∥ y - x ∥$ and $g^{'} (c) \leq M ∥ y - x ∥$ .

Proof

technique · direct

1.1

If $x = y$ the estimate reads $∥ f (x) - f (x) ∥ \leq 0$ and holds; if $w : = f (y) - f (x) = 0$ it reads $0 \leq M ∥ y - x ∥$ and holds. Hence we may assume $x \neq y$ and $w \neq 0$ , and then [L3] provides a norming functional for this nonzero $w$ .

givenL7algebra

1.2

By [L3] fix $φ \in Y^{*}$ with $∥ φ ∥ = 1$ and $φ (w) = ∥ w ∥$ ; then $φ (f (y)) - φ (f (x)) = ∥ f (y) - f (x) ∥ \geq 0$ .

L3L7algebra

1.3

Define $γ : R \to X$ by $γ (t) : = x + t (y - x)$ and put $J : = γ^{- 1} [U] = {t \in R : γ (t) \in U} .$ By [L2], $[0, 1] \subseteq J$ . The set $J$ is open: if $t_{0} \in J$ , openness of $U$ gives $ρ > 0$ with $B (γ (t_{0}), ρ) \subseteq U$ , and $∣ t - t_{0} ∣ < ρ / ∥ y - x ∥$ implies $∥ γ (t) - γ (t_{0}) ∥ = ∣ t - t_{0} ∣ ∥ y - x ∥ < ρ$ (recall that $x \neq y$ ). Define the well-typed function $g : J \to R$ by $g (t) : = φ (f (γ (t)))$ .

L2def-metric-topologyL7construct

1.4

The map $γ$ is differentiable at every $t_{0} \in R$ with constant derivative $D γ (t_{0}) = (s \mapsto s (y - x))$ , because $γ (t_{0} + s) - γ (t_{0}) = s (y - x)$ exactly, so the remainder vanishes.

givenL7algebra

1.5

The restriction $γ ∣_{J} : J \to U$ is differentiable at every $t_{0} \in J$ with the derivative in [step 1.4]. Since $J$ and $U$ are open and $γ [J] \subseteq U$ , two applications of [L4], first to $f \circ γ ∣_{J}$ and then to the bounded linear functional $φ$ , show that $g$ is differentiable on $J$ with $g^{'} (t_{0}) = φ (D f (γ (t_{0})) (y - x)) .$ In particular $g$ is continuous on $J$ , hence its restriction to $[0, 1]$ is continuous there and differentiable on $(0, 1)$ .

L1L4step 1.3step 1.4algebra

2.1

For every $t_{0} \in [0, 1]$ the bound in the statement gives $∥ D f (γ (t_{0})) ∥ \leq M$ because $γ (t_{0}) \in [x, y]$ , so by [step 1.5], [L6] and $∥ φ ∥ = 1$ , $∣ g^{'} (t_{0}) ∣ = ∣ φ (D f (γ (t_{0})) (y - x)) ∣ \leq ∥ D f (γ (t_{0})) ∥ \cdot ∥ y - x ∥ \leq M ∥ y - x ∥ .$

step 1.5L6L7algebra

2.2

By [L5] applied to $g$ on the interval $[0, 1]$ , whose hypotheses were verified in [step 1.5], there is $c \in (0, 1)$ with $g (1) - g (0) = g^{'} (c)$ .

step 1.5L5

3.1

Combining [step 1.2], [step 1.3] and [step 1.5], $∥ f (y) - f (x) ∥ = φ (f (y)) - φ (f (x)) = g (1) - g (0) = g^{'} (c) = ∣ g^{'} (c) ∣ \leq M ∥ y - x ∥,$ the second equality because $g (1) - g (0) = ∥ f (y) - f (x) ∥ \geq 0$ forces $g^{'} (c) \geq 0$ .

step 1.2step 1.3step 2.1step 2.2L7algebra

4.1

The estimate is [step 3.1] under the assumptions made there, and [step 1.1] disposes of the two degenerate cases; hence it holds in general.

step 1.1step 3.1∎

Depends on

Used by

Inverse function theorem for Banach spaces Theorem

Dependency tree · two levels

43 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Zuoqin Wang, Lecture 6 — §2.2.3 (mean value theorem via Hahn–Banach) (standard reference, not scraped)