TheoremStatement: AI-adaptedProof: AI-generated

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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.

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On a convex open set, a uniform bound $∥ D f (z) v ∥_{2} \leq M ∥ v ∥_{2}$ implies $∥ f (y) - f (x) ∥_{2} \leq M ∥ y - x ∥_{2}$

Statement

Let $U \subseteq R^{m}$ be convex and open, and let $f : U \to R^{n}$ be totally differentiable at every point. If $M \geq 0$ satisfies $∥ D f (z) v ∥_{2} \leq M ∥ v ∥_{2}$ for every $z \in U$ and $v \in R^{m}$ , then

$∥ f (y) - f (x) ∥_{2} \leq M ∥ y - x ∥_{2} (x, y \in U) .$

Facts & Assumptions

Given: The stated convex open domain, total differentiability, and uniform derivative bound.

[L1]

A convex subset contains every line segment between two of its points (A convex subset of $R^{m}$ contains every line segment between two of its points).

[L2]

The chain rule for total derivatives is $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ).

[L3]

The vector mean-value inequality gives $∥ f (b) - f (a) ∥_{2} \leq M (b - a)$ when the derivative norm is bounded by $M$ (The mean value inequality: if $f : [a, b] \to R^{m}$ is continuous and differentiable on $(a, b)$ with $∥ f^{'} ∥_{2} \leq M$ , then $∥ f (b) - f (a) ∥_{2} \leq M (b - a)$ ).

[L4]

Total differentiability implies continuity at the point of total differentiability (Total differentiability gives a local $O (∥ h ∥_{2})$ increment bound and therefore continuity).

Proof

technique · direct

1.1L1L2L3

If $n = 0$ , both function values are the unique zero vector, so the left-hand side is $0$ and the right-hand side is nonnegative. If $x = y$ the conclusion is also immediate. For the remaining case $n \geq 1$ and $x \neq y$ , the source dimension is necessarily positive; put $γ (t) = x + t (y - x)$ for $0 \leq t \leq 1$ ; [L1] keeps $γ ([0, 1])$ in $U$ .

2.1step 1.1L2algebra

The chain rule gives $(f \circ γ)^{'} (t) = D f (γ (t)) (y - x)$ for $0 < t < 1$ , whose norm is at most $M ∥ y - x ∥_{2}$ by hypothesis.

3.1step 1.1step 2.1L3L4∎

By [L4] the curve $f \circ γ$ is continuous at the endpoints, so [L3] applied on $[0, 1]$ yields $∥ f (y) - f (x) ∥_{2} \leq M ∥ y - x ∥_{2}$ .

Depends on

Dependency tree · 2 levels

Sources

J. Lebl, Basic Analysis I, §8.4