Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02
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Total differentiability gives a local O(∥h∥2) increment bound and therefore continuity

Statement

If f:U→Rn is totally differentiable at a∈U, then some C,δ>0 satisfy ∥f(a+h)−f(a)∥2≤C∥h∥2 whenever ∥h∥2<δ and a+h∈U. In particular f is continuous at a.

Facts & Assumptions

Given: A total derivative Df(a) for f at a.

[L1]

The normalized remainder in the total-derivative definition tends to zero as h tends to zero (The total (Fréchet) derivative Df(a) as the linear first-order approximation with o(∥h∥2) remainder).

[L2]

Every Euclidean linear map has a norm bound ∥Lh∥2≤K∥h∥2 for some K≥0 (Every Euclidean linear map has a unique matrix and satisfies ∥Lh∥2≤K∥h∥2 for some K≥0).

Proof

technique · direct
1.1

By [L1], choose δ>0 such that the remainder r(h)=f(a+h)−f(a)−Df(a)h satisfies ∥r(h)∥2≤∥h∥2 whenever 0<∥h∥2<δ.

L1L2
2.1

If K bounds Df(a) as in [L2], the triangle inequality gives ∥f(a+h)−f(a)∥2≤(K+1)∥h∥2 for those h, and it also holds at h=0.

step 1.1L2algebra
3.1

Given ε>0, take ∥h∥2<min⁡{δ,ε/(K+1)}; step 2.1 is the metric continuity condition at a.

step 1.1step 2.1∎

Depends on

Used by

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Sources