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TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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Total differentiability gives a local O(h2)O(\|h\|_2) increment bound and therefore continuity

Statement

If f:URnf:U\to\mathbb R^n is totally differentiable at aUa\in U, then some C,δ>0C,\delta>0 satisfy f(a+h)f(a)2Ch2\|f(a+h)-f(a)\|_2\le C\|h\|_2 whenever h2<δ\|h\|_2<\delta and a+hUa+h\in U. In particular ff is continuous at aa.

Facts & Assumptions

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

[L1]

The normalized remainder in the total-derivative definition tends to zero as hh tends to zero (The total (Fréchet) derivative Df(a)Df(a) as the linear first-order approximation with o(h2)o(\|h\|_2) remainder).

[L2]

Every Euclidean linear map has a norm bound Lh2Kh2\|Lh\|_2\le K\|h\|_2 for some K0K\ge0 (Every Euclidean linear map has a unique matrix and satisfies Lh2Kh2\|Lh\|_2\le K\|h\|_2 for some K0K\ge0).

Proof

technique · direct
1.1

By [L1], choose δ>0\delta>0 such that the remainder r(h)=f(a+h)f(a)Df(a)hr(h)=f(a+h)-f(a)-Df(a)h satisfies r(h)2h2\|r(h)\|_2\le\|h\|_2 whenever 0<h2<δ0<\|h\|_2<\delta.

L1L2
2.1

If KK bounds Df(a)Df(a) as in [L2], the triangle inequality gives f(a+h)f(a)2(K+1)h2\|f(a+h)-f(a)\|_2\le(K+1)\|h\|_2 for those hh, and it also holds at h=0h=0.

step 1.1L2algebra
3.1

Given ε>0\varepsilon>0, take h2<min{δ,ε/(K+1)}\|h\|_2<\min\{\delta,\varepsilon/(K+1)\}; step 2.1 is the metric continuity condition at aa.

step 1.1step 2.1

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 72 results over 17 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