TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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Sums and scalar multiples of totally differentiable maps are totally differentiable with the expected derivatives

Statement

If $f,g:U\to\mathbb R^n$ are totally differentiable at $a$ and $c\in\mathbb R$ , then $f+g$ and $cf$ are totally differentiable at $a$ , with

$D(f+g)(a)=Df(a)+Dg(a),\qquad D(cf)(a)=cDf(a).$

Facts & Assumptions

Given: Total first-order expansions for $f$ and $g$ at $a$ .

[L1]

In the total-derivative definition, the normalized remainder 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]

A norm satisfies the triangle inequality and $\|cw\|=|c|\,\|w\|$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

Proof

technique · direct

1.1

Add the two expansions and multiply the first by $c$ to obtain remainders $r_f+r_g$ for $f+g$ and $cr_f$ for $cf$ , with the displayed candidate linear maps.

L1L2

2.1

By [L2], $\|r_f(h)+r_g(h)\|_2/\|h\|_2$ is bounded by the sum of two quantities tending to zero, and $\|cr_f(h)\|_2/\|h\|_2=|c|\,\|r_f(h)\|_2/\|h\|_2$ tends to zero (also when $c=0$ ).

step 1.1L2algebra

3.1

Sums and scalar multiples of linear maps are linear, so step 2.1 verifies the definition with exactly the two stated derivatives.

step 1.1step 2.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 65 results over 19 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

J. Lebl, Basic Analysis I, §8.3 (standard reference, not scraped)