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

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AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The Fréchet derivative is unique

Statement

Let $U \subseteq X$ be open in a real Banach space $X$ , let $f : U \to Y$ map to a real Banach space $Y$ , and let $x \in U$ . If $A, B \in B (X, Y)$ both satisfy the Fréchet remainder condition for $f$ at $x$ , that is

$\lim_{\begin{array}{c} h \to 0 \\ h \neq 0 \end{array}} \frac{∥ f (x + h) - f (x) - A h ∥}{∥ h ∥} = 0 and \lim_{\begin{array}{c} h \to 0 \\ h \neq 0 \end{array}} \frac{∥ f (x + h) - f (x) - B h ∥}{∥ h ∥} = 0,$

then $A = B$ .

Facts & Assumptions

Given: An open $U$ in a real Banach space $X$ , a map $f : U \to Y$ , a point $x \in U$ , and $A, B \in B (X, Y)$ satisfying the two remainder conditions. Write $r_{A} (h) : = f (x + h) - f (x) - A h$ and $r_{B} (h) : = f (x + h) - f (x) - B h$ .

[L1]

The remainder condition means: for every real $ε > 0$ there is a real $δ > 0$ such that $∥ r_{A} (h) ∥ \leq ε ∥ h ∥$ and $∥ r_{B} (h) ∥ \leq ε ∥ h ∥$ whenever $∥ h ∥ < δ$ and $x + h \in U$ (Fréchet derivative between Banach spaces).

[L2]

The norm satisfies the triangle inequality $∥ u + v ∥ \leq ∥ u ∥ + ∥ v ∥$ , absolute homogeneity $∥ λ u ∥ = ∣ λ ∣ ∥ u ∥$ for real $λ$ , and separation $∥ u ∥ = 0 \Rightarrow u = 0$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[L3]

Since $U$ is open and $x \in U$ , there is a real $ρ > 0$ such that $x + h \in U$ whenever $∥ h ∥ < ρ$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

Proof

technique · direct

1.1L3algebra

Fix $v \in X$ with $v \neq 0$ ; by [L3] there is $ρ > 0$ such that $x + h \in U$ whenever $∥ h ∥ < ρ$ , so $x + t v \in U$ for every real $t$ with $∣ t ∣ ∥ v ∥ < ρ$ , that is for every $t$ with $∣ t ∣ < ρ / ∥ v ∥$ .

2.1step 1.1L2algebra

For every real $t \neq 0$ with $∣ t ∣ < ρ / ∥ v ∥$ the two remainders at $h = t v$ satisfy $r_{A} (t v) - r_{B} (t v) = - (t A v - t B v)$ , hence $t (A - B) v = r_{B} (t v) - r_{A} (t v), ∥ (A - B) v ∥ \leq \frac{∥ r_{A} (t v) ∥ + ∥ r_{B} (t v) ∥}{∣ t ∣} .$

3.1L1step 2.1algebra

For real $t$ with $0 < ∣ t ∣ < ρ / ∥ v ∥$ one has $∥ t v ∥ = ∣ t ∣ ∥ v ∥$ , so the right-hand side of [step 2.1] equals $∥ v ∥ (∥ r_{A} (t v) ∥ / ∥ t v ∥ + ∥ r_{B} (t v) ∥ / ∥ t v ∥)$ , and the two quotients tend to $0$ as $t \to 0$ by [L1] applied with $h = t v$ , since $∥ t v ∥ \to 0$ .

3.2L1step 2.1L2choose

Given a real $ε > 0$ , [L1] supplies $δ > 0$ with $∥ r_{A} (h) ∥ \leq ε ∥ h ∥$ and $∥ r_{B} (h) ∥ \leq ε ∥ h ∥$ for $∥ h ∥ < δ$ ; applying [step 2.1] to $h = t v$ with $0 < ∣ t ∣ < \min {ρ / ∥ v ∥, δ / ∥ v ∥}$ gives $∥ (A - B) v ∥ \leq 2 ∥ v ∥ ε$ . As $ε > 0$ was arbitrary, $∥ (A - B) v ∥ = 0$ , and [L2] gives $(A - B) v = 0$ .

4.1step 3.2algebra∎

The vector $v \neq 0$ was arbitrary, so $(A - B) v = 0$ for every nonzero $v$ ; for $v = 0$ linearity of $A - B$ gives $(A - B) 0 = 0$ as well. Hence $A - B = 0$ , that is $A = B$ .

Depends on

Dependency tree · 2 levels

Sources

Zuoqin Wang, Lecture 6 — §2.1 (Exercise 6)