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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Chain sum product and composition rules for Banach derivatives

Statement

Let $U \subseteq X$ be open in a real Banach space $X$ , and let $Y$ , $Z$ , $W$ be real Banach spaces. Then:

Sum rule. If $f_{1}, f_{2} : U \to Y$ are Fréchet differentiable at $x \in U$ and $a, b \in R$ , then $a f_{1} + b f_{2}$ is Fréchet differentiable at $x$ with $D (a f_{1} + b f_{2}) (x) = a D f_{1} (x) + b D f_{2} (x)$ .
Bounded-bilinear product rule. If $f : U \to Y$ and $g : U \to Z$ are Fréchet differentiable at $x$ , and $B : Y \times Z \to W$ is bounded bilinear, then $U ∋ u \mapsto B (f (u), g (u))$ is Fréchet differentiable at $x$ and

$D (B (f, g)) (x) h = B (D f (x) h, g (x)) + B (f (x), D g (x) h) (h \in X) .$

Chain rule. If $f : U \to Y$ is Fréchet differentiable at $x$ , if $W_{0} \subseteq Y$ is an open set with $f [U] \subseteq W_{0}$ , and if $g : W_{0} \to Z$ is Fréchet differentiable at $f (x)$ , then $g \circ f$ is Fréchet differentiable at $x$ and $D (g \circ f) (x) = D g (f (x)) D f (x) .$

No continuity of any derivative map is assumed; these are pointwise statements about one $x$ at a time.

Facts & Assumptions

Given: An open $U \subseteq X$ in a real Banach space $X$ , real Banach spaces $Y, Z, W$ , and $x \in U$ . The three claims have separate map data:

For claim 1, $f_{1}, f_{2} : U \to Y$ are differentiable at $x$ and $a, b \in R$ .
For claim 2, $f : U \to Y$ and $g : U \to Z$ are differentiable at $x$ , and $B : Y \times Z \to W$ is bounded bilinear with a constant $C \geq 0$ as in [L3].
For claim 3, $f : U \to Y$ is differentiable at $x$ , $W_{0} \subseteq Y$ is open with $f [U] \subseteq W_{0}$ , and $g : W_{0} \to Z$ is differentiable at $f (x)$ .

The symbols $f, g$ are local to their respective claims. Throughout the proof, source increments $h$ satisfy $x + h \in U$ ; in claim 3 this guarantees $y + k (h) = f (x + h) \in W_{0}$ .

[L1]

Fréchet differentiability at $x$ with derivative $T$ means that for every real $ε > 0$ there is a real $δ > 0$ such that $∥ f (x + h) - f (x) - T h ∥ \leq ε ∥ h ∥$ for every $h$ with $∥ h ∥ < δ$ and $x + h \in U$ (Fréchet derivative between Banach spaces).

[L2]

The norm satisfies the triangle inequality $∥ u + v ∥ \leq ∥ u ∥ + ∥ v ∥$ , absolute homogeneity $∥ λ u ∥ = ∣ λ ∣ ∥ u ∥$ , 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]

A bounded bilinear map $B$ has a real constant $C \geq 0$ with $∥ B (u, v) ∥ \leq C ∥ u ∥ ∥ v ∥$ for all $u, v$ , and is jointly continuous (A bounded bilinear map between normed spaces, For a bilinear map, boundedness is equivalent to joint continuity).

[L4]

Linear combinations of bounded linear operators with a common source and target are bounded linear. A composite of bounded linear operators is bounded linear, and $∥ S T ∥ \leq ∥ S ∥ ∥ T ∥$ ; the operator norm satisfies $∥ T u ∥ \leq ∥ T ∥ ∥ u ∥$ (Composition satisfies |ST|\le|S|,|T|, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators).

[L5]

If two bounded linear operators satisfy the Fréchet remainder condition for the same map at the same point, they are equal (The Fréchet derivative is unique).

Proof

technique · direct

1.1L1L2L4algebra

For claim 1, write $T : = a D f_{1} (x) + b D f_{2} (x)$ , a bounded linear operator by [L4], and $r (h) : = a f_{1} (x + h) + b f_{2} (x + h) - a f_{1} (x) - b f_{2} (x) - T h$ , which equals $a r_{1} (h) + b r_{2} (h)$ with $r_{j} (h) : = f_{j} (x + h) - f_{j} (x) - D f_{j} (x) h$ . Then $∥ r (h) ∥ / ∥ h ∥ \leq ∣ a ∣ ∥ r_{1} (h) ∥ / ∥ h ∥ + ∣ b ∣ ∥ r_{2} (h) ∥ / ∥ h ∥$ for $h \neq 0$ , and both terms tend to $0$ by [L1]; hence $a f_{1} + b f_{2}$ is differentiable at $x$ with derivative $T$ , which is claim 1.

1.2L3L4algebra

For claim 2 define $L (h) : = B (D f (x) h, g (x)) + B (f (x), D g (x) h)$ for $h \in X$ . Since $D f (x)$ , $D g (x)$ are linear and $B$ is bilinear, $L$ is linear, and $∥ L (h) ∥ \leq C (∥ D f (x) ∥ ∥ g (x) ∥ + ∥ f (x) ∥ ∥ D g (x) ∥) ∥ h ∥$ by [L3] and [L4], so $L$ is a bounded linear operator $X \to W$ .

1.3L1L4L5algebra

For claim 3 let $T : = D f (x)$ and $S : = D g (f (x))$ , write $y : = f (x)$ , $R (k) : = g (y + k) - g (y) - S k$ for $k \in Y$ with $y + k \in W_{0}$ , and $r (h) : = f (x + h) - f (x) - T h$ . Put $k (h) : = f (x + h) - f (x) = T h + r (h)$ ; then $g (f (x + h)) - g (f (x)) - S T h = S r (h) + R (k (h))$ identically in $h$ . Given a real $η > 0$ , apply [L1] for $g$ at $y$ with $ε : = η / (2 (∥ T ∥ + 1))$ to get $δ_{1} > 0$ with $∥ R (k) ∥ \leq \frac{η}{2 (∥ T ∥ + 1)} ∥ k ∥$ for $∥ k ∥ < δ_{1}$ , and apply [L1] for $f$ at $x$ with $ε : = 1$ and with $ε : = η / (2 (1 + ∥ S ∥))$ to get a single $δ_{2} > 0$ such that for $∥ h ∥ < δ_{2}$ both $∥ k (h) ∥ \leq (∥ T ∥ + 1) ∥ h ∥$ and $∥ r (h) ∥ \leq \frac{η}{2 (1 + ∥ S ∥)} ∥ h ∥$ hold (take the smaller of the two thresholds). Then for $∥ h ∥ < \min {δ_{2}, δ_{1} / (∥ T ∥ + 1)}$ one has $∥ R (k (h)) ∥ \leq \frac{η}{2} ∥ h ∥$ and $∥ S r (h) ∥ \leq ∥ S ∥ ∥ r (h) ∥ \leq \frac{η}{2} ∥ h ∥$ , so $∥ g (f (x + h)) - g (f (x)) - S T h ∥ \leq η ∥ h ∥$ . Hence the bounded linear operator $S T$ of [L4] satisfies the remainder condition for $g \circ f$ at $x$ , and by [L5] it is the derivative, which is claim 3.

2.1step 1.2L3algebra

For claim 2 put $Δ_{f} (h) : = f (x + h) - f (x)$ , $Δ_{g} (h) : = g (x + h) - g (x)$ , so that $Δ_{f} (h) = D f (x) h + r_{f} (h)$ and $Δ_{g} (h) = D g (x) h + r_{g} (h)$ with remainders $r_{f}, r_{g}$ as in [L1]. By bilinearity, expanding $B (f (x) + Δ_{f}, g (x) + Δ_{g})$ gives $B (f (x + h), g (x + h)) - B (f (x), g (x)) - L (h) = B (r_{f} (h), g (x)) + B (f (x), r_{g} (h)) + B (Δ_{f} (h), Δ_{g} (h)),$ because $B (Δ_{f}, g (x)) = B (D f (x) h, g (x)) + B (r_{f}, g (x))$ and likewise in the second variable, while the cross term is $B (Δ_{f}, Δ_{g})$ .

3.1step 1.2L1L3L5algebra

For $∥ h ∥$ small, [L1] with $ε = 1$ gives $∥ Δ_{f} (h) ∥ \leq (∥ D f (x) ∥ + 1) ∥ h ∥$ and $∥ Δ_{g} (h) ∥ \leq (∥ D g (x) ∥ + 1) ∥ h ∥$ . Combining this with [step 2.1] and [L3], $∥ B (f (x + h), g (x + h)) - B (f (x), g (x)) - L (h) ∥ \leq C ∥ r_{f} (h) ∥ ∥ g (x) ∥ + C ∥ f (x) ∥ ∥ r_{g} (h) ∥ + C (∥ D f (x) ∥ + 1) (∥ D g (x) ∥ + 1) ∥ h ∥^{2} .$ Dividing by $∥ h ∥$ for $h \neq 0$ and letting $h \to 0$ , every term tends to $0$ by [L1], so the left-hand side is $o (∥ h ∥)$ and the bounded linear operator $L$ of [step 1.2] satisfies the remainder condition for $B (f, g)$ at $x$ ; by [L5] it is the derivative, which is claim 2.

4.1step 1.1step 1.3step 3.1∎

Claim 1 is [step 1.1], claim 2 is [step 3.1], and claim 3 is [step 1.3]; this is exactly the conjunction stated.

Depends on

Dependency tree · 2 levels

Sources

Zuoqin Wang, Lecture 6 — §§2.1.1–2.1.4