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Second derivative formula for a composed cycle map

Statement

Let $X_{0}, X_{1}, X_{2}, X_{3}$ be finite-dimensional real Euclidean spaces and let $U_{i} \subseteq X_{i}$ be open for $i = 0, 1, 2$ . Suppose $S : U_{0} \to U_{1}$ , $C : U_{1} \to U_{2}$ , and $T : U_{2} \to X_{3}$ are $C^{2}$ , meaning that each is totally differentiable with continuous derivative and that its derivative map is itself totally differentiable. Write $L (X, Y)$ for the finite-dimensional space of linear maps $X \to Y$ with the induced Euclidean operator norm. Put $E = T \circ C \circ S$ . For a $C^{2}$ map $f$ , define $D^{2} f (x) [h, k] : = (D (D f) (x) h) k$ . At $x \in U_{0}$ , set $y = S (x)$ and $z = C (y)$ . Then for all $h, k \in X_{0}$ , $D^{2} E (x) [h, k] = D^{2} T (z) [D C (y) D S (x) h, D C (y) D S (x) k] + D T (z) (D^{2} C (y) [D S (x) h, D S (x) k] + D C (y) D^{2} S (x) [h, k]) .$

Facts & Assumptions

Given: The finite-dimensional Euclidean spaces, compatible open domains, $C^{2}$ maps $S, C, T$ , and the stated definition of $D^{2} f$ .

[F1]

Chain sum product and composition rules for Banach derivatives: Fréchet derivatives obey the composition rule and the product rule for bounded bilinear maps.

[F2]

Every finite-dimensional normed space is Banach: a finite-dimensional normed space with a finite basis is complete, hence Banach.

Proof

technique · direct

1.1givenF1F2algebra

The spaces $X_{i}$ and their matrix spaces $L (X_{i}, X_{j})$ have finite bases, so [F2] makes them Banach and [F1] applies. Put $G = C \circ S$ . The composition rule gives $D G (x) = D C (y) D S (x)$ . Composition of bounded linear maps is the bounded bilinear map $(A, B) \mapsto A \circ B$ , since $∥ A \circ B ∥_{o p} \leq ∥ A ∥_{o p} ∥ B ∥_{o p}$ . Applying its product rule and the composition rule in [F1] to this derivative identity, then evaluating on $k$ , gives $D^{2} G (x) [h, k] = D^{2} C (y) [D S (x) h, D S (x) k] + D C (y) D^{2} S (x) [h, k]$ .

2.1givenF1step 1.1algebra∎

Since $E = T \circ G$ , [F1] gives $D E (x) = D T (G (x)) D G (x)$ . Differentiate this bounded-operator product in direction $h$ and evaluate on $k$ : the derivative of $D T \circ G$ contributes $D^{2} T (z) [D G (x) h, D G (x) k]$ , and the derivative of $D G$ contributes $D T (z) D^{2} G (x) [h, k]$ . Substituting $D G (x) = D C (y) D S (x)$ and step 1.1 yields the stated formula.

Depends on

Used by

Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Chin-Lung Wang, Banach Calculus, §§2.1.1–2.1.4 and §2.3 (standard reference, not scraped)