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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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Continuous mixed partials of order kk are invariant under permutations

Statement

Let k2k\ge2. If fCk(U)f\in C^k(U) for an open URmU\subseteq\mathbb R^m, then every iterated derivative of ff of order kk is unchanged by any permutation of its coordinate differentiations.

Facts & Assumptions

Given: A CkC^k scalar field on UU and a word of kk coordinate indices.

[L1]

Adjacent second coordinate derivatives commute under the C2C^2 hypotheses (Clairaut--Schwarz theorem for continuous second partial derivatives).

[L2]

A CkC^k field has every ordered iterated partial derivative through length kk, continuously on UU (CkC^k maps and multi-index derivative notation in Euclidean space).

Proof

technique · induction
1.1

Label the kk differentiation positions and count inversions of a permutation of these labels. A permutation with zero inversions is the identity, so it leaves the derivative unchanged.

basealgebra
1.2

Assume every reordering with at most nn inversions leaves the derivative unchanged.

ih
2.1

A reordering with n+1n+1 inversions has an adjacent inverted pair; exchanging that pair reduces its inversion count by one. If that pair occupies positions r,r+1r,r+1 in the sequence of differentiation operations, first apply only the operations in positions 1,,r11,\ldots,r-1 and call the resulting field gg. Every ordered partial of gg through order two is an ordered partial of ff of length at most kk, hence is continuous by [L2]; thus gC2(U)g\in C^2(U) and [L1] swaps precisely the operations in positions r,r+1r,r+1. Apply the remaining outer operations in positions r+2,,kr+2,\ldots,k to this equality; their existence is again supplied by [L2].

step 1.2L1L2algebra
3.1

The induction hypothesis applies after the swap in step 2.1, so the original reordering leaves the derivative unchanged. Induction on inversion number proves the claim for every finite permutation.

step 1.1step 1.2step 2.1discharge-induction

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 43 results over 15 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