Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generated
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.

The closure Hölder spaces are Banach spaces

Statement

Assume Countable Choice. Let n≥1, let k≥0 be an integer, 0<α<1, K∈{R,C} and let Ω⊆Rn be open and nonempty. Then Cbk,α(Ω;K) with the norm of Hölder spaces Ck,α, closure and interior scaled norms, and Ck,α domains is a Banach space over K: every Cauchy sequence in ∥⋅∥Ck,α(Ω) converges in that norm to a limit whose k-th partial derivatives are α-Hölder on Ω. If in addition Ω is a bounded Ck,α domain with nonempty boundary, then the subspace X0:={u∈Cbk,α(Ω):u extends continuously to Ωˉ with u∣∂Ω=0} is closed in Cbk,α(Ω) and hence is a Banach space. For every bounded open nonempty Ω, the boundary-extension class Ck,α(Ωˉ;K) is also a closed subspace of Cbk,α(Ω;K) and hence a Banach space, as is its zero-boundary subspace. Completeness here is for the full finite Hölder norm. A boundary Hölder seminorm alone does not define a norm on this function space, since it vanishes on nonzero constant functions.

Facts & Assumptions

Given: ACω, n≥1, integers k≥0, 0<α<1, K∈{R,C}, an open nonempty Ω⊆Rn, and a Cauchy sequence (uj) in Cbk,α(Ω;K).

[A1]

The only choice assumption is Countable Choice ACω, used through the sequential completeness of R and C and the cited metric-space completeness conventions. No full Axiom of Choice is used. (The Axiom of Countable Choice (ACω))

[F1]

The norm is ∥u∥Ck,α(Ω)=∑j=0ksup⁡Ωmax⁡∣β∣=j∣Dβu∣+[u]k,α;Ω with [u]k,α;Ω=∑∣β∣=k[Dβu]0,α;Ω, and Cbk,α consists of the Ck functions with finite norm; Dβ are the canonical-order derivatives. (Hölder spaces Ck,α, closure and interior scaled norms, and Ck,α domains, Ck maps and multi-index derivative notation in Euclidean space)

[F2]

Uniformly Cauchy sequences of real- (or complex-) valued functions converge uniformly; a uniform limit of continuous functions is continuous; and if gm→g uniformly on an interval and the derivatives gm′ converge uniformly with gm(t0)→g(t0) at one point, then g′=lim⁡mgm′. (A sequence of real-valued functions converges uniformly if and only if it is uniformly Cauchy, The uniform limit of continuous real-valued functions on a metric space is continuous, A uniform limit of continuous complex-valued functions is continuous, If continuously differentiable functions converge at one point and their derivatives converge uniformly on a closed interval, then the functions converge uniformly to a differentiable function whose derivative is the derivative limit)

[F3]

A Banach space is a complete normed space; a subspace of a complete metric space is complete if and only if it is closed, under Countable Choice. (Banach space, Complete metric space: every Cauchy sequence converges in the space, Closed subspaces of complete metric spaces are complete; the converse under countable choice)

[F4]

The real mean value theorem bounds the increment of a differentiable function on a segment by the supremum of its derivative times the length of the segment. (The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a))

Proof

technique · direct
1.1givenF1F2algebraA1

Uniform limits of the derivative fields. Since (uj) is Cauchy in ∥⋅∥Ck,α(Ω), for every multi-index β with ∣β∣≤k the sequence (Dβuj) is uniformly Cauchy on Ω: for j,l and every x∈Ω, ∣Dβuj(x)−Dβul(x)∣≤∥uj−ul∥Ck,α(Ω). By [F2] there is a bounded function vβ with Dβuj→vβ uniformly on Ω, and vβ is continuous. Moreover sup⁡Ω∣vβ∣=lim⁡jsup⁡Ω∣Dβuj∣≤lim inf⁡j∥uj∥Ck,α(Ω)<∞, the last bound holding because a Cauchy sequence is bounded.

2.1step 1.1F1algebra

The limits are Hölder. For ∣β∣=k and x≠y in Ω, ∣vβ(x)−vβ(y)∣=lim⁡j∣Dβuj(x)−Dβuj(y)∣≤lim inf⁡j[Dβuj]0,α;Ω∣x−y∣α; hence [vβ]0,α;Ω≤lim inf⁡j[Dβuj]0,α;Ω≤lim inf⁡j∥uj∥Ck,α(Ω)<∞ and vβ is α-Hölder on Ω. Consequently v0 (whose finiteness and continuity is step 1.1, including k=0, where no derivative is involved) satisfies ∥v0∥Ck,α(Ω)≤lim inf⁡j∥uj∥Ck,α(Ω)<∞ as soon as vβ=Dβv0 for all ∣β∣≤k, which is proved next.

2.2step 1.1F1F2algebraF4

Identification of the limits with the derivatives of v0. Proceed by induction on ∣β∣. For β=0, v0 is the limit. Suppose vβ=Dβv0 is known on Ω for some ∣β∣<k; fix i and a ball B⋐Ω (every point of Ω lies in such a ball). On B, all uj are Ck; for orders at least two, Continuous mixed partials of order k are invariant under permutations identifies their derivative words with the canonical-order fields. Thus Dβuj→vβ uniformly while Dβ+eiuj→vβ+ei uniformly; by [F2] applied to the restrictions to each coordinate segment inside B (as in the one-variable theorem on a closed interval, at a fixed base point where Dβuj converges), the limit vβ is differentiable in direction ei with ∂ivβ=vβ+ei on B; Since every point lies in such a ball, this gives ∂iDβv0=Dβ+eiv0=vβ+ei on Ω.

3.1step 2.1step 2.2F1F2F3algebra

Convergence in the Hölder norm. Let ε>0 and choose J with ∥uj−ul∥Ck,α(Ω)≤ε for j,l≥J. Fixing l≥J and passing to the limit in the componentwise bounds of steps 1.1 and 2.1 gives sup⁡Ω∣vβ−Dβul∣≤ε for all ∣β∣≤k and [vβ−Dβul]0,α;Ω≤ε for ∣β∣=k; hence ∥v0−ul∥Ck,α(Ω)≤Ckε for every l≥J with a dimensional factor Ck. So the Cauchy sequence converges in the norm to v0∈Cbk,α(Ω), and Cbk,α(Ω;K) is complete: it is a Banach space over K (the vector-space operations are the pointwise ones and the norm is by [F1]). The complex case follows from the real case applied to real and imaginary parts, using [F2]'s complex uniform limit statement.

4.1step 3.1F1F2F3algebra

The boundary-condition subspace. Assume now Ω is a bounded Ck,α domain with nonempty boundary, and let (uj) be a sequence in X0 converging to u in Cbk,α(Ω). Each uj has a continuous extension uˉj to Ωˉ with uˉj=0 on ∂Ω. Since Ωˉ is compact and the extensions are uniformly Cauchy on the dense set Ω, they are uniformly Cauchy on Ωˉ (for x∈Ωˉ and y∈Ω near x, ∣uˉj(x)−uˉl(x)∣=lim⁡y→x∣uˉj(y)−uˉl(y)∣≤sup⁡Ω∣uj−ul∣); hence uˉj converges uniformly on Ωˉ to a continuous uˉ with uˉ∣Ω=u and uˉ=0 on the closed set ∂Ω, so u∈X0. Thus X0 is closed in the Banach space Cbk,α(Ω), and [F3] makes it complete, hence a Banach space.

5.1step 3.1F1F2F3algebra∎

The closure class. Let Ω be any bounded open nonempty set and let uj∈Ck,α(Ωˉ;K) converge to u in Cbk,α(Ω;K). For every ∣β∣≤k, let wj,β be the continuous extension of Dβuj. Density gives sup⁡Ωˉ∣wj,β−wl,β∣=sup⁡Ω∣Dβuj−Dβul∣, so [F2] yields a continuous uniform limit wβ on Ωˉ. Its restriction is Dβu, by convergence in the full norm. Thus u belongs to the boundary-extension class, which is a closed linear subspace of Cbk,α(Ω;K) and is Banach by step 3.1 and [F3], with the same norm by [F1]. Its zero-boundary subspace is closed because the uniform limit w0 of extensions vanishing on ∂Ω also vanishes there, hence is Banach as well. No identification of the closure class with all of Cbk,α(Ω) is required.

Remarks

  • The interior completeness assertion holds for arbitrary open nonempty Ω. The closure-class assertion assumes boundedness to match its definition; its proof and the closedness of the zero-boundary subspace require no boundary regularity. On an arbitrary bounded open set the closure class can be a proper closed subspace of Cbk,α(Ω) when k≥1.
  • The local class Clock,α(Ω) may contain functions with infinite full-domain norm; the displayed norm defines a Banach space on its finite-norm class Ck,α(Ω)=Cbk,α(Ω). No assertion about completeness for a boundary pseudometric is made.

Depends on

Used by

Dependency tree · two levels

64 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