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Positive Hölder cubic map is continuously differentiable

Statement

Assume Countable Choice. Let X and Y=C0,α(Ω‾) be the Hölder spaces with the norms specified in Bounded-domain charged conformal problem, and let S∈Y be its source. The inclusion X↪Y is bounded. The set U:={u∈Y:min⁡Ω‾(1+u)>0} is open, and

NS:U→Y,NS(u):=S(1+u)−3,

is continuously Fréchet differentiable with DNS(u)[h]=−3S(1+u)−4h for every u∈U and h∈Y.

Facts & Assumptions

Given: Countable Choice, the bounded smooth domain Ω, 0<α<1, the Hölder norm on Y=C0,α(Ω‾), and S∈Y.

[A1]

Countable Choice is used for completeness of the Hölder spaces in [F2]. The Axiom of Countable Choice (ACω)

[F1]

Bounded-domain charged conformal problem: ∥g∥Y=∥g∥∞+[g]0,α, with [g]0,α=sup⁡x≠y∣g(x)−g(y)∣/∣x−y∣α.

[F2]

The closure Hölder spaces are Banach spaces: under Countable Choice, C0,α(Ω‾) with its full Hölder norm is a Banach space, as is the closed zero-boundary subspace X of C2,α(Ω‾).

[F3]

Fréchet derivative between Banach spaces and C k map between Banach spaces: Fréchet differentiability means a bounded linear first-order approximation with remainder o(∥h∥), and C1 means this derivative is continuous.

Proof

Given: The hypotheses above.

1.1givenF1algebra

Bounded inclusion. Compactness of Ω‾ and the smooth boundary give a finite cover by interior balls and flattened boundary boxes whose smaller concentric patches still cover Ω‾. In an interior patch, sufficiently close points can be joined by a segment; in a boundary patch, flattening identifies the domain locally with one side of a graph, and the inverse graph map and its inverse are Lipschitz on the smaller box. Therefore there are δ,C0>0 such that any x,y∈Ω‾ with ∣x−y∣<δ can be joined in Ω‾ by a curve of length at most C0∣x−y∣: take a Lebesgue number for the finite smaller-patch cover and pull back the straight segment in the corresponding ball or half-box. For boundary endpoints, approximate both endpoints by interior points in the same patch and pass to the limit using the continuous extensions of u and its first derivatives. The fundamental theorem of calculus along these curves gives ∣u(x)−u(y)∣≤C0∥∇u∥∞∣x−y∣ for such pairs. For ∣x−y∣≥δ, the quotient is at most 2δ−α∥u∥∞. Consequently [u]0,α≤C(∥u∥∞+∥∇u∥∞), so ∥u∥Y≤C∥u∥X.

1.2givenF1F4algebra

The Hölder norm is an algebra norm. For f,g∈Y, the identity f(x)g(x)−f(y)g(y)=f(x)(g(x)−g(y))+g(y)(f(x)−f(y)) gives [fg]0,α≤∥f∥∞[g]0,α+∥g∥∞[f]0,α, hence ∥fg∥Y≤∥f∥Y∥g∥Y. By [F4], the positive minimum m=min⁡Ω‾(1+u) exists; if ∥h∥Y<m/2, then 1+u+h≥m/2>0, so U is open.

2.1F1F5step 1.2algebra

Hölder composition and remainder. On a sufficiently small Y-neighborhood of a fixed u∈U, the values of 1+u lie in a compact interval contained in (0,∞). For j=3,4,5, the scalar derivative of s↦s−j is bounded on that interval; the real mean-value theorem [F5] gives [(1+u)−j]0,α≤C[u]0,α, and these compositions lie in Y. Put z=1+u and r=h/z. Since z≥m>0, ∣z(x)−1−z(y)−1∣≤m−2∣z(x)−z(y)∣, so z−1∈Y and step 1.2 gives ∥r∥Y≤C∥h∥Y. Direct algebra gives (z+h)−3−z−3+3z−4h=z−3(6r2+8r3+3r4)/(1+r)3. For small ∥h∥Y, z+h≥m/2; the same reciprocal estimate gives (1+r)−1=z/(z+h) bounded in Y. The algebra estimate of step 1.2 therefore bounds the norm of this remainder after multiplication by S by C∥h∥Y2=o(∥h∥Y), proving DNS(u)[h]=−3S(1+u)−4h.

3.1A1F1F2F3step 1.1step 2.1algebra∎

Continuity of the derivative. For u,v in a smaller neighborhood with q:=1+u, p:=1+v, and q,p≥m/2, the factorization q−4−p−4=(p−q)(p+q)(p2+q2)/(p4q4) and the reciprocal estimates of step 2.1 give ∥(1+u)−4−(1+v)−4∥Y≤C∥u−v∥Y. Thus ∥DNS(u)−DNS(v)∥Y→Y≤3∥S∥YC∥u−v∥Y, proving continuity of DNS. By [A1] and [F2], X and Y are Banach, so step 2.1 and the derivative continuity just proved establish that NS is continuously Fréchet differentiable as a map between Banach spaces; by step 1.1, its restriction along the bounded inclusion X↪Y is also C1.

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