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Differentiation of an integral functional under growth domination

Statement

Assume Countable Choice (The Axiom of Countable Choice (ACω)). Let Ω⊆Rn be a bounded C1 domain (Bounded C^k domains and boundary charts), 1<p<∞, and let f:Ω×R×Rn→R be a Caratheodory integrand (A Caratheodory integrand composed with measurable functions is measurable) whose classical partial derivatives fs, fξ exist and are continuous in (s,ξ) for almost every x, with a constant C≥0 and functions G∈L1(Ω), g∈Lp′(Ω), where p′=p/(p−1), such that for almost every x and all (s,ξ) ∣f(x,s,ξ)∣≤C(1+∣s∣p+∣ξ∣p)+G(x),∣fs(x,s,ξ)∣+∣fξ(x,s,ξ)∣≤C(1+∣s∣p−1+∣ξ∣p−1)+g(x). Then I(u):=∫Ωf(x,u(x),Du(x)) dx is well defined and finite on W1,p(Ω) (Integer-order Sobolev spaces and their norms), and for all u,v∈W1,p(Ω) δI(u;v)=∫Ω(fs(x,u,Du) v+fξ(x,u,Du)⋅Dv) dx. In particular I is Gateaux differentiable at every u, with bounded Gateaux derivative δI(u)∈W1,p(Ω)∗ (Gateaux and Frechet derivatives of a functional).

Facts & Assumptions

Given: Countable Choice; a bounded C1 domain Ω⊆Rn, 1<p<∞ with Holder conjugate p′=p/(p−1), and a Caratheodory integrand f:Ω×R×Rn→R whose classical partials fs,fξ exist and are continuous in (s,ξ) for almost every x, with C≥0, G∈L1(Ω), g∈Lp′(Ω) and, for almost every x and all (s,ξ), ∣f(x,s,ξ)∣≤C(1+∣s∣p+∣ξ∣p)+G(x),∣fs(x,s,ξ)∣+∣fξ(x,s,ξ)∣≤C(1+∣s∣p−1+∣ξ∣p−1)+g(x).

[F1]

For a Caratheodory integrand and measurable u,w, the composition x↦f(x,u(x),w(x)) is measurable (A Caratheodory integrand composed with measurable functions is measurable).

[F2]

The proof of Integer-order Sobolev spaces are Banach uses only Countable Choice after AC supplies it, so the Countable Choice assumed here supplies that same completeness argument and makes W1,p a Banach space. On a bounded domain, the class u∈W1,p(Ω) has u,Du∈Lp(Ω) and finite norm ∥u∥W1,p(Ω), with ∥u∥p≤∥u∥W1,p and ∥∣Du∣∥p≤n∥u∥W1,p, since ∣Du∣≤∑i∣Diu∣; the constant 1 lies in L1(Ω)∩Lp′(Ω), ∣G∣∈L1(Ω), and ∣u∣p−1,∣Du∣p−1 (and the analogous monomials in v,Dv) lie in Lp′(Ω) (Integer-order Sobolev spaces and their norms, Bounded C^k domains and boundary charts).

[F3]

Holder's inequality: ∫∣hw∣≤∥h∥p′∥w∥p for h∈Lp′(Ω) and w∈Lp(Ω) (Holder's inequality for integrals, including the endpoint cases).

[F4]

Dominated convergence: if measurable qk→q almost everywhere and ∣qk∣≤H almost everywhere for a single H∈L1(Ω), then ∫qk→∫q (Dominated convergence).

[F5]

If the classical partial derivatives of (s,ξ)↦f(x,s,ξ) are continuous at a point, then that map is totally differentiable there (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative); the chain rule identifies the total derivative of t↦f(x,s+tv,ξ+tw) as fs(x,s+tv,ξ+tw)v+fξ(x,s+tv,ξ+tw)⋅w (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)); the one-variable mean value theorem applies to t↦f(x,s+tv,ξ+tw) on a compact interval (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)).

[F6]

I is Gateaux differentiable at u with Gâteaux derivative δI(u)∈W1,p(Ω)∗ precisely when the limits δI(u;v) exist for all v and v↦δI(u;v) is a bounded linear functional (Gateaux and Frechet derivatives of a functional).

Proof

technique · direct, by difference quotients, the mean value theorem, a uniform integrable dominator and dominated convergence
1.1F1F2given

Well-definedness and finiteness. For u∈W1,p(Ω) the map x↦f(x,u(x),Du(x)) is measurable by [F1], and the first bound of the hypothesis, together with G≤∣G∣, gives ∣f(x,u(x),Du(x))∣≤C(1+∣u(x)∣p+∣Du(x)∣p)+∣G(x)∣, whose integral is finite by [F2]. Hence I(u)=∫Ωf(x,u(x),Du(x)) dx is a well-defined real number for every u∈W1,p(Ω).

1.2F5given

Difference quotients and their pointwise limit. Fix u,v∈W1,p(Ω) and put qε(x):=ε−1(f(x,u(x)+εv(x),Du(x)+εDv(x))−f(x,u(x),Du(x))) for ε≠0. For almost every x the map (s,ξ)↦f(x,s,ξ) is C1, so [F5] applies to gx(t):=f(x,u(x)+tv(x),Du(x)+tDv(x)): by the mean value theorem there is θ=θ(x,ε)∈(0,1) with qε(x)=gx′(θε), and gx′(t)=fs(x,u+tv,Du+tDv)v(x)+fξ(x,u+tv,Du+tDv)⋅Dv(x). As ε→0 the arguments (u(x)+θεv(x),Du(x)+θεDv(x)) tend to (u(x),Du(x)), so continuity of the partials gives the pointwise limit qε(x)→fs(x,u(x),Du(x))v(x)+fξ(x,u(x),Du(x))⋅Dv(x) for almost every x.

2.1F2F3step 1.2

A single integrable dominator. For almost every x and every 0<∣ε∣≤1, the representation of step 1.2 and the second bound of the hypothesis give, with θ=θ(x,ε) and the elementary estimate (a+b)p−1≤2p−1(ap−1+bp−1), ∣qε(x)∣≤h(x) (∣v(x)∣+∣Dv(x)∣),h:=C′(1+∣u∣p−1+∣v∣p−1+∣Du∣p−1+∣Dv∣p−1)+∣g∣, for a constant C′ depending only on C and p. Since u,Du,v,Dv∈Lp(Ω), the monomials ∣u∣p−1,…,∣Dv∣p−1 lie in Lp′(Ω), and ∣g∣∈Lp′(Ω), while the constant term is integrable on the bounded domain; hence h∈Lp′(Ω), and Hölder's inequality [F3] gives h(∣v∣+∣Dv∣)∈L1(Ω), uniformly in ε.

3.1F4step 2.1

Dominated convergence identifies the limit. Let εk→0, εk≠0, and choose K such that ∣εk∣≤1 for every k≥K. By step 1.2 the functions qεk converge pointwise almost everywhere to L(x):=fs(x,u,Du)v+fξ(x,u,Du)⋅Dv, and by step 2.1 the tail (qεk)k≥K is dominated by the single L1 function h(∣v∣+∣Dv∣). Applying [F4] to this tail gives ∫Ωqεk→∫ΩL; removing a finite prefix does not change the limit. Since the sequence εk→0 was arbitrary, lim⁡ε→0ε−1(I(u+εv)−I(u))=∫Ω(fs(x,u,Du)v+fξ(x,u,Du)⋅Dv)dx.

4.1F3F6step 3.1∎

Boundedness of the derivative, and conclusion. The map v↦δI(u;v):=∫Ω(fs(x,u,Du)v+fξ(x,u,Du)⋅Dv)dx is linear in v by linearity of the integral, and the pointwise estimate ∣fs(x,u,Du)v+fξ(x,u,Du)⋅Dv∣≤hu(x)(∣v∣+∣Dv∣) with hu:=C(1+∣u∣p−1+∣Du∣p−1)+∣g∣∈Lp′(Ω) gives, by [F3], ∣δI(u;v)∣≤(1+n)∥hu∥p′∥v∥W1,p(Ω); hence δI(u)∈W1,p(Ω)∗. By [F6] the functional I is Gateaux differentiable at u with derivative δI(u) and the displayed formula, as claimed.

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