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Arbitrary L2 boundary data need not have an H1 lifting

Statement refuted

Assume the Axiom of Choice. Let Ω⊂R2 be a bounded C1 domain, let a boundary chart contain the closed straight segment [0,1] strictly inside its patch, and let g:=1(0,1) on that segment, extended by zero. Then g∈L2(∂Ω), but g∉H1/2(∂Ω)=W1/2,2(∂Ω): the Slobodeckij seminorm of the line jump at exponent θ=1−1/2 diverges logarithmically, and by the sharp trace theorem the trace range of W1,2(Ω) is exactly W1/2,2(∂Ω). Consequently no u∈H1(Ω) has Tu=g, so the boundary-value problem with this L2 datum is not solvable in H1: the lifting hypothesis of the weak Dirichlet formulation cannot be relaxed to arbitrary L2 boundary data. No claim is made about the range 1<p<2, where the same jump function does lie in the trace space.

Facts & Assumptions

Given: The Axiom of Choice together with Countable Choice; a bounded C1 domain Ω⊂R2 with a boundary chart containing the closed straight segment [0,1] strictly inside its patch; the jump function g=1(0,1) on that segment, extended by zero; the surface measure on ∂Ω; and the exponent θ=1−1/2=1/2, so that q:=pθ=1 at p=2. (Bounded C^k domains and boundary charts, Surface integration on compact C1 hypersurfaces, The Axiom of Choice, The Axiom of Countable Choice (ACω))

[F1]

The boundary norm of The fractional Sobolev space on a compact C1 boundary is a sum over a finite boundary atlas of the Euclidean Slobodeckij norms of the localised representations (χjg)∘Ψj−1, where χj is a subordinate finite ambient partition; the Euclidean norm is that of The Gagliardo--Slobodeckij space on Euclidean space, the sum of the Lp norm and the extended seminorm [h]s,p=(∫∫∣h(x)−h(y)∣p∣x−y∣−1−spdx dy)1/p, and the set Ws,p(∂Ω) and its topology are independent of the atlas (Chart independence of the fractional boundary norm).

[F2]

Assume Countable Choice. For nonnegative measurable functions on a product of sigma-finite measure spaces the double integral equals the iterated integrals. (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability, The Axiom of Countable Choice (ACω))

[F3]

A bounded measurable function supported in a set of finite surface measure is an Lp(∂Ω) class for every p. (The space Lp(μ) as the quotient by null functions, Surface integration on compact C1 hypersurfaces)

[F4]

Sharp trace theorem: for 1<p<∞ the trace operator T:W1,p(Ω)→Lp(∂Ω) of The Lp trace operator on a bounded C1 domain has range exactly W1−1/p,p(∂Ω) (The sharp trace theorem: boundedness and range in the fractional space, The fractional Sobolev space on a compact C1 boundary).

Proof

1.1F3given

The datum is an L2 class: the indicator of the straight segment (0,1) is bounded and is supported in a set of finite surface measure, so by [F3] it is an Lp(∂Ω) class for every p, in particular g∈L2(∂Ω). At p=2 the exponent is θ=1/2 and q=pθ=1.

1.2F1F2algebra

The line-jump seminorm: for h=1(0,1) on R and q=pθ>0 the integrand ∣h(x)−h(y)∣p is nonzero exactly when one of x,y lies in (0,1) and the other does not. By symmetry and [F2] the double integral equals twice its part with x∈(0,1), y∉(0,1), and the elementary antiderivative ∫u−1−qdu=−u−q/q gives ∫−∞0(x−y)−1−qdy+∫1∞(y−x)−1−qdy=x−q+(1−x)−qq for x∈(0,1). Hence [h]θ,pp=2q∫01(x−q+(1−x)−q)dx=4q∫01x−qdx, which is finite exactly when q<1; translating and scaling the interval (0,1) to another interval (a,b) changes this value by the finite factor (b−a)1−q, so finiteness is intrinsic to the interval indicator.

2.1step 1.2algebra

Divergence at p=2: at q=1 the integral in step 1.2 is ∫01x−1dx=+∞, diverging logarithmically at the endpoint x=0, so [h]1/2,2=+∞.

3.1F1step 2.1

The boundary norm is infinite: fix the finite atlas of [F1] so that it contains the given straight chart with a cutoff equal to one on the closed segment — possible because the segment lies strictly inside the patch — so that this chart's localised representation is the interval indicator h of step 1.2. Then the corresponding summand of the boundary norm is +∞ while every other summand is nonnegative, so ∥g∥W1/2,2(∂Ω)=+∞; by the atlas independence in [F1] the space W1/2,2(∂Ω) is the same set for every atlas, so g∉W1/2,2(∂Ω)=H1/2(∂Ω).

4.1F4step 3.1∎

No H1 lifting: at p=2 the sharp trace theorem identifies the range of T:W1,2(Ω)=H1(Ω)→L2(∂Ω) with W1/2,2(∂Ω); since g lies outside this range, no u∈H1(Ω) satisfies Tu=g, and the inhomogeneous problem with this L2 datum is not solvable in H1.

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