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The trace of an affine function on a ball is its classical restriction

Example

Assume the Axiom of Choice. Let Ω=BR(0)⊂Rn, n≥2, 1≤p<∞, c∈Rn, d∈K, and u(x)=c⋅x+d. Then u∈C∞(Ω‾)∩W1,p(Ω) and, by The trace agrees with classical restriction for continuous Sobolev functions, Tu=(c⋅x+d)∣∂Ω; moreover Tu∈W1−1/p,p(∂Ω) for 1<p<∞ with ∥Tu∥W1−1/p,p(∂Ω)≤C(R,n,p)(∣c∣R+∣d∣) by The sharp trace theorem: boundedness and range in the fractional space. At p=1 only the L1 statement is made; the space is not renamed W0,1(∂Ω). On the sphere ∂BR(0) the boundary Lp norm is the classical surface integral ∥Tu∥Lp(∂Ω)p=Rn−1∫Sn−1∣c⋅Rω+d∣p dσ(ω), with σ the surface measure on the unit sphere.

Facts & Assumptions

Given: The Axiom of Choice; n≥2, R>0, 1≤p<∞, c∈Rn, d∈K; the affine function u(x)=c⋅x+d; and the trace T of The Lp trace operator on a bounded C1 domain.

[F1]

If a W1,p(Ω) class has a continuous representative on Ω‾, its trace is the classical restriction of that representative. (The trace agrees with classical restriction for continuous Sobolev functions)

[F2]

For 1<p<∞ the trace maps W1,p(Ω) onto Wθ,p(∂Ω) with θ=1−1/p and is bounded: ∥Tu∥Wθ,p(∂Ω)≤C(Ω,p)∥u∥W1,p(Ω). (The sharp trace theorem: boundedness and range in the fractional space, The fractional Sobolev space on a compact C1 boundary)

[F3]

The surface integral on the sphere is ∫∂BR(0)h dS=Rn−1∫Sn−1h(Rω) dσ(ω), and the boundary integral is finite for continuous h on the compact boundary. (Surface integration on compact C1 hypersurfaces)

[F4]

The Euclidean ball has finite Lebesgue measure, and ∫BR(0)∣x∣pdx≤Rp∣BR(0)∣. (A box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai), whichever of its faces are included)

Verification

1.1F1F4algebragiven

The trace is the classical restriction, with a controlled norm. The affine function is smooth on Ω‾; its gradient is the constant c, so ∫Ω∣u∣pdx≤2p−1(∣c∣p∫Ω∣x∣pdx+∣d∣p∣Ω∣)≤C(R,n,p)(∣c∣R+∣d∣)p and ∫Ω∣∂iu∣pdx=∣ci∣p∣Ω∣≤C(R,n,p)(∣c∣R+∣d∣)p by [F4]. Hence u∈W1,p(Ω) with ∥u∥W1,p(Ω)≤C′(R,n,p)(∣c∣R+∣d∣), and u∈C(Ω‾), so Tu=(c⋅x+d)∣∂Ω by [F1].

2.1F2F3step 1.1algebra

Fractional membership and the boundary norm. For 1<p<∞ put θ:=1−1/p. By [F2] and step 1.1, Tu∈Wθ,p(∂Ω) with ∥Tu∥Wθ,p(∂Ω)≤C(R,n,p)(∣c∣R+∣d∣); at p=1 the same computation gives only Tu∈L1(∂Ω), and no space W0,1 is introduced. For the Lp value, parametrise the sphere by x=Rω; by [F3] and the definition of the surface integral, ∥Tu∥Lp(∂Ω)p=∫∂Ω∣c⋅x+d∣pdS=Rn−1∫Sn−1∣c⋅Rω+d∣pdσ(ω), which is the classical sphere integral.

3.1step 1.1step 2.1algebragiven∎

Conclusion. Steps 1.1 and 2.1 prove that the affine class lies in W1,p(Ω) with trace its classical restriction, that the trace lies in the fractional boundary space for 1<p<∞ with the stated bound, and that its boundary Lp norm is the displayed surface integral.

Source notes

Teschl's Theorem 9.18 (printed p. 209) is the statement Tf=f∣∂U for continuous functions; Laugesen's Theorem 3.14 (printed pp. 62-64) records the classical boundary values, and Gagliardo's Teorema [1.I] (printed p. 289) the inverse estimate behind the fractional bound. The example keeps the endpoint p=1 out of the fractional notation, as required by the page conventions.

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