Sobolev inequalities for fractional Neumann Laplacians on half spaces

Lemma A.1.

Let p∈(1,∞), t∈(-1p′,2⁢s+1p) and let α be an exponent satisfying

If u∈Lp⁢(R+n;|x1|-t⁢p⁢d⁢x), then Ps⁢u∈Lp⁢(R-n;|x1|-t⁢p⁢d⁢x) and

Proof.

We use Hölder’s inequality and Fubini’s theorem to estimate

The proof is complete. ∎

Corollary A.2.

The linear transform Ps:Lp⁢(R+n)→Lp⁢(R-n) is continuous for any exponent p∈(1,∞].

Proof.

If p=∞, it trivially holds that 𝒫s:L∞⁢(ℝ+n)→L∞⁢(ℝ-n) is nonexpansive. To handle the case p∈(1,∞), take t=0 in Lemma A.1 and conclude. ∎

Remark A.3.

The assumption p>1 is necessary. Let E⊂ℝ+n be any bounded measurable set of positive measure. Since χE∈Lp⁢(ℝ+n) for any p∈[1,∞], we have 𝒫s⁢χE∈Lp⁢(ℝ-n) for any p∈(1,∞], by Corollary A.2. Now, for x∈ℝ-n and R>0 large enough, we estimate

which readily implies 𝒫s⁢χE∉L1⁢(ℝ-n).

Proof of Lemma 4.2.

To check (i), note that the set

is convex, closed and not empty. Thus, the minimization problem (4.6) has a unique solution Ps⁢u∈𝒟Srs⁢(ℝ+n). Further, we have that

because the polynomial t↦ℰs⁢(Ps⁢u+t⁢φ;ℝ2⁢n∖(ℝ-n)2) attains its minimum at t=0. Take φ∈𝒞0∞⁢(ℝ-n). Using (A.2), (4.4) and recalling that φ⁢(x)-φ⁢(y)≡0 for x,y∈ℝ+n, we find

Since φ was arbitrarily chosen, identity (4.5) follows.

Now we prove (ii). The operator u↦Ps⁢u is clearly linear. From KPs⁢u=Ku, we infer that Ps is projector. Since

we see from (A.2) that Ker⁡(Ps)⊥Im⁡(Ps), thus Ps is an orthoprojector.

Finally, statement (iii) is the Pythagorean theorem for orthoprojectors. ∎

Proof of Lemma 4.3.

By Corollary A.2 and thanks to Lemma 4.1, for any u∈𝒟Srs⁢(ℝ+n), we have

Thus, Ps:𝒟Srs⁢(ℝ+n)→L2s*⁢(ℝn) is continuous. Since Ps coincides with the identity on ℛSrs⁢(ℝ+n), the continuity of the embedding ℛSrs⁢(ℝ+n)↪L2s*⁢(ℝn) follows for free.

To prove (ii), take an exponent p∈[1,2s*) and a sequence uh∈𝒟Srs⁢(ℝ+n) such that uh→0 weakly in 𝒟Srs⁢(ℝ+n). We have to show that Ps⁢uh→0 in Lp⁢(Br) for any r>0.

For arbitrary ρ>2⁢r, we write

We estimate 𝒰h∞⁢(x), for x∈Br-, as follows:

We infer that

Trivially,

and Corollary A.2 gives ∥𝒰h0∥Lp⁢(Br-)≤c⁢(p)⁢∥uh∥Lp⁢(Bρ+), so we arrive at

Since uh→0 weakly in 𝒟Srs⁢(ℝ+n), we have that uh is bounded in L2s*⁢(ℝ+n). Thus, given any ε>0, we can find a large ρ=ρ⁢(ε)>0 such that the last term in (A.3) is smaller than ε. Hence,

as, by Lemmata 2.1 and 4.1, we have uh→0 in Lp⁢(Bρ+). Since ε>0 was arbitrarily chosen, we are done. ∎

Proof of Lemma 4.4.

Let ℱ be the Fourier transform in ℝn. It is well known that ℱ⁢[(-Δ)s⁢φ]⁢(ξ)=|ξ|2⁢s⁢ℱ⁢[φ] for any φ∈𝒞0∞⁢(ℝn). Since u∈L2s*⁢(ℝn), we can define the distribution (-Δ)s⁢u via

Next, if φ∈𝒞0∞⁢(ℝ+n), then φ⁢(x)-φ⁢(y)=0 for x,y∈ℝ-n, hence

which concludes the proof of (i).

Next, since u=Ps⁢u and 𝒞0∞⁢(ℝ-n)⊂𝒟Srs⁢(ℝ+n), (A.2) gives (-Δℝ+nN)Srs⁢u=0 on ℝ-n immediately. Using again u=Ps⁢u and the explicit expression for Ps in (4.5), for any x∈ℝ-n, we obtain

and the lemma is proved. ∎

Theorem B.1 (Limits as s→1-).

If u∈H1⁢(Rn), then

Proof.

The conclusion for the restricted quadratic form should be known, at least for bounded domains. We cite, for instance, [4] for related results. We furnish here a complete proof for the convenience of the reader.

We denote by cn any constant possibly depending on the dimension n but not on s; in particular, we have Cn,s≤cn⁢(1-s) for s∈(0,1). We start by proving that

Note that the proof in [7, Subsection 5.1] of a similar result on bounded domains Ω⊂ℝn contains a defect, precisely in the proof of formula (5.5). However, the statement of [7, Proposition 5.1] is correct.

We introduce the notation Πδ={x∈ℝn||x1|<δ} and estimate

We have

To handle ℐ2, we estimate

so that

Thus, for s close to 1 and small δ, we obtain

Formula (B.2) readily follows, because for any ε>0, we can find δ=δ⁢(ε)>0 such that cn⁢∥∇⁡u∥L2⁢(Π2⁢δ)2<ε, and thus

From (B.2) we first infer that ℰs⁢(u;ℝn×ℝn)=ℰs⁢(u;ℝ+n×ℝ+n)+ℰs⁢(u;ℝ-n×ℝ-n)+o⁢(1). Further, by replacing u by u^, the symmetric extension of u|ℝ+n, and using (B.1), we obtain

Thus, ∫ℝ+n|∇u|2dx=ℰs(u;ℝ+n×ℝ+n)+o(1). The conclusion for ℰs⁢(u;ℝ2⁢n∖(ℝ-n)2) readily follows from (4.1) and (B.2). ∎

Theorem B.2 (Limit as s→0+, restricted Laplacian).

If u∈H1⁢(Rn), then

Proof.

For s∈(0,12), we have χℝ+n⁢u∈Hs⁢(ℝn), see [25, Section 2.10.2]. Thus, via (2.1), we can compute

By (B.1) we have

Next, from (2.2), we see that lims→0+⁡μs=12, and the conclusion readily follows. ∎

Theorem B.3 (Limit as s→0+, semirestricted Laplacian).

If u∈H1⁢(Rn), then

Proof.

Identity (B.3) readily follows from (4.2) and Theorem B.2, since

To prove (B.4) we first notice that u∈L2⁢(ℝ+n;x1-2⁢s⁢d⁢x) for 0≤s<12. Choosing p=2, t=s and α=1-s in (A.1), we get that Ps⁢u∈L2⁢(ℝ-n;|x1|-2⁢s⁢d⁢x) and

Since Γ⁢(1-s+2⁢s)=1+o⁢(1) and Γ⁢(s)Γ⁢(2⁢s)=O⁢(s), we infer that

Using also (4.7), we obtain

The conclusion follows, thanks to (B.3) and (B.5). ∎