The Helmholtz Decomposition of a BMO Type Vector Field in a Slightly Perturbed Half Space

We introduce a space of L2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$L^2$$\end{document} vector fields with bounded mean oscillation whose “normal” component to the boundary is well-controlled. We establish its Helmholtz decomposition in the case when the domain is a perturbed C3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C^3$$\end{document} half space in Rn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\textbf{R}}^n$$\end{document}(n≥3)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(n \ge 3)$$\end{document} with small perturbation.


Introduction
This is a continuation of our paper [14]. It is well known that the Helmholtz decomposition plays a key role in analyzing the Stokes and the Navier-Stokes equations [25]. Such decomposition is well studied for L p space with 1 < p < ∞. It is a topological direct sum decomposition L p (Ω) n = L p σ (Ω) ⊕ G p (Ω) of the L p vector fields in a domain Ω ⊆ R n . Here, L p σ (Ω) denotes the L p -closure of the space of all smooth div-free vector fields that are compactly supported and G p denotes the space of all L p gradient fields. If p = 2, such decomposition holds for any domain Ω. It is an orthogonal decomposition and often called Weyl's decomposition. For 1 < p < ∞, the decomposition still holds for various domain including the whole space R n , the half space R n + , a perturbed half space, a bounded smooth domain [10] and an exterior smooth domain. However, there are smooth unbounded domains which do not admit L p -Helmholtz decomposition; see e.g. a nice book of Galdi [11].
If p = ∞, such a decomposition does not hold even when Ω = R n since the projection operator is a kind of the Riesz operator which is unbounded in L ∞ , though it is bounded in L p (1 < p < ∞). We replace L ∞ by BM O space. It turns out that it is convenient to consider a subspace vBM O of BM O to have the Helmholtz decomposition, at least for a half space [12] and a bounded domain [14]. Our goal is to extend such a result to a perturbed half space. Unfortunately, it seems that a direct extension is difficult because the behavior at space infinity is not well controlled. Thus we consider the L 2 intersection of this space. For L p space, Farwig, Kozono, and Sohr [7] established the Helmholtz decomposition of L p ∩ L 2 (p ≥ 2) and L p + L 2 (1 < p < 2) for arbitrary uniformly C 2 smooth domain in R 3 . (Later, it is extended to arbitrary dimension [8].) Although we consider vBM O ∩ L 2 in a slightly perturbed half space in the present paper, our results extend to any uniformly C 3 domain. This will be discussed in a separate forthcoming paper.
Let us recall the BM O space of vector fields introduced in [13], [14]. For µ ∈ (0, ∞], we recall the BM O seminorm. For a locally integrable function f in Ω, i.e., f ∈ L 1 loc (Ω), we set is essentially introduced in [2] and well studied in [5]. The Stokes semigroup in such spaces was studied [2], [4] and it is useful to prove that the analyticity of the Stokes semigroup still holds in some unbounded domains which do not admit L p -Helmholtz decomposition [3].
Our space vBM O requires a control only on the normal component. Let d Γ denote the distance function from Γ. We set where · denotes the standard inner product. This is a seminorm. If Ω = R n + , this is not a norm unless n = 1, ν = ∞. However, if Γ has a fully curved part in the sense of [13,Definition 7], then [·] vBM O µ,ν (Ω) becomes a norm [13,Lemma 8]. In particular, when Ω is a bounded C 2 domain, this is a norm. In this paper, we consider the case where Ω is a perturbed C k half space where h ≡ 0 is in C k c (R n−1 ), i.e., h is a compactly supported C k function in R n−1 ; here x ′ = (x 1 , . . . , x n−1 ) for x ∈ R n . A perturbed C k (k ≥ 2) half space R n h is said to be type (K) if where ∇ ′ := (∂ 1 , ∂ 2 , . . . , ∂ n−1 ). We note that the perturbed C k half space has a fully curved part so that [v] vBM O µ,ν (Ω) is a norm. By definition, there always exists R h > 0 such that the support supp h ⊆ B R h (0 ′ ). We say that the perturbed C k half space R n h has small perturbation if where C * (n) is a specific constant depending only on the space dimension n, To simplify the behavior near the infinity, we consider vBM OL 2 (Ω) := vBM O µ,ν (Ω) ∩ L 2 (Ω).
Note that this space is independent of the choice of µ, ν. The purpose of this paper is to establish the Helmholtz decomposition for the space vBM OL 2 R n h in the case where R n h is a perturbed C 3 half space that has small perturbation with n ≥ 3. Here is our main theorem. Theorem 1. Let R n h be a perturbed C 3 half space of type (K) that has small perturbation with n ≥ 3 and Γ = ∂R n h . Then for any v ∈ vBM OL 2 R n h , there exists a unique decomposition v = v 0 + ∇q with v 0 ∈ vBM OL 2 σ R n h := f ∈ vBM OL 2 R n h div f = 0 in R n h , f · n = 0 on Γ , ∇q ∈ GvBM OL 2 R n h := ∇p ∈ vBM OL 2 R n h p ∈ L 2 loc R n h satisfying the estimate where C(K, R * , R h ) is a constant that depends only on the constant K which controls the second order derivative of h, the reach of the boundary R * and the size R h which characterizes the support of h. In particular, the Helmholtz projection P vBM OL 2 , defined by P vBM OL 2 (v) = v 0 , is a bounded linear map on vBM OL 2 R n h with range vBM OL 2 σ R n h and kernel GvBM OL 2 R n h .
Roughly speaking, the reach R * represents the size of a small neighborhood of the boundary within which every point has a unique projection on the boundary. Here we would like to direct the readers to subsection 2.1 for its precise definition.
Our strategy to prove Theorem 1 follows from the potential theoretic strategy we used to establish the Helmholtz decomposition in a bounded C 3 domain [14]. Let E represents the fundamental solution of −∆ in R n . By [21], we see that as long as the boundary Γ is uniformly C 2 , the space BM O ∞ (Ω) ∩ L 2 (Ω) allows the standard cut-off, i.e., we can decompose v into two parts v = v 1 + v 2 with v 2 = ϕv and v 1 = v − v 2 with some ϕ ∈ C ∞ (R n ) that is supported within a small neighborhood of Γ. Thus, the support of v 2 lies in a small neighborhood of Γ whereas the support of v 1 is away from Γ. For v 1 , by extending v 1 as zero outside Ω, we just set q 1 1 = E * div v 1 . Then, the L ∞ bound for ∇q 1 1 is well controlled near Γ, which yields a bound for b ν semi-norm. To estimate v 2 , we use a normal coordinate system near Γ and reduce the problem to the half space. We extend v 2 to R n so that the normal part (∇d · v 2 )∇d is odd and the tangential part v 2 − (∇d · v 2 )∇d is even in the direction of ∇d with respect to Γ. In such type of coordinate system, the minus Laplacian can be transformed as where η n is the normal direction to the boundary so that {η n > 0} is the half space. We then use a freezing coefficient method to construct volume potential q tan 1 and q nor 1 , which corresponds to the contribution from the tangential part v 2 tan and the normal part v 2 nor , respectively. Since the leading term of div v 2 nor in normal coordinate consists of the differential of η n only, if we extend the coefficient b ij even in η n , q nor 1 is constructed so that the leading term of ∇d · ∇q nor 1 is odd in the direction of ∇d. On the other hand, as the leading term of div v 2 tan in normal coordinate consists of the differential of η ′ = (η 1 , ..., η n−1 ) only, the even extension of b ij in η n gives rise to q tan 1 so that the leading term of ∇d · ∇q tan 1 is also odd in the direction of ∇d. Disregarding lower order terms and localization procedure, q tan 1 and q nor 1 are constructed as One is able to arrange BA −1 small by working in a small neighborhood of a boundary point. Then (I − BA −1 ) −1 is given as the Neumann series ∞ m=0 (BA −1 ) m . The BM O-BM O estimates for ∇q tan 1 and ∇q nor 1 follow from [9]. Since the leading term of ∇d · (∇q tan 1 + ∇q nor 1 ) is odd in the direction of ∇d with respect to Γ, the BM O bound implies b ν bound. The L 2 estimates for ∇q tan 1 and ∇q nor 1 hold as in the localization procedure, the partition of unity we consider for a small neighborhood of Γ is locally finite. As a result, setting q 1 = q 1 1 + q tan 1 + q nor 1 would give us our desired volume potential corresponding to div v. Some results needed for the construction of volume potential q 1 have already been established in [14,Section 3] and [21], for these parts we omit their proofs and recall them directly.
Theorem 2 (Construction of a suitable volume potential). Let Ω ⊂ R n be a uniformly C 3 domain of type (α, β, K) with n ≥ 2. Let R * be the reach of the boundary Γ = ∂Ω. Then, there exists a bounded linear operator v −→ q 1 from vBM OL 2 (Ω) to L ∞ (Ω) such that −∆q 1 = div v in Ω and that there exists a constant C = C(α, β, K, R * ) > 0 satisfying In particular, the operator v −→ ∇q 1 is a bounded linear operator in vBM OL 2 (Ω).
Here (α, β, K) are parameters that characterize the regularity of Γ. We would like to direct the readers to subsection 2.1 for their precise definitions. Although the construction of the suitable volume potential works for arbitrary uniformly C 3 domain in R n with n ≥ 2, for the rest of the theory we need to focus back on perturbed half space in R n with n ≥ 3. For v ∈ vBM OL 2 R n h , we observe that w = v − ∇q 1 is divergence free in R n h . Unfortunately, this w may not fulfill the trace condition w · n = 0 on the boundary Γ = ∂R n h . We construct another potential q 2 by solving the Neumann problem We then set q = q 1 + q 2 . Since ∂q 2 /∂n = ∇q 2 · n, v 0 = v − ∇q gives the Helmholtz decomposition.
To complete the proof of Theorem 1, it suffices to control Lemma 3 (Estimate of the normal trace). Let R n h be a perturbed C 2+κ half space of type (K) with κ ∈ (0, 1), n ≥ 3 and Γ = ∂R n h . Then, there is a constant HereḢ − 1 2 (Γ) is a Hilbert space that is isomorphic toḢ − 1 2 (R n−1 ), which turns out to be the dual space of the homogeneous fractional Sobolev spaceḢ 1 2 (R n−1 ). Here, the homogeneous Sobolev space of order s ∈ R is defined aṡ where S ′ (R n−1 ) denotes the space of Schwartz's tempered distributions and f denotes the Fourier transform of f , see e.g. [1,Section 1.3]. Unfortunately, the spaceḢ 1 2 (R n−1 ) is complete if and only if n ≥ 3. Thus, we assume that n ≥ 3 when theḢ − 1 2 norm appears. The basic idea to establish Lemma 3 is as follows. The L ∞ estimate of w ·n follows directly from the trace theorem established in [13,Theorem 22]. For theḢ − 1 2 estimate for w · n, we split the boundary into the straight part and the curved part. Since we have the L ∞ estimate for w · n and the curved part is compact, the contribution in theḢ − 1 2 estimate for w · n that comes from the curved part can be estimated by the L ∞ norm of w · n. For the contribution in theḢ − 1 2 estimate of w · n that comes from the straight part, we invoke theḢ − 1 2 estimate of w · n in the case of the half space. We finally need the estimate for the Neumann problem.
Lemma 4 (Estimate for the Neumann problem). Let R n h be a perturbed C 2 half space of type (K) that has small perturbation with n ≥ 3. Let R * be the reach of Γ = ∂R n h . For any g ∈ L ∞ (Γ) ∩Ḣ − 1 2 (Γ), there exists a unique (up to constant) solution u ∈ L 2 loc R n h to the Neumann problem ∆u = 0 in R n h ∂u ∂n = g on Γ (2) such that the operator g −→ u is linear. Moreover, there exists a constant C = C(K, .
To establish Lemma 4, the basic strategy is the same as in [14], we firstly show that represents the single layer potential for g. Since the boundary Γ = ∂R n h is curved and not compact, we appeal a perturbation argument. For g ∈ L ∞ (Γ), we decompose g into the curved part g 1 and the straight part g 2 by setting g 1 y ′ , h(y ′ ) : Since g 2 vanishes in the curved part of Γ, we can define g H 2 ∈ L ∞ (R n−1 ) ∩Ḣ − 1 2 (R n−1 ) by setting g H 2 (y ′ , 0) = g 2 y ′ , h(y ′ ) for any y ′ ∈ R n−1 . Note that for any x ∈ R n . By setting g R n 2 (y ′ , y n ) := g H 2 (y ′ , 0) for any (y ′ , y n ) ∈ R n , we can deduce the BM O estimate of ∇E * δ Γ ⊗ g 2 by applying the L ∞ − BM O estimate for singular integral operator [19,Theorem 4.2.7] to ∇∂ xn E * 1 R n + g R n 2 . Since g 1 · ′ , h(· ′ ) is compactly supported in R n−1 , we may extend g 1 to some g e 1,c ∈ L ∞ (R n ) such that g e 1,c is compactly supported and ∇d · ∇g e 1,c = 0 within a small neighborhood of Γ. Then, we can rewrite ∇E * δ Γ ⊗ g 1 as with some compactly supported continuous function f θ,ρ 0 /4 (see Proof of Lemma 25 (i)) that is independent of g. The BM O estimate for the first term on the right hand side of (3) can be controlled using the L ∞ − BM O estimate for singular integration operator. The second term on the right hand side of (3) can be controlled by the L ∞ norm as ∇E(x) is a locally integrable kernel and f θ,ρ 0 /4 has compact support. We thus obtain the BM O estimate for ∇E * δ Γ ⊗ g for g ∈ L ∞ (Γ).
For the b ν estimate of the normal component of ∇E * δ Γ ⊗ g with g ∈ L ∞ (Γ) ∩Ḣ − 1 2 (Γ), we also decompose g into the curved part g 1 and the straight part g 2 . It is possible to show that is compactly supported in R n−1 , estimate (4) allows us to show that for any x ∈ Γ R n ρ 0 with some constant C = C(K, R h , R * ) > 0. On the other hand, for x close to the curved part of Γ, ∇d(x) is not necessarily (0, ..., 0, 1), in this case would contain contributions from ∇ ′ E * (δ Γ ⊗ g 2 ) , which cannot be estimated by the L ∞ bound of g 2 (see Proposition 26). As a result, we introduce theḢ − 1 2 bound of g 2 . Since can be viewed as an element ofḢ 1 2 (Γ) for any x close to the curved part of Γ. Hence, by theḢ − 1 2 −Ḣ 1 2 duality we can deduce that We thus obtain an L ∞ estimate for the normal component of ∇E * δ Γ ⊗ g within a small neighborhood of Γ. The b ν estimate naturally follows.
For g ∈ L ∞ (Γ) ∩Ḣ − 1 2 (Γ), we prove that the trace of with C * (n) denoting a specific fixed constant which depends on dimension n only. Therefore, if R n h is a perturbed C 2 half space that has small perturbation with n ≥ 3, the inverse of I − 2S is well-defined as a bounded linear map from L ∞ (Γ)∩Ḣ − 1 2 (Γ) to L ∞ (Γ)∩Ḣ − 1 2 (Γ) by the Neumann series The solution to the Neumann problem (2) is formally given by We finally need the L 2 estimate for ∇u in R n h . In the case of a half space, for g ∈Ḣ − 1 2 (R n−1 ), the single layer potential E * δ ∂R n + ⊗g is exactly half of the solution u to the Neumann problem. By directly considering the partial Fourier transform of E(x ′ , x n ) with respect to x ′ , we could easily deduce that .
In the case that R n h is a perturbed C 2 half space with n ≥ 3, for g ∈ L ∞ (Γ) ∩Ḣ − 1 2 (Γ), we still decompose g into the curved part g 1 and the straight part g 2 . Since L 2n−2 n (Γ) is continuously embedded inḢ − 1 2 (Γ) and the curved part g 1 has compact support in Γ, g ∈ L ∞ (Γ) would imply that both g 1 , g 2 ∈ L ∞ (Γ) ∩Ḣ − 1 2 (Γ). Since for any x ∈ R n we have that and for any x = (x ′ , x n ) ∈ R n + we have that .
For the curved part g 1 , we recall the argument which establishes the vBM O ∞,ν norm for ∇E * δ Γ ⊗ g . We extend g 1 to g e 1,c ∈ L ∞ (R n ) and consider equation (3). Since ∇ div E is L p integrable for any 1 < p < ∞, see e.g. [18, Theorem 5.2.7 and Theorem 5.2.10], the L 2 norm of the first term on the right hand side of (3) can be estimated by the L ∞ norm of g. Whereas ∇E(x) is an integration kernel that is dominated by a constant multiple of |x| −(n−1) , by the famous Hardy-Littlewood-Sobolev inequality [1, Theorem 1.7], we can also control the L 2 norm of the second term on the right hand side of (3) by the L ∞ norm of g. Combine with the L 2 estimate for the contribution from the straight part g 2 , we finally obtain our desired L 2 estimate . This completes the proof of Lemma 4. When taking the normal trace and solving the Neumann problem, the reason why we need to require the dimension n to be greater than or equal to 3 is because when n ≥ 3, we indeed have the fact thatḢ . Based on these facts, we can show that in the case of any perturbed C 2 half space R n h with boundary Γ = ∂R n h ,Ḣ 1 2 (R n−1 ) is isomorphic toḢ . More importantly, we can estimate theḢ − 1 2 norm of the trace operator S by its L 2n−2 n norm and do cut-offs to boundary data g ∈ L ∞ (Γ) ∩Ḣ − 1 2 (Γ) to decompose it into the curved part g 1 and the straight part g 2 . In the case where n = 2, the spaceḢ 1 2 (R) is no longer complete andḢ − 1 2 (R) is not necessarily the dual space ofḢ 1 2 (R). The completion ofḢ 1 2 (R) cannot be embedded in the space of Schwartz's tempered distributions. Moreover, as a limit case,Ḣ 1 2 (R) is not continuously embedded in L ∞ (R). As a result, at the present we lack of tools to estimate theḢ − 1 2 norm of the trace operator S and we cannot do cut-offs to decompose a boundary data into the curved part and the straight part any more. This is why we focus on the case where the dimension n ≥ 3 in this paper. The problem ofḢ 1 2 (R) is similar toḢ 1 (R 2 ). The spaceḢ 1 (R 2 ) is not complete. Its completion should be the quotient space {u ∈ L 1 loc (R 2 ) | ∇u ∈ L 2 (R 2 ) n }/R as discussed in [11] sinceḢ 1 (R 2 ) includes all smooth compactly supported functions. We have to study an appropriate dual space as in [16]. This paper is organized as follows. In Section 2, we recall results from [21] to localize the problem and results from [14] to construct a suitable volume potential corresponding to div v. Theorem 2 is proved in this section. In Section 3, we take the normal trace inḢ − 1 2 sense by considering isomorphisms betweenḢ − 1 2 (Γ) andḢ − 1 2 (R n−1 ). In Section 4, we establish estimates for the trace operator S of Qg. We show that S is bounded from L ∞ ∩Ḣ − 1 2 to L ∞ ∩Ḣ − 1 2 . In Section 5, we solve Neumann problem (2) by considering the single layer potential with 2(I − 2S) −1 g and establish the vBM OL 2 estimate for its gradient in R n h .

Volume potential construction in a uniformly C 3 domain
For v ∈ vBM OL 2 (Ω), we shall construct a suitable potential q 1 so that v −→ ∇q 1 is a bounded linear operator in vBM OL 2 (Ω) as stated in Theorem 2. The construction in the case where Ω is a uniformly C 3 domain basically follows from the theory in [14], where Ω is a bounded C 3 domain.

Localization tools
Let us recall some uniform estimates established in [21]. Let Ω be a uniformly C k domain in R n with k ∈ N and n ≥ 2. Let Γ = ∂Ω denotes the boundary of domain Ω. There exists α, β > 0 such that for each z 0 ∈ Γ, up to translation and rotation, there exists a function denotes the open ball in R n−1 of radius α with center 0 ′ , that satisfies the following properties: We say that Ω is of type (α, β, K).
Let d denote the signed distance function from Γ which is defined by For a uniformly C k domain Ω, there is R Ω > 0 such that for x ∈ Ω with |d(x)| < R Ω , there is a unique point πx ∈ Γ such that |x − πx| = |d(x)|. The supremum of such R Ω is called the reach of Γ in Ω, we denote this supremum by R Ω * . Let Ω c be the complement of Ω in R n . Similarly, there is R Ω c > 0 such that for x ∈ Ω c with |d(x)| < R Ω c , we can also find a unique point πx ∈ Γ such that |x − πx| = |d(x)|. The supremum of such R Ω c is called the reach of Γ in Ω c , we denote this supremum by R Ω c * . We then define which we call it the reach of Γ. Moreover, d is C k in the ρ-neighborhood of Γ for any ρ ∈ (0, R * ), i.e., d ∈ C k Γ R n ρ for any ρ ∈ (0, R * ) with see e.g. [20, Chap. 14, Appendix], [24,Section 4.4].

Cut-off and extension
In this subsection, we assume that Ω ⊂ R n is a uniformly C 2 domain of type (α, β, K) with n ≥ 2. Let ρ ∈ (0, ρ 0 /2). For a function f defined in Γ R n ρ ∩ Ω, we define f even to be the even extension of f to Γ R n ρ with respect to the boundary Γ, i.e., and f odd to be the odd extension of f to Γ R n ρ with respect to the boundary Γ, i.e., For x ∈ Γ R n ρ , we further define that It is not hard to see that P (x) represents the normal projection to the direction ∇d whereas Q(x) represents the tangential projection to the direction ∇d.

Decomposition of volume potential corresponding to v
In this subsection, we assume that Ω ⊂ R n is a uniformly C 3 domain of type (α, β, K) with n ≥ 2. Let us recall some results that have already been established in [14]. There exists a constant ρ * = C(ρ 0 , K) > 0 such that all theories in this subsection hold for every ρ ∈ (0, ρ * ]. Let ρ ∈ (0, ρ * ] and θ ρ be the cut-off function defined in subsection 2.2. Since now we assume that Γ is uniformly To construct the mapping v → q 1 in Theorem 2, we localize v 2 by using the partition of the unity The corresponding volume potential to v 1 can be estimated directly. Proposition 7. There exists a constant C(ρ) > 0 such that for q 1 1 = E * div v 1 and v ∈ vBM OL 2 (Ω). In particular, for any ν < ρ/4.
Proof. By the BM O-BM O estimate [9] and Proposition 6, we have the estimate .
, the mean value property for harmonic functions implies that i.e., we can estimate |∇q 1 Since the convolution with ∇ 2 E is bounded in L p for any 1 < p < ∞, see e.g. [18, Theorem 5.2.7 and Theorem 5.2.10], we have that Therefore, the estimate Indeed, this extension is independent of the choice of For i ∈ N, we have that ϕ i ∈ C 2 (U ρ,i ) as in this case Γ is of regularity uniformly C 3 . For simplicity of notation, we denote (r Ω ϕ i ) even (v 2 ) even by v 2,i . By Proposition 5 and 6, we can easily deduce that for any We further denote Q v 2,i by w tan i . Now, we are ready to construct the suitable potential corresponding to with some p > n.
Having the estimate for the volume potential near the boundary regarding its tangential component, we are left to handle the contribution from v nor For simplicity of notations, for every i ∈ N we denote ∇d · (r Ω ϕ i ) even (v 2 ) odd by f 2,i . In this case, for any i ∈ N we have that f 2,i ∈ BM OL 2 (R n ) with supp f 2,i ⊂ U ρ,i satisfying the estimate We further denote f 2,i ∇d by w nor i . With a similar idea of proof, we can establish the suitable potential corresponding to v nor 2 .
with some p > n.

Remark 10. Specifically speaking, Proposition 8 is indeed [14, Proposition 4] whose proof is in [14, Section 3.4], Proposition 9 is indeed [14, Proposition 5] whose proof is in [14, Section 3.5].
For Proposition 8, in the local normal coordinate system at x i ∈ P Γ , p tan i,1 is constructed as [9] and in L p (1 < p < ∞) for any k, ℓ = 1, 2, ..., n, see e.g. [ is controlled by the BM O ∩ L 2 norm of v 2,i . On the other hand, p tan i,2 is constructed as the convolution of the Newton potential E with some function of v 2,i , we can directly estimate the L ∞ norm for ∇p tan i,2 by Hölder's inequality as ∇E is locally L p -integrable for p sufficiently close to 1. Similarly, for Proposition 9, p nor i,1 is constructed as If Ω is a general uniformly C 3 domain, then it is not necessary that vBM O ∞,∞ (Ω) is continuously embedded in L 1 (Ω) n . This is why we state Proposition 8 and Proposition 9 in a different form from [14].

Volume potential corresponding to v
We admit Proposition 8 and 9. The corresponding volume potential to v 2 can be constructed by summing up the cut-off of p tan i and p nor Proof of Theorem 2. Let i ∈ N. We firstly consider the contribution from the tangential part. Since Γ is uniformly C 3 , there exists a cut-off function θ i ∈ C 2 c (U 2ρ,i ) such that θ i = 1 in U ρ,i , 0 ≤ θ i ≤ 1 and moreover, the estimate holds for some constant C(ρ) > 0 independent of i. We next set

Note that Proposition 8 ensures that
Since our partition of unity for Γ R n ρ is locally finite, see subsection 2.1, we can deduce that where N * is the constant which characterizes the local finiteness of Q Γ in the sense that any element of Q Γ can intersect for at most N * other elements of Q Γ . Suppose that B is a ball of radius r(B) < ρ. If B does not intersect Γ R n 2ρ , then Lemma 11]. Hence in this case, we have that Hence, by estimate (10) we deduce that Note that supp θ i p tan i,2 ⊂ U 2ρ,i for any i ∈ N and for every x ∈ Γ R n 2ρ , x belongs to at most N * elements of Q Γ . By Proposition 8 again we can deduce that In addition, as Hence, by Proposition 5 we obtain that Let f tan By the same argument above which proves the estimate (11), we can show that holds for any 1 < s < ∞. Since ∇E is in L p ′ B 6ρ (0) for any 1 < p ′ < n/(n − 1), it follows that for all p > n. Thus, we have that Since the convolution kernel ∇E(x) is dominated by a constant multiple of |x| −(n−1) , by the well-known Hardy-Littlewood-Sobolev inequality, see e.g. [1, Theorem 1.7], we deduce that Hence by estimate (12), we see that Combine with estimate (11), we finally obtain that By Proposition 8, the control on the boundary with respect to q tan 1 is estimated by as the partition Q Γ is a locally finite open cover of Γ R n ρ . For the contribution coming from the normal component, we set in a similar way that By making use of Proposition 9 and repeating the whole argument above that treats the case for q tan 1 , we can prove that in the same way. Then the volume potential q 1 corresponding to v can be constructed as where q 1 1 is the volume potential defined in Proposition 7 corresponding to v 1 . By our construction, it can be easily seen that in Ω. We are done.

✷
3 Normal trace for a L 2 vector field with bounded mean oscillation Let R n h be a perturbed C 2+κ half space of type (K) with κ ∈ (0, 1) and n ≥ 3. Let R * > 0 denote the reach of boundary Γ = ∂R n h . We have already shown that there exists a constant C = C(α, β, K, R * ) > 0 such that the estimate [13,Theorem 22]. In this section, we would like to further estimate theḢ − 1 2 norm of the normal trace of v on Γ for w ∈ vBM OL 2 R n h . For simplicity of notations, from now on we define that for any δ > 0, Moreover, for h ∈ C 1 c (R n−1 ), we define the constant C s (h) := 1 + h C 1 (R n−1 ) .

Isomorphism betweenḢ
We would like to firstly give a characterization to the homogeneous fractional Sobolev spacė H 1 2 (R n−1 ) before we defineḢ 1 2 (Γ). Let us recall that if f ∈Ḣ 1 2 (R n−1 ), then f ∈ L 2 loc (R n−1 ) and the Gagliardo seminorm More precisely, if f ∈Ḣ 1 2 (R n−1 ), then it holds that with some constant C(n) that depends only on dimension n; see e.g. [ does not imply that f ∈Ḣ 1 2 (R n−1 ). Constant functions in R n−1 are typical counterexamples. Since we are considering the case where n ≥ 3, we have a very important property thatḢ 1 2 (R n−1 ) can be continuously embedded in L 2n−2 n−2 (R n−1 ), i.e., there exists a constant C(n) > 0 such that the estimate holds for any f ∈Ḣ 1 2 (R n−1 ), see e.g. [1, Theorem 1.38]. As a result, we expect thatḢ 1 2 (R n−1 ) can be identified with the spacė Fortunately, this expectation is indeed true. The Gagliardo seminorm [·]Ḣ 1 2 (R n−1 ) turns out to be a norm onĠ and it are equivalent for any f ∈ C ∞ c (R n−1 ). Hence, the spaceĠ , i.e., it holds as ρ → ∞ (This claim will be established within the proof of Proposition 17 which appears later in subsection 3.3). Therefore, by recalling that the space S 0 (R n−1 ) is dense inḢ 1 2 (R n−1 ) where S 0 (R n−1 ) denotes the subspace of S(R n−1 ) whose Fourier transform vanishes near the origin, see e.g. [1, Proposition 1.35], we can deduce thaṫ With respect to a function f defined on R n−1 , we could define a corresponding function f h that is defined on Γ by setting f h y ′ , h(y ′ ) := f (y ′ ), for all y ′ ∈ R n−1 .
Let the mapping f → f h be denoted by T h , i.e., f h = T h (f ). Since it is not that natural to consider the Fourier transform on a surface, we follow the characterization ofḢ 1 2 (R n−1 ) above to define the homogeneous fractional Sobolev spaceḢ The spaceḢ .
Without causing any ambiguity, we simply use the norm notation · Ḣ 1 .
Finally, we would like to note that Lemma 11. The mapping T h :Ḣ Proof. Let f ∈Ḣ 1 2 (R n−1 ). We naturally have that Hence, theḢ Let f h ∈Ḣ 1 2 (Γ). With respect to f h , we define that f (y ′ ) := f h y ′ , h(y ′ ) for any y ′ ∈ R n−1 . By the mean value theorem, we see that the estimate |h(x ′ ) − h(y ′ )| ≤ ∇ ′ h L ∞ (R n−1 ) |x ′ − y ′ | holds for any x ′ , y ′ ∈ R n−1 . As a result, for any x, y ∈ Γ we have that Hence, we deduce that Therefore, we obtain that ✷ As an application of Lemma 11, we have the following embedding result. Proof. Let f h ∈Ḣ 1 2 (Γ) and f = T −1 h (f h ). SinceḢ 1 2 (R n−1 ) is continuously embedded in L 2n−2 n−2 (R n−1 ), by estimate (15) we see that Let us further recall that a tempered distribution g is said to belong to the homogeneous Sobolev Proposition 13. Let ρ > 0 and ζ ∈ C 1 (R n−1 ) satisfies that ζ is identically a constant in B ρ (0 ′ ) c .
By the mean value theorem, it holds that |ζ( Therefore, by Hölder's inequality and estimate (13) once more, we finally have that .

Hence, by estimate (15) we have that
.
Proof. For f ∈Ḣ 1 2 (R n−1 ), we set for (x ′ , x n ) ∈ R n , i.e., u f is the inverse Fourier transform of e −|xn||ξ ′ | f (ξ ′ ) with respect to ξ ′ . By the Fourier-Plancherel formula, see e.g. [1, Theorem 1.25], we have that On the other hand, by the Fourier-Plancherel formula again, we see that which further implies that Therefore, we obtain that .
Letting ℓ n (f ) = u f for any f ∈Ḣ 1 2 (R n−1 ) completes the proof of Proposition 16. ✷ We start with the half space problem. If w ∈ L 2 (R n + ) n satisfies div w = 0 in R n + , then the normal trace w · n can be taken in theḢ − 1 2 sense.
Proposition 17. There exists a constant C(n) > 0 such that the estimate holds for any w ∈ L 2 (R n + ) n with div w = 0 in R n + .

.3], it holds that
for any ρ > 0 and w ∈ L 2 (R n + ) n with div w = 0 in R n + . Since we have already shown above that θ 2ρ (· ′ , 0)f converges to f inḢ 1 2 norm as ρ → ∞, by Proposition 16 we conclude that for any f ∈Ḣ 1 2 (R n−1 ) and w ∈ L 2 (R n + ) with div w = 0 in R n + , Remark 18. Instead, if the domain Ω ⊂ R n that we are considering is bounded C 2 , then for w ∈ L 2 (Ω) n satisfying div w = 0 in Ω, the normal trace w · n can be taken in the H − 1 2 sense, i.e., there exists a constant C, independent of w, such that for any w ∈ L 2 (Ω) n with div w = 0 in Ω; see e.g. [26], [28]. Now we are ready to consider the perturbed half space problem. For w ∈ vBM OL 2 R n h with div w = 0 in R n h , we show that the normal trace w · n can be taken in the L ∞ ∩Ḣ − 1 2 sense.
h . Then we set w 1 := ϕ * w and w 2 := w − w 1 . By Proposition 6, we see that w 1 , w 2 ∈ vBM OL 2 R n h satisfying Thus, for any y ∈ B ′ Γ (R h ), by picking an arbitrary x ′ ∈ R n−1 such that |x ′ | = R h and considering the mean value theorem, we have that As a result, we deduce that On the other hand, we define that Since w 2 = 0 in R n h ∩ B 1+τ 1 (h) (0), we have that w 2,H = 0 in R n + ∩ B 1+τ 1 (h) (0). By Proposition 17, we have that where n ∂R n + denotes the outward normal on the boundary ∂R n + of the half space R n + . Since T h w 2,H · n ∂R n + = w 2 · n Γ with n Γ denoting the outward normal on the boundary Γ of the perturbed half space R n h , by estimate (16) we obtain that

Estimates for some boundary integrals
Let E denotes the fundamental solution of −∆ in R n , i.e., E(x) := − log|x|/2π (n = 2), where b 1 (n) denotes the volume of the unit ball B 1 (0) in R n . In this section, we assume that R n h is a perturbed C 2 half space of type (K) with boundary Γ = ∂R n h . The purpose of this section is to establish several estimates for the trace operator of where ∂/∂n x denotes the exterior normal derivative with respect to x-variable.
Let us recall that for a perturbed C 2 half space R n h with h ∈ C 2 c (R n−1 ) that is not identically zero, R h > 0 represents the smallest positive real number such that supp h ⊆ B R h (0 ′ ).

Estimate for the normal derivative in y of E
To be specific, we give an estimate for the boundary integral of ∂E ∂ny (x − ·) for x ∈ R n h that is close to the boundary Γ. We have a compatible result with [14,Lemma 6], which deals with the case where the domain is bounded. For ρ ∈ (0, ρ 0 ], we define that Γ Ω ρ := Γ R n ρ ∩ Ω.
Lemma 19. Let R n h be a perturbed C 2 half space of type (K) with boundary Γ = ∂R n h , n ≥ 2 and ρ ∈ (0, ρ 0 ]. Then, it holds that Proof. (i) This follows from the Gauss divergence theorem. For a bounded piecewise C 1 domain D ⊂ R n , we have that We consider R > R h + |x ′ |. By applying the Gauss divergence theorem in D R , we deduce that The last term tends to zero naturally as R → ∞. For the first term, since n y is pointing straightly upward but x is located below (y ′ , y n ) y n = h L ∞ (R n−1 ) + 2|x n | , the kernel ∂E/∂n y (x − y) in this case is exactly the half of the Poisson kernel P δ h,xn (x ′ − y ′ ) with δ h,xn := h L ∞ (R n−1 ) + 2|x n | − x n . Hence, the first integral on the right hand side tends to − 1 2 as R → ∞. We therefore obtain (i).
(ii) Let us observe that where ω(y ′ ) = 1 + |∇ ′ h(y ′ )| 2 1/2 and ∇ ′ is the gradient in y ′ variables. This implies that We set that By the Taylor expansion, for |x ′ − y ′ | < 1 we have that We obtain that We decompose K into the sum of a leading term and a remainder term The term R is estimated as for any y ′ ∈ R n−1 , we have that y∈Γ, Since supp h ⊆ B R h (0 ′ ), the first two terms of estimate (25) can be estimated by the constant C(n)R n−1 h h C 1 (R n−1 ) . On the other hand, since the estimate The third term of estimate (25) can be controlled by the constant C(n) (ρK + ρ + 1) h C 1 (R n−1 ) + ρ . By (i), we observe that The term K 0 is very singular but it is positive for x ∈ Γ Ω ρ . Hence, we have that y∈Γ, where the constant C 19 (K, h, ρ) has the explicit expression Therefore, we finally obtain the estimate which holds for any x ∈ Γ Ω ρ . This completes the proof of Lemma 19. ✷ Before we end this subsection, we would like to give an estimate on the difference between gradients of the signed distance function near the boundary with explicit dependency on the boundary function h. This estimate plays an important role in later estimations of various boundary integrals.
Proposition 21. Let n ≥ 2 and ρ > 0. Suppose that f ∈ C 1 (R n−1 ) satisfies with some constants c 1 and c 2 independent of x ′ ∈ R n−1 . Then the estimate ρ n holds with some constant C(n) > 0 depending on n only.
Proof. By a direct calculation, we see that For y ′ ∈ B ρ (0 ′ ) and x ′ ∈ B 2ρ (0 ′ ) c , we have that |x ′ − y ′ | ≥ ρ. Hence, we can deduce that By symmetry, we also have that Hence, it is sufficient to estimate We then follow the similar idea that proves [15,Lemma 3.2]. Assume that |x ′ | ≤ |y ′ | and connect x ′ and y ′ by a geodesic curve in B |x ′ | (0 ′ ) c . Since the curve length is less than (π/2)|x ′ −y ′ |, by a fundamental theorem of calculus, we observe that Since the integrand of I is symmetric with respect to x ′ and y ′ , we now estimate To estimate I 1 , we observe that Thus, .

L ∞ estimate for the trace operator of Qg
For ρ ∈ (0, ∞), we let 1 ′ Bρ(0 ′ ) to be the characteristic function associated with the open ball B ρ (0 ′ ) in R n−1 , i.e., we define that For g ∈ L ∞ (Γ), we decompose g into the sum of the curved part g 1 and the straight part g 2 where for any x ′ ∈ R n−1 . Note that g 1 , g 2 ∈ L ∞ (Γ). With respect to g 2 , we further define g H 2 ∈ L ∞ (∂R n + ) by setting for x ′ ∈ R n−1 .
Theorem 22. Let R n h be a perturbed C 2 half space with boundary Γ = ∂R n h and n ≥ 3. Moreover, let us assume that h ∈ C 2 c (R n−1 ) satisfies the smallness condition Then, the boundary trace of Qg is of the form with some specific constant C * 0 (n) depending only on n and Proof. For x ∈ Γ Ω ρ 0 , we decompose g into the straight part g 2 and the curved part g 1 , i.e., Moreover, we further decompose I 2 (x) as ∇d(x) − ∇d(y) · ∇E(x − y) g 1 (y) dH n−1 (y) where P xn denotes the Poisson kernel. Let x tends x 0 on the boundary, in this case we have that I 1 (x) tends to 1 2 g H 2 (x 0 ), which is indeed 1 2 g(x 0 ). We then estimate I 2,1 (x 0 ) for x 0 ∈ Γ with |x ′ 0 | ≥ 2R h . By Proposition 20, I 2,1 (x 0 ) can be estimated as If |x ′ 0 | ≥ 3R h , then we have that |x ′ 0 − y ′ | ≥ R h . In this case, If |x ′ 0 | < 3R h , then in this case we have the estimate Hence, for x 0 ∈ Γ with |x ′ 0 | ≥ 2R h , we obtain that Next, we estimate Thus, for x 0 ∈ Γ with |x ′ 0 | ≥ 2R h , ∂E ∂n y (x 0 − y)g 1 (y) dH n−1 (y).
By estimate (24), we have that Hence, we obtain the estimate for |I 2,2 (x 0 )| for the case where |x ′ 0 | ≥ 2R h , i.e., we get that Suppose now that x ∈ Γ Ω ρ 0 with |x ′ | < 2R h . There exists a bounded C 2 domain Ω c ⊂ R n h such that ∂Ω c ∩Γ = B ′ Γ (2R h ). Let us recall a standard result concerning the double layer potential, see e.g. [22,Lemma 6.17]. Let f ∈ L ∞ (∂Ω c ), then the boundary trace of the double layer potential for w ∈ ∂Ω c . We define g c ∈ L ∞ (∂Ω c ) by letting Thus, for any z ∈ Ω c we have that I 2 (z) = I 2,1 (z) + P g c (z).
We next seek to control I 1,1 (x 0 ) for x 0 ∈ Γ with |x ′ 0 | < 2R h . Let θ 2 ∈ C ∞ c (R n−1 ) be a cut-off function such that 0 ≤ θ 2 ≤ 1 in R n−1 , θ 2 = 1 in B 1 (0 ′ ) and supp θ 2 ⊆ B 2 (0 ′ ). We then set that for y ′ ∈ R n−1 . Since we are assuming that 2R h < r h , it holds that Note that for any x 0 , y ∈ Γ such that x 0 = y, we have that Through some simple calculations, we can deduce by the mean value theorem that the estimate holds for any x 0 , y ∈ Γ and the estimate holds for any x 0 , y ∈ Γ with x 0 = y. In addition, for y ∈ Γ such that r h < |y ′ − x ′ 0 | < 2r h , we have that Hence, we see that the estimate holds for any x 0 , y ∈ Γ. By the duality relation betweenḢ 1 2 (R n−1 ) andḢ − 1 2 (R n−1 ), we see that I 1,1 (x 0 ) follows the estimate . By Proposition 13 and Proposition 21, we have that Since L 2n−2 n (R n−1 ) is continuously embedded inḢ − 1 2 (R n−1 ), by estimate (18) and estimate (20) we see that Therefore, the estimate for I 1,1 (x 0 ) reads as Since for I 1,2 (x 0 ) and I 1,3 (x 0 ), the integration region is bounded, I 1,2 (x 0 ) and I 1,3 (x 0 ) can be estimated in exactly the same way as I 2,1 (x 0 ) + I 2,2 (x 0 ) in the case where |x ′ 0 | < 2R h . As a result, here we directly give the estimate for I 1,2 (x 0 ) and I 1,3 (x 0 ) without going through what have already been done again. The estimate for I 1,2 (x 0 ) and I 1,3 (x 0 ) reads as This completes the proof of Theorem 22. ✷ 4.4Ḣ − 1 2 estimate for the trace operator S In this subsection, we assume that R n h is a perturbed C 2 half space with boundary Γ = ∂R n h and n ≥ 3. We shall derive theḢ − 1 2 estimate for the trace operator S from its L 2n−2 n estimate. We begin with the L p estimate for S.
Then, it holds that Sg ∈ L p (Γ) for any 1 ≤ p < ∞. For 1 < p < ∞, Sg satisfies the estimate with some specific constant C * 1 (n, p) > 0 that depends on n and p only. For p = 1, Sg satisfies the estimate with some specific constant C * 2 (n) > 0 that depends on n only and Proof. We firstly consider x ∈ Γ with |x ′ | < 3R h . Since we already have the L ∞ estimate for Sg on Γ according to Theorem 22, the estimate follows naturally, where the surface area Λ 3R h is estimated by Hence, for any 1 ≤ p < ∞, it holds that . Let 1 < p < ∞. We then consider x ∈ Γ with |x ′ | ≥ 3R h . For y ∈ Γ with |y ′ | < 2R h , the triangle inequality implies that |x ′ − y ′ | ≥ |x ′ | − 2R h . In this case, we deduce that Hence, we have that where the integral on the right hand side can be estimated as Therefore, we obtain that For x ∈ Γ with |x ′ | ≥ 3R h , we indeed have that ∇d(x) = (0, ..., 0, 1). Thus, Sg(x) has the form Since |x ′ − y ′ | ≥ |x ′ | − 2R h for any |y ′ | < 2R h , |Sg(x)| can thus be estimated as

✷
We are now ready to state theḢ − 1 2 estimate for the trace operator S.
with some specific constant C * 3 (n) > 0 that depends on n only.
Proof. Since L 2n−2 n (Γ) is continuously embedded inḢ − 1 2 (Γ), by considering estimate (20) and Lemma 23 with p = 2n−2 n , we obtain Corollary 24. ✷ We define constants and C * (n) := C * 0 (n) + C * 3 (n). We would like to emphasize that C * (n) is a specific constant that depends on dimension n only and C * (h) is a constant that depends on the boundary function h only. Theorem 22 and Corollary 24 guarantee that the trace operator S : Moreover, we can require C * (h) to be arbitrarily small by taking R h to be sufficiently small.
5 Neumann problem with bounded data in a perturbed C 2 half space with small perturbation We consider the Neumann problem for the Laplace equation in a perturbed C 2 half space in R n with L ∞ -initial data for n ≥ 3. We shall begin with the half space problem. It is well-known that a solution of the Neumann problem ∆u = 0 in R n is formally given by where N denotes the Neumann-Green function Its exterior normal derivative ∂N/∂n x for y n = 0 is nothing but the Poisson kernel with the parameter x n . By symmetry we observe that as x n > 0 tends to zero. Thus u gives a solution to (29) formally. The function is called the single layer potential of g.
For g ∈ L ∞ (R n−1 ), we let g(x ′ , x n ) := g(x ′ , 0) for any x ∈ R n . Natrually, g ∈ L ∞ (R n ). Let 1 R n + be the characteristic function associated with the half space R n + . In this case, we have that ∇E * (δ ∂R n + ⊗ g) = ∇∂ xn E * 1 R n + g.
Moreover, since −∂ xn (E * (δ ∂R n + ⊗ g)) is the half of the Poisson integral, i.e., holds for any g ∈ L ∞ (R n−1 ), see [14,Lemma 7]. We are able to to establish similar estimates for the case where the domain is a perturbed C 2 half space.
Lemma 25. Let R n h be a perturbed C 2 half space of type (K) with boundary Γ = ∂R n h and n ≥ 3. Then, holds with holds with For g ∈ L ∞ (Γ), the notation E * (δ Γ ⊗ g) in Lemma 25 means that

BMO estimate
We follow the idea of the proof for [14, Lemma 5 (i)], which establishes the same BM O estimate in the case where the domain is a bounded C 2 doamin.
A key observation is that div(g e 1,c ∇d) = g e 1,c ∆d + ∇d · ∇g e 1,c = g e 1,c ∆d + Thus, . By the L ∞ -BM O estimate for the singular integral operator [19,Theorem 4.2.7], the first term is estimated as for x ∈ R n with d(x, U c,ρ 0 /2 ) = inf y∈U c,ρ 0 /2 |x − y| < 1 we have that where For x ∈ R n with d(x, U c,ρ 0 /2 ) = inf y∈U c,ρ 0 /2 |x − y| ≥ 1, same estimate above for |I 2 (x)| holds trivially as |x − y| −(n−1) ≤ 1 for any y ∈ U c,ρ 0 /2 . The proof of the first part of Lemma 25 is now complete. ✷

L ∞ estimate for normal component
The L ∞ estimate to the normal component of ∇E * (δ Γ ⊗ g) within a small neighborhood of Γ for g ∈ L ∞ (Γ) ∩Ḣ − 1 2 (Γ) can be derived by almost the same argument as establishing the boundedness of the trace operator S from L ∞ (Γ) ∩Ḣ − 1 2 (Γ) to L ∞ (Γ) in Theorem 22.
By Proposition 20, we have that On the other hand, by Lemma 19 we have that Then, we deal with the modified straight part g * 2 . Let θ * ∈ C ∞ c (R n−1 ) be a cut-off function such that 0 ≤ θ * ≤ 1, θ * = 1 in B 1 2 +3R h (0 ′ ) and supp θ * ⊆ B 1+3R h (0 ′ ). Let x ∈ Γ Ω ρ 0 with |x ′ | < 3R h . We define that Note that ✷ We would like to emphasize that it is insufficient to obtain an L ∞ estimate for the normal component of ∇E * (δ Γ ⊗ g) in a small neighborhood of Γ if we only assume that g ∈ L ∞ (Γ). Let B be a ball centered at 0 with radius r B such that B ′ Γ (2R h ) ⊂ B/2. By almost the same argument as in the proof of Lemma 25 (ii), we can see that if x ∈ Γ Ω ρ 0 with |x ′ | ≥ r B /2, then we have the estimate ∇d(x) · ∇E * (δ Γ ⊗ g) (x) ≤ C(K, R * , R h ) g L ∞ (Γ) .
The main barrier comes from the contribution of g s in the case that |x ′ | < r B /2.
We assume the contrary of Proposition 26. Suppose that there exist a constant C ′ > 0 such that for any |x ′ | < r B /2 and g ∈ L ∞ (R n−1 ) with supp g ⊆ B r B (0 ′ ) c . As a consequence, the estimate holds for any g ∈ L ∞ (R n−1 ) where With respect to f ∈ L 1 (R n−1 ), we can define the adjoint operator of P by P * (f )(x ′ ) := φ r B (x ′ ) R * j * ψ r B /4 f (x ′ ), x ′ ∈ R n−1 .
Hence, by the Bolzano-Weierstrass theorem, there exists a sequence {t m } m∈N which converges to zero so that the sequence P * (f tm ) L 1 (R n−1 ) m∈N is convergent. By Fatou's lemma, we can then conclude that However, for |x ′ | ≥ 2r B we have that φ r B (x ′ )R * j (x ′ ) = x j /|x ′ | n , which is clearly not L 1 integrable in the region x ′ ∈ R n−1 |x ′ | ≥ 2r B . We reach a contradiction. ✷

L 2 estimate for the gradient of the single layer potential
We begin with the half space case.
Proof. We consider the partial Fourier transform of E with respect to x ′ , i.e., we let Since E(x ′ , x n ) is radial symmetric in R n−1 for any fixed x n > 0, E ′ (ξ ′ , x n ) can be calculated by the Hankel transform of order n−3 2 of the function r is the Bessel function of the first kind of order n−3 2 , see e.g. [17,Formula 6.565.2]. Then by the Fourier-Plancherel formula, we obtain that and ∂ xn E * δ ∂R n + ⊗ g 2 L 2 (R n .

✷
We then generalize this result to arbitrary perturbed C 2 half space R n h .
where The L 2 estimate of ∇E * δ Γ ⊗ g 2 in R n + follows directly from Proposition 27 and estimate (37). We have that .

Solution to the Neumann problem
Let R n h be a perturbed C 2 half space with boundary Γ = ∂R n h and n ≥ 3. We further assume that R n h has small perturbation, i.e., we require that the boundary function h ∈ C 2 c (R n−1 ) satisfies where C * (n) is a specific constant that depends on dimension n only. Under this setting, we are able to construct a solution to Neumann problem (2).
A simple check ensures that u satisfies Neumann problem (2) formally. It is sufficient to establish the vBM OL 2 estimate for ∇u. The vBM O ∞,ρ 0 -norm for ∇u in R n h is guaranteed by Lemma 25. By estimate (33) and estimate (34), we have that where C 25 (K, h, ρ 0 ) := C 25,i (h, ρ 0 ) + C 25,ii (K, h, ρ 0 ). The L 2 estimate of ∇u follows directly from Lemma 28, we have that .
This completes the proof of Lemma 4. ✷