Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 22, 1997, 445–464

UNIFORM DOMAINS OF HIGHER ORDER III

Pekka Alestalo and Jussi Väisälä

Helsingin yliopisto, Matematiikan laitos PL 4 (Yliopistonkatu 5), FIN-00014 Helsinki, Finland pekka.alestalo@helsinki.fi; jussi.vaisala@helsinki.fi

Abstract. We continue the study of (p, c)-uniform domains. Special emphasis is on (p, c)- NUD sets. The ambient space is assumed to be a Hilbert space.

1. Introduction

This paper is a continuation to [Al] and [AV], and we assume that the reader is familiar with these papers. They will be cited as I and II. For example, I.6.11 means the result 6.11 of [Al]. For basic notation and terminology, see I, p. 6, and II.1.4. However, we assume throughout this paper that the space E is a Hilbert space instead of a general normed space. This simplifies several proofs and also gives better estimates for various constants. The finite-dimensional case E = R^{n} is probably the most interesting.

The main emphasis will be on the null-sets (NUD) for homotopically (htop) and homologically (hlog) (p, c)-uniform domains in E. In Section 2 we give elementary estimates for the uniformity constants of certain standard domains. Section 3 deals with cartesian products of NUD sets. The case where the set is contained in a linear subspace of E is studied in Section 4. In Section 5 we consider compact sets in infinite-dimensional spaces and an application of our theory to bilipschitz spheres.

We let x · y denote the inner product of two vectors x, y ∈ E. For A ⊂ E and r > 0 we write

\overline{B}(A, r) = \{x : d(x, A) ≤ r\}, S(A, r) = \{x : d(x, A) = r\}.

Part of this article was written while the first author was visiting the University of Bielefeld in 1995–96. The hospitality of SFB 343 is hereby acknowledged.

1991 Mathematics Subject Classification: Primary Primary 30C65.

2. Estimates for uniformity constants

2.1. Summary of Section 2. We give estimates for the uniformity constants of various standard domains. All proofs are elementary. Remember that E is a Hilbert space. If E is infinite-dimensional, Theorem 5.2 will in certain cases give constants better than those obtained in this section.

2.2. Lemma. Each bounded set A ⊂ E is contained in a ball \overline{B}(a, r) with r = d(A)/\sqrt{2}.

Proof. If dim E = n < ∞, this follows from Jung’s theorem [Fe, 2.10.41]. Indeed, one can choose r = d(A)\sqrt{n/2(n + 1)}. The infinite-dimensional case was proved by J. Daneš [Da, Th. 2].

2.3. Theorem. Each ball and half space in E is htop and hlog (p, c)-uniform for all 0 ≤ p < dim E − 1 with c = \sqrt{3/2} = 1.22474 ….

Proof. Suppose first that G is a ball, which can be assumed to be the unit ball B(1). We show that G is htop (p, c)-uniform. Let f: S^{p} → G be continuous with d = d(|f|) > 0. If d ≥ 1, we let g: \overline{B}^{p+1} → E be the cone extension of f with g(0) = 0. The uniformity conditions are clearly satisfied with the constant 1. Assume that d < 1. By 2.2, there is a ball B = \overline{B}(a, d/\sqrt{2} ) containing |f|. We may choose a so that |a| ≤ 1.

Case 1. B ⊂ G. Let g: \overline{B}^{p+1} → E be the cone extension of f with g(0) = a. Each pair of points in |g| lies in a triangle with side lengths at most d, and hence d(|g|) ≤ d(|f|). If x ∈ |g|, then x ∈ [a, y] for some y ∈ |f|, and we get

d(x, |f|) ≤ |x − y| ≤ d(x, ∂B) ≤ d(x, ∂G).

Again, the uniformity conditions hold with the constant 1.

Case 2. B ⊄ G. Choose e ∈ ∂G with a ∈ [0, e], and set b = (1 − d)e. Then |b| = 1 − d < |a| and B ∩ G ⊂ \overline{B}(b, cd) with c = \sqrt{3/2}. Let g: \overline{B}^{p+1} → E be the cone extension of f with g(0) = b. Each pair of points in |g| lies in a triangle with side lengths at most cd, and hence d(|g|) ≤ cd.

To prove the lens condition, let x ∈ |g|. Then x = (1 − t)y + tb for some y ∈ |f| and 0 ≤ t ≤ 1. We have

|x| ≤ (1 − t)|y| + t|b| ≤ 1 − t + t(1 − d) = 1 − td,

and hence d(x, ∂G) = 1 − |x| ≥ td. This implies that

d(x, |f|) ≤ |x − y| = t|y − b| ≤ tdc ≤ cd(x, ∂G),

which is the lens condition.

Since the proof made use of solely cone extensions, it is valid with obvious modifications also in the homological case. The case where G is a half space follows from the above, since each compact set in G is contained in a ball B ⊂ G.

2.4. Theorem. Let B be a closed ball in E. Then the domain G = E \ B is htop and hlog (p, 5)-uniform for all 0 ≤ p < dim E − 1.

Proof. We may assume that B is the unit ball \overline{B}(1). We show that G is htop (p, 5)-uniform. Let f: S^{p} → G be continuous with d = d(|f|) > 0. We consider two cases.

Case 1. d ≥ 1. Set r = max{|fx| : x ∈ S^{p}}. If d < r − 1, then |f| is contained in a ball in G, and the desired extension of f is given by 2.3. Assume that d ≥ r − 1. Set R = r + \frac{1}{2}d, and let P: G → S(R) be the radial projection P x = Rx/|x|. Since p < dim E − 1, the map P f: S^{p} → S(R) is null-homotopic. This homotopy and the segmental homotopy from f to P f give an extension g: \overline{B}^{p+1} → G of f such that |g| is contained in the union of S(R) and all line segments [x, P x], x ∈ |f|. Since

d(|g|) ≤ 2R = 2r + d ≤ 2(d + 1) + d ≤ 5d,

the turning condition holds with the constant 5. To prove the lens condition let y ∈ |g|. If y ∈ [x, P x] for some x ∈ |f|, then

d(y, |f|) ≤ |y − x| ≤ |y| − 1 = d(y, B).

If y ∈ S(R), then

d(y, |f|) ≤ |y| + r = 2r + \frac{1}{2}d.

Since d(y, B) = r + \frac{1}{2}d − 1, d ≥ 1 and r ≥ 1, we get d(y, |f|) ≤ 5d(y, B), which is the lens condition.

Case 2. d ≤ 1. Applying 2.2 we choose a ball \overline{B}(a, d/\sqrt{2} ) containing |f|. In view of 2.3, we may assume that this ball meets B. Let L be the ray {ta : t ≥ 0}, and write

R = 1 + d\sqrt{2},  C = \{x : d(x, L) ≤ d|x|/\sqrt{2} \},  F = C ∩ S(R).

Let P: G → S(R) be the radial projection. There is an extension g: \overline{B}^{p+1} → G of f such that |g| is contained in the union of F and all line segments [x, P x], x ∈ |f|. It is easy to see that

|g| ⊂ C ∩ (\overline{B}(R) \ B) ⊂ \overline{B}(w, 2d),

where w is the center of the sphere S(R) ∩ ∂C. Hence d(|g|) ≤ 4d, which is the turning condition with the constant 4. To prove the lens condition let y ∈ |g|. The case where y ∈ [x, P x] is treated as in Case 1. If y ∈ F, then

d(y, |f|) ≤ d(|g|) ≤ 4d = 2\sqrt{2} d(y, B).

Hence the lens condition holds with the constant 2\sqrt{2} < 5. We have proved that G is htop (p, 5)-uniform.

The proof for the homological case is essentially the same. Let z be a pcycle in G and assume, for example, that d(|z|) = d ≤ 1. Choose again a ball \overline{B}(a, d/\sqrt{2} ) containing |z| and assume that this ball meets B. With the notation of Case 2 above, |z| is contained in the union A of F and all line segments [x, P x], x ∈ |z|. Since A is contractible, z = ∂g for some chain g in A. As in Case 2 above we see that g satisfies the uniformity conditions with the constant 5.

2.5. Theorem. The complement of a point x_{0} ∈ E is htop and hlog (p, 2)- uniform for all 0 ≤ p < dim E − 1.

Proof. We may assume that x_{0} = 0. We prove only the homotopical case. Let f: S^{p} → G = E \ {0} be continuous with d = d(|f|) > 0, and set r = max{|fx| : x ∈ S^{p}}. We may assume that d ≥ r, since otherwise |f| is contained in a ball in G and we can apply 2.3. Let P: G → S(r) be the radial projection. Then f has an extension g: \overline{B}^{p+1} → G such that |g| is contained in the union of S(r) and all line segments [x, P x], x ∈ |f|. We have d(|g|) ≤ 2r ≤ 2d, which is the turning condition. Let y ∈ |g|. If y ∈ [x, P x] for some x ∈ |f|, then d(y, |f|) ≤ |x − y| ≤ |y| = d(y, ∂G). If |y| = r, then d(y, |f|) ≤ 2r = 2d(y, ∂G).

2.6. Weak uniformity. We recall that an open set U ⊂ E is weakly hlog (p, c)-uniform if the uniformity conditions hold for every null-homologous p-cycle in U. Similarly, U is weakly htop (p, c)-uniform if each null-homotopic f: S^{p} → U has an extension g: \overline{B}^{p+1} → U satisfying the uniformity conditions.

2.7. Theorem. Let T be a closed proper linear subspace of E. Then U = E \ T is weakly htop and hlog (p, \sqrt{5} )-uniform for all 0 ≤ p < dim E − 1.

Proof. We prove the homotopical case. Let P: E → T and N: E → T^{⊥} denote the orthogonal projections. Suppose that f: S^{p} → U is null-homotopic. We may assume that 0 ∈ P |f|. We may also assume that |Nfx| ≤ d for all x ∈ S^{p}, since otherwise |f| is contained in a ball in U, and the desired extension is given by 2.3. Let S(d) be the sphere |x| = d in T^{⊥}, and let Q: T^{⊥} \{0} → S(d) be the radial projection Qx = dx/|x|. For x ∈ U consider the line segments I_{x} = [x, P x + QNx] and J_{x} = [P x + QNx, QNx]. Then U deformation retracts along these segments onto S(d). Thus QNf: S^{p} → S(d) is null-homotopic, and we obtain an extension g: \overline{B}^{p+1} → U of f such that |g| is contained in the union of S(d) and all I_{x} and J_{x}, x ∈ |f|.

Suppose that y, y′ ∈ |g|. Then P y ∈ [P x, 0] and P y′ ∈ [P x′, 0] for some x, x′ ∈ |f|. Since 0 ∈ P |f|, we get |P y − P y′| ≤ d(P |f|) ≤ d. Since |Ny − Ny′| ≤ 2d, this implies the turning condition d(|g|) ≤ d\sqrt{5}.

To prove the lens condition let y ∈ |g|. If y ∈ I_{x} for some x ∈ |f|, the lens condition holds with the constant 1. If y ∈ J_{x}, then

d(y, |f|) ≤ |x − y| ≤ d\sqrt{2} = \sqrt{2} d(y, T).

If y ∈ S(d) and x ∈ |f|, then

|x − y|^{2} = |P x − P y|^{2} + |Nx − Ny|^{2} ≤ d^{2} + 4d^{2} = 5d(y, T)^{2}.

Hence the lens condition holds with the constant \sqrt{5}.

2.8. Theorem. Let T be a closed linear subspace of E with dim T ≥ 1, and let S = S(1) ∩ T be the unit sphere of T. Then G = E \ S is weakly htop and hlog (p, 5)-uniform for all 0 ≤ p < dim E − 1.

Proof. We remark that the result follows with a larger constant from 2.7 and from the Möbius invariance of uniformity.

Let P: E → T and N: E → T^{⊥} be the orthogonal projections. For x ∈ E \ T^{⊥} we have x = P x + Nx = re + Nx for some r > 0 and e ∈ S, which are uniquely determined by x. We consider T^{⊥} × R as a Hilbert space in the natural way, and let H denote the half space T^{⊥} × [0, ∞). Then u = (Nx, r) ∈ H. The pair (u, e) gives the polar coordinates of x with respect to T. If x ∈ T^{⊥}, then u ∈ ∂H but e is undefined. We write u_{0} = (0, 1) ∈ H. In polar coordinates we have S = {(u, e) : u = u_{0}}.

We prove that G is htop (p, 5)-uniform. Let f: S^{p} → G be null-homotopic with d = d(|f|) > 0. We may assume that |f| ⊂ \overline{B}(S, d) = \{x : d(x, S) ≤ d\} and that |f| meets \overline{B}(1), since otherwise we can apply either 2.3 or 2.4. We consider three cases.

Case 1. d ≥ 1. Now |f| ⊂ \overline{B}(1 + d) ⊂ \overline{B}(2d). For x ∈ B(S, 1) \ S let Qx be the point in S(S, 1) closest to x. Explicitly, we can write in polar coordinates

Q(u, e) = (u_{0} + \frac{u − u_{0}}{|u − u_{0}|}, e).

We extend Q to E \ S by setting Qx = x for x ∈ A = \overline{B}(2d) \ B(S, 1) and Qx = 2dx/|x| for |x| ≥ 2d. Then E \ S deformation retracts onto A along the line segments [x, Qx]. Hence Qf is null-homotopic in A, and we obtain an extension g: \overline{B}^{p+1} → G of f such that |g| is contained in the union of A and all segments [x, Qx], x ∈ |f|. Then d(|g|) ≤ d(\overline{B}(2d)) = 4d, which gives the turning condition.

To prove the lens condition let y ∈ |g|. The case y ∈ [x, Qx] is again clear. Suppose that y ∈ A. Since |f| meets \overline{B}(1), we have d(y, |f|) ≤ |y| + 1. If |y| ≤ 2, this implies that d(y, |f|) ≤ 3 ≤ 3d(y, S). If |y| ≥ 2, we also get d(y, |f|) ≤ |y| + 1 ≤ 3(|y| − 1) ≤ 3d(y, S).

Case 2. 1/\sqrt{2} ≤ d ≤ 1. We use the same idea as in Case 1, but we let now A = \overline{B}(1+d)\B(S, d). We get an extension g with d(|g|) ≤ 2+2d ≤ (2\sqrt{2}+2)d < 5d. If y ∈ A, then

d(y, |f|) ≤ |y| + 1 ≤ 2 + d ≤ (2\sqrt{2} + 1)d < 4d ≤ 4d(y, S).

Case 3. d ≤ 1/\sqrt{2}. Applying 2.2 we choose a ball \overline{B}(a, d/\sqrt{2} ) containing |f|. We may assume that there is a point b ∈ S ∩ \overline{B}(a, d/\sqrt{2} ), since otherwise |f| is contained in a ball in G. Then |f| ⊂ \overline{B}(b, d\sqrt{2} ). Set α = arcsin d\sqrt{2} and

Z = {(u, e) : |u − u_{0}| ≤ d, ang(e, b) ≤ α}

in polar coordinates; here ang(e, b) is the angle between the unit vectors e and b. In the case E = R^{3} , T = R^{2} , Z is obtained from the solid torus \overline{B}(S, d) by cutting it by two half planes whose boundary is the x_{3} -axis and which touch the ball \overline{B}(b, d\sqrt{2} ).

We have |f| ⊂ (\overline{B}(S, d) ∩ \overline{B}(b, d\sqrt{2} )) \ S ⊂ Z \ S. Since Ė \ S deformation retracts onto Z \S, f is null-homotopic in Z \S. Let Q: \overline{B}(S, d)\S → S(S, d) be as in the previous cases. Then Z \ S deformation retracts onto W = S(S, d) ∩ Z along the segments [x, Qx], and Qf is null-homotopic in W. Hence we obtain an extension g: \overline{B}^{p+1} → G of f such that |g| is contained in the union of W and all line segments [x, Qx], x ∈ |f|. We show that g satisfies the uniformity conditions.

The set S ∩ Z is a cap of S with diameter 2 sin α = 2d\sqrt{2}. Since Z ⊂ \overline{B}(S ∩ Z, d), this gives the turning condition d(|g|) ≤ d(Z) ≤ 2d\sqrt{2} + 2d < 5d.

To prove the lens condition, it suffices to consider the case y ∈ W, and then d(y, |f|) ≤ d(Z) < 5d = 5d(y, S).

2.9. Theorem. Let Z be the infinite cylinder B(1) × R ⊂ E × R. Then Z is htop and hlog (p, \sqrt{2} )-uniform for all 1 ≤ p < dim E − 1.

Proof. Of course, we use the natural inner product in E × R; see 3.5. Let f: S^{p} → Z be continuous with d = d(|f|) > 0. If d ≤ 1, we choose a ball B = B(a, d/\sqrt{2} ) in E × R containing |f|. We may assume that B does not lie in Z and that a ∈ E × {0} with |a| ≤ 1. The desired extension of f is obtained by coning with vertex at (1 − d)a. If d ≥ 1, let Q: E × R → {0} × R be the projection. The extension is now obtained by using a segmental homotopy from f to Qf and a homotopy of Qf to a point in the segment Q|f|. We omit the estimates for the uniformity constants, because they are rather similar to those in the previous theorems. In the homological case we can apply similar constructions to each component of the carrier of a given p-cycle.

2.10. Remark. Observe that Z is not (0)-uniform.

3. Product sets

3.1. Summary of Section 3. Recall that an open set U is p-acyclic if it has trivial (reduced) homology groups H_{k}(U) for 0 ≤ k ≤ p. Suppose that A ⊂ R^{m} and B ⊂ R^{n} are closed sets without interior points. Assume that R^{m} \ A is pacyclic and that R^{n} \ B is q -acyclic for some p ≤ m − 2 and q ≤ n − 2. It follows from the Alexander duality and from the Künneth formula for cohomology that R^{m+n} \ (A × B) is (p + q + 2)-acyclic. In fact, one can show that it is also simply connected and hence, by the Hurewicz theorem, it is even (p + q + 2)-connected. In this section we shall consider quantitative versions of this and related results.

3.2. The homotopical case. Recall that a closed set A ⊂ E is htop (p, c)- porous if for all r > 0 and for all maps f: Δ^{p} → E there is a map g: Δ^{p} → E such that ∥g − f∥ ≤ r and d(|g|, A) ≥ r/c. This implies that A is htop (q, c)-porous for all 0 ≤ q ≤ p; see I.3.3.3.

We shall prove in 3.5 that the product of a htop (p)-porous set and a htop (q)-porous set is htop (p + q + 1)-porous. We need the following result, which shows that one can replace Δ^{p} in the definition of porosity by any p-dimensional polyhedron if the constant c is allowed to change.

3.3. Lemma. Suppose that A ⊂ E is htop (p, c)-porous and that P is a p-dimensional polyhedron. Suppose also that f: P → E is continuous and that r > 0. Then there is a map g: P → E such that

(1) ∥g − f∥ ≤ r,

(2) d(|g|, A) ≥ r/c_{1} , where c_{1} = c_{1}(c, p).

Proof. Choose a triangulation K of P such that d(fΔ) ≤ r′ = r/6^{p} for every Δ ∈ K. Let K^{k} be the k -skeleton of K and set f_{k} = f | |K^{k}|. Since A is htop (0, c)-porous, we can define g_{0}: |K^{0}| → E such that |g_{0}v − fv| ≤ r′ and d(g_{0}v, A) ≥ r′/c for each vertex v of K. Proceeding inductively, assume that 0 ≤ k < p and that we have defined a map g_{k}: |K^{k}| → E satisfying the following two conditions:

(α_{k})  ∥g_{k} − f_{k}∥ ≤ 6^{k}r′,
(β_{k})  d(|g_{k}|, A) ≥ r′/c(c + 2)^{k}.

Let Δ ∈ K be a (k + 1)-simplex. Using the cone construction we extend g_{k} | ∂Δ to a map u_{Δ}: Δ → E such that d(|u_{Δ}|) = d(g_{k}∂Δ). Applying I.3.4 and (β_{k}) we find an extension g_{Δ}: Δ → E of g_{k} | ∂Δ such that

∥g_{Δ} − u_{Δ}∥ ≤ r′/(c + 2)^{k+1}, d(|g_{Δ}|, A) ≥ r′/c(c + 2)^{k+1}.

Setting g_{k+1} | Δ = g_{Δ} we obtain a map g_{k+1}: |K^{k+1}| → E satisfying (β_{k+1}). To prove (α_{k+1}), observe that

\begin{aligned}d(|g_{Δ}|) & ≤ d(|u_{Δ}|) + 2∥g_{Δ} − u_{Δ}∥ ≤ d(g_{k}∂Δ) + 2r′/(c + 2)^{k+1} \\ & ≤ d(f∂Δ) + 2∥f_{k} − g_{k}∥ + r′ ≤ r′ + 2 · 6^{k}r′ + r′ ≤ 4 · 6^{k}r′. \end{aligned}

Let x ∈ Δ and choose a vertex y of Δ. Then

\begin{aligned}|g_{k+1}x − fx| & ≤ |g_{k+1}x − g_{k+1}y| + |g_{k+1}y − fy| + |fy − fx| \\ & ≤ d(|g_{Δ}|) + ∥g_{0} − f_{0}∥ + d(fΔ) ≤ 4 · 6^{k}r′ + r′ + r′ ≤ 6^{k+1}r′. \end{aligned}

Hence (α_{k+1}) is true.

The last step gives a map g = g_{p}: P → E satisfying (1) and (2) with c_{1} = 6^{p}c(c + 2)^{p}.

3.4. Remark. We consider the special case of 3.3 where dim E ≥ p + 1 and A consists of a single point, say A = {0}. Then g can be constructed directly as follows: Let 0 < ε < r. Choose a piecewise linear map f_{1}: P → E with ∥f_{1} − f∥ < ε. By general position, we may assume that 0 ∉ |f_{1}|. Let P be the radial retraction of E \ {0} onto E \ B(r − ε). Then g = P f_{1} satisfies (1) and (2) of 3.3 with a constant c_{1} arbitrarily close to one. In particular, c_{1} does not depend on p.

3.5. Preparations. In the product E_{1} × E_{2} of two inner products spaces we use the inner product

(x, y) · (x′, y′) = x · x′ + y · y′.

We recall the concept of a dual skeleton. Let K be a finite simplicial complex of dimension n. For 0 ≤ p ≤ n, the dual skeleton D(K^{p}) of the p-skeleton K^{p} of K is the subcomplex of the barycentric subdivision Sd K of K consisting of all simplexes that do not meet |K^{p}|. The dimension of D(K^{p}) is n − p − 1. Thus D(K^{n}) = ∅, and D(K^{n−1}) is the finite set of the barycenters b(Δ) of all n-simplexes Δ ∈ K. If Δ is an n-simplex of K, the (n − p − 1)-simplexes of D(K^{p}) contained in Δ are of the form [b(σ_{p+1}), …, b(σ_{n})], where σ_{n} = Δ and each σ_{i} is an i-face of σ_{i+1} . The most important property of the dual skeleton is that given 0 ≤ p ≤ n, each n-simplex Δ′ ∈ Sd K can be uniquely written as Δ′ = σ ∗ τ, where σ ∈ Sd K^{p} and τ ∈ D(K^{p}). Here σ ∗ τ denotes the join of two joinable simplexes σ and τ; see [Ru, p. 10].

3.6. Theorem. Let A_{1} ⊂ E_{1} be htop (p, c)-porous and let A_{2} ⊂ E_{2} be htop (q, c)-porous. Then A_{1} × A_{2} is htop (p + q + 1, c_{1})-porous in E = E_{1} × E_{2} with c_{1} = c_{1}(c, p, q).

Proof. Let r > 0 and let f: Δ^{p+q+1} → E be a map. Choose a triangulation K of Δ^{p+q+1} such that d(fΔ) ≤ \frac{1}{4}r for all Δ ∈ K. Write f = (f_{1}, f_{2}) with f_{i}: Δ^{p+q+1} → E_{i}, P_{1} = |K^{p}| and P_{2} = |D(K^{p})|. By 3.3 we find maps h_{i}: P_{i} → E_{i} such that

∥h_{i} − f_{i}|P_{i}∥ ≤ r/4, d(|h_{i}|, A_{i}) ≥ r/4c_{2}

for i = 1, 2, where c_{2} = c_{2}(c, p, q).

Let Δ be a (p + q + 1)-simplex of Sd K. Then Δ can be uniquely expressed as σ ∗ τ, where σ ∈ SdK^{p} and τ ∈ D(K^{p}). Each point z ∈ Δ can be written as z = (1 − t)x + ty with x ∈ σ, y ∈ τ, 0 ≤ t ≤ 1. The representation is unique except for z ∈ σ ∪ τ, which occurs only if t = 0 or t = 1. We define a map g_{Δ}: Δ → E as follows: For x ∈ σ, y ∈ τ we set

g_{Δ}x = (h_{1}x, f_{2}x),  g_{Δ}y = (f_{1}y, h_{2}y),  g_{Δ}(\frac{1}{2}(x + y)) = (h_{1}x, h_{2}y).

On the line segments [x, \frac{1}{2}(x + y)] and [\frac{1}{2}(x + y), y], g_{Δ} is defined to be affine. Explicitly, for z = (1 − t)x + ty we set

g_{Δ}z = \{ \begin{matrix}(h_{1}x, (1 − 2t)f_{2}x + 2th_{2}y) & if 0 ≤ t ≤ \frac{1}{2}, \\ ((2 − 2t)h_{1}x + (2t − 1)f_{1}y, h_{2}y) & if \frac{1}{2} ≤ t ≤ 1. \end{matrix}

The maps g_{Δ} clearly define a map g: Δ^{p+q+1} → E. We show that this is the desired map.

Let z = (1 − t)x + ty be as above. If 0 ≤ t ≤ \frac{1}{2} , then

|h_{1}x − f_{1}z| ≤ |h_{1}x − f_{1}x| + |f_{1}x − f_{1}z| ≤ \frac{1}{4}r + d(fΔ) ≤ \frac{1}{2}r,
|(1 − 2t)f_{2}x + 2th_{2}y − f_{2}z| ≤ (1 − 2t)|f_{2}x − f_{2}z| + 2t|h_{2}y − f_{2}z|
≤ (1 − 2t)d(fΔ) + 2t|h_{2}y − f_{2}y| + 2t|f_{2}y − f_{2}z|
≤ (1 − 2t)d(fΔ) + \frac{1}{4}r + 2td(fΔ) ≤ \frac{1}{2}r.

These inequalities imply that |gz − fz| ≤ r/\sqrt{2} < r. Similar arguments show that this is true also if \frac{1}{2} ≤ t ≤ 1. Thus ∥g − f∥ ≤ r.

Consider again the point z = (1 − t)x + ty ∈ Δ = σ ∗ τ. If 0 ≤ t ≤ \frac{1}{2} , we have

d(gz, A_{1} × A_{2}) ≥ d(h_{1}x, A_{1}) ≥ r/4c_{2}.

If \frac{1}{2} ≤ t ≤ 1, then

d(gz, A_{1} × A_{2}) ≥ d(h_{2}y, A_{2}) ≥ r/4c_{2}.

Hence d(|g|, A_{1} × A_{2}) ≥ r/4c_{2} , and we have proved the theorem with c_{1} = 4c_{2} = 4 · 6^{s}c(c + 2)^{s}, s = p ∨ q.

Recall from II.1.4 that a set A ⊂ E has a property Q(p) involving an integer p ≥ 0 completely if A has property Q(k) for all 0 ≤ k ≤ p. For example, A is completely htop (p, c)-NUD if A is htop (k, c)-NUD for all 0 ≤ k ≤ p. By I.4.9, this is quantitatively equivalent to htop (p + 1, c)-porosity. Hence Theorem 3.6 has the following corollary:

3.7. Theorem. Let A_{1} ⊂ E_{1} be completely htop (p, c)-NUD and let A_{2} ⊂ E_{2} be completely htop (q, c)-NUD. Then A_{1}×A_{2} is completely htop (p+q+2, c_{1})- NUD in E_{1} × E_{2} with c_{1} = c_{1}(c, p, q).

3.8. Remark. We consider the special case of 3.6 where dim E_{2} ≥ q + 1 and A_{2} consists of a single point, say A_{2} = {0}. Let c_{0} > 1. By 3.4 we can choose the map h_{2} in the proof of 3.6 so that d(|h_{2}|, 0) ≥ r/c_{0} . It follows that A_{1} × {0} is htop (p + q + 1, c_{1})-porous in E with c_{1} = c_{1}(c, p) = 4 · 6^{p}c(c + 2)^{p}.

3.9. The homological case. It is natural to conjecture that 3.6 and 3.7 are true if htop is replaced by hlog. In fact, we believe that they are true in the stronger form where the hlog properties of A_{1} and A_{2} imply the corresponding htop property of A_{1} × A_{2} . We prove this for finite-dimensional spaces:

3.10. Theorem. Let A_{1} ⊂ R^{m} be completely hlog (p, c)-NUD and let A_{2} ⊂ R^{n} be completely hlog (q, c)-NUD. Then A_{1} × A_{2} is completely htop (p + q + 2, c_{1})-NUD in R^{m+n} with c_{1} = c_{1}(c, m, n).

Proof. The proof is based on compact families of sets, and it does not give an explicit estimate for the constant c_{1} . Since the proof is rather long, we give only a sketch. The topological fact behind the proof is the inequality dim(A_{1} × A_{2}) ≤ dim A_{1} + dim A_{2} for the topological dimension.

Part 1. The compactness theory of [Vä_{2} ] cannot be directly applied, because we cannot allow rotations in R^{m} × R^{n}. We must therefore rewrite part of this theory replacing the group sim of all similarities of R^{n} by the subgroup sim^{∗} consisting of all maps f: R^{n} → R^{n} of the form fx = λx + a, λ > 0, a ∈ R^{n}.

Fix an integer n ≥ 1. As before, we let K^{n} denote the family of all nonempty compact subsets of Ṙ^{n}, and we write K_{∞}^{n} = {A ∈ K^{n} : ∞ ∈ A}. If L ⊂ K^{n}, we set L^{∗} = {A ∈ L : 0 ∈ ∂A and S^{n−1} ∩ ∂A ≠ ∅}. We say that L is ∗-stable if sim^{∗}L = L and if L^{∗} is compact. For H ⊂ K^{n}, we let σ^{∗}(H) denote the union of all ∗-stable subfamilies of H. Compared with the theory of [Vä_{2} ], a stable family is always ∗-stable, and we have σ(H) ⊂ σ^{∗}(H). A ∗-filtration of σ^{∗}(H) is a function c ↦ L_{c}, defined for c ≥ 1, such that

(1) c < d implies L_{c} ⊂ L_{d},

(2) each L_{c} is contained in a ∗-stable subfamily of H,

(3) each ∗-stable subfamily of H is contained in some L_{c}.

In this case, σ^{∗}(H) = ∪{L_{c} : c ≥ 1}.

We next give the following modification of I.6.4: Let n ≥ 2, 0 ≤ p ≤ n − 1, and let M ⊂ K_{∞}^{n} be a ∗-stable family such that the reduced Čech cohomology groups H^{n−p−1}(A) and H^{n−1}(A) are trivial for all A ∈ M. Then for each c ≥ 1 there is c_{1} = c_{1}(c, M) ≥ 1 such that R^{n} \A satisfies the condition HT^{∗}(p, c, c_{1}) of II.3.4 for all A ∈ M. The proof of this is obtained by modifying the proof of I.6.4.

Using this result and II.3.10, we can show as in I.6.6 that, with the notation of II.3.16, we have σ^{∗}(D_{p}) = HU(p) and that c ↦ HU(p, c) is a ∗-filtration of σ^{∗}(D_{p}). Furthermore, with the notation of II.3.16 we have σ^{∗}(L_{n−p−2}) = HN(p), and c ↦ HN(p) gives a ∗-filtration of σ^{∗}(L_{n−p−2}). Indeed, I.6.10 implies that HN(p, c) ⊂ σ^{∗}(L_{n−p−2}), and the opposite inclusion follows rather easily from [HW, VIII 4F]. With the convention of II.3.10, the statement is also valid for p = −1; then one may have n = 1.

Part 2. We turn to the proof of 3.10. Since htop and hlog (0)-uniformity coincide, the case p ≤ 0, q ≤ 0 is a special case of 3.7. We may thus assume that p ≥ 1. From 3.7 it follows that A_{1} × A_{2} is htop (1, c_{1})-NUD. By the quantitative Hurewicz theorem II.4.5, it suffices to prove that A_{1} × A_{2} is completely hlog (p + q + 2, c_{2})-NUD with c_{2} depending only on (c, m, n).

Let M be the family of all sets A_{1} × A_{2} where A_{1} and A_{2} satisfy the hypotheses of 3.10. It is rather easy to see that \overline{M} is ∗-stable. We next show that dim F ≤ m + n − p − q − 4 for all F ∈ \overline{M}. Indeed, F is of the form B_{1} × B_{2} , where B_{1} and B_{2} are limits of sets satisfying the hypotheses of 3.10. From I.6.3 and from I.6.10 it follows that dim B_{1} ≤ m − p − 2 and dim B_{2} ≤ n − q − 2. By [HW, p. 33] this implies that dim F ≤ dim B_{1} + dim B_{2} ≤ m + n − p − q − 4.

We have proved that \overline{M} is a ∗-stable family in L_{m+n−p−q−4} . By Part 1, M is contained in some HN(p + q + 2, c_{2}), and the theorem is proved.

Theorem 3.10 can also be expressed in terms of hlog porosity, defined in II.6.2. By II.6.15, the complete hlog (c, p)-NUD property is quantitatively equivalent to complete hlog (c, p + 1)-porosity. In view of I.4.9, Theorem 3.10 has the following corollary:

3.11. Theorem. Let A_{1} ⊂ R^{n} be completely hlog (p, c)-porous and let A_{2} ⊂ R^{n} be completely hlog (q, c)-porous. Then A_{1} × A_{2} is htop (p + q + 1, c_{1})- porous in R^{m+n} with c_{1} = c_{1}(c, m, n).

4. Subsets and changing dimension

4.1. Summary of Section 4. We first show that the htop and hlog porosity and complete uniformity properties of a set A ⊂ E are inherited by closed subsets. Next we consider the case where A is contained in a linear subspace E_{1} of E, and we compare the properties of A in E_{1} and in E. As a corollary we obtain results on planar sections of A.

4.2. Theorem. Suppose that A is htop or hlog (p, c)-porous in E. Then each closed subset of A has the same property.

Proof. This follows at once from the definition of porosity.

4.3. Theorem. Suppose that A is completely htop or hlog (p, c)-NUD in E. Then each closed B ⊂ A is completely htop or hlog (p, c′)-NUD for all c′ > c, respectively.

Proof. We remark that in view of I.4.9 and II.6.15, we obtain from 4.2 the weaker result where c′ depends on p and c.

We first consider the homotopical case. If suffices to show that B is htop (p, c′)-NUD. Let f: S^{p} → E \ B be a map and let ε > 0. Since A is completely htop (p, c)-NUD, A is LC^{p} rel E; see II.5.2 and II.5.3.1. By II.5.4 there is a map f_{1}: S^{p} → E \ A with ∥f_{1} − f∥ < ε. Since A is htop (p, c)-NUD, there is an extension g_{1}: \overline{B}^{p+1} → E \ A of f_{1} satisfying the uniformity conditions in E \ A. Let e ∈ S^{p} and 0 ≤ t ≤ 1. Setting

g(te) = \{ \begin{matrix}(2t − 1)f(e) + (2 − 2t)f_{1}(e) & for \frac{1}{2} ≤ t ≤ 1, \\ g_{1}(2te) & for 0 ≤ t ≤ \frac{1}{2}, \end{matrix}

we obtain an extension g: \overline{B}^{p+1} → E \ B of f. It is easy to see that g satisfies the uniformity conditions in E \ B with a constant c′ = c′(c, ε) such that c′ → c as ε → 0. This proves the homotopical case of the theorem.

The homological case is proved similarly, using II.5.10 instead of II.5.4. Indeed, let z be a p-cycle in E and let ε > 0. Then II.5.10 gives z_{1} ∈ Z_{p}(\overline{B}(|z|, ε) \ A) and g_{1} ∈ S_{p+1}(\overline{B}(|z|, ε)) such that ∂g_{1} = z − z_{1} . Since A is hlog (p, c)- NUD, z_{1} = ∂g_{2} for some g_{2} satisfying the uniformity conditions in E \ A. Then z = ∂(g_{1} + g_{2}), and g_{1} + g_{2} satisfies the uniformity conditions in E \ B with a constant c′ arbitrarily close to c.

4.4. Raising dimension. Suppose that E_{1} is a closed linear subspace of E and that A is a closed subset of E_{1} . We want to compare the properties of A in E_{1} and in E. It turns out that, roughly speaking, raising the dimension by one improves the order of porosity and NUD by one. An early example of this phenomenon was given in 1982 by S. Granlund, P. Lindqvist and O. Martio [GLM, 4.18], who observed that a closed set A ⊂ S^{n} is (0)-porous if and only if it is (0)-NUD in R^{n+1} .

The converse problem is discussed in 4.6.

4.5. Theorem. Suppose that E is a Hilbert space and that E_{1} is a closed linear subspace of E of codimension at least k ≥ 1. Let A be closed in E_{1} .

(1) If A is htop (p, c)-porous in E_{1} , then A is htop (p + k, c_{1})-porous in E with c_{1} = c_{1}(c, p).

(2) If A is completely htop (p, c)-NUD in E_{1} , then A is completely htop (p + k, c_{1})-NUD in E with c_{1} = c_{1}(c, p, k).

Suppose also that dim E = n < ∞.

(3) If A is completely hlog (p, c)-porous in E_{1} , then A is htop (p + k, c_{1})- porous in E with c_{1} = c_{1}(c, n).

(4) If A is completely hlog (p, c)-NUD in E_{1} , then A is completely htop (p + k, c_{1})-NUD in E with c_{1} = c_{1}(c, n).

Proof. Writing E_{2} = E_{1}^{⊥} we can identify E with E_{1} × E_{2} and A with A × {0}. Now (1) follows directly from Remark 3.8, and (2) follows from (1) and from I.4.9. The statements (3) and (4) follow from 3.10 and 3.11, respectively, since {0} is completely hlog (k − 1)-porous in R^{k} by 2.5 and II.6.15.

4.6. Lowering dimension. Let us consider the situation A ⊂ E_{1} ⊂ E as in 4.4 and in 4.5. We ask whether the properties of A in E imply lower-order properties of A in E_{1} . It turns out that in the homotopical case the answer is negative. This will be shown by counterexamples 4.12 and 4.13, but a partially positive answer will be given in 4.14. The next result gives a positive answer in the homological case. In view of an application in 5.5, we give the result also in terms of weak uniformity; see 2.6.

By a hyperplane in a Hilbert space E we mean the orthogonal complement b^{⊥} = {x : x · b = 0} of any nonzero vector b ∈ E.

4.7. Theorem. Suppose that T is a hyperplane in E and that A ⊊ T is [weakly] hlog (p, c)-NUD in E with p ≥ 0. If p ≥ 1, then T \ A is [weakly] hlog (p − 1, c_{1})-uniform with any c_{1} > c\sqrt{2}; if p = 0, then A is 2c^{2} -porous in T.

Proof. Write T = e^{⊥} with |e| = 1 and H_{1} = {x ∈ E : x · e > 0}, H_{2} = {x ∈ E : x · e < 0}. As in [Do_{2} , Prop. 1] we can make use of a Mayer–Vietoris sequence to obtain an isomorphism H_{p−1}(T \ A) → H_{p}(E \ A). Hence it suffices to prove the weak version.

Suppose first that p = 0, and let x ∈ T, r > 0. We may assume that x = 0. Writing t = \sqrt{4c^{2} − 1}, a = re/t, b = −re/t, we can find an arc γ ⊂ E\A joining a to b and satisfying the c-uniformity conditions in E\A. Choose a point y ∈ γ∩T. Since |y|^{2} + (r/t)^{2} ≤ (2rc/t)^{2} by the turning condition, we have |y| ≤ r. On the other hand, the cigar condition implies that d(y, A) ≥ |y − a|/c ≥ r/ct > r/2c^{2} . Thus A is 2c^{2} -porous in T.

Suppose then that p ≥ 1. Assume that z ∈ Z_{p−1}(T \ A) with d(|z|) = d > 0 and that z ∼ 0 in T \A. By 2.2, |z| is contained in a ball \overline{B}(a, s) with a ∈ T and s = d/\sqrt{2}. Setting b_{1} = se and b_{2} = −se we consider the cones z_{1} = b_{1} · z and z_{2} = b_{2} · z; see [Do_{1} , III.(4.7)]. Then ∂z_{1} = ∂z_{2} = z by [Do_{1} , III.(4.8)]. Since z ∼ 0 in T \ A, we have z = ∂h for some h ∈ S_{p}(T \ A). Setting z′ = z_{1} − z_{2} and h′ = b_{2} · h − b_{1} · h we get ∂h′ = h − b_{2} · z − h + b_{1} · z = z′, and thus z′ ∼ 0 in E \ A.

Since A is weakly hlog (p, c)-NUD, we have z′ = ∂g′ for some g′ ∈ S_{p+1}(E \ A) satisfying the uniformity conditions

(4.8)  \begin{aligned}d(|g′|) & ≤ cd(|z′|) ≤ 2cs = cd\sqrt{2}, \\ d(x, |z′|) & ≤ cd(x, A) \end{aligned}

for all x ∈ |g′|.

Let ε > 0 and set B = \overline{B}(|g| ∩ T, ε) ∩ T. We first show that z bounds in B. Choose a positive integer j such that d(|σ|) < ε for all σ < Sd^{j}g′; for notation, see II.1.4. Next choose 0 < δ < ε such that d(|σ|, T) > δ for all σ < Sd^{j}g′ with |σ| ∩ T = ∅. Finally, choose an integer k ≥ j such that d(|τ|) ≤ δ for all τ < Sd^{k}g′.

Define a map φ: R → R by φ(t) = 0 for |t| ≤ δ and by φ(t) = t − δsgn t for |t| ≥ δ. For y ∈ T and t ∈ R we write r(y + te) = y + φ(t)e. We obtain a map r: E → E, which retracts the layer T + \overline{B}(δ) onto T.

Let Sd^{k}g′ = ∑_{σ∈J} n_{σ}σ be the normal representation. Write

g_{1} = ∑\{n_{σ}σ : σ ∈ J, |σ| ∩ \overline{H}_{1} ≠ ∅\}, g_{2} = Sd^{k}g′ − g_{1}, g = r_{#}(Sd^{k}z_{1} − ∂g_{1}).

Then |g| ⊂ \overline{H}_{1} . Since

∂g_{1} + ∂g_{2} = ∂Sd^{k}g′ = Sd^{k}z_{1} − Sd^{k}z_{2},

we have Sd^{k}z_{1}−∂g_{1} = ∂g_{2}−Sd^{k}z_{2} , and hence g = r_{#}(∂g_{2}−Sd^{k}z_{2}), which implies that |g| ⊂ \overline{H}_{2} . Thus |g| ⊂ \overline{H}_{1} ∩ \overline{H}_{2} = T.

Since r | T = id, we have

∂g = r_{#}Sd^{k}∂z_{1} = r_{#}Sd^{k}z = Sd^{k}z.

Hence z bounds in |g| ∪ |z|; cf. I.1.3.1. Since |z| = |z′| ∩ T ⊂ |g′| ∩ T ⊂ B, it suffices to show that |g| ⊂ B. Let x ∈ |g|. Then x = ry for some y = x + te ∈ |∂g_{1}| ∪ |Sd^{k}z_{1}| with |t| ≤ δ. Since |∂g_{1}| ∪ |Sd^{k}z_{1}| ⊂ |Sd^{k}g′| ⊂ |Sd^{j}g′|, there is σ < Sd^{j}g′ with y ∈ |σ|. Then |σ| meets T by the choice of δ. Choosing a point x_{1} ∈ |σ| ∩ T we have x_{1} ∈ |g′| ∩ T. Since |x − x_{1}| ≤ |y − x_{1}| ≤ d(|σ|) ≤ ε, we have x ∈ B. We have proved that z bounds in B.

Let c_{1} > c\sqrt{2}. It suffices to show that for sufficiently small ε, B satisfies the uniformity conditions

\begin{aligned}d(B) & ≤ c_{1}d, \\ d(x, |z|) & ≤ c_{1}d(x, A) \end{aligned}

for all x ∈ B. Since d(B) ≤ d(|g′|) + 2ε ≤ cd\sqrt{2} + 2ε by (4.8), the first condition is clear. To prove the second one, it suffices to show that d(x, |z|) ≤ cd(x, A)\sqrt{2} for all x ∈ |g′| ∩ T. This follows from (4.8) and from the elementary lemma (4.9) below.

4.9. Lemma. Let T = b^{⊥} be a hyperplane in E, b ≠ 0, and let Y ⊂ \overline{B}(|b|) ∩ T. Then d(x, Y) ≤ d(x, b ∗ Y)\sqrt{2} for all x ∈ T. Here b ∗ Y is the cone of Y with vertex b.

Proof. Let x ∈ T, let y ∈ Y , and let L be the line through y and b. It suffices to show that |x − y| ≤ d(x, L)\sqrt{2}. Write x = y′ + v with y′ ∈ span(y) and v ⊥ y. Let P be the orthogonal projection of E onto L. Set z = P x = P y′. From similar triangles we obtain

\frac{|z − y|}{|y′ − z|} = \frac{|y|}{|b|} ≤ 1.

Hence

|x − y|^{2} = |v|^{2} + |y′ − z|^{2} + |z − y|^{2} ≤ 2|v|^{2} + 2|y′ − z|^{2} = 2|x − z|^{2} = 2d(x, L)^{2}.

4.10. Theorem. Suppose that E_{1} ⊂ E is a closed linear subspace of finite codimension k. Let A ⊊ E_{1} be [weakly] hlog (p, c)-NUD in E with p ≥ k − 1. If p ≥ k, then E \ A is [weakly] hlog (p − k, c_{1})-uniform in E_{1} for all c_{1} > 2^{k/2}c, and hence with c_{1} = (\frac{3}{2})^{k}c. If p = k − 1, then A is 2^{k}c^{2} -porous in E_{1} .

Proof. This follows from 4.7 by induction. To obtain the required constant in the case p = k −1, notice that the proof of 4.7 actually gives the porosity constant ct = c\sqrt{4c^{2} − 1} in the case p = 0.

4.11. Remark. There are several obvious corollaries of Theorem 4.10. For example, suppose that A is completely hlog (p, c)-NUD in E with p ≥ k. If A ⊂ E_{1} , then A is completely hlog (p − k, c_{1})-NUD in E_{1} . If A ⊄ E_{1} , then still A ∩ E_{1} is completely hlog (p − k, c_{1})-NUD in E_{1} by the monotonicity result 4.3.

4.12. Example. We show that 4.7 is not true if hlog is replaced by htop. Indeed, we construct a set A ⊂ R^{3} such that A is completely htop (2)-NUD in R^{4} but not htop (1)-NUD in R^{3} . This set is the famous Antoine’s necklace constructed in a self-similar manner. Related considerations appear in the recent work of S. Semmes [Se].

Let S be a circle of radius r in R^{3} . For 0 < t < r, the set T = \overline{B}(S, t) is a solid 3-torus with core S. We say that a line L ⊂ R^{3} is a piercing line of T if L meets S in two diametrically opposite points.

The building block of the Antoine set A is a solid 3-torus T together with a collection of solid 3-tori T_{1}, …, T_{m}, contained in T and linked with each other in the well-known manner. For details, see [Mo, Section 18] and [Se, Section 3]. We assume that the sets T_{1}, …, T_{m} are mutually congruent and similar to T.

We show that A is completely htop (2)-NUD in R^{4} . By I.2.2 it suffices to find a quasiconformal map f: R^{4} → R^{4} carrying A to a completely htop (2)-NUD set. We show that fA can be chosen to be a porous subset of a line. This will prove the assertion by I.4.15 or by 4.5(2).

Fix a piercing line L of T. We fatten the solid 3-tori T and T_{j} to solid 4-tori C and C_{j} in the natural way. Since the sets C_{j} are no longer linked in R^{4} , there is a bilipschitz homeomorphism f_{0}: R^{4} → R^{4} such that

(1) f_{0} = id outside C,

(2) f_{0} | C_{j} is a similarity for 1 ≤ j ≤ m,

(3) L is a piercing line of f_{0}C_{j} for 1 ≤ j ≤ m. The map f will agree with f_{0} outside the sets C_{j}. In the sets C_{j}, we iterate

the construction in the obvious manner. A limiting process gives a quasiconformal homeomorphism f, which maps the Antoine set A onto a Cantor type subset of L. It is easy to see that fA is porous in L. We have thus proved that A is completely htop (2)-NUD in R^{4} . However, it is not htop (1)-NUD in R^{3} , since R^{3} \ A is not simply connected. Observe that A is hlog (1)-NUD in R^{3} by 4.7.

We can modify this example to get a set A′ ⊂ R^{3} such that (1) A′ is completely htop (2)-NUD in R^{4} , (2) A′ is not htop (1)-NUD in R^{3} , (3) R^{3} \ A′ is simply connected. This set A′ is a subset of A, obtained by removing a thin “slice of sausage” from each of the infinite number of solid 3-tori in the construction of A. Then A′ is no longer topologically wild. But if the slices become relatively thinner and thinner along the construction, A′ cannot be htop (1)-NUD in R^{3} . On the other hand, A′ is completely htop (2)-NUD in R^{4} by 4.3.

4.13. Example. Assume that A is completely hlog (p)-NUD in R^{n}. Then A is completely htop (p + 1)-NUD in R^{n+1} by 4.5(4), but A need not be htop (p)-NUD in R^{n}. This is the case in Example 4.12 with n = 3, p = 1. We give another example with n = 4, p = 1.

Let A ⊂ R^{4} be a BT arc such that R^{4} \ A is not simply connected. Such an arc exists by [Vä_{1} , 6.3] and [Bl, Th. 3E]. Then A is not htop (1)-NUD in R^{4} . However, it is completely hlog (1)-NUD in R^{4} by I.5.6 and [MV, 3.9]. Thus A is completely htop (2)-NUD in R^{5} .

However, the following homotopical version of 4.7 was pointed to us by the referee.

4.14. Theorem. Suppose that T is a hyperplane in R^{n} and that A ⊊ T is completely htop (p, c)-NUD in R^{n} and htop (1, c)-NUD in T with p ≥ 1. Then A is completely htop (p − 1, c_{1})-NUD in T with c_{1} = c_{1}(c, n).

Proof. By II.4.2, the set A is completely hlog (p, c_{2})-NUD in R^{n} with c_{2} = c_{2}(c, p). From 4.7 it follows that A is completely hlog (p − 1, 2c_{2})-NUD in T. Since A is also htop (1, c)-NUD in T, the theorem follows from II.4.5.

5. Miscellaneous results

5.1. Summary of Section 5. This section consists of two parts. In the first part we consider the case where dim E = ∞ and A ⊂ E is compact or, more generally, boundedly compact, which means that A meets every ball \overline{B}(r) in a compact set. It turns out that A has all NUD and porosity properties with universal constants. We prove in detail the htop NUD case in 5.2 and consider the other cases in 5.3. We thank E. Saksman for useful discussions concerning this part.

In the second part we consider L-bilipschitz maps f: S^{k} → R^{n} and give a new proof for the fact that fS^{k} is completely weakly hlog (n − 2, c)-NUD with c = c(L, n).

5.2. Theorem. Let dim E = ∞ and let A ⊂ E be boundedly compact. Then A is htop (p, c)-NUD for each p ≥ 0 and for each c > \sqrt{3/2} = 1.22474 ….

Proof. Let I^{p+1} be the cube [−1, 1]^{p+1} , and let f : ∂I^{p+1} → E \ A be continuous. We may assume that f is piecewise linear and that d(|f|) = d > 0. By Lemma 2.2 there is a ∈ E with |f| ⊂ \overline{B}(a, d/\sqrt{2} ). We may assume that a = 0. Let ε > 0. Since the set A′ = A ∩ \overline{B}(2d) is compact, there is a finite-dimensional linear subspace E_{1} of E such that |f| ⊂ E_{1} and A′ ⊂ \overline{B}(E_{1}, ε). Let b be a vector perpendicular to E_{1} with |b| = d. Let g: I^{p+1} → E be the cone of f with g(0) = b. We show that g satisfies the uniformity conditions in E \ A if ε is small enough.

Let x_{1}, x_{2} ∈ |g|. Then x_{j} ∈ [b, y_{j}] for some y_{j} ∈ |f|. Since |b − y_{j}|^{2} = d^{2} + |y_{j}|^{2} ≤ 3d^{2}/2 and |y_{1} − y_{2}| ≤ d, we obtain |x_{1} − x_{2}| ≤ d\sqrt{3/2}, and hence d(|g|) ≤ d\sqrt{3/2}, which is the turning condition.

To prove the lens condition let c > \sqrt{3/2} and let x ∈ |g|, z ∈ A. Then x ∈ [b, y] for some y ∈ |f|, and hence d(x, |f|) ≤ |x − y| ≤ d\sqrt{3/2}. If z ∉ A′, we have |x − z| ≥ |z| − |x| ≥ 2d − d = d, and hence d(x, |f|) ≤ |x − z|\sqrt{3/2}. Assume that z ∈ A′, set r = d(|f|, A), and write x = (1 − t)y + tb with 0 ≤ t ≤ 1. Then |x − y| = t|y − b| ≤ td\sqrt{3/2} and d(x, E_{1}) = td(b, E_{1}) = td. If td\sqrt{3/2} ≤ \frac{1}{2}r, then

2|x − y| ≤ r ≤ |y − z| ≤ |y − x| + |x − z|,

and hence |x − y| ≤ |x − z|. If td\sqrt{3/2} ≥ \frac{1}{2}r, then for small ε we get

\frac{|x − y|}{|x − z|} ≤ \frac{td\sqrt{3/2}}{d(x, E_{1}) − ε} = \frac{\sqrt{3/2}}{1 − ε/td} ≤ \frac{\sqrt{3/2}}{1 − ε\sqrt{6}/r} < c.

These estimates yield d(x, |f|) ≤ cd(x, A).

5.3. Variations. Theorem 5.2 remains true if htop is replaced by hlog. The proof is almost the same. Furthermore, the set A is htop and hlog (p, c)-porous for all p ≥ 0 and c > 1. The proof is easier than that of 5.2, since we can make use of a translation in a direction perpendicular to a suitable finite-dimensional subspace.

5.4. Bilipschitz spheres. Suppose that f: S^{n−1} → R^{n} is θ-quasimöbius. It is well known but not easy to prove that the components of R^{n} \ fS^{n−1} are cuniform domains with c = c(θ, n). A proof based on compactness was announced in [Vä_{2} , 5.10], and a direct proof with c independent of n was recently given by P. MacManus [Ma_{2} ]. He gives c = 322θ(8), but his definition for the uniformity constant is different from ours. A third proof is given in [Vä_{3} , 5.25]. It gives an explicit estimate depending on n. On the other hand, it shows that these domains are completely hlog (n − 2, c)-uniform.

We shall give a fourth proof, which is valid only in the case where f is bilipschitz. On the other hand, it is valid for maps f: S^{k} → R^{n}, 0 ≤ k ≤ n − 1, and gives weak hlog (p)-uniformity for all relevant p. It was inspired by the proof of A. Dold [Do_{2} ] for the Jordan–Brouwer separation theorem. The basic idea is to extend f to a bilipschitz map R^{n+k} → R^{n+k}. This is possible by the following Lipschitz version of the well-known Klee trick; see [Ru, p. 74] and [Ma_{1} , 1.3].

5.5. Theorem. Let A ⊂ R^{k} and let f: A → R^{n} be M -bilipschitz. Set A_{1} = A×{0} ⊂ R^{k}×R^{n} = R^{k+n}, and define f_{1}: A_{1} → R^{k+n} by f_{1}(x, 0) = (0, fx). Then f_{1} extends to an M′ -bilipschitz map F: R^{k+n} → R^{k+n} with M′ = M^{2}\sqrt{7}.

Proof. By the Kirszbraun theorem [Fe, 2.10.43], there are M -Lipschitz extensions g: R^{k} → R^{n} and h: R^{n} → R^{k} of f and f^{−1} , respectively. Define G, H: R^{k+n} → R^{k+n} by G(x, y) = (x, y + gx) and H(x, y) = (x − hy, y). The map F = HG is an extension of f_{1} . We show that it is M′ -bilipschitz. Let z = (x, y), z′ = (x′, y′) ∈ R^{k+n}. Since

F (x, y) = (x − h(y + gx), y + gx),

we obtain by the Schwarz inequality

|F z − F z′|^{2} ≤ (|x − x′| + |h(y + gx) − h(y′ + gx′)|)^{2} + (|y − y′| + |gx − gx′|)^{2}

≤ (|x − x′| + M(|y − y′| + M|x − x′|))^{2} + (|y − y′| + M|x − x′|)^{2}

≤ ((1 + M^{2})^{2} + M^{2})|z − z′|^{2} + (1 + M^{2})|z − z′|^{2}
= (2 + 4M^{2} + M^{4})|z − z′|^{2} ≤ 7M^{4}|z − z′|^{2}.

Hence F is M′ -bilipschitz. Since F^{−1}(x, y) = (x + hy, y − g(x + hy)), similar estimates show that F^{−1} is M′ -Lipschitz.

Uniform domains of higher order III 463

5.6. Question. Is there a quasisymmetric version of 5.5? A direct analogue is false, since a quasisymmetric map f: A → R^{n}, A ⊂ R^{k}, need not be Hölder continuous. However, this cannot happen if, for example, A is connected.

5.7. Theorem. Let 0 ≤ k ≤ n − 1 and let f: S^{k} → R^{n} be M -bilipschitz. Then fS^{k} is completely weakly hlog (n − 2, c)-NUD with c = 35 · (\frac{3}{2})^{k+1}M^{4} .

Proof. By 2.8, the set S^{k} is completely weakly hlog (k + n − 1, 5)-NUD in R^{k+1+n}. Since S^{k} ⊂ R^{k+1} , 5.5 gives an M^{2}\sqrt{7}-bilipschitz extension F: R^{k+1+n} → R^{k+1+n} of f_{1} . Since an L-bilipschitz map increases uniformity constants at most by a factor L^{2} , fS^{k} is completely weakly hlog (k+n−1, 35M^{4})-NUD in R^{k+1+n}. Applying Theorem 4.10, with E = R^{k+1+n}, E_{1} = {0} × R^{n}, p = k + n − 1, gives the theorem.

5.8. Remarks. 1. Theorem 5.7 remains true if S^{k} is replaced by R^{k} and c by 16 · (\frac{3}{2})^{k}M^{4} . In the proof we replace 2.8 by 2.7.

2. In these results, the set fS^{k} or fR^{k} is hlog (p, c)-NUD for p ≠ n − k − 1, 0 ≤ p ≤ n − 2.

5.9. Corollary. If f: S^{n−1} → R^{n} is M -bilipschitz, the components of R^{n} \ fS^{n−1} are c-uniform domains with c = 35 · (\frac{3}{2})^{n}M^{4} .

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Received 19 February 1996