On Bilipschitz Extensions in Real Banach Spaces

Definition 1. A domain D in E is called c-uniform in the norm metric, provided there exists a constant c with the property that each pair of points z₁, z₂ in D can be joined by a rectifiable arc α in D satisfying

• (1)

  min _j=1,2 ℓ(α[z_j, z]) ≤ cd_D(z) for all z ∈ α, and

• (2)

  ℓ(α) ≤ c|z₁ − z₂|,

where ℓ(α) denotes the length of α, α[z_j, z] the part of α between z_j and z, and d_D(z) the distance from z to the boundary ∂D of D.

Definition 2. Suppose GE  , G^′E^′, and M ≥ 1. We say that a homeomorphism f : G → G^′ is M-bilipschitz if

for all x, y ∈ G, and M-QH if for all x, y ∈ G.

Theorem A (see [11], Theorem  7.)Suppose that D is a K-quasidisk in ℝ², that D^′ is a Jordan domain in ℝ², and that ϕ : ∂D → ∂D^′ is L₁-bilipschitz. Then there exist L-bilipschitz $$ and $$ such that f = f^⋆ = ϕ on ∂D and L depends only on K and L₁, where $$ and $$.

Theorem B (see [12], Theorem  2.12.)Suppose that X is a closed set in ℝⁿ, n ≠ 4, and that f : ℝⁿ → ℝⁿ is a K-QC map such that f∣_X is L-bilipschitz. Then there is an L₁-bilipschitz map g : ℝⁿ → ℝⁿ such that

• (1)

  g∣_X = f∣_X;

• (2)

  g(D) = f(D) for each component D of ℝⁿ∖X;

• (3)

  L₁ depends only on K, L, and n.

Open Problem 1. Suppose that D is a Jordan domain in $$ and that f∣_∂D is M-bilipschitz. Characterize mappings f having M^′-bilipschitz extension to D with M^′ = M^′(c, M).

Open Problem 2. Suppose that D is a Jordan domain in $$. For which domains D does each M-bilipschitz f in the ∂D have M^′-bilipschitz extension to D with M^′ = M^′(c, M)?

Theorem C (see [13], Theorem  2.11.)Suppose that D and D^′ are Jordan domains in $$ and that ∞ ∈ D^′ if and only if ∞ ∈ D. Suppose also that f : D → D^′ is a K-quasiconformal mapping and that f extends to a homeomorphism $$ such that f∣_∂D is M-bilipschitz. Then there exists an M-bilipschitz map $$ with g∣_∂D = f∣_∂D, where M^′ = M^′(M, K).

Theorem D (see [13], Theorem  4.9.)Suppose that D and D^′ are Jordan domains in $$. Then each M-bilipschitz f in ∂D has an M^′-bilipschitz extension g : D → D^′ with g∣_∂D = f∣_∂D if and only if D is a K-quasidisk, where M^′ = M^′(M, K) and K = K(M).

Open Problem 3. Suppose that D and D^′ are bounded domains with connected boundaries in E and E^′. Suppose also that f : D → D^′ is M-QH and that f extends to a homeomorphism $$ such that f∣_∂D is M-bilipschitz. Is it true that f  M^′-bilipschitz with M^′ = M^′(c, M)?

Theorem 3. Suppose that D and D^′ are bounded domains with connected boundaries in E and E^′, respectively. Suppose also that f : D → D^′ is M-QH and that f extends to a homeomorphism $$ such that f∣_∂D is M-bilipschitz. If D^′ is a c-uniform domain, then f is M^′-bilipschitz with M^′ = M^′(c, M).

Theorem E (see [5], Theorem  6.16.)For a domain D, the following are quantitatively equivalent:

• (1)

  D is a c-uniform domain;

• (2)

  k_D(z₁, z₂) ≤ c^′log   (1 + |z₁ − z₂|/min   {d_D(z₁), d_D(z₂)}) for all z₁, z₂ ∈ D;

• (3)

  $$ for all z₁, z₂ ∈ D.

Definition 4. Let α be an arc in E. The arc may be closed, open, or half open. Let $$, n ≥ 1, be a finite sequence of successive points of α. For h ≥ 0, we say that $$ is h-coarse if k_D(x_j−1, x_j) ≥ h for all 1 ≤ j ≤ n. Let Φ_k(α, h) be the family of all h-coarse sequences of α. Set

with the agreement that ℓ_k(α, h) = 0 if Φ_k(α, h) = ∅. Then the number ℓ_k(α, h) is the h-coarse quasihyperbolic length of α.

Definition 5. Let D be a domain in E. An arc α ⊂ D is (ν, h)-solid with ν ≥ 1 and h ≥ 0 if

for all x, y ∈ α. A (ν, 0)-solid arc is said to be a ν-neargeodesic, that is, an arc α ⊂ D is a ν- neargeodesic if and only if ℓ_k(α[x, y]) ≤ νk_D(x, y) for all x, y ∈ α.

Obviously, a ν-neargeodesic is a quasihyperbolic geodesic if and only if ν = 1.

In [19], Väisälä got the following property concerning the existence of neargeodesic in E.

Theorem F (see [19], Theorem  3.3.)Let {z₁, z₂} ⊂ D and ν > 1. Then there is a ν-neargeodesic in D joining z₁ and z₂.

Theorem G (see [5], Theorem  4.15.)For domains D ≠ E and D^′ ≠ E^′, suppose that f : D → D^′ is M-QH. If γ is a c-neargeodesic in D, then the arc γ^′ is c₁-neargeodesic in D^′ with c₁ depending only on c and M.

Let G ≠ E and G^′ ≠ E^′ be metric spaces, and let φ :  [0, ∞)   → [0, ∞) be a growth function, that is, a homeomorphism with φ(t) ≥ t. We say that a homeomorphism f : G → G^′ is φ-semisolid if

for all x, y ∈ G, and φ-solid if both fand f⁻¹ satisfy this condition.

We say that f is fully φ-semisolid (resp. fully φ-solid) if f is φ-semisolid (resp. φ-solid) on every subdomain of G. In particular, when G = E, corresponding subdomains are taken to be proper ones. Fully φ-solid mapsare also called freely φ-quasiconformal maps, or briefly φ-FQC maps.

Theorem H (see [5], Theorem  7.18.)Let D and D^′ be domains in E and E^′, respectively. Suppose that D is a c-uniform domain and that f : D → D^′ is φ-FQC (see Section 2 for the definition). Then the following conditions are quantitatively equivalent:

• (1)

  D^′ is a c₁-uniform domain;

• (2)

  f is η-quasimöbius.

Theorem I (see [20], Theorem  1.1.)Suppose that D is a c-uniform domain and that f : D → D^′ is (M, C)-CQH, where DE and D^′E^′. Then the following conditions are quantitatively equivalent:

• (1)

  D^′ is a c₁-uniform domain;

• (2)

  f extends to a homeomorphism $$ and $$ is η-QM rel ∂D.

Theorem 6. Suppose that DE and D^′E^′, that D is a c-uniform domain, and that f : D → D^′ is φ-FQC. Then the following conditions are quantitatively equivalent:

• (1)

  D^′ is a c₁-uniform domain;

• (2)

  f is θ-quasimöbius;

• (3)

  f extends to a homeomorphism $$ and $$ is θ₁-QM rel ∂D.

Theorem J (see [1], Theorem  2.44.)Suppose that GE and G^′E^′ is a c-uniform domain, and that f : G → G^′ is M-QH. If D ⊂ G is a c-uniform domain, then D^′ = f(D) is a c^′-uniform domain with c^′ = c^′(c, M).

Theorem K (see [5], Theorem  6.19.)Suppose that DE is a c-uniform domain and that γ is a c₁-neargeodesic in D with endpoints z₁ and z₂. Then there is a constant b = b(c, c₁) ≥ 1 such that

• (1)

  min _j=1,2 ℓ(γ[z_j, z]) ≤ bd_D(z) for all z ∈ α, and

• (2)

  ℓ(γ) ≤ b|z₁ − z₂|.

Theorem L (see [21], Theorem  1.2.)Suppose that D₁ and D₂ are convex domains in E, where D₁ is bounded and D₂ is c-uniform for some c > 1, and that there exist z₀ ∈ D₁∩D₂ and r > 0 such that 𝔹(z₀, r) ⊂ D₁ ∩ D₂. If there exist constants R₁ > 0 and c₀ > 1 such that R₁ ≤ c₀r and $$, then D₁ ∪ D₂ is a c^′-uniform domain with c^′ = (c + 1)(2c₀ + 1) + c.

Lemma 7. There is a constant M₀ = M₀(M) > M such that if the points z₁, z₂ ∈ D satisfies dist (z₁, ∂D) ≤ ε and dist (z₂, ∂D) ≤ ε for sufficiently small ε > 0, then

Proof. Let x₁, x₂ ∈ ∂D be such that |z₁ − x₁| = (4/3)dist (z₁, ∂D), |z₂ − x₂| ≤ (4/3)dist (z₂, ∂D) and |x₁ − x₂| ≤ max   {|z₁ − x₁|, |z₂ − x₂|} < 3|x₁ − x₂| for sufficiently small ε > 0. It follows from “f being M-QH in D and homeomorphic in $$” that H(x, f) ≤ K (cf. [1]) for each x ∈ D, where K depends only on M. Hence,

If |z₁ − z₂| ≤ (1/4K²M)max   {|z₁ − x₁|, |z₂ − x₂|}, then for each z ∈ [z₁, z₂],

and so we have which shows that

If |z₁ − z₂| > (1/4K²M)max   {|z₁ − x₁|, |z₂ − x₂|}, then by the assumption “f being M-bilipschitz in ∂D,”

The same discussion as the above shows that

Lemma 8. There is a constant M₁ = M₁(c, M) such that if the points x ∈ D and $$ satisfies dist (z, ∂D) ≤ ε for sufficiently small ε > 0, then

Proof. Let $$ such that dist (x₀, ∂D) ≤ ε for sufficiently small ε > 0, and let x₂ be the intersection point of 𝕊(x₀, (1/2)d_D(x)) with [x₀, x]. Then we have

which implies that Hence, and so

Let T be a 2-dimensional linear subspace of E which contains x₀ and x₂, and we use τ to denote the circle T∩𝕊(x₀, (1/2)d_D(x)). Take w₁ ∈ τ∩∂D such that τ(x₂, w₁) ⊂ D and ℓ(τ[x₂, w₁]) ≤ 2d_D(x). Let $$ and denote τ(x₁, w₁) by τ₁.

Claim 1. There must exist a 2³²-uniform domain D₁ in D and $$ satisfying dist (x₃, ∂D) ≤ ε for sufficiently small ε > 0 such that x₀, $$ and (1/12)d_D(x) ≤ |x₃ − x₀| ≤ (11/12)d_D(x).

If d_D(x₁) = 0, then we take D₁ = 𝔹(x, d_D(x)) and x₃ = x₁. Obviously, |x₃ − x₀| = (1/2)d_D(x). Hence Claim 1 holds true in this case.

If d_D(x₁) > 0, we divide the proof of Claim 1 into two parts.

Case 1. (d_D(x₁) ≤ (5/12)d_D(x)). Then we take D₁ = 𝔹(x, d_D(x)) ∪ 𝔹(x₁, d_D(x₁)) and $$ such that dist (x₃, ∂D) ≤ ε for sufficiently small ε > 0. It follows from Theorem L that D₁ is a 29-uniform domain and

from which we see that Claim 1 is true.

Case 2. (d_D(x₁) > (5/12)d_D(x)). Obviously, d_D(x₁) > (5/6)|x₁ − x₀|. We let w₂ ∈ τ₁ be the first point along the direction from x₁ to w₁ such that

If |w₂ − x₁| ≤ (1/3)d_D(x), then we take D₁ = 𝔹(x, d_D(x))  ∪  𝔹(w₂, d_D(w₂)),and let $$ such that dist (x₃, ∂D) ≤ ε for sufficiently small ε > 0. Then

It follows from Theorem L that D₁ is a 677-uniform domain, which shows that Claim 1 is true.

If |w₂ − x₁| > (1/3)d_D(x), then we first prove the following subclaim.

Subclaim  1. There exists a simply connected domain $$ in D, where t = 1 or 2, such that

• (1)

  x₀, $$;

• (2)

  for each i ∈ {0, …, t}, (5/12)d_D(x) ≤ r_i ≤ d_D(x);

• (3)

  if t = 2, then |x − w₂| − r₀ − r₂ ≥ (1/144)d_D(x);

• (4)

  r_i + r_i+1 − |v_i − v_i+1| ≥ (1/144)d_D(x), where i ∈ {0,1} if t = 2 or i = 0 if t = 1.

Here B_i = 𝔹(v_i, r_i), v_i ∈ τ[x₂, w₂], v₁ ∉ B₀, and v₂ ∉ τ[x₂, v₁].

To prove this subclaim, we let y₂ ∈ τ₁ be such that |x₁ − y₂| = (1/3)d_D(x) and let C₀ = 𝔹(x, d_D(x)) and C₁ = 𝔹(y₂, d_D(y₂)). Since d_D(y₂) > (5/12)d_D(x), we have

Next, we construct a ball denoted by C₂.

•  

   If $$, then we let C₂ = 𝔹(w₂, d_D(w₂)).

•  

  If $$, then we let y₃ be the intersection of 𝕊(y₂, d_D(y₂)) with τ₁[y₂, w₁]. Since ℓ(τ₁) ≤ 2d_D(x) and d_D(z) ≥ (5/12)d_D(x) for all z ∈ τ₁(x₁, x₂), we have

  $$ ()

which implies that We take C₂ = 𝔹(w₂, d_D(w₂)). Then (26) implies

Now we are ready to construct the needed domain D₁.

If d_D(w₂) + d_D(x) − |w₂ − x| ≥ (1/48)d_D(x), then we take B₀ = C₀, B₁ = C₂, and D₁ = B₀ ∪ B₁ with v₀ = x, v₁ = w₂, r₀ = d_D(x), and r₁ = d_D(w₂). Obviously, D₁ satisfies all the conditions in Subclaim 1. In this case, t = 1.

If d_D(w₂) + d_D(x) − |w₂ − x| < (1/48)d_D(x), then we take B₀ = 𝔹(x, (35/36)d_D(x)) with r₀ = (35/36)d_D(x) and v₀ = x, B₁ = C₁ with r₁ = d_D(y₂) and v₁ = y₂, and B₂ = C₂ with r₂ = d_D(w₂) and v₂ = w₂. Then Inequalities (24) and (27) show that $$ satisfies all the conditions in Subclaim 1. In this case, t = 2.

Hence, the proof of Subclaim 1 is complete.

The following follows from a similar argument as in the proof of [22, Theorem 1.1].

Corollary  9. The domain D₁ constructed in Subclaim 1 is a 2³²-uniform domain.

Let $$ such that dist (x₃, ∂D) ≤ ε for sufficiently small ε > 0. Then

Then the proof of Claim 1 easily follows from (28), Subclaim 1, and Corollary 9.

We come back to the proof of Lemma 8. It follows from (28) and Lemma 7 that

Then it follows from Theorem J that $$ is an M^′-uniform domain, where M^′ = M^′(c, M). Hence, we know from Theorem 6 that f⁻¹ is a θ-Quasimöbius in $$, where θ = θ(c, M), and so (19), (20), (28), and (29) imply that which, together with (20), shows Thus, the proof of Lemma 8 is complete.

Lemma 10. For all x ∈ D, if $$ such that dist (x, ∂D) ≤ ε for sufficiently small ε > 0, then $$, where M₁ = M₁(c, M).

Proof. Suppose on the contrary that there exist points x₁ ∈ D and $$ with dist (y₁, ∂D) ≤ ε for sufficiently small ε > 0 such that

We take $$ such that dist (y₂, ∂D) ≤ ε for sufficiently small ε > 0. From Lemma 7 we know that

Let T₁ be a 2-dimensional linear subspace of E determined by x₁, y₁ and y₂, and ω the circle T₁∩𝕊(y₁, d_D(x₁)). We take y₃ ∈ ω∩∂D which satisfies ω(x₁, y₃) ⊂ D and ℓ(ω[x₁, y₃]) ≤ 4d_D(x₁). Let ω₁ = ω(x₁, y₃) and w₁ be the first point along the direction from x₁ to y₃ such that

Let $$ such that dist (w₁, ∂D) ≤ ε for sufficiently small ε > 0. Then it follows from Lemma 8 that

which, together with Lemmas 7 and 8 and (32), implies that Hence, we infer from (32) that

Since ℓ(ω₁) ≤ 4d_D(x₁), by the choice of w₁, one has

whence which contradicts with (37). The proof of Lemma 10 is complete.

Lemma 11. For x₁ ∈ D and x₂ ∈ ∂D, we have

where M₂ = 2M₀ + M₁.

Proof. For x₁ ∈ D, we let $$ such that dist (y₁, ∂D) ≤ ε for sufficiently small ε > 0. Then it follows from Lemma 8 that

For x₂ ∈ ∂D, if |y₁ − x₂| ≤ 2|x₁ − y₁|, then by Lemma 7, we have

If |y₁ − x₂| > 2|y₁ − x₁|, then we have

Hence, by Lemma 7 and (41),

from which the proof follows.

Lemma 12. For x₁ ∈ D and x₂ ∈ ∂D, one has

where $$.

Proof. We begin with a claim.

Claim 2. For all z ∈ D, we have $$.

To prove this claim, we let w₂ ∈ [z, y₁] be such that $$. It follows from [18] that

By Lemma 8, we have Hence, Lemma 10 implies $$, whence which shows that Claim 2 is true.

Now we are ready to finish the proof of Lemma 12. For x₁ ∈ D and x₂ ∈ ∂D, if |x₁ − x₂| ≤ 2M₀M₁d_D(x₁), then by Claim 2,

If |x₁ − x₂| > 2M₀M₁d_D(x₁), then we take $$ such that dist (w₃, ∂D) ≤ ε for sufficiently small ε > 0, and so

whence Lemmas 7 and 8 imply from which the proof is complete.

Lemma 13. D is a c₁-uniform domain, where c₁ = c₁(c, M).

Proof. We first prove that f⁻¹ is θ₁-Quasimöbius rel ∂D^′, where $$, M₂ and M₃ are the same as in Lemmas 11 and 12, respectively. By definition, it is necessary to prove that for $$, $$, $$, $$,

where x₁, x₂ ∈ ∂D^′. Obviously, to prove Inequality (52), we only need to consider the following three cases.

Case 3 ($$). Since f is M-bilipschitz in ∂D, we have

Case 4 ($$, $$, $$ and$$). It follows from Lemmas 11 and 12 that

Case 5 ($$, $$ and $$). We obtain from Lemmas 11 and 12 that

The combination of Cases 3 ~ 5 shows that Inequality (52) holds, which implies that f⁻¹ is a θ₁-Quasimöbius rel ∂D^′. Hence, Theorem 6 shows that D is a c₁-uniform domain, where c₁ depends only on c and M.