The two-phase problem for harmonic measure in VMO Xavier Tolsa 23 - - PowerPoint PPT Presentation

The two-phase problem for harmonic measure in VMO

Xavier Tolsa 23 August 2019

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 1 / 22

Rectifiability

E ⊂ Rn+1 is n-rectifiable if it is Hn-a.e. contained in a countable union of C 1 (or Lipschitz) n-dimensional manifolds.

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Two-phase problem for harmonic measure 23 August 2019 2 / 22

Rectifiability

E ⊂ Rn+1 is n-rectifiable if it is Hn-a.e. contained in a countable union of C 1 (or Lipschitz) n-dimensional manifolds. A Borel measure µ is n-rectifiable if it is of the form µ = g Hn|E, with E n-rectifiable and g ∈ L1

loc(Hn|E).

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Two-phase problem for harmonic measure 23 August 2019 2 / 22

Uniform rectifiability

E ⊂ Rd is n-AD-regular if Hn(E ∩ B(x, r)) ≈ r n for all x ∈ E, 0 < r ≤ diam(E).

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Two-phase problem for harmonic measure 23 August 2019 3 / 22

Uniform rectifiability

E ⊂ Rd is n-AD-regular if Hn(E ∩ B(x, r)) ≈ r n for all x ∈ E, 0 < r ≤ diam(E). E is uniformly n-rectifiable if it is n-AD-regular and there are M, θ > 0 such that for all x ∈ E, 0 < r ≤ diam(E), there exists a Lipschitz map g : Rn ⊃ Bn(0, r) → Rd, ∇g∞ ≤ M, such that Hn E ∩ B(x, r) ∩ g(Bn(0, r))

• ≥ θ r n.
• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 3 / 22

Uniform rectifiability

E ⊂ Rd is n-AD-regular if Hn(E ∩ B(x, r)) ≈ r n for all x ∈ E, 0 < r ≤ diam(E). E is uniformly n-rectifiable if it is n-AD-regular and there are M, θ > 0 such that for all x ∈ E, 0 < r ≤ diam(E), there exists a Lipschitz map g : Rn ⊃ Bn(0, r) → Rd, ∇g∞ ≤ M, such that Hn E ∩ B(x, r) ∩ g(Bn(0, r))

• ≥ θ r n.

A Borel measure µ is n-AD-regular if it is of the form µ = g Hn|E, with E n-AD-regular and g ≈ 1. It is uniformly n-rectifiable if, additionally, E is uniformly n-rectifiable.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 3 / 22

Harmonic measure

Ω ⊂ Rn+1 open and connected. For p ∈ Ω, ωp is the harmonic measure in Ω with pole in p.

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Two-phase problem for harmonic measure 23 August 2019 4 / 22

Harmonic measure

Ω ⊂ Rn+1 open and connected. For p ∈ Ω, ωp is the harmonic measure in Ω with pole in p. That is, for E ⊂ ∂Ω, ωp(E) is the value at p of the harmonic extension of χE to Ω.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 4 / 22

Harmonic measure

Ω ⊂ Rn+1 open and connected. For p ∈ Ω, ωp is the harmonic measure in Ω with pole in p. That is, for E ⊂ ∂Ω, ωp(E) is the value at p of the harmonic extension of χE to Ω. Questions about the metric properties of harmonic measure: When Hn ≈ ωp? Which is the connection with rectifiability? Dimension of harmonic measure?

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 4 / 22

Harmonic measure

Ω ⊂ Rn+1 open and connected. For p ∈ Ω, ωp is the harmonic measure in Ω with pole in p. That is, for E ⊂ ∂Ω, ωp(E) is the value at p of the harmonic extension of χE to Ω. Questions about the metric properties of harmonic measure: When Hn ≈ ωp? Which is the connection with rectifiability? Dimension of harmonic measure? In the plane if Ω is simply connected and H1(∂Ω) < ∞, then H1 ≈ ωp. (F.& M. Riesz) Many results in C using complex analysis (Carleson, Makarov, Jones, Bishop, Wolff, Garnett,...). Analogue of Riesz theorem fails in higher dimensions (Wu, Ziemer). In higher dimensions, need real analysis techniques.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 4 / 22

Connection between harmonic measure and Riesz transform

Let E(x) be the fundamental solution of the Laplacian in Rn+1: E(x) = cn 1 |x|n−1 .

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 5 / 22

Connection between harmonic measure and Riesz transform

Let E(x) be the fundamental solution of the Laplacian in Rn+1: E(x) = cn 1 |x|n−1 . The kernel of the Riesz transform R is K(x) = x |x|n+1 = c ∇E(x).

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Two-phase problem for harmonic measure 23 August 2019 5 / 22

Connection between harmonic measure and Riesz transform

Let E(x) be the fundamental solution of the Laplacian in Rn+1: E(x) = cn 1 |x|n−1 . The kernel of the Riesz transform R is K(x) = x |x|n+1 = c ∇E(x). The Green function G(·, ·) of Ω is G(x, p) = E(x − p) −

• E(x − y) dωp(y).
• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 5 / 22

Connection between harmonic measure and Riesz transform

Let E(x) be the fundamental solution of the Laplacian in Rn+1: E(x) = cn 1 |x|n−1 . The kernel of the Riesz transform R is K(x) = x |x|n+1 = c ∇E(x). The Green function G(·, ·) of Ω is G(x, p) = E(x − p) −

• E(x − y) dωp(y).

Therefore, for x ∈ Ω: c ∇xG(x, p) = K(x − p) −

• K(x − y) dωp(y).

That is, c ∇xG(x, p) = K(x − p) − Rωp(x).

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 5 / 22

One-phase problems and two-phase problems

In one-phase problems for harmonic measure we have one domain Ω and usually we are interested in the connection between and ω and Hn|∂Ω.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 6 / 22

One-phase problems and two-phase problems

In one-phase problems for harmonic measure we have one domain Ω and usually we are interested in the connection between and ω and Hn|∂Ω. In two-phase problems for harmonic measure we have two domains Ω1, Ω2 with ∂Ω1 ∩ ∂Ω2 = ∅ and usually we are interested in the connection between and ω1, ω2 and Hn|∂Ωi in ∂Ω1 ∩ ∂Ω2. Ω1 Ω2 E

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 6 / 22

Quantitative results: NTA domains

Let Ω ⊂ Rn+1 be open. For x, y ∈ Ω, a curve γ ⊂ Ω from x to y is a C-cigar curve with bounded turning if

min(H1(γ(x, z)), H1(γ(y, z))) ≤ C dist(z, Ωc) for all z ∈ γ, and H1(γ) ≤ C |x − y|.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 7 / 22

Quantitative results: NTA domains

Let Ω ⊂ Rn+1 be open. For x, y ∈ Ω, a curve γ ⊂ Ω from x to y is a C-cigar curve with bounded turning if

min(H1(γ(x, z)), H1(γ(y, z))) ≤ C dist(z, Ωc) for all z ∈ γ, and H1(γ) ≤ C |x − y|.

Ω is NTA if:

• All x, y ∈ Ω are connected by a C-cigar curve with bounded turning.
• It has exterior corkscrews, i.e. for every ball B centered at ∂Ω there

is another ball B′ ⊂ B \ Ω with r(B′) ≈ r(B).

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 7 / 22

Quantitative results: NTA domains

Let Ω ⊂ Rn+1 be open. For x, y ∈ Ω, a curve γ ⊂ Ω from x to y is a C-cigar curve with bounded turning if

min(H1(γ(x, z)), H1(γ(y, z))) ≤ C dist(z, Ωc) for all z ∈ γ, and H1(γ) ≤ C |x − y|.

Ω is NTA if:

• All x, y ∈ Ω are connected by a C-cigar curve with bounded turning.
• It has exterior corkscrews, i.e. for every ball B centered at ∂Ω there

is another ball B′ ⊂ B \ Ω with r(B′) ≈ r(B). A non trivial NTA domain:

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 7 / 22

Quantitative results: NTA domains

Let Ω ⊂ Rn+1 be open. For x, y ∈ Ω, a curve γ ⊂ Ω from x to y is a C-cigar curve with bounded turning if

min(H1(γ(x, z)), H1(γ(y, z))) ≤ C dist(z, Ωc) for all z ∈ γ, and H1(γ) ≤ C |x − y|.

Ω is NTA if:

• All x, y ∈ Ω are connected by a C-cigar curve with bounded turning.
• It has exterior corkscrews, i.e. for every ball B centered at ∂Ω there

is another ball B′ ⊂ B \ Ω with r(B′) ≈ r(B). A non trivial NTA domain:

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 7 / 22

Quantitative results: NTA domains

Let Ω ⊂ Rn+1 be open. For x, y ∈ Ω, a curve γ ⊂ Ω from x to y is a C-cigar curve with bounded turning if

min(H1(γ(x, z)), H1(γ(y, z))) ≤ C dist(z, Ωc) for all z ∈ γ, and H1(γ) ≤ C |x − y|.

Ω is NTA if:

• All x, y ∈ Ω are connected by a C-cigar curve with bounded turning.
• It has exterior corkscrews, i.e. for every ball B centered at ∂Ω there

is another ball B′ ⊂ B \ Ω with r(B′) ≈ r(B). A non trivial NTA domain:

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 7 / 22

Quantitative results: NTA domains

Let Ω ⊂ Rn+1 be open. For x, y ∈ Ω, a curve γ ⊂ Ω from x to y is a C-cigar curve with bounded turning if

min(H1(γ(x, z)), H1(γ(y, z))) ≤ C dist(z, Ωc) for all z ∈ γ, and H1(γ) ≤ C |x − y|.

Ω is NTA if:

• All x, y ∈ Ω are connected by a C-cigar curve with bounded turning.
• It has exterior corkscrews, i.e. for every ball B centered at ∂Ω there

is another ball B′ ⊂ B \ Ω with r(B′) ≈ r(B). A non trivial NTA domain:

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 7 / 22

Quantitative results: NTA domains

Let Ω ⊂ Rn+1 be open. For x, y ∈ Ω, a curve γ ⊂ Ω from x to y is a C-cigar curve with bounded turning if

min(H1(γ(x, z)), H1(γ(y, z))) ≤ C dist(z, Ωc) for all z ∈ γ, and H1(γ) ≤ C |x − y|.

Ω is NTA if:

• All x, y ∈ Ω are connected by a C-cigar curve with bounded turning.
• It has exterior corkscrews, i.e. for every ball B centered at ∂Ω there

is another ball B′ ⊂ B \ Ω with r(B′) ≈ r(B). A non trivial NTA domain:

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 7 / 22

Quantitative results: NTA domains

Let Ω ⊂ Rn+1 be open. For x, y ∈ Ω, a curve γ ⊂ Ω from x to y is a C-cigar curve with bounded turning if

min(H1(γ(x, z)), H1(γ(y, z))) ≤ C dist(z, Ωc) for all z ∈ γ, and H1(γ) ≤ C |x − y|.

Ω is NTA if:

• All x, y ∈ Ω are connected by a C-cigar curve with bounded turning.
• It has exterior corkscrews, i.e. for every ball B centered at ∂Ω there

is another ball B′ ⊂ B \ Ω with r(B′) ≈ r(B). A non trivial NTA domain:

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 7 / 22

One phase quantitative results for harmonic measure

Denote σ = Hn|∂Ω.

Theorem (Dahlberg)

If Ω is a Lipschitz domain, then dωp

dσ ∈ B2(σ), and thus ωp ∈ A∞(Hn|∂Ω).

Remark: dωp

dσ ∈ B2(σ) means that for any ball B centered at ∂Ω,

• −
• B

dωp dσ 2 dσ 1/2 ≤ C ωp(B) σ(B) .

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 8 / 22

One phase quantitative results for harmonic measure

Denote σ = Hn|∂Ω.

Theorem (Dahlberg)

If Ω is a Lipschitz domain, then dωp

dσ ∈ B2(σ), and thus ωp ∈ A∞(Hn|∂Ω).

Remark: dωp

dσ ∈ B2(σ) means that for any ball B centered at ∂Ω,

• −
• B

dωp dσ 2 dσ 1/2 ≤ C ωp(B) σ(B) .

Theorem (David, Jerison / Semmes, 1990)

If Ω is NTA and ∂Ω is n-AD-regular (and thus uniformly n-rectifiable), then ωp ∈ A∞(Hn|∂Ω).

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 8 / 22

One phase quantitative results for harmonic measure

Denote σ = Hn|∂Ω.

Theorem (Dahlberg)

If Ω is a Lipschitz domain, then dωp

dσ ∈ B2(σ), and thus ωp ∈ A∞(Hn|∂Ω).

Remark: dωp

dσ ∈ B2(σ) means that for any ball B centered at ∂Ω,

• −
• B

dωp dσ 2 dσ 1/2 ≤ C ωp(B) σ(B) .

Theorem (David, Jerison / Semmes, 1990)

If Ω is NTA and ∂Ω is n-AD-regular (and thus uniformly n-rectifiable), then ωp ∈ A∞(Hn|∂Ω). A related result for semiuniform domains by J. Azzam (2018).

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Two-phase problem for harmonic measure 23 August 2019 8 / 22

A quantitative two-phase result

Theorem (Azzam, T., Mourgoglou, 2018)

Let Ω1 ⊂ Rn+1 be an NTA domain such that Ω2 = Rn+1 \ Ω1 is also NTA. The following conditions are equivalent: (a) ω2 ∈ A∞(ω1).

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Two-phase problem for harmonic measure 23 August 2019 9 / 22

A quantitative two-phase result

Theorem (Azzam, T., Mourgoglou, 2018)

Let Ω1 ⊂ Rn+1 be an NTA domain such that Ω2 = Rn+1 \ Ω1 is also NTA. The following conditions are equivalent: (a) ω2 ∈ A∞(ω1). (b) both ω1 and ω2 have very big pieces of uniformly n-rectifiable measures.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 9 / 22

A quantitative two-phase result

Theorem (Azzam, T., Mourgoglou, 2018)

Let Ω1 ⊂ Rn+1 be an NTA domain such that Ω2 = Rn+1 \ Ω1 is also NTA. The following conditions are equivalent: (a) ω2 ∈ A∞(ω1). (b) both ω1 and ω2 have very big pieces of uniformly n-rectifiable measures. (c) either ω1 or ω2 has big pieces of uniformly n-rectifiable measures.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 9 / 22

A quantitative two-phase result

Theorem (Azzam, T., Mourgoglou, 2018)

Let Ω1 ⊂ Rn+1 be an NTA domain such that Ω2 = Rn+1 \ Ω1 is also NTA. The following conditions are equivalent: (a) ω2 ∈ A∞(ω1). (b) both ω1 and ω2 have very big pieces of uniformly n-rectifiable measures. (c) either ω1 or ω2 has big pieces of uniformly n-rectifiable measures. We say that µ has big pieces of uniformly n-rectifiable measures if there exists some ε ∈ (0, 1) such that, for every ball B centered at supp µ with r(B) ≤ diam(supp µ), there exist a uniformly n-rectifiable set E and G ⊂ E such that µ(B \ G) ≤ ε µ(B) and µ|G is comparable to µ(B)

r(B)n Hn|G.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 9 / 22

A quantitative two-phase result

Theorem (Azzam, T., Mourgoglou, 2018)

Let Ω1 ⊂ Rn+1 be an NTA domain such that Ω2 = Rn+1 \ Ω1 is also NTA. The following conditions are equivalent: (a) ω2 ∈ A∞(ω1). (b) both ω1 and ω2 have very big pieces of uniformly n-rectifiable measures. (c) either ω1 or ω2 has big pieces of uniformly n-rectifiable measures. We say that µ has very big pieces of uniformly n-rectifiable measures if for every ε ∈ (0, 1) and for every ball B centered at supp µ with r(B) ≤ diam(supp µ), there exist a uniformly n-rectifiable set E and G ⊂ E such that µ(B \ G) ≤ ε µ(B) and µ|G is comparable to µ(B)

r(B)n Hn|G.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 9 / 22

Previous contributions

Some relevant contributions to the (non-quantitative) 2-phase problem: Bishop, Carleson, Garnett, Jones: Jordan arcs in the plane. Bishop: general domains in the plane. Kenig-Preiss-Toro: NTA domains in Rn+1. Azzam, Mourgoglou, T.: CDC domains in Rn+1. Azzam, Mourgoglou, T., Volberg: general domains in Rn+1.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 10 / 22

An example

Ω2 Ω1

Picture by J. Azzam

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Two-phase problem for harmonic measure 23 August 2019 11 / 22

An example

Ω2 Ω1

Picture by J. Azzam

Ω1 and Ω2 are NTA domains such that H1|∂Ω1 = H1|∂Ω2 is not locally finite. However, ω1, ω2 are 1-rectifiable and ω2 ∈ A∞(ω1).

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 11 / 22

Reifenberg flat domains

For δ ∈ (0, 1), R > 0, we say that Ω ⊂ Rn+1 is (δ, R)-Reifenberg flat if, for any ball B centered at ∂Ω with r(B) ≤ R,

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 12 / 22

Reifenberg flat domains

For δ ∈ (0, 1), R > 0, we say that Ω ⊂ Rn+1 is (δ, R)-Reifenberg flat if, for any ball B centered at ∂Ω with r(B) ≤ R, (a) infL(distH(∂Ω ∩ B, L ∩ B) ≤ δ R.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 12 / 22

Reifenberg flat domains

For δ ∈ (0, 1), R > 0, we say that Ω ⊂ Rn+1 is (δ, R)-Reifenberg flat if, for any ball B centered at ∂Ω with r(B) ≤ R, (a) infL(distH(∂Ω ∩ B, L ∩ B) ≤ δ R. (b) For the optimal n-plane L, one of the connected components of B ∩

• x ∈ Rn+1 : dist(x, L) ≥ 2δ r
• is contained in Ω and the other in Rn+1 \ Ω.
• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 12 / 22

Reifenberg flat domains

For δ ∈ (0, 1), R > 0, we say that Ω ⊂ Rn+1 is (δ, R)-Reifenberg flat if, for any ball B centered at ∂Ω with r(B) ≤ R, (a) infL(distH(∂Ω ∩ B, L ∩ B) ≤ δ R. (b) For the optimal n-plane L, one of the connected components of B ∩

• x ∈ Rn+1 : dist(x, L) ≥ 2δ r
• is contained in Ω and the other in Rn+1 \ Ω.

If for all δ ∈ (0, 1) there exists R > 0 such that ∂Ω is (δ, R)-Reifenberg flat, then Ω is called vanishing Reifenberg flat.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 12 / 22

Reifenberg flat domains

For δ ∈ (0, 1), R > 0, we say that Ω ⊂ Rn+1 is (δ, R)-Reifenberg flat if, for any ball B centered at ∂Ω with r(B) ≤ R, (a) infL(distH(∂Ω ∩ B, L ∩ B) ≤ δ R. (b) For the optimal n-plane L, one of the connected components of B ∩

• x ∈ Rn+1 : dist(x, L) ≥ 2δ r
• is contained in Ω and the other in Rn+1 \ Ω.

If for all δ ∈ (0, 1) there exists R > 0 such that ∂Ω is (δ, R)-Reifenberg flat, then Ω is called vanishing Reifenberg flat. If Ω is (δ, R)-Reifenberg flat with δ small enough, then it is also NTA (Kenig-Toro).

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 12 / 22

The one-phase problem in Reifenberg flat domains

Theorem (Kenig-Toro, 97,99,03)

Let Ω ⊂ Rn+1 be a bounded δ-Reifenberg flat domain for some δ ∈ (0, 1) small enough, with n-AD-regular boundary. The following are equivalent:

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 13 / 22

The one-phase problem in Reifenberg flat domains

Theorem (Kenig-Toro, 97,99,03)

Let Ω ⊂ Rn+1 be a bounded δ-Reifenberg flat domain for some δ ∈ (0, 1) small enough, with n-AD-regular boundary. The following are equivalent: (a) log dω

dσ ∈ VMO(σ).

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 13 / 22

The one-phase problem in Reifenberg flat domains

Theorem (Kenig-Toro, 97,99,03)

Let Ω ⊂ Rn+1 be a bounded δ-Reifenberg flat domain for some δ ∈ (0, 1) small enough, with n-AD-regular boundary. The following are equivalent: (a) log dω

dσ ∈ VMO(σ).

(b) The inner unit normal NΩ ∈ VMO(σ).

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 13 / 22

The one-phase problem in Reifenberg flat domains

Theorem (Kenig-Toro, 97,99,03)

Let Ω ⊂ Rn+1 be a bounded δ-Reifenberg flat domain for some δ ∈ (0, 1) small enough, with n-AD-regular boundary. The following are equivalent: (a) log dω

dσ ∈ VMO(σ).

(b) The inner unit normal NΩ ∈ VMO(σ). (c) The inner unit normal NΩ ∈ VMO(σ) and Ω is vanishing Reifenberg flat.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 13 / 22

The one-phase problem in Reifenberg flat domains

Theorem (Kenig-Toro, 97,99,03)

Let Ω ⊂ Rn+1 be a bounded δ-Reifenberg flat domain for some δ ∈ (0, 1) small enough, with n-AD-regular boundary. The following are equivalent: (a) log dω

dσ ∈ VMO(σ).

(b) The inner unit normal NΩ ∈ VMO(σ). (c) The inner unit normal NΩ ∈ VMO(σ) and Ω is vanishing Reifenberg flat. Recall that f ∈ VMO(σ) if lim

r→0 sup x∈∂Ω

−

• B(x,r)
• f − −
• B(x,r)

f dσ

• dσ = 0.
• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 13 / 22

The one-phase problem in Reifenberg flat domains

Theorem (Kenig-Toro, 97,99,03)

Let Ω ⊂ Rn+1 be a bounded δ-Reifenberg flat domain for some δ ∈ (0, 1) small enough, with n-AD-regular boundary. The following are equivalent: (a) log dω

dσ ∈ VMO(σ).

(b) The inner unit normal NΩ ∈ VMO(σ). (c) The inner unit normal NΩ ∈ VMO(σ) and Ω is vanishing Reifenberg flat. Recall that f ∈ VMO(σ) if lim

r→0 sup x∈∂Ω

−

• B(x,r)
• f − −
• B(x,r)

f dσ

• dσ = 0.

The result is false without the δ-Reifenberg flatness assumption.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 13 / 22

The one-phase problem in Reifenberg flat domains

Theorem (Kenig-Toro, 97,99,03)

Let Ω ⊂ Rn+1 be a bounded δ-Reifenberg flat domain for some δ ∈ (0, 1) small enough, with n-AD-regular boundary. The following are equivalent: (a) log dω

dσ ∈ VMO(σ).

(b) The inner unit normal NΩ ∈ VMO(σ). (c) The inner unit normal NΩ ∈ VMO(σ) and Ω is vanishing Reifenberg flat. Recall that f ∈ VMO(σ) if lim

r→0 sup x∈∂Ω

−

• B(x,r)
• f − −
• B(x,r)

f dσ

• dσ = 0.

The result is false without the δ-Reifenberg flatness assumption. Previous result by Jerison and Kenig in C 1 domains.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 13 / 22

The two-phase problem in Reifenberg flat domains

Theorem (Prats-Tolsa, 2019)

Let Ω1 ⊂ Rn+1 be a bounded δ-Reifenberg flat domain for some δ ∈ (0, 1) small enough and let Ω2 = Rn+1 \ Ω1. The following are equivalent:

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 14 / 22

The two-phase problem in Reifenberg flat domains

Theorem (Prats-Tolsa, 2019)

Let Ω1 ⊂ Rn+1 be a bounded δ-Reifenberg flat domain for some δ ∈ (0, 1) small enough and let Ω2 = Rn+1 \ Ω1. The following are equivalent: (a) log dω2

dω1 ∈ VMO(ω1).

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 14 / 22

The two-phase problem in Reifenberg flat domains

Theorem (Prats-Tolsa, 2019)

Let Ω1 ⊂ Rn+1 be a bounded δ-Reifenberg flat domain for some δ ∈ (0, 1) small enough and let Ω2 = Rn+1 \ Ω1. The following are equivalent: (a) log dω2

dω1 ∈ VMO(ω1).

(b) Ω1 is vanishing Reifenberg flat, NΩ1 ∈ VMO(ω1), and

dω2 dω1 ∈ B3/2(ω1).

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 14 / 22

The two-phase problem in Reifenberg flat domains

Theorem (Prats-Tolsa, 2019)

Let Ω1 ⊂ Rn+1 be a bounded δ-Reifenberg flat domain for some δ ∈ (0, 1) small enough and let Ω2 = Rn+1 \ Ω1. The following are equivalent: (a) log dω2

dω1 ∈ VMO(ω1).

(b) Ω1 is vanishing Reifenberg flat, NΩ1 ∈ VMO(ω1), and

dω2 dω1 ∈ B3/2(ω1).

The result is false without the δ-Reifenberg flatness assumption.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 14 / 22

The two-phase problem in Reifenberg flat domains

Theorem (Prats-Tolsa, 2019)

Let Ω1 ⊂ Rn+1 be a bounded δ-Reifenberg flat domain for some δ ∈ (0, 1) small enough and let Ω2 = Rn+1 \ Ω1. The following are equivalent: (a) log dω2

dω1 ∈ VMO(ω1).

(b) Ω1 is vanishing Reifenberg flat, NΩ1 ∈ VMO(ω1), and

dω2 dω1 ∈ B3/2(ω1).

The result is false without the δ-Reifenberg flatness assumption. We do not assume Hn|∂Ωi to be locally finite.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 14 / 22

The two-phase problem in Reifenberg flat domains

Theorem (Prats-Tolsa, 2019)

Let Ω1 ⊂ Rn+1 be a bounded δ-Reifenberg flat domain for some δ ∈ (0, 1) small enough and let Ω2 = Rn+1 \ Ω1. The following are equivalent: (a) log dω2

dω1 ∈ VMO(ω1).

(b) Ω1 is vanishing Reifenberg flat, NΩ1 ∈ VMO(ω1), and

dω2 dω1 ∈ B3/2(ω1).

The result is false without the δ-Reifenberg flatness assumption. We do not assume Hn|∂Ωi to be locally finite. Previously Kenig and Toro had shown that (a) ⇒ Ω1 is vanishing Reifenberg flat.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 14 / 22

The two-phase problem in Reifenberg flat domains (II)

Theorem (Prats-Tolsa, 2019)

Let Ω1 ⊂ Rn+1 be a bounded δ-Reifenberg flat domain for some δ ∈ (0, 1) small enough and let Ω2 = Rn+1 \ Ω1. The following are equivalent: (a) log dω2

dω1 ∈ VMO(ω1).

(b) Ω1 is vanishing Reifenberg flat, NΩ1 ∈ VMO(ω1), and

dω2 dω1 ∈ B3/2(ω1).

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 15 / 22

The two-phase problem in Reifenberg flat domains (II)

Theorem (Prats-Tolsa, 2019)

Let Ω1 ⊂ Rn+1 be a bounded δ-Reifenberg flat domain for some δ ∈ (0, 1) small enough and let Ω2 = Rn+1 \ Ω1. The following are equivalent: (a) log dω2

dω1 ∈ VMO(ω1).

(b) Ω1 is vanishing Reifenberg flat, NΩ1 ∈ VMO(ω1), and

dω2 dω1 ∈ B3/2(ω1).

(c) Ω1 is vanishing Reifenberg flat, ω2 ∈ A∞(ω1), and lim

ρ→0 sup r(B)≤ρ

−

• B

|NΩ1 − NB| dω1 = 0, where NB is the normal to the n-plane minimizing β∞,∂Ω(B).

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 15 / 22

The two-phase problem in Reifenberg flat domains (II)

Theorem (Prats-Tolsa, 2019)

Let Ω1 ⊂ Rn+1 be a bounded δ-Reifenberg flat domain for some δ ∈ (0, 1) small enough and let Ω2 = Rn+1 \ Ω1. The following are equivalent: (a) log dω2

dω1 ∈ VMO(ω1).

(b) Ω1 is vanishing Reifenberg flat, NΩ1 ∈ VMO(ω1), and

dω2 dω1 ∈ B3/2(ω1).

(c) Ω1 is vanishing Reifenberg flat, ω1 has big pieces of uniformly rectifiable measures, and lim

ρ→0 sup r(B)≤ρ

−

• B

|NΩ1 − NB| dω1 = 0, where NB is the normal to the n-plane minimizing β∞,∂Ω(B).

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 15 / 22

Question

Is (c) is equivalent to the following? (c’) Ω1 is vanishing Reifenberg flat, ω1 has big pieces of uniformly rectifiable measures, and NΩ1 ∈ VMO(ω1).

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 16 / 22

Question

Is (c) is equivalent to the following? (c’) Ω1 is vanishing Reifenberg flat, ω1 has big pieces of uniformly rectifiable measures, and NΩ1 ∈ VMO(ω1). Equivalently, is the following true? lim

ρ→0 sup r(B)≤ρ

• NB − −
• B

NΩ1 dω1

• = 0.
• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 16 / 22

Perhaps not...

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 17 / 22

Perhaps not...

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 17 / 22

Perhaps not...

Here, NB is “essentially” vertical ↑,

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 17 / 22

Perhaps not...

Here, NB is “essentially” vertical ↑, while

−

• NΩ1 dω1 is horizontal →.
• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 17 / 22

Perhaps not...

Here, NB is “essentially” vertical ↑, while

−

• NΩ1 dω1 is horizontal →.

In fact, NΩ1 is constant ω1-a.e. and horizontal.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 17 / 22

More remarks

Other related result by Kenig-Toro, Bortz-Hofmann, and Bortz-Engelstein-Goering-Toro-Zhao assuming that Ω1 is a chord-arc domain and considering log dω1

dσ , log dω2 dσ ∈ VMO(σ).

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 18 / 22

More remarks

Other related result by Kenig-Toro, Bortz-Hofmann, and Bortz-Engelstein-Goering-Toro-Zhao assuming that Ω1 is a chord-arc domain and considering log dω1

dσ , log dω2 dσ ∈ VMO(σ).

Results of Engelstein assuming that log dω2

dω1 ∈ C α, for some α > 0.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 18 / 22

More remarks

Other related result by Kenig-Toro, Bortz-Hofmann, and Bortz-Engelstein-Goering-Toro-Zhao assuming that Ω1 is a chord-arc domain and considering log dω1

dσ , log dω2 dσ ∈ VMO(σ).

Results of Engelstein assuming that log dω2

dω1 ∈ C α, for some α > 0.

Without the Reifenberg flatness condition, Badger, Engelstein and Toro have studied the singular set of ∂Ωi, showing uniqueness of the blowup when log dω2

dω1 ∈ C α. Previous contributions of Badger.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 18 / 22

Ideas for (a) ⇒ (b)

The vanishing Reifenberg flat condition was proved by Kenig-Toro.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 19 / 22

Ideas for (a) ⇒ (b)

The vanishing Reifenberg flat condition was proved by Kenig-Toro. To show that NΩ1 ∈ VMO(ω1): We use jump identities for the Riesz transform valid for arbitrary n-rectifiable sets [T.]: R+ω1(x) − R−ω1(x) = cnΘn(x, ω1) NΩ1(x) for ω1-a.e. x, where Θn(x, ω1)(x) = lim

r→0

ω1(B(x, r)) (2r)n .

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 19 / 22

Ideas for (a) ⇒ (b)

The vanishing Reifenberg flat condition was proved by Kenig-Toro. To show that NΩ1 ∈ VMO(ω1): We use jump identities for the Riesz transform valid for arbitrary n-rectifiable sets [T.]: R+ω1(x) − R−ω1(x) = cnΘn(x, ω1) NΩ1(x) for ω1-a.e. x, where Θn(x, ω1)(x) = lim

r→0

ω1(B(x, r)) (2r)n . We show that R+ω1(x) − R−ω1(x) is small using the fact that ω1 is close to c ω2 at small scales and the connection between Riesz transform and harmonic measure.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 19 / 22

Ideas for (a) ⇒ (b)

The vanishing Reifenberg flat condition was proved by Kenig-Toro. To show that NΩ1 ∈ VMO(ω1): We use jump identities for the Riesz transform valid for arbitrary n-rectifiable sets [T.]: R+ω1(x) − R−ω1(x) = cnΘn(x, ω1) NΩ1(x) for ω1-a.e. x, where Θn(x, ω1)(x) = lim

r→0

ω1(B(x, r)) (2r)n . We show that R+ω1(x) − R−ω1(x) is small using the fact that ω1 is close to c ω2 at small scales and the connection between Riesz transform and harmonic measure. We show that Θn(x, ω1) is far from 0 in big sets using a criterion for rectifiability of general Radon measures of Girela-Sarri´

• n and T.

To show that it is not too big we apply the Alt-Caffarelli-Friedman formula.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 19 / 22

Ideas for (b) ⇒ (a)

To show that log dω2

dω1 ∈ VMO(ω1):

(1) We approximate Ω1, Ω2 by Reifenberg flat chord-arc domains Ω1, Ω2 using a technique from Azzam-Mourgoglou-T.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 20 / 22

Ideas for (b) ⇒ (a)

To show that log dω2

dω1 ∈ VMO(ω1):

(1) We approximate Ω1, Ω2 by Reifenberg flat chord-arc domains Ω1, Ω2 using a technique from Azzam-Mourgoglou-T. (2) We show that the approximating domains Ω1, Ω2 have normal in VMO( σ), where σ = Hn|∂

Ω1.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 20 / 22

Ideas for (b) ⇒ (a)

To show that log dω2

dω1 ∈ VMO(ω1):

(1) We approximate Ω1, Ω2 by Reifenberg flat chord-arc domains Ω1, Ω2 using a technique from Azzam-Mourgoglou-T. (2) We show that the approximating domains Ω1, Ω2 have normal in VMO( σ), where σ = Hn|∂

Ω1.

(3) By Kenig-Toro, log d

ωi d σ ∈ VMO(

σ). By applying the maximum principle and some arguments from Korey, we deduce that log dω2

dω1 ∈ VMO(ω1).

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 20 / 22

Ideas for (b) ⇒ (a)

To show that log dω2

dω1 ∈ VMO(ω1):

(1) We approximate Ω1, Ω2 by Reifenberg flat chord-arc domains Ω1, Ω2 using a technique from Azzam-Mourgoglou-T. (2) We show that the approximating domains Ω1, Ω2 have normal in VMO( σ), where σ = Hn|∂

Ω1.

(3) By Kenig-Toro, log d

ωi d σ ∈ VMO(

σ). By applying the maximum principle and some arguments from Korey, we deduce that log dω2

dω1 ∈ VMO(ω1).

The most delicate step is (2): At some point we have to show that

−

• B NΩ1 dω1 is close to NB, the unit

normal of the best approximating n-plane for B ∩ ∂Ω1.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 20 / 22

Where is used the condition B3/2?

We apply again the jump identity for Rω. In some step we need to estimate

• B

Θ(x, ω1) dω1(x).

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 21 / 22

Where is used the condition B3/2?

We apply again the jump identity for Rω. In some step we need to estimate

• B

Θ(x, ω1) dω1(x). By the ACF formula, we know that Θ(x, ω1) Θ(x, ω2) ω1(B) r(B)n ω2(B) r(B)n =: CB.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 21 / 22

Where is used the condition B3/2?

We apply again the jump identity for Rω. In some step we need to estimate

• B

Θ(x, ω1) dω1(x). By the ACF formula, we know that Θ(x, ω1) Θ(x, ω2) ω1(B) r(B)n ω2(B) r(B)n =: CB. Then

• B

Θ(x, ω1) dω1(x) C 1/2

B

• B

Θ(x, ω1)1/2 Θ(x, ω2)1/2 dω1(x) =

• B

dω1 dω2 1/2 dω1(x) =

• B

dω1 dω2 3/2 dω2(x) < ∞.

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 21 / 22

Thank you! Congratulations, John and Don!

• X. Tolsa (ICREA / UAB)

Two-phase problem for harmonic measure 23 August 2019 22 / 22