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 ⊂ R n +1 is n -rectifiable if it is H n -a.e. contained in a countable union of C 1 (or Lipschitz) n -dimensional manifolds. X. Tolsa (ICREA / UAB) Two-phase problem for harmonic measure 23 August 2019 2 / 22
Rectifiability E ⊂ R n +1 is n -rectifiable if it is H n -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 H n | E , with E n -rectifiable and g ∈ L 1 loc ( H n | E ). X. Tolsa (ICREA / UAB) Two-phase problem for harmonic measure 23 August 2019 2 / 22
Uniform rectifiability E ⊂ R d is n -AD-regular if H n ( E ∩ B ( x , r )) ≈ r n for all x ∈ E , 0 < r ≤ diam( E ) . X. Tolsa (ICREA / UAB) Two-phase problem for harmonic measure 23 August 2019 3 / 22
Uniform rectifiability E ⊂ R d is n -AD-regular if H n ( 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 : R n ⊃ B n (0 , r ) → R d , �∇ g � ∞ ≤ M , such that H n � � ≥ θ r n . E ∩ B ( x , r ) ∩ g ( B n (0 , r )) X. Tolsa (ICREA / UAB) Two-phase problem for harmonic measure 23 August 2019 3 / 22
Uniform rectifiability E ⊂ R d is n -AD-regular if H n ( 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 : R n ⊃ B n (0 , r ) → R d , �∇ g � ∞ ≤ M , such that H n � � ≥ θ r n . E ∩ B ( x , r ) ∩ g ( B n (0 , r )) A Borel measure µ is n -AD-regular if it is of the form µ = g H n | 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 ⊂ R n +1 open and connected. For p ∈ Ω, ω p is the harmonic measure in Ω with pole in p . X. Tolsa (ICREA / UAB) Two-phase problem for harmonic measure 23 August 2019 4 / 22
Harmonic measure Ω ⊂ R n +1 open and connected. For p ∈ Ω, ω p is the harmonic measure in Ω with pole in p . That is, for E ⊂ ∂ Ω, ω p ( E ) is the value at p of the harmonic extension of χ E to Ω. X. Tolsa (ICREA / UAB) Two-phase problem for harmonic measure 23 August 2019 4 / 22
Harmonic measure Ω ⊂ R n +1 open and connected. For p ∈ Ω, ω p is the harmonic measure in Ω with pole in p . That is, for E ⊂ ∂ Ω, ω p ( E ) is the value at p of the harmonic extension of χ E to Ω. Questions about the metric properties of harmonic measure: When H n ≈ ω p ? 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 ⊂ R n +1 open and connected. For p ∈ Ω, ω p is the harmonic measure in Ω with pole in p . That is, for E ⊂ ∂ Ω, ω p ( E ) is the value at p of the harmonic extension of χ E to Ω. Questions about the metric properties of harmonic measure: When H n ≈ ω p ? Which is the connection with rectifiability? Dimension of harmonic measure? In the plane if Ω is simply connected and H 1 ( ∂ Ω) < ∞ , then H 1 ≈ ω p . (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 R n +1 : 1 E ( x ) = c n | 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 R n +1 : 1 E ( x ) = c n | x | n − 1 . The kernel of the Riesz transform R is x K ( x ) = | x | n +1 = c ∇E ( x ) . 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 R n +1 : 1 E ( x ) = c n | x | n − 1 . The kernel of the Riesz transform R is x K ( x ) = | x | n +1 = c ∇E ( x ) . The Green function G ( · , · ) of Ω is � E ( x − y ) d ω p ( y ) . G ( x , p ) = E ( x − p ) − 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 R n +1 : 1 E ( x ) = c n | x | n − 1 . The kernel of the Riesz transform R is x K ( x ) = | x | n +1 = c ∇E ( x ) . The Green function G ( · , · ) of Ω is � E ( x − y ) d ω p ( y ) . G ( x , p ) = E ( x − p ) − Therefore, for x ∈ Ω: � K ( x − y ) d ω p ( y ) . c ∇ x G ( x , p ) = K ( x − p ) − c ∇ x G ( x , p ) = K ( x − p ) − R ω p ( x ) . That is, 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 H n | ∂ 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 H n | ∂ Ω . In two-phase problems for harmonic measure we have two domains Ω 1 , Ω 2 with ∂ Ω 1 ∩ ∂ Ω 2 � = ∅ and usually we are interested in the connection between and ω 1 , ω 2 and H n | ∂ Ω i in ∂ Ω 1 ∩ ∂ Ω 2 . Ω 1 Ω 2 E X. Tolsa (ICREA / UAB) Two-phase problem for harmonic measure 23 August 2019 6 / 22
Quantitative results: NTA domains Let Ω ⊂ R n +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( H 1 ( γ ( x , z )) , H 1 ( γ ( y , z ))) ≤ C dist( z , Ω c ) for all z ∈ γ , and H 1 ( 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 ⊂ R n +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( H 1 ( γ ( x , z )) , H 1 ( γ ( y , z ))) ≤ C dist( z , Ω c ) for all z ∈ γ , and H 1 ( 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 ⊂ R n +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( H 1 ( γ ( x , z )) , H 1 ( γ ( y , z ))) ≤ C dist( z , Ω c ) for all z ∈ γ , and H 1 ( 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
b b Quantitative results: NTA domains Let Ω ⊂ R n +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( H 1 ( γ ( x , z )) , H 1 ( γ ( y , z ))) ≤ C dist( z , Ω c ) for all z ∈ γ , and H 1 ( 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
b b Quantitative results: NTA domains Let Ω ⊂ R n +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( H 1 ( γ ( x , z )) , H 1 ( γ ( y , z ))) ≤ C dist( z , Ω c ) for all z ∈ γ , and H 1 ( 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
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