aa r X i v : . [ m a t h . C A ] J un Champagne subregions of the unit disc
Joanna Pres ∗ Abstract
This paper concerns harmonic measure on the domains that arise when infinitelymany disjoint closed discs are removed from the unit disc. It investigates which con-figurations of discs are unavoidable for Brownian motion, and obtains refinements ofrelated results of Akeroyd, and of Ortega-Cerd`a and Seip.
Keywords harmonic measure, capacity.
Let B ( x, r ) denote the open ball of centre x and radius r > R n ( n ≥ B = B (0 , B ( x k , r k ) such that | x k | → r k / (1 − | x k | ) <
1, and use them to form a champagne subregion
Ω = B \ E of the unit ball, where E = S ∞ k =1 B ( x k , r k ). We assume for convenience that 0 ∈ Ω. Wesay that E is unavoidable if it carries full harmonic measure with respect to Ω, that is, theharmonic measure of ∂B relative to Ω is zero. This can be viewed as the case where theprobability that Brownian motion starting from the origin first exits Ω through ∂B is zero.Akeroyd [3] has shown that the following phenomenon occurs. Theorem A
Let ε > . There is a champagne subregion of the unit disc B such that P k r k < ε and yet E is unavoidable. Ortega-Cerd`a and Seip [7] provided a description of unavoidable configurations of discs in B and improved Akeroyd’s result. Theorem B (a)
Let ( x k ) be a sequence in B ⊂ R satisfying inf j = k | x j − x k | − | x k | > and B ( x, a (1 − | x | )) ∩ { x k : k ∈ N } 6 = ∅ ( x ∈ B ) (1.2) ∗ School of Mathematical Sciences, University College Dublin, Belfield, Dublin 4, Ireland;[email protected], [email protected] or some a ∈ (0 , . Let r k = (1 − | x k | ) φ ( | x k | ) for some decreasing function φ : [0 , → (0 , . Then E is unavoidable if and only if Z − t ) log(1 /φ ( t )) dt = ∞ . (b) For any ( x k ) satisfying (1.1) and (1.2) , and for any ε > , there exists a champagnesubregion of B ⊂ R such that P k r k < ε and E is unavoidable. Recently Gardiner and Ghergu [6] obtained the following result for more general champagnesubregions when n ≥
3, where the separation condition (1.1) is relaxed. Let σ denotenormalised surface area measure on ∂B (normalised arclength on ∂B , if n = 2). Theorem C
Let Ω be a champagne subregion of the unit ball in R n ( n ≥ ). (a) If E is unavoidable, then X k (1 − | x k | ) | y − x k | r n − k = ∞ f or σ − almost every y ∈ ∂B. (1.3)(b) If (1.3) holds together with the separation condition inf j = k | x j − x k | r − /nk (1 − | x k | ) > , (1.4) then E is unavoidable. If we substitute n = 2 into (1.4) we obtain (1.1), so it would be natural to assume thatthe latter is the appropriate separation condition to use in the case of the disc. However, amore careful analysis of the plane case yields a stronger, and less obvious, result. This will,in turn, lead to interesting refinements of Theorems A and B. Theorem 1.1.
Let
Ω = B \ E be a champagne subregion of the unit disc B .(a) If E is unavoidable, then X k (1 − | x k | ) | y − x k | (cid:26) log 1 − | x k | r k (cid:27) − = ∞ for σ − almost every y ∈ ∂B. (1.5) In particular, X k (1 − | x k | ) | y − x k | = ∞ f or σ − almost every y ∈ ∂B. b) Conversely, if (1.5) holds together with the separation condition inf j = k | x k − x j | n log −| x k | r k o / − | x k | > , (1.6) then E is unavoidable. Using Theorem 1.1 we can describe the unavoidable configurations of discs for which r k = (1 − | x k | ) φ ( | x k | ), where φ : [0 , → (0 ,
1) is decreasing. Let M : [0 , → [1 , ∞ ) be anincreasing function satisfying M (1 − t/ ≤ cM (1 − t ) (0 < t ≤ c >
1. The number of centres in a given disc will be denoted by N a ( x ) = B ( x, a (1 − | x | )) ∩ { x k : k ∈ N } ] for some a ∈ (0 , . We can now present our refinement of Theorem B(a).
Theorem 1.2.
Let
Ω = B \ E be a champagne subregion of the unit disc, where r k =(1 − | x k | ) φ ( | x k | ) .(a) If E is unavoidable and there are constants a ∈ (0 , and b > such that N a ( x ) ≤ bM ( | x | ) for all x ∈ B , then Z M ( t )(1 − t ) log(1 /φ ( t )) dt = ∞ . (1.7) (b) Conversely, if (1.7) holds together with inf j = k | x k − x j | n log φ ( | x k | ) o / − | x k | > , (1.8) and if there are constants a ∈ (0 , and b > such that N a ( x ) ≥ bM ( | x | ) for all x ∈ B ,then E is unavoidable. From this we deduce a further strengthening of Theorem A.
Corollary 1.3.
Let ε > .(a) For any α > there is a champagne subregion of the unit disc such that P k r αk < ε and E is unavoidable.(b) For any α > there is a champagne subregion of the unit disc such that X k (cid:26) log 1 r k (cid:27) − α < ε and E is unavoidable.
3e observe that for every α > c ( α ) > r α ≤ c ( α ) { log(1 /r ) } − for all r ∈ (0 , Remark 1.4.
The conclusion of Corollary 1.3(b) fails when α = 1. For details see the endof Section 3.We will establish the above results by combining the methods of Gardiner and Ghergu[6] with some new ideas. From now on we will assume that the dimension n is 2. We referto [4] for the relevant background material on potential theory. For a positive superharmonic function u on B we define the reduced function of u relative toa subset E of BR Eu = inf { v : v is positive and superharmonic on B and v ≥ u on E } . The Poisson kernel for B is given by P ( x, y ) = 1 − | x | | x − y | for y ∈ ∂B and x ∈ B. In view of [4, Theorem 6.9.1], E is unavoidable if and only if R E (0) = 1. By the use of thesame theorem and the fact that R ∂B P ( · , y ) dσ ( y ) = 1 it can be seen that E is unavoidableif and only if R EP ( · ,y ) (0) = 1 = P (0 , y ) for σ -almost every y ∈ ∂B . Since Ω is connectedand P ( · , y ) − R EP ( · ,y ) ( y ∈ ∂B ) is a nonnegative harmonic function on Ω, it follows from themaximum principle that E is unavoidable if and only if R EP ( · ,y ) ≡ P ( · , y ) for σ − almost every y ∈ ∂B. (2.1)Now for n ∈ N and m ∈ Z such that 0 ≤ m < n +4 let S m,n = (cid:26) re iθ : 2 − n − ≤ − r ≤ − n and 2 πm n +4 ≤ θ ≤ π ( m + 1)2 n +4 (cid:27) and z m,n = (1 − − n ) exp(2 πim/ n +4 ) . It is easy to see that the diameter diam( S m,n ) of S m,n satisfies2 − n ≤ diam( S m,n ) ≤ − n √ . (2.2)Theorem 1 of [5] (cf. [2, Corollary 7.4.4]) tells us that R EP ( · ,y ) ≡ P ( · , y ) if and only if X m,n (cid:26) − n | z m,n − y | (cid:27) (cid:26) log 2 − n c ( E ∩ S m,n ) (cid:27) − = ∞ , (2.3)4here c ( · ) denotes logarithmic capacity, and n log − n c ( E ∩ S m,n ) o − is interpreted as 0 whenever E ∩ S m,n is polar.Before commencing the proof of Theorem 1.1 we state the following lemma. Lemma 2.1.
There is a constant c > such that, for any S m,n and any B ( x k , r k ) whichintersects S m,n : c − − n ≤ − | x k | ≤ c − n , and c − ≤ | z m,n − y || x k − y | ≤ c for all y ∈ ∂B. The constant c depends on the value of sup { r k / (1 − | x k | ) } . The proof of Lemma 2.1 is straightforward and hence omitted.We shall also make use of the elementary observation that, if p > f ( t ) = log( pt ) / log t ( t >
1) is decreasing.
Proof of Theorem 1.1(a).
Suppose that E is unavoidable. Applying [8, Theorem 5.1.4 (a)],concerning a relation between the logarithmic capacity and unions, and Lemma 2.1 we obtain X m,n (cid:26) − n | z m,n − y | (cid:27) (cid:26) log 2 − n c ( E ∩ S m,n ) (cid:27) − ≤ X k X m,n (cid:26) − n | z m,n − y | (cid:27) (cid:26) log 2 − n c ( B ( x k , r k ) ∩ S m,n ) (cid:27) − ≤ c X k (1 − | x k | ) | x k − y | X m,n (cid:26) log 2 − n c ( B ( x k , r k ) ∩ S m,n ) (cid:27) − ≤ C ( c ) X k (1 − | x k | ) | x k − y | X m,n (cid:26) log c − n c ( B ( x k , r k ) ∩ S m,n ) (cid:27) − . The last inequality can be deduced from the observation following Lemma 2.1 and (2.2)combined with the fact that c ( B ( x k , r k ) ∩ S m,n ) ≤ diam( S m,n ). Since a given disc B ( x k , r k )intersects at most c of the sets S m,n (where c is independent of k and of m, n ) and by thefact that c ( B ( x k , r k ) ∩ S m,n ) ≤ c ( B ( x k , r k )) = r k , we obtain X m,n (cid:26) − n | z m,n − y | (cid:27) (cid:26) log 2 − n c ( E ∩ S m,n ) (cid:27) − ≤ C ( c ) c X k (1 − | x k | ) | y − x k | (cid:26) log 1 − | x k | r k (cid:27) − . In view of (2.1) and (2.3) we see that (1.5) holds if E is unavoidable. Following [1], we define the capacity C by C ( A ) = inf (cid:26) µ ( R ) : Z log + | x − y | dµ ( y ) ≥ A (cid:27) . C ( B ( x, r )) is comparable with (cid:26) r Z r t log + t dt (cid:27) − , from which it follows that there exists a constant c > c − ( log 1 r ) − ≤ C ( B ( x, r )) ≤ c ( log 1 r ) − (0 < r ≤ / . (2.4)Let | A | denote the Lebesgue measure of a planar set A , and for α > αA = { αx : x ∈ A } .We write A ◦ to denote the interior of A . We recall the following quasiadditivity property ofthe capacity C (see [1, Theorem 3]). Theorem D
For k ∈ N and ρ k > let η ( ρ k ) be such that | B ( y k , η ( ρ k )) | = C ( B ( y k , ρ k )) .Let η ∗ ( ρ k ) = max { η ( ρ k ) , ρ k } . If the discs { B ( y k , η ∗ ( ρ k )) } are pairwise disjoint and F is ananalytic subset of S k B ( y k , ρ k ) , then there is a constant c > such that C ( F ) ≥ c X k C ( F ∩ B ( y k , ρ k )) . (2.5)Using (2.4) we deduce that if F is an analytic subset of S k B ( y k , ρ k ), where ρ k ≤ / (4 πc )for all k , and if the discs n B (cid:16) y k , [ πc log(1 /ρ k )] − / (cid:17) : k ∈ N o are pairwise disjoint, then (2.5) holds. Lemma 2.2.
Suppose that | x j − x k | (cid:16) log −| x k | r k (cid:17) / − | x k | ≥ √ πc c ( j = k ) , (2.6) where c is as in Lemma 2.1 and c as in (2.4) . Then, for each , C ( α n [ ∪ k B ( x k , r k ) ∩ S ◦ m,n ]) ≥ c X k C ( α n [ B ( x k , r k ) ∩ S ◦ m,n ]) , (2.7) where α n = (4 πc c ) − n and c is as in Theorem D.Proof. Let F = ∪ k [ B ( y k , ρ k ) ∩ α n S ◦ m,n ], where y k = α n x k and ρ k = α n r k . We observe that ρ k = α n r k ≤ α n (1 − | x k | ) ≤ α n c − n = 14 πc . k such that B ( x k , r k ) ∩ S m,n = ∅ we also havelog 1 ρ k = log 4 πc c − n r k ≥ log 1 − | x k | r k . For j = k , we thus see from (2.6) that | y k − y j | { πc log(1 /ρ k ) } − / = α n | x k − x j |{ log(1 /ρ k ) } / πc ) − / ≥ α n | x k − x j | n log −| x k | r k o / πc ) − / ≥ √ πc c (1 − | x k | ) α n πc ) − / ≥ πc c / − n α n = c / > . Hence the required inequality (2.7) holds by (2.5).
Remark 2.3. (a) Suppose that E ∩ S m,n is non-polar. We know by [4, Lemma 5.8.1] thatthere exists a unit measure ν on ∂ ( α n [ E ∩ S m,n ]) such that − log c ( α n [ E ∩ S m,n ]) = Z log 1 | x − y | dν ( y )for each x ∈ α n ( E ◦ ∩ S ◦ m,n ). Since, for x, y ∈ α n ( E ∩ S m,n ), we have | x − y | ≤ α n diam( S m,n ) ≤ πc c < , it follows that Z log + | x − y | dν ( y ) ≥ − log c ( α n [ E ∩ S m,n ]) > x ∈ α n ( E ◦ ∩ S ◦ m,n ) . Hence C ( α n [ E ◦ ∩ S ◦ m,n ]) ≤ ν ( R ) − log c ( α n [ E ∩ S m,n ]) = 1log c ( α n [ E ∩ S m,n ]) . (2.8)Furthermore, since c ( α n [ E ∩ S m,n ]) = α n c ( E ∩ S m,n ), we have C ( α n [ E ◦ ∩ S ◦ m,n ]) ≤ ( log 2 − n c ( E ∩ S m,n ) ) − . (2.9)7b) Let k ∈ N . Suppose that r k ≤ (2 c ) − (1 − | x k | ), that is, the diameter of B ( x k , r k )is not greater than the length of the shortest line segment joining points of S m,n , where S m,n ∩ B ( x k , r k ) = ∅ . Then B ( x k , r k ) intersects at most four sets S m,n , and one of themcontains a disc of radius r k /
4. Recall that α n = { πc c } − n . Since 0 < α n r k < /
2, we canuse (2.4) to obtain C ( α n [ B ( x k , r k ) ∩ S ◦ m,n ]) ≥ c − (cid:26) log 4 α n r k (cid:27) − = c − (cid:26) log 16 πc c − n r k (cid:27) − ≥ c − (cid:26) log 16 πc c (1 − | x k | ) r k (cid:27) − ≥ C ( c , c ) (cid:26) log 1 − | x k | r k (cid:27) − . In general, if r k ≤ − | x k | , then (2 c ) − r k ≤ (2 c ) − (1 − | x k | ) and there exists S m,n thatintersects B ( x k , r k ) and contains a disc of radius r k / (2 c ). Thus C ( α n [ B ( x k , r k ) ∩ S ◦ m,n ]) ≥ c − (cid:26) log 128 c α n r k (cid:27) − ≥ C ( c , c ) (cid:26) log 4 α n r k (cid:27) − ≥ C ( c , c ) (cid:26) log 1 − | x k | r k (cid:27) − . Proof of Theorem 1.1(b).
Suppose that (1.5) holds, together with separation condition (1.6).In view of (2.1) and (2.3), in order to prove that E is unavoidable it is enough to show that,for σ -almost every y ∈ ∂B , X m,n (cid:26) − n | z m,n − y | (cid:27) (cid:26) log 2 − n c ( E ∩ S m,n ) (cid:27) − = ∞ . Let δ ∈ (0 ,
1) be small enough so that for all k , | x k − x j | n log −| x k | δr k o / − | x k | ≥ √ πc c ( j = k ) . E δ = ∪ k B ( x k , δr k ). We see from Lemma 2.2, Lemma 2.1 and Remark 2.3(b) that X m,n (cid:26) − n | z m,n − y | (cid:27) C ( α n [ E δ ∩ S ◦ m,n ]) ≥ c X k X m,n (cid:26) − n | z m,n − y | (cid:27) C ( α n [ B ( x k , δr k ) ∩ S ◦ m,n ]) ≥ C ( c , c , c ) X k (1 − | x k | ) | y − x k | (cid:26) log 1 − | x k | δr k (cid:27) − . Moreover, by the observation following Lemma 2.1, we have log −| x k | δr k ≤ c δ − log −| x k | r k , where c depends only on the value of sup { r k / (1 − | x k | ) } . Hence X m,n (cid:26) − n | z m,n − y | (cid:27) C ( α n [ E δ ∩ S ◦ m,n ]) ≥ C ( c , c , c , c ) δ X k (1 − | x k | ) | y − x k | (cid:26) log 1 − | x k | r k (cid:27) − . Now, the fact that E δ ⊆ E ◦ and (1.5) yield X m,n (cid:26) − n | z m,n − y | (cid:27) C ( α n [ E ◦ ∩ S ◦ m,n ]) = ∞ for σ − almost every y ∈ ∂B. Therefore, by (2.9), X m,n (cid:26) − n | z m,n − y | (cid:27) (cid:26) log 2 − n c ( E ∩ S m,n ) (cid:27) − = ∞ , and we conclude that E is unavoidable. Proof of Theorem 1.2(a).
Suppose that E is unavoidable and that N a ( x ) ≤ bM ( | x | ) ( x ∈ B ).It follows from Theorem 1.1(a) that X k (1 − | x k | ) | y − x k | (cid:26) log 1 φ ( | x k | ) (cid:27) − = ∞ for σ − almost every y ∈ ∂B. (3.1)Reasoning as in the proof of [6, Theorem 2] with { φ ( t ) } n − replaced by { log(1 /φ ( t )) } − wesee that Z B M ((3 | x | − + )(1 − | x | ) log { /φ ((3 | x | − + ) } dx = ∞ .
9t follows that ∞ = Z / M (3 t − t (1 − t ) log { /φ (3 t − } dt ≤ Z / M (3 t − − t ) log { /φ (3 t − } dt, and so (1.7) holds. Proof of Theorem 1.2(b).
Again, by adapting the method from the proof of [6, Theorem2(b)], it can be shown that if Ω is a champagne subregion with r k = (1 − | x k | ) φ ( | x k | ) forwhich (1.7) holds, and if there are constants a ∈ (0 ,
1) and b > N a ( x ) ≥ bM ( | x | )for all x ∈ B , then X k (1 − | x k | ) | y − x k | (cid:26) log 1 − | x k | r k (cid:27) − = ∞ ( y ∈ ∂B ) . Since (1.8) corresponds to (1.6), it follows from Theorem 1.1(b) that E is unavoidable. Proof of Corollary 1.3.
Since (b) implies (a) in Corollary 1.3 we only prove part (b). Let ε > α >
1. Let β > / ( α −
1) and φ ( t ) = exp (cid:16) − c (1 − t ) β (cid:17) for t ∈ [0 ,
1) and some c ∈ (0 , p n be the integer part of 2 nβ/ . We divide each “square” S m,n into p n “subsquares”. Let { x k : k ∈ N } be the collection of centres of those “subsquares”, and let r k = (1 − | x k | ) φ ( | x k | ). Then the following observations can be made.(a) Each x k belongs to some S m,n and so 1 −| x k | ≤ − n . Moreover, since | x k − x j | ≥ γ − n /p n ( j = k ) for some universal constant γ >
0, for j = k we have | x k − x j | n log φ ( | x k | ) o / − | x k | = | x k − x j |{ c (1 − | x k | ) β } / (1 − | x k | ) ≥ C ( c , γ ) 2 − n p n − | x k | ) β/ ≥ C ( c , γ ) 2 − n nβ/ − n ) β/ = C ( c , γ ) . (b) Let M : [0 , → [1 , ∞ ) be given by M ( t ) = (1 − t ) − β . Then M satisfies M (1 − t/
2) = 2 β M (1 − t ) ( t ∈ (0 , , and clearly (1.7) holds.(c) If 0 < a < N a ( x ) ≥ M ( | x | ) for all x ∈ B .(d) If c is sufficiently small, the closed discs B ( x k , r k ) are pairwise disjoint.10t follows from Theorem 1.2(b) that E is unavoidable.Finally, since r k ≤ exp (cid:16) − c (1 −| x k | ) β (cid:17) , we deduce that X k (cid:16) log r k (cid:17) α ≤ c α X k (1 − | x k | ) αβ ≤ c α X n (2 − n ) αβ p n n +4 ≤ c α X n (cid:8) − β ( α − (cid:9) n . Since − β ( α −
1) + 1 <
0, the above geometric series converges and so does P k { log(1 /r k ) } − α .By omitting a finite number of the discs we can arrange that P k { log(1 /r k ) } − α < ε and yetthe collection of discs is unavoidable. Details of Remark 1.4 . To see that the conclusion of Corollary 1.3(b) fails when α = 1,suppose that there is a champagne subregion Ω of the unit disc such that P k { log r k } − converges. Since | x k | →
1, by omitting a finite number of discs we can have B ( x k , r k ) ⊂ B \ B (0 , /
2) for each k and P k { log r k } − ≤ / (2 log 4). Let A be the union of all the remainingdiscs B ( x k , r k ). Since the Green capacity of B ( x k , r k ) relative to B (0 ,
2) is dominated by1 / log(1 /r k ) (see [4, (5.8.5)]), it follows that the value of the capacitary Green potential of A on B (0 ,
2) is not bigger than 1 / A is avoidable and so is thewhole collection { B ( x k , r k ) : k ∈ N } . References [1] H. Aikawa and A. A. Borichev. Quasiadditivity and measure property of capacity andthe tangential boundary behavior of harmonic functions.
Trans. Amer. Math. Soc. ,348(3):1013–1030, 1996.[2] H. Aikawa and M. Ess´en.
Potential theory—selected topics . Lecture Notes in Mathematics,Volume 1633. Springer, Berlin, 1996.[3] J. R. Akeroyd. Champagne subregions of the disk whose bubbles carry harmonic measure.
Math. Ann. , 323(2):267–279, 2002.[4] D. H. Armitage and S. J. Gardiner.
Classical potential theory . Springer Monographs inMathematics. Springer, London, 2001.[5] M. Ess´en. On minimal thinness, reduced functions and Green potentials.
Proc. EdinburghMath. Soc. (2) , 36(1):87–106, 1992.[6] S. J. Gardiner and M. Ghergu. Champagne subregions of the unit ball with unavoidablebubbles.
Ann. Acad. Sci. Fenn. Math. , 35(1):321–329, 2010.[7] J. Ortega-Cerd`a and K. Seip. Harmonic measure and uniform densities.
Indiana Univ.Math. J. , 53(3):905–923, 2004.[8] T. Ransford.