GROWTH AND ASYMPTOTIC SETS OF SUBHARMONIC FUNCTIONS II

We study the relation between the growth of a subharmonic function in the half space R + and the size of its asymptotic set. In particular, we prove that for any n ≥ 1 and 0 < α ≤ n, there exists a subharmonic function u in the R + satisfying the growth condition of order α : u(x) ≤ x−α n+1 for 0 < xn+1 < 1, such that the Hausdorff dimension of the asymptotic set ⋃ λ6=−∞ A(λ) is exactly n−α. Here A(λ) is the set of boundary points at which f tends to λ along some curve. This proves the sharpness of a theorem due to Berman, Barth, Rippon, Sons, Fernández, Heinonen, Llorente and Gardiner cumulatively. Research partially supported by the National Science Foundation.

satisfying the growth condition of order α : u(x) ≤ x −α n+1 for 0 < x n+1 < 1, such that the Hausdorff dimension of the asymptotic set Here A(λ) is the set of boundary points at which f tends to λ along some curve.This proves the sharpness of a theorem due to Berman, Barth, Rippon, Sons, Fernández, Heinonen, Llorente and Gardiner cumulatively.
A function f defined in a domain D is said to have an asymptotic value b ∈ [−∞, ∞] at a point a ∈ ∂D provided that there exists a path γ in D ending at a so that u(p) tends to b as p tends to a along γ.The set of all points on ∂D at which f has an asymptotic value b is denoted by A(f, b) and called the asymptotic set for the value b.
G. R. MacLane [M1], [M2] studied the class of analytic functions in the unit disk having asymptotic values at a dense subset of the unit circle.Hornblower studied the analogous class of subharmonic functions.Since then, many have worked on problems of the following nature: for a subharmonic function u of a certain growth, if A(u, +∞) is a small set, then u has nice boundary behavior on a large set.Denote by R n+1 + = {(x, y) : x = (x 1 , . . ., x n ) ∈ R n , y > 0} the upper half space in R n+1 .For α > 0, denote by M α the class of subharmonic functions u in R n+1 + which satisfy the growth condition: Research partially supported by the National Science Foundation.

Denote by F(u) the Fatou set consisting of points on ∂R n+1
+ at which u has finite vertical limits.For β > 0, denote by H β the β-dimensional Hausdorff content.
Theorem A. Let n ≥ 1, 0 < α ≤ n and u be a subharmonic function in the class M α .Let B be a ball on ∂R n+1 + .If We prove in this note that Theorem B is sharp.
Theorem 1.Given n ≥ 1 and 0 < α ≤ n, there exists a subharmonic function u in M α so that It has been proved ( [FHL], [W]) that for 0 < α ≤ 1, there exists a harmonic function In general, there is more flexibility in constructing subharmonic functions than harmonic functions.In the proof of Theorem 1, we assign values of u on a grid in R n+1 + to force u to have the desired growth, and shift the cumbersome work to the proof of subharmonicity.In order to construct such harmonic functions, we need to assure the harmonicity before regulating the growth.Our attempts have been unsuccessful when 1 < α ≤ n and it is not clear whether n − α is the critical dimension in the harmonic case.
We proceed to prove Theorem 1 for n ≥ 2, using ideas from [FHL] and [W].
From now on, assume n ≥ 2 and 0 < α ≤ n; and denote by C, C 1 , C 2 . . .positive constants depending at most on n and α, with actual values of C varying from line to line.

Two lemmas.
Let L be a cylindrical set of the form {(x, y) : x ∈ S and c < y < d} or {(x, y) : x ∈ S and c ≤ y ≤ d}, with S ⊆ R n and c, d ∈ R 1 .Denote by L t , L s and L b , the top ∂L ∩ {y = d}, the side ∂L ∩ {c < y < d} and the bottom ∂L ∩ {y = c} of L respectively.Lemma 1.Let r > 0 and D, Q be two cubes in R n+1 defined by Let G be the Green function for Q, and ω (x,y) (S, Q) be the harmonic measure of a set S on ∂Q with respect to Q evaluated at the point (x, y) ∈ Q.Then there exist constants C 1 , C 2 and C such that for sup |x j | < 2r, 4r < y < 8r and (x , 0) ∈ D b , and n the unit inward normal at (x , 0).Both lemmas are elementary.The precise statement of Lemma 2 can be found in [D].

A partition of R n+1
+ .Choose and fix an odd integer R: where C 1 and C 2 are constants from Lemma 1, C 3 and C 4 are to be specified later.Choose for k ≥ 1, δ k so that Denote by A the collection of all integer lattice points on R n and by and let 1 is an odd integer, the inequality follows.
For each k ≥ 0, denote by H k the set consisting of Ω k and those Γ j,a (j > k and a ∈ A j ) that can be connected to Ω k by paths not intersecting Ω k for any k = k.That is, Γ j,a ⊆ H k if and only if j > k and the line segment {x = a, Note from the comment in the last paragraph, the interiors of H k 's are mutually disjoint and that k≥0 Size of the asymptotic sets.
Suppose that u is a function in M α that satisfies (9) lim Then any asymptotic path γ, along which u has an asymptotic value b = −∞, does not meet ∪ j≥0 ∂H j {0 < y < t} for some t > 0; therefore γ ∩ {0 < y < t/2} is contained in a certain H k .From this, it follows that where Γ * j is the projection of Γ j onto ∂R n+1 + .Let T be a unit cube on R n and k a positive integer.Given K > k, the set T ∩ (∩ j≥k Γ * j ) can be covered by at most which approaches 0 as K → ∞.This implies that dim(A (u)) ≤ n − α.
In view of (1), dim(A (u)) = n − α. J.-M.Wu Construction of the function u.Now we are ready to construct a subharmonic u in M α that has the property (9).Let for k ≥ 0, ( 10) and let for λ > 0, and for some constant C.
Extend u to be continuous on R n+1 + , bounded on Ω 0 , harmonic in each Ω k (k ≥ 0) and harmonic in the interior of each Γ k (k ≥ 1).By the maximum principle, ( 13) Note from the definition of u that u is negative on ∪∂H j and Since {m k } is unbounded, u satisfies (9).Therefore (2) holds for u.

Subharmonicity.
The subharmonicity is proved by induction.Note that u is harmonic in {y > r 1 }, and suppose that u is subharmonic in {y > r k } for some k ≥ 1.In order to prove that u is subharmonic in {y > r k+1 }, we need to verify the submean value property on Γ s k ∪ {y = r k }.We shall prove that u has a local minimum at each point in Γ s k ∪ 5 4 Γ t k \ 3 4 Γ t k , and compare the normal derivatives from both sides on the remaining part and then use Lemma 2.
First we give estimates of u on some subsets of Ω k and Γ k .
For k ≥ 1 and a ∈ A k , let Note from ( 6) and ( 8) that {Q k,a : a ∈ A k } are mutually disjoint and that and , where ω is the harmonic measure.In view of (3), ( 6) and ( 10) and by the maximum principle.It follows from ( 4), ( 6) and (10) that This completes the proof of Lemma 3.
From ( 13), ( 15) and the monotonicity of Because of (15 We claim that ( 18) for some constant C 3 > 0 depending only on n.To this end, fix a ∈ A k and let g be a C 2 function in a neighborhood of Q k,a , with values g(x, r k ) = u(x, r k ) and g(x, y) . This is possible because u satisfies ( 11) and ( 12).Let h be a function continuous on Q k,a , harmonic in Q k,a with boundary values h(x, y) = 0 on ∂Q k,a ∩ {y = r k } and h(x, y) = u(x, y) , a boundary estimate of derivatives [PW,p. 144] shows that in From these, estimate (18) follows.
To obtain an upper bound for ∂u ∂y − if necessary, we shall replace C 3 in ( 18) by a larger number.Recall from (15) that u ≥ −m k in {5r k+1 < y < r k }, thus v is nonnegative; and note from (14 Combining estimates ( 17) through ( 20), we conclude that on {y = which is negative in view of ( 6), ( 8) and ( 10).This proves ( 16).Hence u is subharmonic in R n+1 + .
u has a local minimum −m k , thus the submean value property.In view of Lemma 2, to prove the subharmonicity on {y = r k }\