Growth and asymptotic sets of subharmonic functions II

Publicacions Matem`atiques, Vol 42 (1998), 449–460. GROWTH AND ASYMPTOTIC SETS OF SUBHARMONIC FUNCTIONS II Jang-Mei Wu Abstract We study the relation between the growth of a subharmonic function in the half space Rn+1 +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 uin the Rn+1 +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 S λ6=−∞ A(λ) is exactly n−α. Here A(λ) is the set of boundary points at which ftends to λalong some curve. This proves the sharpness of a theorem due to Berman, Barth, Rippon, Sons, Fern´andez, Heinonen, Llorente and Gardiner cumulatively. A function fdefined in a domain Dis said to have an asymptotic value b∈[−∞,∞]atapointa∈∂D provided that there exists a path γ in Dending at aso that u(p) tends to bas ptends to aalong γ. The set of all points on ∂D at which fhas an asymptotic value bis 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 uof a certain growth, if A(u, +∞) is a small set, then uhas nice boundary behavior on a large set. Denote by Rn+1 += {(x, y):x=(x 1 ,... ,x n)∈R n,y>0}the upper half space in Rn+1. For α>0, denote by Mαthe class of subharmonic functions uin Rn+1 + which satisfy the growth condition: u(x, y)≤C(u)y−αfor 0 <y<1 for some constant C(u)>0. Research partially supported by the National Science Foundation. 450 J.-M. Wu Denote by F(u) the Fatou set consisting of points on ∂Rn+1 +at which uhas finite vertical limits. For β>0, denote by Hβthe β-dimensional Hausdorff content. The following theorem is due to Berman, Barth, Rippon, Sons, Fern´andez, Heinonen, Llorente and Gardiner cumulatively, see [FHL] and [G]. Theorem A. Let n≥1,0<α≤nand ube a subharmonic function in the class Mα.LetBbeaballon∂R n+1 +.If H n−α (A(u, +∞)∩B)<H n−α (B), then Hn(F(u)∩B)>0. Denote by A0(u)= [ −∞<b≤+∞ A(u, b). Theorem A implies the following. Theorem B. Let n≥1,0<α≤nand u∈M α . Then (1) dim(A0(u)) ≥n−α. We prove in this note that Theorem B is sharp. Theorem 1. Given n≥1and 0<α≤n, there exists a subharmonic function uin Mαso that (2) dim(A0(u)) = n−α. It has been proved ([FHL], [W]) that for 0 <α≤1, there exists a harmonic function hin Mαsuch that dim(A0(h)) = n−α. In general, there is more flexibility in constructing subharmonic functions than harmonic functions. In the proof of Theorem 1, we assign values of uon a grid in Rn+1 +to force uto 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≤nand it is not clear whether n−αis 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, C1,C 2... positive constants depending at most on nand α, with actual values of Cvarying from line to line. Growth of subharmonic functions 451 Two lemmas. Let Lbe a cylindrical set of the form {(x, y):x∈Sand c<y<d} or {(x, y):x∈Sand c≤y≤d}, with S⊆Rnand c,d∈R1. Denote by Lt,Lsand Lb, the top ∂L ∩{y=d}, the side ∂L ∩{c<y<d}and the bottom ∂L ∩{y=c}of Lrespectively. Lemma 1. Let r>0and D,Qbe two cubes in Rn+1 defined by D={(x, y) : sup |xj|<2r, 0<y<4r} and Q={(x, y) : sup |xj|<4r, 0<y<8r}, let E={(x, 0) : sup |xj|<r/2}and F={(x, 0) : sup |xj|<3r/2}.Let Gbe the Green function for Q, and ω(x,y)(S, Q)be the harmonic measure of a set Son ∂Q with respect to Qevaluated at the point (x, y)∈Q. Then there exist constants C1,C2and Csuch that ω(x,y)(E,Q)>C 1for (x, y)∈Dt,(3) ω(x,y)(E,Q)>C 2 ω (x,y)(F∪Qt∪Qs,Q)for (x, y)∈Dt∪Ds,(4) C−1(8r−y)r−n−1≤∂G ∂n ((x, y),(x0,0)) ≤C(8r−y)r−n−1 (5) for sup |xj|<2r, 4r<y<8rand (x0,0) ∈Db, and nthe unit inward normal at (x0,0). Lemma 2. Let ube a function continuous in B={(x, y):Σx 2 j+ y 2<r 2 }and harmonic in B\{y=0}with first partials continuous on B∩{y≥0}and also on B∩{y≤0}. If the left and right partials satisfy µ∂u ∂y¶− (x, 0) ≤µ∂u ∂y¶+ (x, 0) on B∩{y=0}, then uis subharmonic in B. Both lemmas are elementary. The precise statement of Lemma 2 can be found in [D]. A partition of Rn+1 +. Choose and fix an odd integer R: (6) R>max{105n,105n/α,2C−1 1,C −1 2,(2C3/C4)2/α}, 452 J.-M. Wu where C1and C2are constants from Lemma 1, C3and C4are to be specified later. Choose for k≥1, δkso that (7) 2δkR2kis an odd integer and (8) (k+R3α/2n)R−2kα/n ≤δk≤2(k+R3α/2n)R−2kα/n. Note from (6) that R−α/n <10−5,thusδ k<1 100 . Let for k≥1, rk=δkR−k2. Denote by Athe collection of all integer lattice points on Rnand by Ak={R−k2a:a∈A}. For k≥1 and a∈Ak, let Γk,a ={(x, y):r k+1 ≤y≤rkand sup 1≤j≤n |xj−aj|≤r k }, Γ k=∪ a Γ k,a, Ωk={(x, y):r k+1 <y<r k }\Γk, and let Ω0={(x, y):y>r 1 }. Note that each Ωkis connected because n≥2, that sets in {Γk}k≥1∪ {Ωk}k≥0have mutually disjoint interiors and that [ k≥1 Γk∪[ k≥0 Ωk=Rn+1 +. The top Γt k,a of each Γk,a is either completely contained in Γb k−1 or completely contained in Ωb k−1. To prove this, we claim that (Γb k−1∩Ωb k−1)∩Γt k=∅. Suppose that (x, rk)∈Γb k−1∩Ωb k−1, then |xj−pR−(k−1)2|=δk−1R−(k−1)2for some integers j(1 ≤j≤n) and p. To see that (x, rk)/∈Γt k, it is enough to show that |xj−qR −k 2|>δ kR −k 2 for all integers q; or equivalently, |pR2k−1±δk−1R2k−1−q|>δ kfor all integers q. Because δk<1 100 and 2δk−1R2k−1is an odd integer, the inequality follows. Growth of subharmonic functions 453 For each k≥0, denote by Hkthe set consisting of Ωkand those Γj,a (j>kand a∈Aj) that can be connected to Ωkby paths not intersecting Ωk0for any k06=k. That is, Γj,a ⊆Hkif and only if j>kand the line segment {x=a, rj 2≤y≤rk 2}is contained in Ωk∪(∪{Γi:i>k}). Note from the comment in the last paragraph, the interiors of Hk’s are mutually disjoint and that [ k≥0 Hk=Rn+1 +. Size of the asymptotic sets. Suppose that uis a function in Mαthat satisfies (9) lim k→∞ sup    u(x, y):(x, y)∈ [ j≥0 ∂Hj \{y≤rk}   =−∞. Then any asymptotic path γ, along which uhas an asymptotic value b6=−∞, does not meet ∪j≥0∂HjT{0<y<t}for some t>0; therefore γ∩{0<y<t/2}is contained in a certain Hk. From this, it follows that A0(u)⊆[ k≥1  \ j≥k Γ∗ j  where Γ∗ jis the projection of Γjonto ∂Rn+1 +. Let Tbe a unit cube on Rnand ka positive integer. Given K>k, the set T∩(∩j≥kΓ∗ j) can be covered by at most N≡C(k)µ1 R−k2·2rk R−(k+1)2...2r K−1 R −K2¶n ≡C(k)(2δk)n(2δk+1)n...(2δK−1)nRK2n cubes in Tof side length 2δKR−K2each. If n≥β>n−α, then in view of (8), N·(2δKR−K2)β ≤C(n, R, α, β, k)4nK ((K+R2)!)nRK2(n−α−β)R2K(n−β)α/n which approaches 0 as K→∞. This implies that dim(A0(u)) ≤n−α. In view of (1), dim(A0(u)) = n−α. 454 J.-M. Wu Construction of the function u. Now we are ready to construct a subharmonic uin Mαthat has the property (9). Let for k≥0, (10) Mk=Rαk2and mk=R−1+αk2, and let for λ>0, λΓt k,a ={(x, rk) : sup j |xj−aj|≤λrk}, and λΓt k=[ a λΓt k,a. Define uon (∪k≥1∂Γk)∪(∪k≥0∂Ωk)≡(∪k≥1Γs k)∪(∪k≥1{y=rk})as follows: for each k≥1 u=−mkon Γs k, and uis C2on {y=rk}with values u=Mkon 1 2Γt k, −mk≤u≤Mkon 3 4Γt k\1 2Γt k, u=−mkon 5 4Γt k\3 4Γt k, −mk≤u≤−m k−1on 3 2Γt k\5 4Γt k, u=−mk−1on {y=rk}\3 2Γt k, with partial derivatives (11) Σj¯¯¯¯ ∂ ∂xj u¯ ¯ ¯ ¯ ≤CM k r −1 k, and (12) Σi,j ¯¯¯¯ ∂2 ∂xi∂xj u¯ ¯ ¯ ¯ ≤CM k r −2 k, for some constant C. Growth of subharmonic functions 455 Extend uto be continuous on Rn+1 +, bounded on Ω0, harmonic in each Ωk(k≥0) and harmonic in the interior of each Γk(k≥1). By the maximum principle, (13) −m1≤u≤M1on {y≥r1}, −mk+1 ≤u≤Mk+1 on {rk+1 ≤y≤rk}(k≥1). Note from the definition of uthat uis negative on ∪∂Hjand u≤−m k−1on  [ j≥0 ∂Hj \{y≤rk}(k≥1). Since {mk}is unbounded, usatisfies (9). Therefore (2) holds for u. Subharmonicity. The subharmonicity is proved by induction. Note that uis harmonic in {y>r 1 }, and suppose that uis subharmonic in {y>r k }for some k≥1. In order to prove that uis subharmonic in {y>r k+1}, we need to verify the submean value property on Γs k∪{y=rk}. We shall prove that uhas a local minimum at each point in Γs k∪¡5 4Γt k\3 4Γt k¢, and compare the normal derivatives from both sides on the remaining part and then use Lemma 2. First we give estimates of uon some subsets of Ωkand Γk. For k≥1 and a∈Ak, let Dk,a ={(x, y) : sup j |xj−aj|<2rk,r k<y<5r k }, Q k,a ={(x, y) : sup j |xj−aj|<4rk,r k<y<9r k }, D k=[ a D k,a and Qk=[ a Qk,a. Note from (6) and (8) that {Qk,a :a∈Ak}are mutually disjoint and that Qk⊆{r k<y<r k−1 }\∂Γk−1. Let for k≥0, Ω0 k=Ω k \D k+1, and for k≥1, Γ0 k=Γ k \D k+1. 456 J.-M. Wu Lemma 3. For k≥1, (14) u>C M kon Dt k, and (15) u>−m kon Ω0 k∪Γ0 k. Proof: Let k≥2 and recall that −mk≤u≤Mkin {rk≤y≤rk−1} and Qk,a ⊆{r k≤y≤r k−1 }. Apply the maximum principle to uon Qk,a,wehave u(x, y)≥Mkω(x,y)µ1 2Γt k,a,Q k,a¶−mkµ1−ω(x,y)µ1 2Γt k,a,Q k,a¶¶ for (x, y)∈Qk,a, where ωis the harmonic measure. In view of (3), (6) and (10) u(x, y)≥MkC1−mk(1 −C1)>M k C 1 /2 for (x, y)∈Dt k,a. This proves (14). Next consider uon Ωk−1, and note that u≥−m k−1on ∂Ωk−1\3 2Γt k, −mk≤u≤Mkon 3 2Γt kand u=Mkon 1 2Γt k. Apply the maximum principle to mk−1+uon Ωk−1, we obtain mk−1+u(x, y)≥Σ0µMkω(x,y)µ1 2Γt k,a,Ωk−1¶ −mkω(x,y)µ3 2Γt k,a\1 2Γt k,a,Ωk−1¶¶ for (x, y)∈Ωk−1, where Σ0sums over those a∈Aksuch that Qk,a ⊆ Ωk−1and ωis the harmonic measure. If Qk,a ⊆Ωk−1and (x, y)∈Qk,a, then ω(x,y)µ1 2Γt k,a,Ωk−1¶≥ω(x,y)µ1 2Γt k,a,Q k,a¶ and ω(x,y)µ3 2Γt k,a\1 2Γt k,a,Ωk−1¶≤ω(x,y)µ3 2Γt k,a ∪Qt k,a ∪Qs k,a,Q k,a¶ by the maximum principle. It follows from (4), (6) and (10) that ω(x,y)µ3 2Γt k,a ∪Qt k,a ∪Qs k,a,Q k,a¶≤Mk mk ω(x,y)µ1 2Γt k,a,Q k,a¶ Growth of subharmonic functions 457 for (x, y)inD t k,a ∪Ds k,a. Hence u≥−m k−1on Dt k,a ∪Ds k,a. Since u≥ −mk−1on ∂Ω0 k−1\Dk,u>−m k−1in Ω0 k−1by the maximum principle. Therefore u>−m kin Ω0 kfor k≥1. The estimate on Γ0 kfollows by a similar argument. This completes the proof of Lemma 3. From (13), (15) and the monotonicity of mk, it follows that u≥−m k on Ω0 k∪Γ0 k∪{y≥r k }. Therefore at each point in Γs k∪¡5 4Γt k\3 4Γt k¢,u has a local minimum −mk, thus the submean value property. In view of Lemma 2, to prove the subharmonicity on {y=rk}\ ¡5 4Γt k\3 4Γt k¢, it suffices to prove (16) µ∂u ∂y¶− ≤µ∂u ∂y¶+ on {y=rk}\ µ5 4Γt k\3 4Γt k¶. Because of (15), u≥−m k−1in {rk≤y≤rk−1}\Dk. Since u= −mk−1on {y=rk}\2Γt k, (17) µ∂u ∂y¶+ ≥0on{y=r k }\2Γt k. We claim that (18) ¯¯¯¯¯µ∂u ∂y¶+¯ ¯ ¯ ¯ ¯ ≤C3Mkr−1 kon 2Γt k for some constant C3>0 depending only on n. To this end, fix a∈ Akand let gbe a C2function in a neighborhood of Qk,a, with values g(x, rk)=u(x, rk) and g(x, y)=−m k−1in Qk,a\Dk,a, |grad g|≤CM k r −1 k and |4g|≤CM k r −2 k on Qk,a. This is possible because usatisfies (11) and (12). Let hbe a function continuous on Qk,a, harmonic in Qk,a with boundary values h(x, y)=0on∂Qk,a ∩{y=rk}and h(x, y)=u(x, y)+mk−1on ∂Qk,a ∩ {y>r k }. Since 4(u−h)=0inQ k,a and u−h=gon ∂Qk,a,a boundary estimate of derivatives [PW, p. 144] shows that in Qk,a, |grad(u−h)|≤max ∂Qk,a |grad g|+ max Qk,a |4g|·diam Qk,a.