/compile/output.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 9, No. 3, 2016, 292-304 ISSN 1307-5543 – www.ejpam.com Hardy Spaces on the Polydisk Khim R. Shrestha University of Great Falls, 1301 20th St S, Great Falls, MT 59405 Abstract. In this paper we will study the boundary values properties of the functions in the Hardy spaces; generalize the F. and M. Riesz theorem to higher dimensions; discuss the existence of boundary values of the functions in H p(Dn) on non-distinguished boundary ∂Dn \Tn and the intersection of the spaces H p u (Dn). 2010 Mathematics Subject Classifications: 32A35, 32A40 Key Words and Phrases: Poisson integral, boundary values, exhaustion function 1. Introduction This paper basically consists of two parts. In the first part, consisting of Sections 2, 3 and 4, we study the properties of the functions on the classical Hardy spaces of n-harmonic functions and the Hardy spaces of holomorphic functions on the polydisk. In Section 2 we will show that the functions in the classical Hardy spaces can be restored by the Poisson integral of its radial limit. In Section 3 we will restate and prove the celebrated F. and M. Riesz theorem to higher dimensions. In Section 4 we will study the boundary values of the functions in H p(D) on the non-distinguished boundary, ∂Dn \Tn. The second part of this paper consists of Section 5. In this section we study the Poletsky– Stessin Hardy spaces H p u (D 2) on bidisk. We mainly establish two things - there are nontrivial Poletsky–Stessin Hardy spaces and the intersection of the Poletsky–Stessin Hardy spaces over all exhaustion functions is H∞(D2), the space of bounded holomorphic functions on D2. 2. Hardy Spaces and Poisson Integral Formula An n-harmonic function u onDn is a function which is harmonic in each variable separately. Denote by hp(Dn) the space of all n-harmonic functions satisfying sup 0≤r<1 ∫ Tn |ur(ζ)| p dm(ζ)<∞ (1) Email address: khim.shrestha@ugf.edu http://www.ejpam.com 292 c© 2016 EJPAM All rights reserved. K. Shrestha / Eur. J. Pure Appl. Math, 9 (2016), 292-304 293 where ur(ζ) = u(rζ) and dm is the normalized Lebesgue measure on Tn. The p -th root of (1) defines a norm on hp(Dn) when p ≥ 1. With this norm hp(Dn) is Banach. We will use the following notations: z =(z1, . . . , zn) ζ =(ζ1, . . . ,ζn) P(z,ζ) =P(z1,ζ1) . . . P(zn,ζn) where P(z,ζ) is the Poisson kernel and P(z j ,ζ j) = Re � ζ j + z j ζ j − z j � = 1− |z j | 2 |ζ j − z j |2 , j = 1, . . . , n. Theorem 1. Let u ∈ hp(Dn), p > 1. Then there exists a function f ∈ Lp(Tn) such that u(z) = ∫ Tn P(z,ζ) f (ζ) dm(ζ). Proof. Take r j ր 1. Then (1) implies that there is a weakly convergent subsequence of ur j . We will write the subsequence ur j just to avoid the sub-subscript. Hence for g ∈ Lq(Tn) g 7→ lim j→∞ ∫ Tn g(ζ)ur j (ζ) dm(ζ) is a linear functional on Lq(Tn). By Riesz theorem there exists an f ∈ Lp(Tn) such that lim j→∞ ∫ Tn g(ζ)ur j (ζ) dm(ζ) = ∫ Tn g(ζ) f (ζ) dm(ζ). Now take g(ζ) = P(z,ζ). Then u(z) = lim j→∞ ur j (z) = lim j→∞ ∫ Tn P(z,ζ)ur j (ζ) dm= ∫ Tn P(z,ζ) f (ζ) dm(ζ). The second equality above follows from [7, Theorem 2.1.2]. What makes the above proof work is the duality of Lp spaces. Since L∞ is the dual of L1, the same result holds with the same proof for p =∞. Of course we have to change the statement accordingly. But unfortunately L1 is not dual of anything, we don’t have the same result for p = 1. Instead, since the space of finite signed measures on Tn is dual of the space of continuous functions C(Tn) we have the following result from [7, Theorem 2.1.3, (e)]. Theorem 2. If the hypothesis of Theorem 1 holds for p = 1 then there exists a finite signed measure µ on Tn with u(z) = ∫ Tn P(z,ζ) dµ(ζ). K. Shrestha / Eur. J. Pure Appl. Math, 9 (2016), 292-304 294 So the function u ∈ hp(Dn), p > 1, is the Poisson integral of some function f ∈ Lp(Tn). Is there any other connection between u and f ? We know, when n = 1, f is the boundary value function of u and when n > 1 the following theorem [7, Theorem 2.3.1] answers this question. Theorem 3. If f ∈ L1(Tn), if σ is a measure on Tn which is singular with respect to dm, and if u= P[ f + dσ], then u∗(ζ) = f (ζ) for almost every ζ ∈ Tn. Recall that u∗(ζ) = limr→1 u(rζ) is the radial limit. Thus any n-harmonic function satisfy- ing the growth condition (1) for p > 1 can be restored by the Poisson integral of its boundary value function. For p = 1 we just saw in Theorem 2 that u(z) = P[dµ](z). By the Lebesgue decomposition theorem dµ= f dm+ dσ where σ is singular with respect to m and f ∈ L1(Tn). Hence we have u∗(ζ) = f (ζ) but u can not be restored by the Poisson integral of its boundary value function unless, of course, P[dσ] = 0. Also in [7] it has been proved that if f ∈ Lp(Tn), 1 ≤ p <∞, and u = P[ f ] then ur converges to f in the Lp-norm as r → 1, i.e. limr→1 ‖ur − f ‖Lp = 0. But when p = 1 we have the weak-∗ convergence. Theorem 4. Let f (z) = P[dµ](z) with µ a finite signed measure on Tn. Then fr dm → dµ weak-∗ as r → 1. Proof. Let ϕ ∈ C(Tn). Then � � � � � ∫ Tn ϕ(ζ) fr(ζ) dm(ζ)− ∫ Tn ϕ(ζ) dµ(ζ) � � � � � = � � � � � ∫ Tn ϕ(ζ) �∫ Tn P(rζ,η) dµ(η) � dm(ζ)− ∫ Tn ϕ(η) dµ(η) � � � � � (∵ P(rζ,η) = P(rη,ζ)) = � � � � � ∫ Tn �∫ Tn P(rη,ζ)ϕ(ζ) dm(ζ) � dµ(η)− ∫ Tn ϕ(η) dµ(η) � � � � � = � � � � � ∫ Tn �∫ Tn P(rη,ζ)ϕ(ζ) dm(ζ)−ϕ(η) � dµ(η) � � � � � →0 because the inner integral goes to zero uniformly on η. Hence fr dm→ dµ weak-∗ as r → 1. We define H p(Dn), 0 < p <∞, to be the class of all holomorphic functions f ∈ Dn for which sup 0≤r<1 ∫ Tn | fr(ζ)| p dm<∞ and H∞(Dn) is the space of all bounded holomorphic functions in Dn. K. Shrestha / Eur. J. Pure Appl. Math, 9 (2016), 292-304 295 Since | f |p is n-subharmonic, sup in the definition can be replaced by lim as r → 1. It is known that if f ∈ H p(Dn), 0 < p <∞, then f has a non-tangential limit at almost all points of Tn [11, Ch. XVII, Theorem 4.8]. We denote this limit by f ∗ as in [7] and call it a boundary value function. Moreover, we have the following results from Rudin (see [7, Theorem 3.4.2 and 3.4.3]). Theorem 5. If f ∈ H p(Dn), 0< p <∞, then f ∗ ∈ Lp(Tn) and (i) limr→1 ∫ Tn | fr | p dm= ∫ Tn | f ∗|p dm (ii) limr→1 ∫ Tn | fr − f ∗|p dm= 0. When p ≥ 1 the function in H p(Dn) can be represented by the Poisson integral of its boundary value function. Theorem 6. If f ∈ H1(Dn), then f (z) = ∫ Tn P(z,ζ) f ∗(ζ) dm. (The case n= 1 can be found in [6, Theorem 17.11].) Proof. Since z ∈ Dn, P(z,ζ) is bounded on Tn and by (ii) of the theorem above � � � � � ∫ Tn P(z,ζ) fr(ζ) dm(ζ)− ∫ Tn P(z,ζ) f ∗(ζ) dm(ζ) � � � � � ≤ ∫ Tn P(z,ζ)| fr(ζ)− f ∗(ζ)| dm(ζ) →0. Now by [7, Theorem 2.1.2] f (z) = lim r→1 fr(z) = lim r→1 ∫ Tn P(z,ζ) fr(ζ) dm(ζ) = ∫ Tn f ∗(ζ) dm(ζ). 3. The F. and M. Riesz Theorem Now we want to generalize the F. and M. Riesz theorem. Theorem 7. Let µ be a complex Borel measure on Tn. If ∫ Tn ei(kθ ) dµ(θ ) = 0 for k = (k1, . . . , kn) ∈ Z n with at least one k j , j = 1,2, . . . , n positive, where (kθ ) = k1θ1 + . . .+ knθn then µ is absolutely continuous with respect to dm. K. Shrestha / Eur. J. Pure Appl. Math, 9 (2016), 292-304 296 (When n= 1 see [6, Theorem 17.13].) Proof. Define f (z) = P[dµ](z). Then, with the notations z =(z1, . . . , zn) with z j = r je iθ j , j = 1, . . . , n r |k| =r |k1| 1 . . . r |kn| n (k · θ ) =k1θ1 + . . .+ knθn (k · t) =k1 t1 + . . .+ kn tn and using the series representation for the Poisson kernel, we get f (z) = ∫ Tn P(z, ei t) dµ(t) = ∫ Tn ∑ k∈Zn r |k|ei(k·θ )e−i(k·t) ! dµ(t) = ∑ k∈Zn �∫ Tn e−i(k·t)dµ(t) � r |k|ei(k·θ ) = ∑ k∈Zn + ckzk where ck = ∫ Tn e−i(k·t) dµ(t) and zk = r |k|ei(k·θ ). Notice that all other integrals in the above sum vanish by the hypothesis. Thus f (z) is holomorphic. For 0≤ r < 1, ∫ Tn | fr(ζ)| dm(ζ) = ∫ Tn � � � � � ∫ Tn P(rζ,η) dµ(η) � � � � � dm(ζ) ≤ ∫ Tn �∫ Tn P(rζ,η) d|µ|(η) � dm(ζ) = ∫ Tn �∫ Tn P(rζ,η) dm(ζ) � d|µ|(η) =‖µ‖. Thus f ∈ H1(Dn) and hence f (z) = P[ f ∗](z), where f ∗ ∈ L1(Tn). Now the uniqueness of the Poisson integral representation shows that dµ= f ∗dm and the proof is completed. K. Shrestha / Eur. J. Pure Appl. Math, 9 (2016), 292-304 297 4. Boundary Values Do the boundary values of functions in H p(Dn) exist on the non-distinguished boundary? Now we want to look into this question. Let { j1, . . . , jk} and {i1, . . . , il} be disjoint sets of indices such that their union is {1, . . . , n} where j1 < j2 < . . .< jk and i1 < i2 < . . .< il . Define the sections of Dn as follows D n z j1 ,...,z jk = {(z1, . . . , zn) ∈ D n : z j1 , . . . , z jk are fixed} and define fz j1 ,...,z jk = f |Dn z j1 ,...,z jk . We will write fz j1 ,...,z jk (zi1 , . . . , zil ) instead of fz j1 ,...,z jk (z1, . . . , zn). We will see below that for f ∈ H p(Dn), 1 ≤ p <∞, the non-tangential limit of fz j1 ,...,z jk exists at almost all points of the distinguished boundary of the section Dn z j1 ,...,z jk which is Tl and the function fz j1 ,...,z jk can be restored by the Poisson integral of this limit. Theorem 8. Let f ∈ H p(Dn), 1≤ p <∞. Then fz j1 ,...,z jk ∈ H p(Dl). Proof. Without loss of generality we suppose that { j1, . . . , jk} = {1, . . . , k}. Let’s use the following notations for the Poisson kernels Pj(ζ j) = ¨ P(z j,ζ j) j = 1, . . . , k P(rξ j ,ζ j) j = k+ 1, . . . , n where |ξ j |= 1. Then, for 0< r < 1, by Theorem 6 fz1,...,zk (rξk+1, . . . , rξn) = ∫ Tn P1(ζ1) . . . Pn(ζn) f ∗(ζ1, . . . ,ζn) dmn. By Hölder and Fubini ∫ Tl | fz1,...,zk (rξk+1, . . . , rξn)| pdml = ∫ Tl � � � � � ∫ Tn P1(ζ1) . . . Pn(ζn) f ∗(ζ1, . . . ,ζn) dmn � � � � � p dml ≤ ∫ Tl �∫ Tn P1(ζ1) . . . Pn(ζn)| f ∗(ζ1, . . . ,ζn)| p dmn � dml = ∫ Tn P1(ζ1) . . . Pk(ζk)| f ∗(ζ1, . . . ,ζn)| p × �∫ Tl Pk+1(ζk+1) . . . Pn(ζn) dml � dmn ≤ 2k (1− |z1|) . . . (1− |zk|) ∫ Tn | f ∗(ζ1, . . . ,ζn)| pdmn. The last quantity above is independent of r and is finite by Theorem 5. Thus the theorem is proved. The following corollary is immediate. K. Shrestha / Eur. J. Pure Appl. Math, 9 (2016), 292-304 298 Corollary 1. If f ∈ H p(Dn), 1 ≤ p <∞, then the non-tangential limit f ∗z j1 ,...,z jk of the function fz j1 ,...,z jk exists almost everywhere on Tl and belongs to Lp(Tl). The following theorems are the direct consequences of Theorems 5 and 6. Theorem 9. If 1≤ p <∞ and f ∈ H p(Dn), then (i) lim r→1 ∫ Tl |( fz j1 ,...,z jk )r | p dml = ∫ Tl | f ∗ z j1 ,...,z jk |p dml (ii) lim r→1 ∫ Tl |( fz j1 ,...,z jk )r − f ∗z j1 ,...,z jk |p dml = 0 where ( fz j1 ,...,z jk )r(ζi1 , . . . ,ζil ) = fz j1 ,...,z jk (rζi1 , . . . , rζil ). Theorem 10. If f ∈ H1(Dn), then fz j1 ,...,z jk (zi1 , . . . , zil ) = ∫ Tl P(zi1 ,ζi1 ) . . . P(zil ,ζil ) f ∗z j1 ,...,z jk (ζi1 , . . . ,ζil ) dml . Theorem 11. Let f be a holomorphic function in Dn. If 1≤ p <∞ and sup (z j1 ,...,z jk ) |z j1 |=...=|z jk | ‖ fz j1 ,...,z jk ‖Hp(Dn−k) = M <∞, then f ∈ H p(Dn). Proof. For simplicity we take { j1, . . . , jk} = {1, . . . , k}. And, of course, this theorem makes sense only when k > 0. Now for 0≤ r < 1, ∫ Tn | f (rζ1, . . . , rζn)| p dmn = ∫ Tk �∫ Tn−k | f (rζ1, . . . , rζn)| p dmn−k � dmk ≤ ∫ Tk � sup 0≤t<1 ∫ Tn−k | f (rζ1, . . . , rζk, tζk+1, . . . , tζn)| p dmn−k � dmk = ∫ Tk ‖ frζ1,...,rζk ‖p Hp(Dn−k) dmk ≤M p. Thus f ∈ H p(Dn). 5. Poletsky–Stessin Hardy Spaces on the Bidisk Let u be a negative continuous plurisubharmonic function on the bidisk D 2 = {(z1, z2) ∈ C 2 : |z1|< 1, |z2|< 1} K. Shrestha / Eur. J. Pure Appl. Math, 9 (2016), 292-304 299 such that u(z1, z2) → 0 as (z1, z2) → (ζ1,ζ2) ∈ ∂D 2. Following Demailly [2], for r < 0 we define Su(r) = � (z1, z2) ∈ D 2 : u(z1, z2) = r Bu(r) ={(z1, z2) ∈ D 2 : u(z1, z2)< r}. For convenience we will write z = (z1, z2). Associated with this u we define the positive measure µu,r called Monge-Ampère measures by µu,r = (dd cur) 2 −χD2\Bu(r) (dd cu)2 where ur = max{u, r}. These measures are supported by the level sets Su(r). Demailly has proved the following [2, Theorem 1.7]. Theorem 12 (Lelong–Jensen Formula). For all r < 0 every plurisubharmonic function ϕ on D2 is µu,r -integrable and µu,r(ϕ) = ∫ Bu(r) ϕ(dd cu)2 + ∫ Bu(r) (r − u)(dd cϕ)∧ (dd cu). Denote by E (D2) the set of all continuous negative plurisubharmonic functions u on D2 and equal to zero on ∂D2 whose Monge–Ampère mass is finite, i.e. ∫ D2 (dd cu)2 <∞ and denote by E1(D 2) the set of those u ∈ E (D2) for which ∫ D2 dd cu= 1. Following [3] we define, what we call, the Poletsky–Stessin Hardy space H p u (D 2), p > 0, as the space of all holomorphic functions on D2 for which lim sup r→0− µu,r(| f | p)<∞. These new spaces are contained in the classical spaces, that is, H p u (D 2) ⊂ H p(D2). Since µu,r(| f | p) is an increasing function of r the lim sup in the definition can be replaced by lim. For p ≥ 1 ‖ f ‖p H p u = lim r→0− µu,r(| f | p) is a norm and with this norm H p u (D 2) is Banach [3, Theorem 4.1]. The Poletsky–Stessin Hardy spaces on the unit disk have been studied in detail in [1, 5, 8–10]. In [4] Poletsky has proved that the intersection of all Poletsky–Stessin Hardy spaces H p u (D), p ≥ 1, where D is a strongly pseudoconvex domain with C2 boundary, is H∞(D), the space of bounded holomorphic functions. Hence it immediately follows that the intersection of all H p u (D) is H∞(D). We will prove this result for the polydisk. It is enough to consider the bidisk. K. Shrestha / Eur. J. Pure Appl. Math, 9 (2016), 292-304 300 Let ζ = (ζ1,ζ2) ∈ T 2 and α = (α1,α2), 0 < α1,α2 < π/2. Following [11] we define the approach region Tα(ζ) as Tα(ζ) = Tα1 (ζ1)× Tα2 (ζ2) where Tα j (ζ j) is the Stolz angle at ζ j ∈ T with vertex angle 2α j . Here we will consider only the congruent symmetric approach regions meaning that the Stolz angles are symmetric with respect to the radius to ζ j and the vertex angles are equal, i.e. α1 = α2. Following [4] we define the Green ball of radius 0< r < 1 and center at w to be the set G(w, r) = {z ∈ D2 : g(z, w)< log r} where g(z, w) is the Green function for D2 with pole at w. The Green function for D2 is explicitly given by g(z, w) = log max �� � � � z1 −w1 1−w1z1 � � � � , � � � � z2 −w2 1−w2z2 � � � � � . Hence it follows that G(w, r) = � z1 ∈ D : � � � � z1 −w1 1−w1z1 � � � � < r � × � z2 ∈ D : � � � � z2 − w2 2−w2z2 � � � � < r � . Lemma 1. Let ζ = (ζ1,ζ2) ∈ T 2 and 0 < r < 1. For any 0 < t < 1 there exists 0 < α < π/2 such that G(tζ, r) ⊂ Tα(ζ) where tζ = (tζ1, tζ2) and Tα(ζ) = Tα(ζ1)× Tα(ζ2). Proof. Observe that ( z j ∈ D : � � � � � z j − tζ j 1− tζ jz j � � � � � < r ) is the image of the disk {|w j |< r} ⊂ C under the conformal map w j 7→ w j + tζ j 1+ tζ jw j which is a disk contained in D with center at t(1− r2) 1− r2 t2 ζ j and radius equal to r(1− t2) 1− r2 t2 . The tangents to this disk that pass through ζ j make an angle of α= arcsin � r(1+ t) 1+ t r2 � K. Shrestha / Eur. J. Pure Appl. Math, 9 (2016), 292-304 301 with the radius to ζ j . Hence ( z j ∈ D : � � � � � z j − tζ j 1− tζ jz j � � � � � < r ) ⊂ Tα(ζ j) for j = 1,2 and G(tζ, r) ⊂ Tα(ζ). Since for fixed 0< r < 1 t 7→ r(1+ t) 1+ t r2 is an increasing function of t ∈ [0,1] we have 0< r(1+ t) 1+ t r2 ≤ 2r 1+ r2 < 1. From this it follows that 0< α≤ arcsin � 2r 1+ r2 � < π 2 . Remark 1. For fixed 0< r < 1, t 7→ r(1− t2) 1− r2 t2 is a decreasing function of t ∈ [0,1] that decreases to zero as t → 1. Therefore we can make the size of the Green ball G(tζ, r) as small as we want simply by choosing t close enough to 1. The plurisubharmonic envelope Eφ of a continuous function φ on a domain Ω ⊂ Cn is the maximal plurisubharmonic function on Ω less than or equal to φ. For a sequence of functions {u j} ⊂ E (D 2), we denote by E{u j} the envelope of inf{u j}. The following Lemma [4, Theorem 3.3] gives the estimate on the Monge–Ampère mass of the envelope. Lemma 2. If Ω is a strongly hyperconvex domain and continuous plurisubharmonic functions {u j} ⊂ E (Ω), then ∫ Ω (dd c E{u j}) n ≤ ∑ ∫ Ω (dd cu j) n. Theorem 13. Let f be a holomorphic function on D2. Suppose that f has non-tangential limits at points {ζ j} ⊂ T 2 and lim j→∞ | f ∗(ζ j)|=∞. Then for any p ≥ 1 there exists u ∈ E1(D 2) such that f /∈ H p u (D 2). The proof that Poletsky gave to this theorem in [4] in the case when D is a strongly pseu- doconvex domain with C2 boundary also works when the domain is a polydisk. We will mimic his proof in our context. Proof. Let us take a sequence {a j} of positive numbers such that ∞ ∑ j=1 a j <∞ and ∞ ∑ j=1 a2 j | f ∗(ζ j)| p =∞. K. Shrestha / Eur. J. Pure Appl. Math, 9 (2016), 292-304 302 For 0 < t j < 1 we write G j = G(t jζ j , e−1). By Lemma 1 there exists 0 < α j < π/2 such that G j ⊂ Tα j (ζ j). Now we inductively construct a sequence {tk}, 0 < tk < 1, satisfying certain conditions. Choose any 0< t1 < 1. Suppose that t1, . . . , tk−1 have already been chosen. Now chose 0< tk < 1 so that the following conditions are satisfied: (i) | f |> | f ∗(ζk)|/2 on Gk (ii) Gk ∩ G j = φ (iii) g(z, tkζk)> −a j/2 k+1 on G j (iv) a j g(z, t jζ j)> −ak/2 j+1 on Gk for 1≤ j ≤ k−1. The conditions (i) and (ii) can be achieved simply by taking tk close enough to 1. Since G j , j < k, and Gk are disjoint, g(z, tkζk)→ 0 uniformly on G j as tk → 1. Hence (iii) can be achieved for tk close enough to 1. Since g(z, t jζ j) = 0 when z ∈ ∂D2, we can choose tk so close to 1 that Gk ⊂ k−1 ⋂ j=1 � z ∈ D2 : a j g(z, t jζ j)> −ak/2 j+1 . Thus (iv) can be achieved. Define u j(z) = a j max{g(z, t jζ j),−2}. Note that if F is an open set in D2 containing G(t jζ j , e−2) then ∫ F (dd cu j) 2 = a2 j . Let u = E{u j}. Since the series v = ∑∞ j=1 u j converges uniformly on D2, v ∈ E (D2). So u ≥ v is a continuous plurisubharmonic function on D2 equal to 0 on ∂D2. By Lemma 2, ∫ D2 (dd cu)2 ≤ ∞ ∑ j=1 ∫ D2 (dd cu j) 2 = ∞ ∑ j=1 a2 j <∞. Hence u ∈ E (D2). Now we evaluate ∫ Gk (dd cu)2. Observe that uk ≥ u ≥ v on D2. By the conditions on the choices of t j , on ∂ Gk we get −ak ≥ u≥ − k−1 ∑ j=1 ak 2 j+1 − ak − ∞ ∑ j=k+1 ak 2 j+1 ≥ − 3 2 ak. Hence u + 3ak/2 ≥ 0 on ∂ Gk and the set Fk = {6(u + 3 2 ak) < uk} compactly belongs to Gk. Moreover, if z ∈ ∂ G(tkζk, e−2) then 6 � u(z) + 3 2 ak � ≤ 6 � uk(z) + 3 2 ak � = −3ak < −2ak = uk(z). K. Shrestha / Eur. J. Pure Appl. Math, 9 (2016), 292-304 303 Thus G(tkζk, e−2) ⊂ Fk. By the comparison principle 36 ∫ Gk (dd cu)2 = ∫ Gk (dd c6(u(z) + 3 2 ak)) 2 ≥ ∫ Fk (dd cuk) 2 = a2 k . Hence by Lelong–Jensen formula ‖ f ‖p H p u ≥ ∫ D2 | f |p(dd cu)2 ≥ ∞ ∑ k=1 ∫ Gk | f |p(dd cu)2 ≥ 1 36 · 2p ∞ ∑ k=0 | f ∗(ζk)| pa2 k =∞. Hence f /∈ H p(D2). The following corollary shows the existence of nontrivial Poletsky–Stessin Hardy spaces on the bidisk. Corollary 2. For every p ≥ 1 there exists a function u ∈ E1(D 2) such that H p u (D 2) 6⊆ H p(D2). Proof. Take f ∈ H p(D2) that is unbounded. Then the non-tangential limit f ∗ on T2 must be unbounded because otherwise f (z) = ∫ T2 P(z,ζ) f ∗(ζ) dm would imply that f (z) is bounded. So there exists a set of points {ζ j} ∈ T 2 such that lim j→∞ | f ∗(ζ j)|=∞. Hence the corollary follows from Theorem 13. Now we prove the most important theorem of this section. Theorem 14. Let p ≥ 1. Then ⋂ u∈E1(D 2) H p u (D 2) = H∞(D2). Proof. Let f ∈ ⋂ u∈E1(D 2)H p u (D 2). Then the non-tangential limit f ∗ on T2 is bounded because otherwise by Theorem 13 there would exist a u ∈ E1(D 2) such that f /∈ H p u (D 2). Thus, since f ∗ is bounded, f (z) = ∫ T2 P(z,ζ) f ∗(ζ) dm implies that f ∈ H∞(D2). ACKNOWLEDGEMENTS I would like to express my sincere gratitude to my Ph. D. advisor Prof. E. A. Poletsky for his immense support and guidance on this work. REFERENCES 304 References [1] M. A. Alan and N. G. Goğuş. Poletsky-Stessin-Hardy spaces in the plane, Complex Analysis and Operator Theory, 8, 975-990. 2014. [2] J. P. Demailly. Mesures de Monge-Ampr̀e et mesures pluriharmoniques, Mathematische Zeitschrift, 194, 519-564. 1987. [3] E. A. Poletsky and M. I. Stessin. Hardy and Bergman Spaces on Hyperconvex Domains and Their Composition Operators, Indiana University Mathematics Journal, 57, 2153-2201. 2008. [4] E. A. Poletsky. Projective Limits of Poletsky–Stessin Hardy Spaces, arXiv:1503.00575. [5] E. A. Poletsky and K. R. Shrestha. On Weighted Hardy Spaces on the Unit Disk, arXiv:1503.00535. [6] W. Rudin. Real and Complex Analysis, Third edition, McGraw Hill, 1987. [7] W. Rudin. Function Theory in Polydiscs, W. A. Benjamin, Inc. New York 1969. [8] S. Şahin. Poletsky-Stessin Hardy spaces on domains bounded by an analytic Jordan curve in C, arXiv:1303.2322. [9] K. R. Shrestha. Boundary Values Properties of Functions in Weighted Hardy Spaces, arXiv:1309.6561. [10] K. R. Shrestha. Weighted Hardy spaces on the unit disk, Complex Analysis and Operator Theory, 9, 1377-1389. 2015. [11] A. Zygmund. Trigonometric Series, Third Edition, Cambridge University Press, 2002.