COMPOSITION OPERATORS ON HARDY ORLICZ SPACES ON PLANAR DOMAINS

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1 Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 42, 207, COMPOSITION OPERATORS ON HARDY ORLICZ SPACES ON PLANAR DOMAINS Michał Rzeczkowski Adam Mickiewicz University, Faculty of Mathematics and Computer Science 6-64 Poznan, Poland; Abstract. We show isomorphic and isometric characterizations of Hardy Orlicz spaces on multiply-connected domains whose boundary consists of finitely many disjoint analytic Jordan curves. We also study composition operators on these spaces. In particular we obtain characterization of compact composition operators on Hardy Orlicz spaces in terms of Carleson s measures defined in the paper.. Introduction LetHΩ denote the space of all holomorphic functions onω, whereωis a domain on the Riemann sphere. For analytic map ϕ HΩ, ϕ: Ω Ω, the composition operator is defined by C ϕ f := f ϕ, f HΩ. This map may act on various quasi-banach spaces X of analytic functions on Ω. Composition operators are fundamental objects of study in analysis that arise naturally in many situations. For example a classical result due to Forelli see [4] states that all surjective isometries of the Hardy space H p D, < p <, p 2 are weighted composition operators. The problem of relating operator theoretic properties e.g., boundedness, compactness, weak compactness, order boundedness, spectral properties of composition operators C ϕ to function theoretic properties of generating function symbol of C ϕ has been a subject of great interest for quite some time. One of the famous problem was to characterize compact composition operators on Hardy spaces H p D. In the eighties MacCluer see [0] or [] proved that C ϕ : H p D H p D is compact if and only if the pullback measure µ ϕ defined by the formula µ ϕ B := mϕ B, where m is normalized Lebesgue measure on D and ϕ is radial limit of ϕ is vanishing Carleson measure, i.e., µ ϕ Wa,h = oh as h 0 for any Carleson window Wa,h := {z D: h < z <, argaz < h}, a D. We note that this problem was also solved by Shapiro [8, 9]. He described compact composition operators in terms of Nevanlinna counting function. It should be emphasized that MacCluer s result together with the famous Carleson Lemma contributes to study another class of operators inclusion operators j µ : H p D L p D,µ, where µ is finite Borel measure. This idea is very useful and often used in contemporary research recently composition operators acting between Hardy-type spaces are Mathematics Subject Classification: Primary 47B33, 46E30. Key words: Hardy Orlicz spaces, composition operators, Hardy spaces on planar domain, Carleson measures.

2 594 Michał Rzeczkowski studied thoroughly. The problems of characterizing compactness, weak compactness, absolute p-summability and other properties are under consideration on various variants of Hardy spaces. We refer to [6, 7, 8, 9] where the authors have extended the results of Shapiro and MacCluer to the case of composition operators on Hardy Orlicz spaces. In this paper we investigate composition operators on Hardy Orlicz spaces on multiply-connected domains Ω whose boundary consists of finitely many disjoint analytic Jordan curves. These spaces are generalizations of classical Hardy spaces H p Ω on multiply-connected domains Ω introduced by Rudin in [6]. Notice that in spite of many similarities there are significant differences between theory of Hardy spaces on unit disk and multiply connected domains we refer to the paper of Sarason [7] and book of Fisher [2], where H p Ω spaces are studied. Our goal is to show isomorphic and isometric characterizations of Hardy Orlicz spaces on Ω as well as to provide a complete description of compact composition operators C ϕ : H Φ Ω H Φ Ω in terms of Φ-Carleson measure on Ω, where Φ is an Orlicz function. 2. Preliminaries In this section we set out some prerequisites which occur in this paper. In particular we recall some basic facts about Hardy Orlicz spaces and harmonic measures. Hardy Orlicz spaces on disc. Let Φ: [0, [0, be an Orlicz function, i.e., continuous and nondecreasing function such that lim t Φt = and Φt = 0 if and only if t = 0. The Orlicz function Φ satisfies the 2 -condition Φ 2 if for some constant β > and for some t 0 > 0, one has Φβt 2βΦt, for t t 0. Given a measure space Ω,Σ,µ the Orlicz space L Φ Ω := L Φ Ω,Σ,µ is the space of all equivalence classes of Σ-measurable functions f: Ω C for which there is a constant λ > 0 such that Φλ f dµ <. Ω It is easy to check that if there exists C > 0, such that Φt/C Φt/2 for all t > 0, then L Φ Ω is a quasi-banach lattice equipped with the quasi-norm { f } f Φ := inf λ > 0; Φ dµ. λ It is well known that Φ is a norm in the case when Φ is a convex function. For Φt = t p, p 0, ], we have L Φ Ω = L p Ω and the norms coincide. We refer the reader to [4] for more complete information about Orlicz spaces. Let D be the unit disc of the complex plane. Throughout the paper we identify D with T = [0,2π. In the same way Hardy spaces H p D are defined out of the Lebesgue spaces L p T, we define Hardy Orlicz spaces H Φ := H Φ D from the Orlicz spaces L Φ T. For f HD and r 0, denote by f r : T C the function given by f r e it = fre it. Following [] H Φ consists of analytic functions f: D C such that f H Φ := sup f r L Φ T <. 0 r< The formula defines a quasi-norm in H Φ and it is a norm when Φ is a convex function. We note that for every f H Φ the radial limit f t := lim r freit, t T, Ω

3 Composition operators on Hardy Orlicz spaces on planar domains 595 exists a.e. and f H Φ = f L Φ T. Recall that see [8] the inverse is also true: for given f L Φ T such that its Fourier coefficients f n vanish for n < 0, the analytic extension fz = P[f ]z := n=0 f nz n, z D, belongs to H Φ and f H Φ = f L T. Φ We denote by HM Φ the subspace of finite elements of H Φ, i.e., the space of all f HD such that for every λ > 0 we have sup 0 r< T Φλ fre it dt <. Harmonic measures. Let Ω be a domain on the Riemann sphere and let u: Ω R be a continuous function on = Ω. The Dirichlet problem is to find if there exists a function ũ: Ω R which is continuous and satisfies two conditions:. ũ is harmonic on Ω, 2. ũ = u on. The Dirichlet problem can be solved for many domains. For our considerations the following result is sufficient see [2, Corollary.4.5.]. Proposition 2.. If each component of Ω is nontrivial, then the Dirichlet problem is solvable in Ω. In particular if Ω is a multiply-connected domain whose boundary consists of finitely many disjoint analytic Jordan curves then the Dirichlet problem is solvable for Ω. Recall that an analytic arc is the image ψ, of the open interval, under a map which is one-to-one and analytic on a neighborhood in C of, to C and an analytic Jordan curve is a Jordan curve that is finite union of open analytic arcs. Let Ω be a domain on the sphere for which the Dirichlet problem is solvable we write Ω SDP and let p Ω. If u C Ω then the map u ũp is linear and bounded by the Maximum Modulus Principle. The Riesz representation theorem implies that there is a unique real measure ω p on = Ω such that ũp = udω p. This measure is called harmonic measure on for p. Note that ω p is probability measure which has no atoms. Now we recall important properties of harmonic measures. First notice that ω p depends of the point p Ω, but it can be shown that for p and q Ω, ω p and ω q are boundedly mutually absolutely continuous. Further, if K is a compact subset of Ω we write K Ω, then there is a constant M such that ω q E Mω p E for all q K and for all measurable set E Ω. Suppose now that Ω and Ω 2 are two domains and f is holomorphic function which maps Ω onto Ω 2 homeomorphically. If Ω SDP, then also Ω 2 SDP. Let p Ω and p 2 = fp. Denote by ω the harmonic measure on Ω for p and define measure µ on Ω 2 as follows: µe = ω f E, E Ω 2.

4 596 Michał Rzeczkowski Then µ is harmonic measure on Ω 2 for p 2. Let Ω,Ω 2 SDP, Ω Ω 2. Let p Ω and let ω and ω 2 be harmonic measure on Ω and Ω 2 respectively, for p. Then for each compact set E Ω Ω 2 we have ω E ω 2 E. For next properties we need the Green s function of a domain. Assume that Ω is a domain on the Riemann sphere and Ω SDP, p Ω. A function g,p is a Green s function for Ω with a pole or singularity at p p, if z gz;p is harmonic on Ω\{p}, gz; p + log z p is harmonic in a neighbourhood of p, limgz;p = 0 for all ζ Ω. z ζ Assume now that Ω consist of m+ disjoint analytic Jordan curves. Let p Ω and let gz;p be Green s function for Ω at p. Denote by hz = hz;p the harmonic conjugate ofgz;p note thathis multivalued. Then we have that locallyq = g+ih is analytic and its derivative is single-valued on Ω. The following results see [2] show relationships between harmonic measure arc length and Green s function. Theorem 2.2. [2, Theorem.6.4.] Suppose Ω is bounded by a finite number of disjoint analytic Jordan curves. Then for each z Ω we have dω z = 2π n g,zds, where g,p is the Green s function for Ω with pole at z, is the derivative in the n direction of outwards normal at Ω and ds is arc length. The function P z ζ := dωzζ = gζ,z, where ζ Ω and z Ω is called ds 2π n Poisson kernel for Ω. It satisfies inequalities for positive constants c, c 2. Lebesgue measure on Ω. c dω z ds c 2 For convienience we assume that s is normalized Theorem 2.3. [2, Proposition.6.5.] dω p ζ = i 2π Q ζdζ. 3. Hardy Orlicz spaces on planar domains In this section we define Hardy Orlicz spaces on planar domains and prove basic properties of these spaces. We start from the study of certain properties of domains and subharmonic functions. Let Ω be a domain on the Riemann sphere. Recall that a regular exhaustion of Ω is a sequence {Ω n } n= of subdomains of Ω which satisfy the following conditions: Ω n Ω n+ for n N, 2 n= = Ω, 3 every component of Ω n, is nontrivial for each n N. It can be proved that each domain has a regular exhaustion see [2, Proposition.5.3]. Recall that upper semicontinuous function u: Ω [, is called subharmonic we write u subhω, if for every compact subset K Ω and every continuous function h harmonic on interior of K with h u on K, we have h u on K as well. In this case h is called harmonic majorant of u. It can be proved that if u has harmonic majorant, then it has the least harmonic majorant and it is unique. It is well known that subharmonic functions satisfy Maximum Modulus Principle. Using this fact and the Harnack Theorem we can obtain the following result.

5 Composition operators on Hardy Orlicz spaces on planar domains 597 Theorem 3.. Let Ω SDP and u subhω, p Ω. Then u has harmonic majorant if and only if for each regular exhaustion {Ω n } of Ω there exist a constant C such that Ω n udω p,n C, where ω p,n is harmonic measure on Ω n for p. In fact infimum over those constants is equal tovp, wherevis the least harmonic majorant of u. Now we can define Hardy Orlicz spaces on general domains. Let Φ: [0, [0, be convex Orlicz function. Suppose that Ω SDP, {Ω n } is regular exhaustion of Ω and z 0 Ω. For g: Ω C denote g n := g Ωn. We define Hardy Orlicz space on Ω by the following condition: H Φ Ω := {f HΩ: f H Φ Ω < }, where { f H Φ Ω := lim f n Φ = lim inf ε > 0: n n Ω n Φ fn ε } dω n,z0. Notice that by Theorem 3. we can also describe H Φ Ω in terms of harmonic majorant; it is a set of all holomorphic functions f, for which there exists λ > 0, such that subharmonic function Φλ f has harmonic majorant. Moreover f H Φ Ω = inf{ε > 0: v f,ε z 0 }, where v f,ε is the least harmonic majorant of Φ f ε. It s clear that H Φ Ω is a Banach space. We denote by HM Φ Ω the subspace of finite elements of H Φ Ω, i.e., closure of H Ω in H Φ Ω. For further work we need additional assumption on domain Ω. Let Ω be a bounded domain whose boundary consists of m+ disjoint analytic Jordan curves, i.e. m := Ω = k, where k is analytic Jordan curve and k j = for k j. Assume that 0 is the boundary of the unbounded component of the complement of Ω. Denote by E 0 bounded component of S 2 \ 0 and for k {,2,...,m}, denote by E k unbounded component of S 2 \ k, where S 2 is the Riemann sphere. From now Ω will be always a set of this type. We also define by H Φ 0E k the subspace of H Φ E k which consists those functions which vanish at. The first result shows that H Φ Ω can be represented as a certain direct sum: Theorem 3.2. For each f H Φ Ω we have the following decomposition k=0 2 fz = f 0 z+f z+...+f m z, z Ω, where f 0 H Φ E 0 andf k H Φ 0 E k for each k m. Moreover, the mapf f 0 is a bounded linear projection of H Φ Ω onto H Φ E 0 and f f k is a bounded linear projection of H Φ Ω onto H Φ 0 E k.

6 598 Michał Rzeczkowski Proof. Fix f H Φ Ω, z Ω and let C 0,...,C m be a smooth Jordan curves so close to 0,..., m respectively, that z is exterior to C,...,C m and interior to C 0. Now for each 0 k m f k z := fζ 2πi C k ζ z dζ. It is clear that f = f 0 + f f m and f f k is a linear operator. Further f k HE k and it is independent of the choice of C k, 0 k m. We have also f k = 0 for each k m. It s easily seen that if f H0E Φ k for some k then f l = 0 for l k. Hence f = f k. We need to show that f k H Φ E k. Note that for all l k the function f l is bounded pointwise on some neighborhood U of k. From the formula 2 and the fact that H Φ Ω is a linear space we conclude that f H Φ U, and so this implies that f H Φ E k. Note that interpolation theorem for general variants of Hardy spaces on planar domains which are defined as a similar direct sums has been recently proved in [3]. Recall that polynomials are dense in HM Φ. Denote by RΩ the set of rational functions whose poles are off Ω. We formulate analogue of this result for HM Φ Ω. Proposition 3.3. RΩ is dense in HM Φ Ω. Proof. Fixk {0,...,m}. SinceΩ E k thenω z,ω A ω z,ek A for every point z Ω and every subset A k. Thus the H Φ E k norm is larger then H Φ Ω. We need to show that f k is a limit in HM Φ E k of a sequence of functions holomorphic in a neighborhood of Ē k. Let η k be a Riemann mapping of D onto E k. Since D is analytic, this mapping can be extended to holomorphic and one-to-one in some neighborhood of D. The same is true for the inverse function η k. Put g k := f k η k, then g k HM Φ and therefore, by Runge s theorem, there is a function G analytic on a neighborhood of D and such that G g k H Φ < ε and it is equivalent to G η k f k H Φ E k < ε, and G η k is analytic in a neighborhood of Ē k. Applying again Runge s theorem we can approximate G η k uniformly on Ēk by elements of RΩ. Proposition 3.4. [2, Proposition ] If u L,ds and uζ 0 = dζ, z / ζ z then u = 0 a.e. on. Now we formulate the main result of this section. Let us remark that ω := ω z for a point z Ω. Theorem 3.5. Every f H Φ Ω has boundary values f almost everywhere dω on and f L Φ,ω. Moreover fz = f 3 w dw, z Ω, 2πi w z f w 4 0 = dw, z / Ω, w z 5 fz = f ζdω z ζ, z Ω.

7 Composition operators on Hardy Orlicz spaces on planar domains 599 Finally, the mapping f f is an isomorphism of H Φ Ω onto closed subspace of L Φ,ω and it is an isometry of HM Φ Ω onto closed subspace of L Φ,ω. Proof. By Theorem 3.2 it is enough to show that f k has boundary values a.e. ds on and that this boundary-value functions lies in L Φ,ω. Fix k {0,...,m}. For l k the function f k is analytic on l. So 3, 4 and 5 hold immediately. Let η k be the Riemann mapping of D onto E k. Note that η k extends to be analytic and conformal on a neighborhood of Ω since k is analytic the same is true for η k : E k D. Further g k = f k η k H Φ so gk exists a.e. on T and g k LΦ T. Then f k = g k η k has boundary values a.e. ds on k and fk = g k η k a.e. so that fk LΦ k,ds and hence fk LΦ,ω. Now if z Ω then f k z = g k η k z = gk ξ 2πi ξ = ξ η dξ. k z Putting ξ = η k ζ we have Now we notice that f k z = 2πi k fk ζ η k ζ η k zη k ζ dζ. η k ζ η k ζ η k z = ζ z +Sz, where S is analytic in a neighborhood of Ω, since the function in the left-hand side of the equality has a simple pole at z with residue equal to. So we have Sζfk ζ = 0. k f If k l then k ζ dζ = 0 and so 3 and 4 hold.to prove 5 recall that dω l ζ z zζ = i 2π Q zζdζ, where Q z ζ = gζ;z+ihζ;z. Then Q zζ = +Rζ, where R is z ζ holomorphic on Ω, hence f krζdζ = 0. Consequently f ζdω z ζ = i 2π f ζ ζ z dζ + i 2π f ζrζdζ = fz. To show that f f is isomorphism from H Φ Ω into L Φ,ω it is enough to use Theorem 3.2 and the facts that f k fk is isomorphism from HΦ E k into L Φ k,ω and for l k the function f k is analytic on l. Now we show that f f is isometry from HM Φ Ω into L Φ,ω. Let q RΩ and let u be the harmonic function on Ω given by 6 uz = qζ Φ dω z, z Ω. q H Φ Ω Then Φ qz uz, and if v is any harmonic majorant of Φ q H Φ Ω qx ux = Φ liminf{vz : z x}. q H Φ Ω qz q H Φ Ω, then It implies that the harmonic functionv u is non-negative atand hence for allz Ω. q Thus, the function u given by 6 is the least harmonic majorant of Φ and q H Φ Ω we conclude that q H Φ Ω = q L Φ,ω for q RΩ. Now take f HM Φ Ω. By

8 600 Michał Rzeczkowski Proposition 3.3 there exist {q n } in RΩ such that q n f in H Φ Ω. Then q n f uniformly on compact subsets of Ω and it s clear that q n q m H Φ Ω = q n q m L Φ. Hence {q n } is a Cauchy sequence in L Φ. So if q n g in L Φ,ω, then fz = gζdω z ζ, z Ω, since all the harmonic measures are boundedly mutually absolutely continuous. Moreover we have fz = gζ dζ, z Ω, 2πi ζ z and 0 = gζ ζ z dζ = f ζ dζ, z / Ω. ζ z Now by Proposition 3.4 we have g = f a.e. dω. Finally q n f L Φ,ω 0 and f L Φ,ω = lim q n L Φ,ω = lim q n H Φ Ω = lim f H Φ Ω. Notice that in the case when Φ 2, i.e., Φ2x KΦx for some constants K, x 0 and x x 0, then H Φ Ω = HM Φ Ω. So from Theorem 3.2 it follows the analogous result proved by Rudin in [6, Theorem 3.2.] for Hardy space H p Ω, p <, since every power function Φt = t p satisfies 2 condition. Recall that a Banach space X of holomorphic functions on an open subset Ω of the complex plane has the Fatou property ifx is continuously embedded inhω, the space of holomorphic functions on Ω, equipped with its natural topology of compact convergence, and if it has the following property: for every bounded sequence {f n } in X which converges uniformly on compact subsets of Ω to a function f, one has f X. Note that H Φ Ω has the Fatou property. We have seen that H Φ Ω is isomorphic to H Φ E 0 H0E Φ H0E Φ m. Since conformal maps are isometries in the class of Hardy Orlicz spaces we may assume that E 0 = D, E i = a i + r i D for i =,...,m, where r i 0,, a i D, a i a j if i j and the circles D,a i +r i D are pairwise disjoint and E i E j =. Spaces of those type on such Ω, which is called circular domain are isomorphic to a general described before so from now Ω denotes this special case. Let us note that conformal maps from E i onto unit disc and inverse which we used before for this particular Ω are of the form: { ri η i z = +a z i dla z D\{0}, dla z = 0, η i z = { ri z a i dla z E i \{ }, 0 dla z =. for i m and we put η 0 = id D. It was proved in [8] that the norm of the evaluation functional δ z : H Φ D C at z D satisfies condition δ z Φ. z Using this fact, Theorem 3.2 and conformal maps it is easy to prove, that for z Ω, the norm of δ z : H Φ Ω C satisfies δ z CΦ, where the constant C distz, depends only on Ω and z.

9 Composition operators on Hardy Orlicz spaces on planar domains 60 At the end of this section we introduce useful family of functions. Recall that see [8] for a D and r 0, r 2, u a,r z = ārz is holomorphic function on D with u a,r H r, u a,r H = and u a,r H Φ Φ. For i m and a and r as above we define r r 2, u i a,r z = u a,r η i z = z Ω, ārr i z a i and u 0 a,r = u a,r Ω. Note that for each 0 i m function u i a,r extends to holomorphic function on E i. It is also clear that we have analogous norm estimation: u i a,r H E i r, u i a,r H E i =, u i a,r H Φ E i Φ, and ui a,r H Ω r C r, u i a,r H Ω, u i a,r H Φ Ω Φ, for some constant C. r 4. Composition operators on Hardy Orlicz spaces of planar domains In this section we study properties of composition operators C ϕ : H Φ Ω H Φ Ω and canonical inclusion maps j µ : H Φ Ω L Φ Ω,µ, where µ is a Borel measure µ on Ω. Recall that if ϕ is analytic map ϕ: Ω Ω in this case we write ϕ Υ := Υ Ω, then for f HΩ we define C ϕ as follow: C ϕ fz := f ϕz, z Ω. First let s see that C ϕ defines bounded operator. Linearity is clear. To show that C ϕ : H Φ Ω H Φ Ω is bounded take f H Φ Ω with f H Φ Ω = λ, where λ = inf{ε > 0: v f,ε z 0 } and v f,ε being the least harmonic majorant of Φ f. ε We have f ϕ Φ v f,ε ϕ, ε andv f,ε ϕ is harmonic function. Further Φ fϕz0 v f,ε ϕz 0. Put w 0 = ϕz 0, then from Harnack s inequality there exists constant c > 0 such that v f,ε w 0 cv f,ε z 0, now taking C = maxc, it is clear that C ϕ f H Φ Ω C f H Φ Ω. In what follows we investigate compactness of composition operators on Hardy Orlicz spaces of planar domains. We will use the following result see [8, Proposition 3.8]. Proposition 4.. Let X,Y be two Banach spaces of analytic functions on an open set Ω C, X having the Fatou property. Let ϕ Υ be such that C ϕ Y whenever f X. Then C ϕ : X Y is compact if and only if for every bounded sequence {f n } in X which converges to 0 uniformly on compact subsets of Ω, one has C ϕ f n Y 0. We can apply Proposition 4. in the case whenx = Y = H Φ Ω getting necessary condition for compactness of C ϕ. Proposition 4.2. If C ϕ : H Φ Ω H Φ Ω is compact operator, then for each 0 i m we have 7 lim sup Φ C ϕ u i s p,s p D s H Φ Ω = 0. ε

10 602 Michał Rzeczkowski Proof. Let {p n } be any sequence in D, and {s n } 0,, with s n. By the fact that u i p,s H Φ Φ, for every s 0, we conclude that the sequence s {fn i } defined by f i nz = Φ s n u i p n,s n z, z Ω, is bounded in H Φ Ω for all 0 i m. Moreover Φ is concave and Φ 0 = 0, so there exists a constant C > 0 such that Φ x Cx for x. But it implies that for i m and We conclude that for 0 i m fnz i C s n 2 s n pnsnr i z a i 2 fn 0 C s n 2, z s n z 2 f i n 0, on compact subsets of Ω and by Proposition 4. we have that C ϕ f n H Φ Ω 0. Therefore 7 clearly holds. Carleson measures. Let ϕ Υ. Since ϕ is bounded by Theorem 3.5, the boundary function ϕ exists and it is in L Φ,ω or what is equivalent ϕ L Φ,ds. For given ϕ Υ and any Borel subset of Ω we define pullback measure as follows: µ ϕ B := sϕ B. Further we have { C ϕ f H Φ Ω f ϕ L Φ,ω = inf ε > 0: { inf ε > 0: { = inf ε > 0: = f L Φ Ω,µ ϕ. Ω, f ϕ } Φ dω ε f ϕ } Φ ds ε f } Φ dµ ϕ ε It shows that some properties of a composition operator C ϕ : H Φ Ω H Φ Ω can be expressed in terms of an inclusion operator j µϕ : H Φ Ω L Φ Ω,µ ϕ. We will consider inclusion operator in general not only for a pullback measures. Lemma 4.3. Suppose that j µ : H Φ Ω L Φ Ω,µ ϕ is a compact operator. Then µ = 0. Proof. First note that lemma is true when Ω = D see [8, Lemma 4.8]. To have general case consider operators T i : H Φ H Φ Ω, defined as follow T i f = f η i Ω, where 0 i m. It is clear that T i is a bounded operator for each 0 i m. Since {e n } given by e n z := z n is weakly null in HM Φ then {T i e n } is weakly null in

11 Composition operators on Hardy Orlicz spaces on planar domains 603 HM Φ Ω for 0 i m. By compactness of j µ we have that ri n dµ < ε, z a i n where i m, and Φ Ω Ω Φ z n dµ < ε for every ε 0, and sufficiently large n. Hence Φµ i < ε for each 0 i m and it implies that µ = 0. Let {Ω n } be arbitrary regular exhaustion of Ω and let µ be Borel measure on Ω. We define operator I n : H Φ Ω L Φ Ω,µ as follow I n f := fχ Ω\Ωn. Theorem 4.4. Let µ be a finite measure on Ω, and Φ an Orlicz function. The following assertions are equivalent: The inclusion operator j µ : H Φ Ω L Φ Ω,µ is well defined and compact. 2 For every bounded sequence {f n } in H Φ Ω converging to 0 uniformly on compact sets, we have f n L Φ Ω,µ 0. 3 H Φ Ω is included in L Φ Ω and lim I n H n Φ Ω L Φ Ω,µ = 0. for any regular exhaustion {Ω n } of Ω. Proof. To show that 2 implies is a standard and it can be found in [5]. We show that implies 2. From lemma we know that µ Ω = 0; take {f n } B H Φ Ω withf n 0 uniformly on compact subsets ofω. In particularf n z 0 for allz Ω. Since µ Ω = 0 we have that f n 0 µ-a.e. on Ω. Suppose that f n L Φ Ω,µ 0 is not true. From the compactness of inclusion operator we have that there exists a subsequence {f nk } such that f nk g 0 in L Φ Ω. But f nk g µ-a.e. and this gives a contradiction. Suppose now that 3 holds. We have that j µ = lim n j µ I n Notice that j µ I n is inclusion operator of H Φ Ω into L Φ Ω,ν for a finite measure ν supported by Ω n. Applying equivalence of and 2 to I n ν := j µ I n we get that I n ν is compact for every n N and also j µ is compact operator as the limit of compact operators. Assume that 2 holds. Notice that the sequence { I n } is decreasing so if 3 were not true, there would exist a constant ε 0 > 0, a sequence {f n } in the unit ball of H Φ Ω and regular exhaustion {Ω n } of Ω such that I n f n L Φ Ω,µ > ε 0, for n N. The sequence {f n } is uniformly bounded on compact subets of Ω. By Theorem 3.2 we can decompose each f n as follow: f n z = f 0,n z+f,n z+ +f m,n z, z Ω, nfi,n with f i,n H0 ΦE r i. For each i n define by g i,n z := i z a i z and g 0,n z := z n f 0,n z. It is clear that g i,n is bounded in H0E Φ i for 0 i m and g i,n 0 uniformly on compact subets of E i, so g n = g 0,n + +g m,n is bounded in

12 604 Michał Rzeczkowski H Φ Ω and g n 0 uniformly on compact subets of Ω. By 2 we should have that if g n L Φ Ω,µ 0. But if dist i,z n, or in other words z An i, where then A n i = { z E i : z = a i +r i +εe iθ, 0 ε, θ [0,2π]}, n A n 0 = D\ D, n r i n g i,n z = fi,n z z a i + n n fi,n z 4 f i,nz for i m and in a similar manner we obtain g 0,n z f 4 0,nz for z A n 0. Now if we renumber {Ω n } to have Ω\Ω n A n = m i=0 An i and in the same way we renumber {f n }, then 4 g n L Φ Ω,µ I n f n L Φ Ω,µ ε 0 > 0, for n big enough and this is a contradiction to the fact g n L Φ Ω,µ 0. Carleson window on i = Ω of center ξ i and radius 0 < h < min i j dist i, j is the set W 0 ξ,h = {z Ω: h < z, argξz < h}, i = 0 W i ξ,h = {z Ω: z a i < r i, argξz < h}, i m. h It is trivial that z W 0 ξ,h if and only if η i z W i η i ξ,h when i. Note also that area of Carleson window satisfies estimation AW i ξ,h 2r 2 i h2, and the constant 2r 2 i is optimal. Let µ be a finite measure on Ω, 0 < h < min i j dist i, j, A > 0, and Φ an Orlicz function, we denote: ρ µ h := max 0 i m sup ξ i µwξ,h, ρ K µ h := sup µt 0<t h, t γ A h :=. Φ AΦ Following [8] we define the following conditions: h R 0 lim ρµh h 0 + γ A = 0 for every A > 0. h K 0 lim Kµhh h 0 + γ A = 0 for every A > 0. h C 0 Inclusion operator j µ : H Φ Ω L Φ Ω,µ is compact. Our purpose is to show that we have the following implications K 0 = C 0 = R 0. We will use the following proposition see [5, Lemma 3.3.] Proposition 4.5. Condition R 0 is equivalent to: h lim h 0 + Φ Φ ρ µh = 0. We note that in [8] or [5] it was proved that C 0 implies R 0 in the case of Ω = D. Based on our Theorem 4.4 we prove below a more general result. Theorem 4.6. Condition C 0 implies R 0.

13 Composition operators on Hardy Orlicz spaces on planar domains 605 Proof. Using Theorem 4.4, in the same way that Proposition 4. was used to prove Proposition 4.2, we deduce that for each 0 i m we have 8 lim sup Φ u i r p D r p,r L Φ Ω,µ = 0. It is easy to check that u 0 p,rz 9, when z W 0p,h, and r = h, thus We obtain analogous inequality Φ 9 u 0 p,r L Ω,µ. Φ µw 0 p,h Φ 9 u i p,r L Ω,µ, Φ µw i η i p,h if i using the fact that z W 0 p,h if and only if η i z W i η i p,h. Further taking for each i 0 the supremum over ξ i = η i p in fact we take m + -times supremum over p D and use the fact that η i is bijection and then maximum over i = 0,...,m we obtain from 8 that lim h 0 + Φ h Φ ρ µh For ξ, and α > consider the following family of sets G α ξ := {z Ω: z ξ < αdistz,}. = 0. Put G ξ := G 3 ξ. The maximal non-tangential function M f will be defined by: M f ξ := sup{ fz : z G ξ }. Using the fact that the maximal non-tangential function M f is weak type, and strong type, see [5, p ] we can extend using the same methods important result which was proved in [8, Proposition 3.5]. Proposition 4.7. Assume that the Orlicz function Φ 2. Then every linear, or sublinear, operator which is of weak-type, and strong type, is bounded from L Φ into itself. In particular, for every f L Φ, the maximal non-tangential function M f L Φ. Theorem 4.8. Let µ be finite borel measure on Ω, f: Ω C continuous function. Then for each t > 0 and for each 0 < h < 4 min i jdist i, j we have 9 µ{z Ω: distz, h, fz > t} C K µ hs{ξ : M f ξ > t}. It was proved in [8] that for D the inequality holds with C =. Using exactly the same methods it is easy to obtain 9 for a circular domain. Recall that the norm of evaluation functional at point z Ω on H Φ Ω is not larger than CΦ for a constant C. Put c 0 = max,c to have c 0 > distz, and denote by c = max,c, where C is the constant from Theorem 4.8. Lemma 4.9. Let µ be a finite Borel measure on Ω. Suppose that Φ 2 and there exists A > 0 and 0 < h A < 4 min i jdist i, j such that for every h 0,h A we have K µ h h γ Ah,

14 606 Michał Rzeczkowski then for every f B H Φ Ω and for every Borel subset E Ω we have A Φ f dµ µeφx A + c ΦM f ds, 2c 0 c 0 E where x A = A 2 Φ h A. Proof. For every w > 0, the inequality fz > w implies that w < c 0 Φ, distz, hence distz, < Φw/c 0. It follows by Theorem 4.8 that µ{z Ω: fz > w} c K µ s{ξ : M f ξ > w}, Φs/c 0 if Φw/c 0 min 4 i jdist i, j. Further A Φ f dµ = Φ tµ { f > 2c 0 t/a} E dt. 2c 0 E By assumption for Φw/c 0 > /h A we have: K µ Φw/c 0 Φw/c 0 ΦAw/c 0. Then we have { µ z Ω fz > 2c } 0t A and E 0 { Φ2t/A c Φ2t s ξ : M f ξ > 2c } 0t. A A xa { Φ f dµ = Φ tµedt+ Φ ts M f ξ > 2c 0t 2c 0 0 x A A { Φ2t/A Φx A µe+c x A Φ2t Φ ts } dt M f ξ > 2c 0t A } dt. Note that one has Φt tφ t Φ2t, for any Orlicz function Φ. Using these inequalities we obtain { Φ2t/A x A Φ2t Φ ts M f ξ > 2c } 0t dt A { Φ2t/A s M f ξ > 2c } 0t dt 0 t A 2 { Φ 2t/As M f ξ > 2c } 0t dt A 0 A { } Φ xs M f > c 0 x dx which implies desired inequality. 2 A 0 = Φ M f ds c 0 c 0 Φ M f ds Theorem 4.0. If Φ 2, then condition K 0 implies C 0.

15 Composition operators on Hardy Orlicz spaces on planar domains 607 Proof. Take ε > 0. Denote by C the norm of maximal operator M f. Set A = 4Cc 0c, where c ε 0 and c are constants from the previous Lemma. Condition K 0 implies existence of 0 < h A < min 4 i jdist i, j such that K µ h 2h γ Ah, when h h A. If f B H Φ Ω then for each n N by Lemma 4.9 we have: Ω\Ω n Φ f dµ = ε 2 2 A f Φ dµ Ω\Ω n 4Cc 0 c 2 µ Ω\Ω n Φx A + c c 0 µ Ω\Ω n Φx A + c 0, A f Φ Ω\Ω n 2Cc 0 c Mf Φ ds Cc for any regular exhaustion {Ω n } of Ω. Since c 0 > and µ = 0 we have µ Ω\ 2 Ω n Φx A + c 0 for a sufficiently large n. This implies by Theorem 4.4 that j µ is compact, i.e., C 0 holds. We are able to show that in the case of µ = µ ϕ that conditions C 0, R 0 and K 0 are equivalent. To prove that we need the following result: Theorem 4.. There exists a constant k > 0 such that for every ϕ Υ, each 0 i m and ξ i i we have 0 where ε 0, and h is small enough. µ ϕ W i ξ i,εh kεµ ϕ W i ξ i,h, Proof. First notice that Theorem is true when Ω = D, ϕ Υ D see [8], in this case h < ϕ0. To prove it in general case fix 0 i m. Let U i,...,ui n i Ω be finite family of domains which satisfies conditions: i for every j n i the set Uj i is simply connected, ii for every j n i the boundary Uj i of Ui j is formed by analytic Jordan curve, iii i n i j= Ui j and every arc contained in i Uj i is free analytic boundary arc for every j n i, iv for every 0 l m and l i we have l n i j= Ui j =. Let τ i,j : Uj i D be conformal map from Ui j onto disc D, then τ i,j extends for each j n i to diffeomorphism τ i,j : Ūi j D. The same is true for the inverse mapτ i,j. Note that there exist finite constants C i,j,c i,j > 0 such that c i,j τ i,j z C i,j. It implies that for every z,z 2 D there exist finite constants C i,j,c i,j > 0 such that c i,j z z 2 τ i,j z τ i,j z 2 C i,j z z 2. Put C = max 0 i m max j ni C i,j and c = min 0 i m min j ni c i,j. Fix ξ k Ω and let h be as small that the preimage ϕ W k ξ,h of Carleson window W k ξ,h is contained in a certain arc of U i j for 0 i m and j n i. For 0 k m define function ψ i,j,k : D D as follow: ψ i,j,k z = η k ϕ τi,j z, z D. dµ

16 608 Michał Rzeczkowski Since ψ i,j,k Υ D then the measure µ ψi,j,k satisfies 0. Thus for sufficiently small h > 0 we have µ ψi,j,k W0 η ξ,h = s {z D: ψ i,j,k z W 0η ξ,h} = s {z D: ϕ τ i,j z W kξ,h}. Using with universal constants C,c we easily see that the last amount is comparable to s {z i : ϕ z W k ξ,h} = µ ϕ Wk ξ,h, for every 0 k m and every ξ k and sufficiently small h > 0. Thus for sufficiently small Carleson windows we have µ ψi,j,k W 0 η ξ,h µ ϕ W k ξ,h for every 0 k m. Then for 0 k m we have µ ϕ W k ξ,εh K µ ψi,j,k W0 η k ξ,εh CK εµ ψi,j,k W0 η k ξ,h CK K 2 εµ ϕ Wk ξ,h. Using this Theorem we obtain characterization of compact composition operators. Theorem 4.2. Let ϕ Υ and let Φ 2 be the Orlicz function. Composition operator C ϕ : H Φ Ω H Φ Ω is compact if and only if for every A > 0 2 ρ ϕ h lim h 0 + γ A h = 0, where ρ ϕ := ρ µϕ. Equivalently C ϕ is compact if and only if h 3 lim h 0 + Φ Φ ρ ϕh = 0. Proof. Equivalence of 2 and 3 was proved in 4.5. For h small enough and 0 < t < h using 0 with ε = t we obtain h µ ϕ W i ξ i,t k t h µ ϕw i ξ i,h for each ξ i, i = 0,...,m. Taking supremum over ξ we get ρ ϕ t k tρ h ϕh. Hence for µ = µ ϕ ρ µ t K µ h = sup ρ µh 0<t h t h, so from Theorem 4.6 and Theorem 4.0 conditionsr 0,K 0 andc 0 are equivalent. Now we prove that condition 7 in Proposition 4.2 is also sufficient to the compactness of composition operator. Corollary 4.3. Let ϕ Υ and let Φ 2 be the Orlicz function. Composition operatorc ϕ : H Φ Ω H Φ Ω is a compact operator if and only if for each0 i m we have 4 lim sup Φ C ϕ u i s p,s p D s H Φ Ω = 0.

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