LINEAR FRACTIONAL COMPOSITION OPERATORS OVER THE HALF-PLANE

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1 LINEAR FRACTIONAL COMPOSITION OPERATORS OVER THE HALF-PLANE BOO RIM CHOE, HYUNGWOON KOO, AND WAYNE SMITH Abstract. In the setting of the Hardy saces or the standard weighted Bergman saces over the unit ball in C n, linear fractional comosition oerators are known to behave quite rigidly in the sense that they cannot form any nontrivial comact differences or, more generally, linear combinations. In this aer, in the setting of the standard weighted Bergman saces over the half-lane, we comletely characterize bounded/comact differences of linear fractional comosition oerators. Our characterization reveals that a linear fractional comosition oerator can ossibly form a comact difference, which is a new half-lane henomenon due to the half-lane not being bounded. Also, we obtain necessary conditions and sufficient conditions for a linear combination to be bounded/comact. As a consequence, when the weights and exonents of the weighted Bergman saces are restricted to a certain range, we obtain a characterization for a linear combination to be bounded/comact. Alying our results, we rovide an examle showing a double difference cancelation henomenon for linear combinations of three linear fractional comosition oerators, which is yet another half-lane henomenon.. Introduction Various asects of comosition oerators acting on classical function saces such as the Hardy saces and the weighted Bergman saces have been studied over the ast several decades; we refer to monograhs by Cowen- MacCluer [7] and Shairo [2] for an overview of the work before the mid- 990s. In articular, initiated by the isolation henomenon on the Hardy sace over the unit disk, which was first observed by Berkson [] and then refined by Sundberg and Shairo [23], the study of comact differences or, more generally, linear combinations of comosition oerators has been a toic of growing interest in the theory of comosition oerators; see, for examle, [0, 2, 8, 23] for the Hardy saces and [3, 4,, 3, 5, 6, 7, 9, 20] for the weighted Bergman saces. Date: May 7, 206; (Revised) January 7, Mathematics Subject Classification. Primary 47B33; Secondary 30H20. Key words and hrases. Linear fractional comosition oerator; Difference; Linear combination; Bounded oerator; Comact oerator; Hilbert-Schmidt oerator; Weighted Bergman sace; Half-lane. B. R. Choe was suorted by NRF(205RDAA ) of Korea and H. Koo was suorted by NRF(207RA2B200255) of Korea.

2 2 B. CHOE, H. KOO, AND W. SMITH In the case of the weighted Bergman saces over the unit disk, comact differences were first characterized in general by Moorhouse [7] in terms of a natural angular derivative cancelation roerty; see also [3] and [4] for extensions to the olydisk and the ball in C n. However, in a quite secialized context, linear fractional comosition oerators (i.e. comosition oerators induced by linear fractional mas), which were first introduced and studied by Cowen and MacCluer [8] in several variables, are known to behave quite rigidly on the weighted Bergman saces over the ball in the sense that they cannot form a nontrivial comact linear combination; see [6]. Such rigid behavior, when restricted to differences, was earlier observed indeendently in [] and [3]. A similar rigidity henomenon is known to hold in the context of the Fock-Sobolev saces; see [2]. The urose of the current aer is to investigate this rigidity henomenon, as well as related roerties, for linear fractional comosition oerators on the weighted Bergman saces over the uer half-lane. Let H be the uer half of the comlex lane C, i.e., H := {z C : Im z > 0} and let S be the class of all holomorhic self-mas of H. Each φ S induces a comosition oerator C φ defined by C φ f = f φ for functions f holomorhic on H. It is clear that C φ takes the sace of holomorhic functions on H into itself. For >, ut da (z) := c (Im z) da(z) where c = 2 (+) π is a normalizing constant and A is the area measure on H. For 0 < <, we denote by A (H) the -weighted Bergman sace consisting of all holomorhic functions f on H such that the norm { } / f A := f da H is finite. As is well known, each sace A (H) is a closed subsace of L (da ). Thus A (H) is a Banach sace for < and, in articular, A 2 (H) is a Hilbert sace. Also, when 0 < <, the sace A (H) is a comlete metric sace under the translation-invariant metric (f, g) f g A. As is well known, in the setting of the unit disk comosition oerators are always bounded on the Hardy saces and weighted Bergman saces by the Littlewood Subordination Princile. However, such boundedness is not guaranteed any more on A (H) due to the domain H not being bounded. For examle, it is not hard to check that the saces A (H) are not Möbius invariant. In fact, Elliott and Wynn [9] obtained a characterization showing that behavior at of an inducing ma is the key to boundedness of the corresonding comosition oerator; see Theorem 2. in 2. Their work also shows that no comosition oerator on A (H) is comact; see [22]

3 LINEAR FRACTIONAL COMPOSITION OPERATORS 3 for a more general result in this direction. Quite recently, the current authors [5] studied bounded/comact differences on A (H). One of our results therein reveals some new half-lane henomena, caused by the domain H being unbounded, that have no analogues over the unit disk. Execting further half-lane henomena, we roceed in this aer to the investigation of bounded/comact differences of linear fractional comosition oerators acting on A (H). We denote by S the class of all linear fractional self-mas of H. As is quite elementary, S contains Aut (H), the class of all automorhisms (=biholomorhic self-mas) of H. It is known that a comosition oerator that is induced by an automorhism but is not bounded behaves quite rigidly in the sense that it cannot form a bounded difference with any other comosition oerator; see [5, Theorem 6.2]. Meanwhile, by Proosition 2.2 below, the only mas in S that induce bounded comosition oerators are linear olynomials. Excluding those two extreme tyes of linear fractional mas, we ut S f : = {φ S : φ( ) and φ / Aut (H)}. Note that every ma in Sf takes the real axis R onto either a circle or a straight line and, moreover, in the latter case, the image line must be arallel to R. So, Sf can be decomosed into the following three disjoint subclasses: S h : = {φ S f : φ(h) = ɛi + H for some ɛ > 0} S t : = {φ S f : φ(h) is a disk tangent to R} S c : = {φ S f : φ(h) is a disk in H}. Our results will turn out to deend on which subclasses the inducing mas under consideration are from. In 2 we collect some basic material that will be needed subsequently. In 3 we obtain comlete characterizations for bounded/comact differences of linear fractional comosition oerators. We first show that a ma φ St induces no comact differences and, when in addition φ( ) R, no bounded differences; see Theorem 3.6 and Theorem 3.3, resectively. We then obtain characterizations for the remaining cases, deending on whether the inducing mas have the same derivative at ; see Theorem 3.9 and Theorem 3.0. As is exected, our results reveal interesting new half-lane henomena demonstrating that the rigidity henomenon sometimes holds and sometimes does not, deending on three factors: (i) the subclasses to which the inducing mas belong; (ii) the derivatives at of the inducing mas; and (iii) the arameters and. In articular, unlike the cases of the disk and the ball, we see that linear fractional comosition oerators can ossibly form comact differences. On the other hand, our characterizations imly another tye of rigid behavior. Namely, we show that the three notions of boundedness, comactness and the Hilbert-Schmidt roerty all

4 4 B. CHOE, H. KOO, AND W. SMITH coincide for differences induced by mas in Sc Sh ; see Theorem 3.2 and Theorem 3.3. In 4 we study general linear combinations of linear fractional comosition oerators. First, we observe behavior of inducing mas at necessary for boundedness and comactness, resectively. For boundedness, we show that an inducing ma involved in a bounded linear combination cannot ma into R; see Theorem 4.2. For comactness, we show that an inducing ma involved in a comact linear combination cannot fix and cannot belong to St ; see Theorem 4.3. Next, for a restricted range of arameters and, we characterize boundedness by means of a certain coefficient cancelation roerty; see Theorem 4.5 and Theorem 4.6. As a consequence when > + 2, we obtain a characterization for boundedness/comactness; see Corollary 4.7. Finally, alying our results, we rovide a concrete examle of linear combinations of three linear fractional comosition oerators, which reveals yet another half-lane henomenon; see Examle 4.8. Constants. Throughout the aer we use the same letter C to denote various ositive constants which may vary at each occurrence but do not deend on the essential arameters. Variables indicating the deendency of constants C will be often secified in arenthesis. For nonnegative quantities X and Y the notation X Y or Y X means X CY for some inessential constant C. Similarly, we write X Y if both X Y and Y X hold. 2. Preliminaries In this section we collect some basic facts and reliminary results to be used throughout the aer. 2.. Bounded Comosition Oerators. Comosition oerators on the weighted Bergman saces over the half-lane, which are not always bounded unlike the disk case, are characterized by Elliott and Wynn as follows. Theorem 2. ([9]). Let > and 0 < <. Let φ S. Then C φ is bounded on A (H) if and only if φ( ) = and φ ( ) exists. In fact the above theorem is stated in [9] only for = 2, but remains true for all by the well-known Carleson measure characterization; see [5, Theorem 2.4]. Note that the derivative condition in Theorem 2. is trivially satisfied by all linear fractional self-mas of H. Thus, when alied to linear fractional comosition oerators, Theorem 2. reduces to the next roosition. Proosition 2.2. Let >, 0 < < and φ S. Then C φ is bounded on A (H) if and only if φ( ) = Linear Fractional Self-mas of H. We need more information on mas in the aforementioned three subclasses Sf. For that urose we first

5 LINEAR FRACTIONAL COMPOSITION OPERATORS 5 recall some elementary facts. A function u M(H, D), the class of all biholomorhic mas taking H onto the unit disk D, is of the form u(z) = ζ z w z w where w = u (0) H and ζ is a unimodular comlex number. Also, a function v Aut (H) is of the form v(z) = az + b cz + d where a, b, c, d are real coefficients with ad bc > 0. Deending on whether the images are bounded or not, general forms of mas in Sf can be described by means of biholomorhic mas in the receding aragrah. First, a ma φ St Sc is of the form (2.) φ(z) = a[u(z) + i] + ξ (i = ) for some a > 0, u M(H, D) and ξ H. In this case note φ Sc if and only if ξ H. On the other hand, a ma φ Sh is of the form (2.2) φ(z) = v(z) + ɛi for some ɛ > 0 and v Aut (H) with v( ). In general a ma φ Sf can be written in the form φ(z) = a + b z w for some b C and a, w H determined by (2.) or (2.2). Note that a and b are comuted by a = φ( ) and b = lim z[φ(z) φ( )]. z The limit above (when φ( ) ) is often regarded as the derivative at and denoted by φ ( ). Note φ ( ) 0. So, φ Sf can be written as (2.3) φ(z) = φ( ) + φ ( ) z w for some w H. In this case note φ S t S c if and only if w H. Also, note φ ( ) < 0 if φ( ) R Integrals of Kernel-tye Functions. For w H, let τ w be the function on H defined by (2.4) τ w (z) := z w. We remark in assing that [iτ w (z)] +2 is the reroducing kernel at w H for the sace A 2 (H). The roof of the next integral equality is quite elementary and thus omitted (or see [5, Lemma 2.5]).

6 6 B. CHOE, H. KOO, AND W. SMITH Lemma 2.3. Given > and s real, the equality { c,s z w +2+s da (Im w) if s > 0 (z) = s if s 0 H holds for w H. Here, c,s is a constant. (2.5) As a consequence, we have Also, when s > + 2, we have (2.6) and thus (2.7) τ s w τ s w A τ s w A (H) s > + 2. τ s w A = (constant) (Im w) s 2 0 uniformly on comact subsets of H as w Ĥ. Here, Ĥ := H { } and lim z g(z) = 0 means that Ĥ su H\K g 0 as the comact set K H exands to the whole of H, or equivalently that g(z) 0 as Im z 0 + and g(z) 0 as z Pseudohyerbolic Distance. The seudo-hyerbolic distance ρ(z, w) between z, w H is given by ρ(z, w) := z w z w. Given φ, ψ S, we will use the notation (2.8) σ(z) = σ φ,ψ (z) := ρ ( φ(z), ψ(z) ) for short. The well-known Schwarz-Pick Lemma asserts ρ ( φ(z), φ(w) ) ρ(z, w) for all φ S and z, w H; see [4, Theorem 2..]. As the limiting version as w z, we have (2.9) for all z H. φ (z) Im z Im φ(z) 2.5. Comact Oerator. It seems better to clarify the notion of comact oerators, since the saces under consideration are not Banach saces when 0 < <. Suose X and Y are toological vector saces whose toologies are induced by comlete metrics. A continuous linear oerator T : X Y is said to be comact if the image of every bounded sequence in X has a subsequence that converges in Y. We have the following convenient comactness criterion for a linear combination of comosition oerators acting on the weighted Bergman saces.

7 LINEAR FRACTIONAL COMPOSITION OPERATORS 7 Lemma 2.4. Let > and 0 < <. Let T be a linear combination of comosition oerators and assume that T is bounded on A (H). Then T is comact on A (H) if and only if T f n 0 in A (H) for any bounded sequence {f n } in A (H) such that f n 0 uniformly on comact subsets of H. A roof can be found in [7, Proosition 3.] for a single comosition oerator over the unit disk and it can be easily modified for a linear combination over the half-lane Joint Carleson Measure Characterization. For a difference of comosition oerators there is a Carleson measure characterization for boundedness/comactness. To describe it, let φ, ψ S. Given > and 0 < <, the so-called joint ullback measure ω, = ω,;φ,ψ is defined by ω, (E) := σ da + σ da φ (E) ψ (E) for Borel sets E H. Here, σ is the function introduced in (2.8). Note that ω, is in fact the sum of two ullback measures (σ da ) φ and (σ da ) ψ. By a standard argument one can verify (2.0) g dω, = [g(φ) + g(ψ)]σ da H H for any ositive Borel function g on H. The Carleson roerty of these joint ullback measures characterize boundedness/comactness of C φ C ψ on A (H) as in the next theorem valid for inducing mas which are not necessarily linear fractional; see [5] for more information. In the next theorem the term locally finite means that the measure under consideration has finite mass on each comact subset of H. Theorem 2.5 ([5]). Let > and 0 < <. Let φ, ψ S and assume that ω, is locally finite. Then C φ C ψ is bounded (comact, res.) on A (H) if and only if the embedding A (H) L (dω, ) is bounded (comact, res.). Considering constants mas, one may check that the assumtion in Theorem 2.5 that ω, is locally finite is essential. We also remark that C φ C ψ being bounded on A (H) imlies, by the Carleson measure characterization [5, Theorem 3.3], that ω, is locally finite. 3. Bounded/Comact Differences In this section we study the most basic case, i.e, the case of difference. We obtain comlete characterizations for bounded/comact differences of linear fractional comosition oerators. It will turn out as a secial case of Theorem 4.3 in Section 4 that neither of the inducing mas in a comact difference can fix. Thus, in view of Proosition 2.2 and the rigidity

8 8 B. CHOE, H. KOO, AND W. SMITH henomenon for automorhisms, we may consider only the inducing linear fractional mas in S f. Before roceeding, we observe a simle condition necessary for bounded difference. The following lemma is taken from [5, Proosition 6.4], which is valid for arbitrary inducing mas. Lemma 3.. Let φ, ψ S and assume that C φ C ψ is bounded on some A (H). Assume φ( ). Then there is a sequence {z n } H such that z n nontangentially and φ(z n ) ψ(z n ) 0 as n. When alied to linear fractional comosition oerators, the above lemma reduces to the next one. Lemma 3.2. Let φ, ψ S f. If C φ C ψ is bounded on some A (H), then φ( ) = ψ( ). We now roceed to characterize bounded/comact differences of comosition oerators induced by mas in Sf. We first show non-boundedness/noncomactness results for the subclass St. Note that to each ma in St corresonds a (unique) boundary oint which is maed to a real number. We remark in assing that such a boundary oint is in fact the oint at which the given ma has finite angular derivative; see [5] for the notion of angular derivatives. It turns out that such boundary behavior is the main cause for athological henomena for the class St. The next theorem shows that no ma in St which takes into R can induce any bounded difference. In fact this result extends to general linear combinations; see Theorem 4.2 in Section 4. Theorem 3.3. Let φ St and ψ Sf be distinct mas and assume φ( ) R. Then C φ C ψ is not bounded on any A (H). Proof. By Lemma 3.2 we may assume φ( ) = ψ( ). Put b = φ ( ) and b 2 = ψ ( ). Recall that b and b 2 are negative real numbers, because φ( ) = ψ( ) := x is real. Note from (2.3) φ(z) = x + b z w and ψ(z) = x + b 2 z w 2 for some w H and w 2 H. We claim (3.) σ(z) c as Re z (with Im z fixed) for some c > 0 deending on Im z; recall that σ as in (2.8) denotes the seudohyerbolic distance between φ and ψ. Since Im φ(z) = b Im z + Im w z w 2, with (3.) granted, we see that [ ] Im z s σ(z) as Re z with Im z fixed Im φ(z)

9 LINEAR FRACTIONAL COMPOSITION OPERATORS 9 for any s > 0. Now, we conclude by [5, Theorem 4.3] that C φ C ψ is not bounded on any A (H). We now rove (3.). If w = w 2, then b b 2 and b b 2 σ(z) = b z w z w b 2 b b 2 = z w z w b b, 2 z w which imlies (3.). On the other hand, if w w 2, then σ(z) = b (z w 2 ) b 2 (z w ) b (z w 2 ) b 2 (z w ) = (b b 2 )z + b 2 w b w 2 b z b 2 z + b 2 w b w 2. Thus, if b = b 2 in addition, the above simlifies to σ(z) = w w 2 2iIm z + (w 2 w ) so that (3.) holds. On the other hand, if b b 2, we have (3.) with c =. This comletes the roof. It remains to investigate the case when a ma in S t takes a finite boundary oint into R. We need the next lemma. Lemma 3.4. Let φ St and assume φ(0) R. Then the following statements hold: (a) φ (0) > 0 and Im φ (0) > 0; Im φ(z) (b) Im z φ (0) as z 0 along the ath Im z = Re z ; Im φ(z) (c) Im z φ (0) + Im φ (0) 2 as z 0 along the ath Im z = z 2. Proof. Note from (2.) that φ is of the form ( φ(z) = a ζ z w ) z w + i + y for some a > 0, w H, y R and a unimodular number ζ. Note that ζ w w +i is real, because φ(0) is. Since ζ w w is unimodular, it follows that ζ w w = i. Accordingly, we have A little maniulation yields Note φ (0) = 2aIm w w 2 (3.2) [ φ(z) = ai ] w(z w) + y. w(z w) φ(z) = 2aIm w wz w 2 z w + y. > 0 and y = φ(0). So, we obtain φ(z) = φ (0) wz z w + φ(0).

10 0 B. CHOE, H. KOO, AND W. SMITH Note φ (0) = 2φ (0) w Also, note and thus Im φ (0) = 2φ (0) Im w w 2 > 0. So, (a) holds. Im φ(z) = φ (0) Im w z 2 + Im z w 2 z w 2. This imlies (b) and (c). The roof is comlete. We now extend Theorem 3.3 by showing that a ma in St in general cannot induce any comact difference. This result also extends to general linear combinations; see Theorem 4.3 in Section 4. We first recall the next result, which is art of [5, Theorem 4.3]. Lemma 3.5. Let > and 0 < <. Let φ, ψ S. If C φ C ψ is comact on A (H), then [ Im z Im φ(z) + Im z ] (+2)/ σ(z) = 0. Im ψ(z) lim z Ĥ Theorem 3.6. Let φ S t and ψ S f be distinct mas. Then C φ C ψ is not comact on any A (H). Proof. We may assume φ( ) = ψ( ) H by Lemma 3.2 and Theorem 3.3. Since φ S t and φ( ) H, there is a (unique) x R such that φ(x) R. For simlicity assume x = 0; otherwise, consider the translated function φ(z + x). Put We claim (3.3) Γ := {z H : Im z = Re z or Im z = z 2 }. lim su σ(z) > 0. z 0, z Γ Once this claim is granted, we see from Lemma 3.4 that [ ] Im z s σ(z) o() as z 0 along Γ Im φ(z) for any s > 0. Consequently, we conclude by Lemma 3.5 that C φ C ψ is not comact on any A (H). We now rove (3.3). We slit the roof into three cases: (i) φ(0) ψ(0); (ii) φ(0) = ψ(0) and φ (0) ψ (0); (iii) φ(0) = ψ(0) and φ (0) = ψ (0). Case (i): This case is easily treated, because φ(0) ψ(0) σ(z) φ(0) ψ(0) > 0 as z 0. In articular, (3.3) holds.

11 LINEAR FRACTIONAL COMPOSITION OPERATORS Case (ii): Since φ(0) = ψ(0), we have φ(z) ψ(z) φ (0) ψ (0) 0 z as z 0. Meanwhile, along the ath Re z = Im z so that z = ( + i)im z, we have by Lemma 3.4 φ(z) ψ(z) z = φ(z) ψ(z) z 2i Im φ(z) i[φ (0) iψ (0)] + i Im z as z 0. Note φ (0) iψ (0); otherwise σ(z) along the ath under consideration, which is imossible. It follows that σ(z) φ (0) ψ (0) φ (0) iψ (0) > 0 as z 0 along the ath Im z = Re z. Thus (3.3) holds. Case (iii): Note that a linear fractional ma is comletely determined by data at a given oint u to second order. Thus, φ (0) ψ (0). Since φ(0) = ψ(0) and φ (0) = ψ (0), we have φ(z) ψ(z) z 2 φ (0) ψ (0) 0 2 as z 0. Also, along the ath Im z = z 2, we have by Lemma 3.4 Im φ(z) z 2 φ (0) + Im φ (0) 2 as z 0. It follows that σ(z) = φ(z) ψ(z) φ(z) ψ(z) 2iIm φ(z) φ(z) ψ(z) φ(z) ψ(z) + 2Im φ(z) φ (0) ψ (0) φ (0) ψ (0) + 4φ (0) + 2Im φ (0) > 0 as z 0 along the ath Im z = z 2. Thus (3.3) holds. The roof is comlete. (3.4) In what follows we use the notation for a C and r > 0. D r (a) = D(a, r) := {z H : z a < r} Lemma 3.7. Let r > 0 and φ Sf. Then there is a constant C = C(r, φ) > 0 such that f φ da C f A D r(0) for functions f A (H) with > and 0 < <.

12 2 B. CHOE, H. KOO, AND W. SMITH Proof. Pick w H such that (2.3) holds. Note We thus have by (2.9) φ (z) = φ ( ) z w 2 φ ( ) ( z + w ) 2. ηim z Im φ(z), z D r (0) where η := φ ( ). Note φ (ξ) = w + φ ( ) (r+ w ) 2 ξ φ( ). So, given f A (H), we have by an elementary change-of-variables D r(0) f φ da η φ(d r(0)) φ ( ) 2 η d 4 f A f(ξ) φ ( ) 2 ( ) 2 da (ξ) ξ φ( ) where d > 0 is the distance between φ ( D r (0) ) and φ( ). comlete. The roof is Lemma 3.8. Let > and 0 < <. Let φ Sc Sh and {f n} be a bounded sequence in A (H) such that f n 0 uniformly on comact subsets of H. Then lim f n φ da = 0 n D r(0) for each r > 0. In the roof below we will use a well-known submean value tye inequality asserting that to each > corresonds a constant C = C() > 0 such that (3.5) f(z) C (Im z) +2 f A, z H for functions f A (H), 0 < <. Proof. The case φ Sc is clear, because φ(h) is relatively comact in H. So, assume φ Sh. Note that φ has a unique singularity in R, say at x, because φ( ) H. Fix r > 0. Given ɛ > 0, decomose the integral under consideration into two ieces (3.6) I n := f n φ da = + D r(0) D r(0) D ɛ(x) D r(0)\d ɛ(x) for each n. Note that the whole image φ(h) stays away from the real axis. φ ( D r (0) \ D ɛ (x) ) is relatively comact in H. It follows that f n φ da 0 D r(0)\d ɛ(x) Thus

13 LINEAR FRACTIONAL COMPOSITION OPERATORS 3 as n (with ɛ fixed). Meanwhile, note from (3.5) f n φ(z) f n A [Im φ(z)] +2 f n A for all n and z H. Thus, setting M := su n f n A <, we obtain f n φ da M A [D ɛ (x)] D r(0) D ɛ(x) for all n and ɛ > 0. Combining the estimates above, we obtain by (3.6) lim su I n M A [D ɛ (x)]; n one may check that the constant suressed above deends only on and. We now conclude the lemma by taking the limits ɛ 0 in the right hand side of the above. The roof is comlete. We now finish comlete descrition of bounded/comact differences by obtaining characterizations for the remaining cases which are not covered in Theorem 3.3 and Theorem 3.6. The characterizations turn out to deend not only on subclasses, but also on whether derivatives at coincide or not. When derivatives at do not coincide, the characterization is as follows. Theorem 3.9. Let > and 0 < <. Let φ, ψ Sf be distinct mas such that φ ( ) ψ ( ). Then the following statements are equivalent: (a) C φ C ψ is bounded on A (H); (b) φ( ) = ψ( ) H and > + 2. If φ, ψ Sc Sh in addition, then the above statements are also equivalent to (c) C φ C ψ is comact on A (H). Proof. We first rove that (a) imlies (b) for φ, ψ Sf. Assume (a). Then φ( ) = ψ( ) H by Lemma 3.2 and Theorem 3.3. Now, since φ( ) = ψ( ), we have and thus (3.7) z[φ(z) ψ(z)] φ ( ) ψ ( ) 0 φ(z) ψ(z) z as z. Also, as z, we have φ(z) ψ(z), because φ(z) ψ(z) 2Im φ( ) > 0. It follows that (3.8) σ(z) z as z. Let f A (H). Pick r > 0 sufficiently large so that 2 f φ(z) f φ( ) for z r; such an r exists by continuity. Let ω := ω,;φ,ψ be the joint ullback measure induced by φ and ψ. Note that ω is locally finite, because

14 4 B. CHOE, H. KOO, AND W. SMITH C φ C ψ is bounded on A (H) by assumtion; see the remark after Theorem 2.5. We thus have by Theorem 2.5, (2.0), and (3.8) that > f φ σ da (z) da f φ( ) z. H\D r(0) H\D r(0) This being true for any f A (H), the last integral must be finite. We thus conclude > + 2, as desired. So, (b) holds. Next, we show that (b) imlies (a). So, assume (b). If φ St, then there is a (unique) real oint which is maed by φ to another real oint, because φ( ) H. The same remark also alies to ψ. Choose R > 0 sufficiently large so that D R (0) contains all of such real oint(s), if any. Then we see that the set Ω := φ ( H \ D R (0) ) ψ ( H \ D R (0) ) stays away from the real axis. It follows that ( ) f φ f ψ da C su f φ ψ da H\D R (0) Ω H\D R (0) (3.9) C f A φ ψ da H\D R (0) for some constants C > 0 indeendent of f; the second inequality comes from (3.5) and the Cauchy Estimates. Note that the estimate (3.7) remains valid, because φ( ) = ψ( ) H. Now, we see from > + 2 that the integral above is finite. Meanwhile, the missing integral over D R (0) is treated exactly by Lemma 3.7. We thus conclude (a), as required. Now, we now further assume φ, ψ Sc Sh and show that (b) imlies (c). So, assume (b). Having already roved the equivalence of (a) and (b), we see that ω is locally finite as mentioned above. So, by Theorem 2.5, it suffices to show that the embedding A (H) L (dω) is comact. Let {f n } be a bounded sequence in A (H). By normal family argument there is a subsequence {f nj } which converges ointwise to some f, a function holomorhic on H, uniformly on comact sets of H. Note f A (H) by Fatou s Lemma. Put g j := f nj f for each j. Now, it is sufficient to show g j 0 in L (dω). By (2.0), we need to show ( g j (φ) + g j (ψ) )σ da 0, H or equivalently by symmetry, (3.0) I j := H g j (φ) σ da 0 as j. Fix any r > 0. Note that {g j } is bounded in A (H) and g j 0 uniformly on comact sets of H. Thus, since σ <, we note from Lemma 3.8 lim su j I j = lim su j H\D r(0) g j (φ) σ da.

15 LINEAR FRACTIONAL COMPOSITION OPERATORS 5 Note that the estimate (3.8) is available, because φ( ) = ψ( ) by assumtion. Also, Im φ stays away from 0 because φ Sc Sh, and so (3.5) yields g j φ M := su j g j A <. Hence g j φ σ da M da (z) z H\D r(0) H\D r(0) for all j and r > 0. Note that the last integral is finite, because > + 2. Combining the estimates above, we obtain lim su I j M da (z) j H\D r(0) z for all j and r > 0; one may check that the constant suressed above deends only on and. We now conclude (3.0) by taking the limit r in the right hand side of the above. So, (c) holds. The imlication (c) = (a) being trivial, this comletes the roof of equivalences of (a), (b) and (c) for φ, ψ Sc Sh. The roof is comlete. Remarks. () For the imlication (b) = (c) in the roof of Theorem 3.9 one may aly Lemma 2.4, as in the roof of Theorem 4.5 later, rather than the joint Carleson measure characterization (Lemma 2.5). (2) When the range > + 2 is given as a standing hyothesis, Theorem 3.9 will be extended in the next section to general linear combinations; see Corollary 4.7. (3) Theorem 3.9 is a henomenon for differences of comosition oerators induced by linear fractional mas. See [24, Corollary 3.9] for a related result regarding a different class of inducing mas. When derivatives at coincide, additional cancelation occurs and we have the following characterization. Theorem 3.0. Let > and 0 < <. Let φ, ψ Sf be distinct mas such that φ ( ) = ψ ( ). Then the following statements are equivalent: (a) C φ C ψ is bounded on A (H); (b) φ( ) = ψ( ) H and > If φ, ψ Sc Sh in addition, then the above statements are also equivalent to (c) C φ C ψ is comact on A (H). Proof. The roof is almost the same as that of Theorem 3.9 excet one difference. We only remark the difference. When φ( ) = ψ( ) H and φ ( ) = ψ ( ) =: d 0, we have ( φ(z) ψ(z) = d z w z w 2 for some distinct w, w 2 H. This imlies ) = [φ(z) ψ(z)]z 2 d(w w 2 ) 0 d(w w 2 ) (z w )(z w 2 )

16 6 B. CHOE, H. KOO, AND W. SMITH as z. So, we have φ(z) ψ(z) z 2 (3.) σ(z) z 2 and hence as z. Now estimate (3.), being better than (3.8), results in the imroved lower bound for. As a consequence of Theorems , we have the next corollary. Corollary 3.. Let >, 0 < < and assume > + 2. Let φ, ψ S f and assume φ( ) = ψ( ) H. Then the following statements hold: (a) C φ C ψ is bounded on A (H). (b) C φ C ψ is comact on A (H) if φ, ψ S c S h. We now close the section with consequences of Theorems concerning Hilbert-Schmidt differences. Recall that a bounded linear oerator T on a searable Hilbert sace X with an orthonormal basis {e β } is called Hilbert-Schmidt if the norm /2 T HS(X) := T e β 2 is finite. As is well known, the value (ossibly ) of the sum on the righthand side of the above does not deend on the choice of orthonormal basis. It is well known that every Hilbert-Schmidt oerator is comact; see [25, Section 6.2] for details. Recall from Theorem 3.6 that φ St can not induce a comact difference. Hence we only consider inducing mas in Sc Sh. It will turn out that three notions of boundedness, comactness and the Hilbert-Schmidt roerty all coincide for those tyes of differences. For >, the following Hilbert-Schmidt norm estimate is roved in [5, Theorem 7.6]: (3.2) C φ C ψ 2 HS(A 2 (H)) H β [ Im z Im φ(z) + Im z ] +2 σ 2 (z) dµ(z) Im ψ(z) for φ, ψ S. Here, dµ(z) := (Im z) 2 da(z) denotes the Möbius-invariant measure. Theorem 3.2. Let >. Let φ, ψ Sc Sh be distinct mas such that φ ( ) ψ ( ). Then the following statements are equivalent: (a) φ( ) = ψ( ) H and < 0; (b) C φ C ψ is bounded on A 2 (H); (c) C φ C ψ is comact on A 2 (H); (d) C φ C ψ is Hilbert-Schmidt on A 2 (H).

17 LINEAR FRACTIONAL COMPOSITION OPERATORS 7 Proof. We already know by Theorem 3.9 (with = 2) that (a), (b) and (c) are equivalent to one another. It is also clear that (d) imlies (c). So, in order to comlete the roof, it suffice to show that (b) imlies (d). Assume (b). Note that the estimate (3.8) is available. Also, since φ, ψ Sc Sh, we have Im φ(z) and Im ψ(z) for all z H. Accordingly, we have by (3.2) C φ C ψ 2 HS(A 2 (H)) (Im z) +2 σ 2 (z) dµ(z) H = (Im z) σ 2 (z) da(z) H (Im z) (Im z) da(z) + z 2 da(z). D (0) H\D (0) The first integral of the above is clearly finite. So is the second one, because < 0. Thus we conclude that (d) holds, which comletes the roof. Similarly, when φ ( ) = ψ ( ), one may use Theorem 3.0 and (3.) in lace of Theorem 3.9 and (3.8) to conclude the next theorem. Theorem 3.3. Let >. Let φ, ψ Sc Sh be distinct mas such that φ ( ) = ψ ( ). Then the following statements are equivalent: (a) φ( ) = ψ( ) H and < 2; (b) C φ C ψ is bounded on A 2 (H); (c) C φ C ψ is comact on A 2 (H); (d) C φ C ψ is Hilbert-Schmidt on A 2 (H). 4. Bounded/Comact Linear Combinations In this section we study bounded/comact linear combinations of linear fractional comosition oerators. Throughout the section we reserve the letter N for the number of oerators which are involved in linear combinations. We first observe some necessary conditions for boundedness and comactness. For a necessary condition for boundedness, we need a reliminary lemma. We ause to set some notation. Put (4.) Γ := {w H : Im w = w 2 }; this is a circle tangent to R at the origin. Given R > 0, ut { } Ω R (w) := z H : R w 2 z + w < 2R w 2 for w Γ with w < R. Using the elementary fact that λ + w = R w 2 λ + w( R 2 w 2 ) = R R 2 w 2,

18 8 B. CHOE, H. KOO, AND W. SMITH we see that (4.2) where Ω R (w) = 2R (w) \ R (w) ( ) R (w) := D w( R 2 w 2 ), R R 2 w 2 ; recall that D(, ) denotes the set defined in (3.4). Note w = Re w i for w Γ. So, the imaginary art of the to w 2 R oint of R (w) is equal to. Thus, setting R 2 w { 2 2R (w) := z 2R (w) : Im z R } R 2 w 2, we have When ζ H is fixed, this yields 2R (w) Ω R (w) 2R (w). [ (4.3) A ΩR (w) + ζ ] R +2, w < 2R as R ; the constants suressed here are indeendent of w. In what follows the notation w Γ 0 means w 0 along the circle Γ. Recall that τ w denotes the test function defined in (2.4). Lemma 4.. Let > and 0 < <. For distinct oints w,..., w N H, ut ψ j (z) = z w j S for each j. Let λ,..., λ N C \ {0} and t > +2. Then to each R > 0 sufficiently large corresonds a constant C > 0 such that N λ j C ψj τw(z) t as w Γ 0. j= C w 2t, z Ω R(w) + w Proof. Let R > be a large number to be secified later. For w Γ with w < R, note (4.4) So, we have (4.5) Ω R (w) + w := { z H : R w 2 ψ (z) w < 2R w 2}. wψ (z) ψ (z) w w 2 + 2R w 3 R w 2 = R + 2 w < 3 R for z Ω R (w) + w. Since ψ j ψ = w j w =: ξ j for each j, we have ψ j w = ψ + ξ j ( w ψ + ξ j ) = + ξ jψ ψ w ξ j wψ ψ w

19 LINEAR FRACTIONAL COMPOSITION OPERATORS 9 so that ( ψj (z) w ) t = ( + O( w ) ( ) ψ (z) w ) wψ t t ξ j ψ w for z Ω R (w) + w. Now, taking the sum over j and setting N ( ) wψ (z) t F (z) := λ j ξ j, ψ (z) w we obtain (4.6) Sτ t w(z) = N j= j= λ j ( ψj (z) w ) t = + O( w ) ( ψ (z) w ) t F (z), z Ω R(w) + w where S := N j= λ jc ψj. In order to estimate F (z), we need some notation. Recall that ξ j s are all distinct. Thus the determinant of the corresonding Vandermonde matrix is nonzero, and so the vectors (, ξ j, ξj 2,..., ξn j ), j N, are linearly indeendent. Since λ j 0 for each j, we see that there exists a smallest integer L 0 such that N β := λ j ξj L 0. Put (4.7) j= L E(λ) := ( λ) t c l λ l, λ <, where c l = c l (t) are the Taylor coefficients of the function ( λ) t. Put M 0 := su 0< λ 2 We now consider R so large that l=0 E(λ) λ L+ < and M := R > 6M 0M and R 6 max ξ j. β c L j N λ j ξ j L+. Note from (4.5) ξ j wψ (z) ψ (z) w 3 ξ j R 2, z Ω R(w) + w for each j. Also, note ( ) wψ (z) L F (z) = βc L + ψ (z) w N j= j ( ) ξj wψ (z) λ j E. ψ (z) w

20 20 B. CHOE, H. KOO, AND W. SMITH Accordingly, for z Ω R (w) + w, we obtain wψ (z) L F (z) βc L wψ (z) L+ ψ (z) w M 0 M ψ (z) w wψ (z) L ( β c L ψ (z) w 3M ) 0M (by (4.5)) β c L R β c L wψ (z) L 2 ψ (z) w. In addition, we have wψ (z) ψ (z) w w 2 2R w 3 2R w 2 = 2R w 6R for z Ω R (w) + w with w 3R. Combining the last two estimates, we obtain F (z) β c L 2(6R) L for z Ω R (w) + w with w 3R. This, together with (4.6), yields Sτw(z) t ψ (z) w t (2R w ) 2t, z Ω R(w) + w as w Γ 0. This is the desired estimate, with the constant C (2R) 2t ; the roof is comlete. Note that the restrictions ψ j ( ) in the next theorem cause no loss of generality by Theorem 2.. Also, note that this result contains Theorem 3.3 as a secial case. Theorem 4.2. Let > and 0 < <. Let ψ,..., ψ N S be distinct mas such that ψ j ( ) for all j. For λ,..., λ N C\{0}, assume that N j= λ jc ψj is bounded on A (H). Then ψ j ( ) H for all j. Proof. Suose not. We will derive a contradiction. We may assume a := ψ ( ) R. Put Q := {j : ψ j ( ) = a}. For j Q, ut b j := ψ j ( ). Then b j ψ j (z) = a z w j for some w j H. Note b j > 0. Now set Q := {j Q : b j = b } and Q 2 := Q \ Q. Notice that Q since Q, but Q 2 = is ossible. Set ψ(z) = b z +a. Then, recalling that a R and b > 0, C ψ is invertible on A (H) with

21 C ψ LINEAR FRACTIONAL COMPOSITION OPERATORS 2 = C ψ. Let ψ j = ψ ψ j, and S := N λ j C ψj = SC ψ, j= Sk := j Q k λ j C ψj for k =, 2. Note that S is bounded on A (H), because S is. Note also that for j Q. Also, c j = for j Q. Pick t > 2(+2) and ut c j ψ j (z) := where c j = b j > 0 z w j b f w := τ t w, w Γ for short; recall that Γ is the circle secified in (4.). Note from (2.5) and (2.6) that we have f w A (H) with f w A (Im w) t 2 = w 2(t 2) for w Γ. We now show how the oerators S, S 2 and S ( S + S 2 ) act on these test functions f w. First, we consider S. Note that the w j s with j Q are distinct, since the mas ψ j are assumed to be distinct. Now ick a large number R = R(w ) > 0 rovided by Lemma 4. (alied to functions ψ j s). We may further assume that estimate (4.3) holds. Then, by Lemma 4., we have lim w 0 Γ f w S f w da =. A Ω R (w)+w Next, we consider S 2. Let j Q 2. Note that (4.4) remains valid when ψ is relaced by ψ. Thus, for z Ω R (w) + w, we have ψ j (z) w = c j + w z w j c j z w + w z w c j c j z w j z w (4.8) c j z w c j w w j 2R w 2 (z w )(z w j ). In conjunction with this, we note from (4.2) that w z w stays bounded (by some ositive constant deending on R) as w Γ 0 with z Ω R (w)+w. Also c j for j Q 2, and hence estimate (4.8) yields a constant C > 0, indeendent of w, such that ψ j (z) w C w, z Ω R (w) + w

22 22 B. CHOE, H. KOO, AND W. SMITH as w Γ 0. Since this is true for each j Q 2, we obtain S 2 f w (z) w t, z Ω R(w) + w as w Γ 0; the constant suressed here is indeendent of w. This, together with (4.3), yields lim w 0 Γ f w S 2 f w da w t 2(+2) = 0; A Ω R (w)+w recall that t > 2( + 2) by our choice of t. Finally, we consider S ( S + S 2 ). For j Q we have ψ j ( ) a, and so f w ψ j remains bounded on Ω R (w) + w as w Γ 0. Hence, recalling (4.3) again, we see that lim w 0 Γ f w [ S S S ] 2 fw da = 0. A Ω R (w)+w It follows from these observations that S is not bounded on A (H), which is the contradiction we sought. The roof is comlete. For comactness we have the next necessary condition. It is quite interesting to see from (a) of the next result and Theorem 2. that, as far as linear fractional comosition oerators are concerned, only unbounded ones can form a comact linear combination; this is not the case if the inducing mas are not necessarily linear fractional; see [5, Examle 5.]. Theorem 4.3. Let > and 0 < <. For distinct mas ψ,..., ψ N S and λ,..., λ N C\{0}, assume that N j= λ jc ψj is comact on A (H). Then the following statements hold: (a) ψ j ( ) for all j; (b) ψ j / S t for all j. Proof. We first rove (a). Suose not. We will derive a contradiction. For each j with ψ j ( ) = we ut a j =: ψ j (0) > 0 and b j = ψ j (0) H so that ψ j (z) = a j z + b j. We may assume ψ ( ) = without loss of generality. Put Pick s > +2 Q := {j : ψ j ( ) = and a j = a }. and consider test functions for ositive integers n. Note f n (z) := τ s i (z + n) = (z + n + i) s f n A = f A <

23 LINEAR FRACTIONAL COMPOSITION OPERATORS 23 for all n. Also, note that f n 0 uniformly on comact subsets of H. Now, we claim that there is a number ɛ > 0 such that N (4.9) Sf n da ɛ where S := λ j C φj H for all large n. This is imossible by comactness of T, which will comlete the roof of (a) once (4.9) is established. For the roof of (4.9), consider j / Q. First, when ψ j ( ), it is clear that ( ) f n ψ j (z) = ψ j (z) + i + n s = O n s as z and n. Next, consider the case φ j ( ) =. Note ψ j (z) + n + i = a jz + b j + i + a ( j n + a ) j n a a a j n a j z x n b j + i a j= where x n = n a. Since a j a, it follows that ( ) f n ψ j (z) = O n s, z D (x n ) as n. Combining these estimates, we obtain ( ) f n ψ j da = O n s D (x n) j / Q as n ; note A [D (x n )] = A [D (x )] for all n. Now, this yields Sf n da T f n da H D (x n) = λ j (f n ψ j ) da + o() D (x n) j Q (4.0) = F da + o() D (0) as n where F (z) := λ j (a z + b j + i) s. j Q Note that the b j s with j Q are all distinct, because the ψ j s are. Thus F cannot be identically zero. So, we conclude (4.9) by (4.0), as claimed, comleting the roof of (a). We now turn to the roof of (b). As above, we assume that the conclusion fails and will derive a contradiction. To describe our test functions, we need some reliminary observations. To begin with, we assume this time ψ St

24 24 B. CHOE, H. KOO, AND W. SMITH without loss of generality. Note ψ ( ) H by (a) and Theorem 4.2. Note also that comactness of the given oerator is invariant under the horizontal translations. So, we may also assume ψ (0) = 0 for simlicity (as in the roof of Theorem 3.5). This time we ut Q := {j : ψ j (0) = 0} so that Q. For each j Q note that ψ j (H) is a disk tangent to R at 0 and hence ψ j (H) = D rj (ir j ) for some r j > 0. By the roof of Lemma 3.4 we have (4.) ψ j(0) > 0 and Im ψ j (0) = [ψ j (0)]2 r j > 0 for each j Q. Thus, for j Q, setting we have (4.2) and (4.3) x j = Re [ψ j (0)] 2 and y j := Im [ψ j (0)] 2 Re ψ j (s) = ψ j(0)s + x j s 2 + O(s 3 ) Im ψ j (s) = y j s 2 + O(s 3 ) for s (0, ) sufficiently small. In articular, we get (4.4) as s 0. Put Also, ut Im ψ j (s) s 2 y j y = min j Q y j and Q y := {j Q : y j = y}. r = max j Q y r j and Q y,r := {j Q y : r j = r}. Since a linear fractional ma is comletely determined by data u to second order at a given oint, we see that x j x k whenever j, k Q y,r and j k. Accordingly, we may further assume Q y,r and (4.5) x > x j for j Q y,r with j. We now introduce our test functions. For t > +2 to be secified later, ut, f ψ (s) := τ ψ t (s) τψ t (s) A

25 LINEAR FRACTIONAL COMPOSITION OPERATORS 25 for s (0, ). Note from Lemma 2.4 and (2.7) (4.6) T f ψ (s) A 0 as s 0. Here, T := N j= λ jc ψj. We will show that this in fact fails to hold, which is the contradiction we seek. Put w j := ψ j (s) for simlicity. We claim there is some ɛ > 0 such that (4.7) w j w w j w + 3ɛ, j Q \ {} for all s sufficiently small. In order to rove this, we consider three cases searately; (i) j Q \ Q y, (ii) j Q y \ Q y,r and (iii) j Q y,r \ {}. First, consider Case (i). In this case we have by (4.3) (4.8) as s 0. Now, since (4.9) Im (w j w ) = (y j y) s 2 + O(s 3 ) w j w w w = i w j w 2Im w, we obtain by (4.4) and (4.8) ( ) wj w Re = Im (w j w ) y j y w w 2Im w 2y as s 0. Note that the last exression is ositive, because j Q \ Q y. So, (4.7) holds for a suitable choice of ɛ > 0. Next, consider Case (ii). In this case we have by (4.) It follows from (4.2) and (4.3) that ψ j(0) = 2r j y and r j < r. w j w = 2y( r j r)s + O(s 2 ), which, together with (4.4), yields w j w w w as s 0. This imlies (4.7) with any choice of ɛ > 0. Finally, consider Case (iii). In this case note from (4.) ψ j(0) = 2ry and Im ψ j (0) = 2y. We thus have by (4.2) and (4.3) again w j w = (x j x )s 2 + O(s 3 ), which, together with (4.4) and (4.9), yields ( ) w j w w w x x j + i 2y as s 0. Thus we see from (4.5) that (4.7) holds for a suitable choice of ɛ > 0, comleting the roof of (4.7).

26 26 B. CHOE, H. KOO, AND W. SMITH Now, using (4.7), we roceed to the lower ointwise estimate of T f w over suitably chosen sets. Put and (4.20) δ := 2 max j Q\{} ψ j(0) { } η := min δ, 4ψ (0). For each j, we claim (4.2) τ w ψ j (z) + 2ɛ w w, z D yɛηs 2(s) for all s sufficiently small. Note that this is clear for j / Q, because ψ j (0) 0. So, assume j Q \ {}. Note ψ j (z) ψ j (s) δ z s, j Q \ {} for all z and s sufficiently close to 0. For such z and s, we have ψ j (z) w w j w ψ j (z) ψ j (s) w j w δ z s and hence by (4.7) (4.22) ψ j (z) w w w w j w w w δ z s w w + 3ɛ δ z s w w. Meanwhile, for z D yɛηs 2(s), note from (4.4) z s w w yɛηs2 w w ɛη 2 as s 0. We thus have by (4.22) ψ j (z) w w w + 3ɛ δɛη + 2ɛ, z D yɛηs 2(s) for all s sufficiently small; the last inequality holds by (4.20). So, (4.2) holds, as asserted. On the other hand, for j =, we claim (4.23) τ w ψ (z) + ɛ w w, z D yɛηs 2(s) for all s sufficiently small. Let z D yɛηs 2(s) for s sufficiently small. Since ψ (z) ψ (s) 2ψ (0) z s < 2ψ (0)yɛηs 2, we have ψ (z) ψ (s) w w 2ψ (0)yɛηs2 2ψ Im w (0)ɛη as s 0. It follows that ψ (z) w w w + ψ (z) ψ (s) w w + 4ψ (0)ɛη + ɛ;

27 LINEAR FRACTIONAL COMPOSITION OPERATORS 27 the last inequality holds by (4.20). So, (4.23) holds, as asserted. Having established estimates in (4.2) and (4.23), we are now ready to show that (4.6) fails to hold. Let z D yɛηs 2(s) with s sufficiently small. Using the original notation ψ (s), rather than w, we have by (4.2) and (4.23) T τψ t (s) (z) λ τψ t (s) ψ (z) λ j τψ t (s) ψ j(z) j ( + ɛ) t λ ψ (s) ψ (s) t Thus, choosing t so large that ( ) + ɛ t λ j λ + 2ɛ 2, j we have T τ t ψ (s) (z) ( + ɛ) t λ 2 and hence by (2.6) and (4.4) (4.24) T f ψ (s)(z) ( ) + ɛ t + 2ɛ j ψ (s) ψ (s) t ψ (s) ψ (s) +2 s 2(+2). λ j. This holds for z D yɛηs 2(s) with s sufficiently small. Now, integrating both sides of (4.24) over D yɛηs 2(s) against the measure da, we obtain T f ψ (s) da A [D yɛηs 2(s)] s 2(+2) D yɛηs 2 (s) for all s sufficiently small; one may check that the constants suressed in these estimates are indeendent of s. This is a contradiction to (4.6). The roof is comlete. We now turn to sufficient conditions for bounded/comact linear combinations. In view of Theorem 4.3 and Theorem 4.2, we may restrict the inducing mas to the class Sf. We ause to introduce some notation to be used in the rest of the section. Let φ, φ 2,..., φ N be distinct mas in Sf. We decomose the index set J := {, 2,..., N} into a disjoint union m J = ν= where sets J ν are the equivalent classes induced by the equivalence relation on J defined by j k if and only if φ j ( ) = φ k ( ). By (2.3) there are J ν

28 28 B. CHOE, H. KOO, AND W. SMITH oints w j H such that φ j (z) = φ j ( ) + φ j ( ) z w j for each j J. Given λ,..., λ N C \ {0}, we ut N (4.25) T := λ j C φj j= for short. For nonnegative integers n, k and l, let c n;k,l be the coefficients determined by the identity (4.26) ( + x + + x n ) l = c n;k,l x k. k=0 Note c n;k,0 = 0 for k and c n;k,l = 0 for k > ln. Finally, ut and E 0 : = {(0, 0)} E n : = {(0, 0)} {(k, l) : k + l n, k n, l 0}. for ositive integers n. These sets are secified so that c n;k,l 0 for (k, l) E n. In the next lemma γ ν denotes the common value of φ j ( ) s for j J ν. Lemma 4.4. Let >, 0 < < and n be a nonnegative integer. For distinct mas φ, φ 2,..., φ N Sf with φ j ( ) H for all j, let T be as in (4.25). Then T f(z) = c n;k,l m f (l) (γ ν ) [φ l! j( )] l k w j z k+l (k,l) E n ν= j J ν ( ) + f A O z n+ for f A (H) as z. The big Oh factor above is indeendent of f. Proof. Let f A (H). Let z be sufficiently large so that w j < z for all j. For each j, ut a j := φ j ( ) and b j := φ j ( ) for short. Then each φ j admits series reresentation which yields (4.27) φ j (z) = a j + b j z φ j (z) a j = b j z n k=0 ( wj z k=0 ( wj z ) k, ) k ( ) + O z n+2.

29 LINEAR FRACTIONAL COMPOSITION OPERATORS 29 In terms of the coefficients introduced in (4.26), we have by (4.27) ( ) l ln ( ) k ( ) (φ j (z) a j ) l bj wj = c n;k,l + O z z z n+2 k=0 for each l 0. Recall a j H for all j by assumtion. Thus, considering the Taylor exansion of f near a j, we obtain n f (l) (a j )( ) l ( f φ j (z) = φj (z) a j + O f A l! φ j(z) a j n+) = l=0 n l=0 f (l) (a j ) l! = f (l) (a j ) c n;k,l l! (k,l) E n ( ) l ln ( bj wj c n;k,l z z k=0 b l j w j k z k+l + O ) k ( ) f A + O ( f A z n+ ). z n+ One may check that the constant suressed in the big oh estimates above is indeendent of f. Also, the factor f A comes from (3.5) and the Cauchy Estimates. Now, summing u these estimates, we obtain T f(z) = c n;k,l N f (l) (a j )λ j b l k l! jw j ( ) z k+l + O f A z n+, (k,l) E n j= which imlies the lemma. Now, based on Lemma 4.4, we introduce the following coefficient cancelation roerty of order n: (CCP λ j [φ j( )] l w k n ) j = 0 for all (k, l) E n and ν m. j J ν This coefficient cancelation roerties will lay an essential role below in our boundedness/comactness characterizations of T. Note that the restrictions φ j ( ) H in the next two theorems cause no loss of generality by Theorem 4.2. Theorem 4.5. Let >, 0 < < and n be a nonnegative integer such that > +2 n+. For distinct mas φ, φ 2,..., φ N Sf with φ j ( ) H for all j, let T be as in (4.25). Assume (CCP n ). Then the following statements hold: (a) T is bounded on A (H); (b) Moreover, T is comact on A (H) when φ j Sc Sh for all j. Proof. Let f A (H). By (CCP n ) and Lemma 4.4 there are constants C > 0 and R > 0, indeendent of f, such that T f(z) C f A z n+, z R

30 30 B. CHOE, H. KOO, AND W. SMITH and hence (4.28) H\D R (0) T f(z) da (z) C f A H\D R (0) da (z) z (n+). Note that the last integral is finite, because (n + ) > + 2 by assumtion. So, we conclude (a) by Lemma 3.7. Now, assume φ j Sc Sh for all j. To establish comactness of T, we use Lemma 2.4. Let {f n } be a bounded sequence in A (H) which converges to 0 uniformly on comact subsets of H. Put M := su n f n A <. Note that (4.28) remains true when R is relaced by any r R. Thus, for r R, we have by Lemma 3.8 lim su T f n (z) da (z) = lim su T f n (z) da (z) n n H C M H\D r(0) H\D r(0) da (z) z (n+) ; This being true for all r R, we take the limit r and obtain T f n (z) da (z) = 0 lim n so that (b) holds. The roof is comlete. H The converse also holds when the arameter is restricted to a certain range as in the next theorem. Theorem 4.6. Let >, 0 < < and n be a nonnegative integer such that +2 n+ < +2 n. For distinct mas φ, φ 2,..., φ N Sf with φ j ( ) H for all j, let T be as in (4.25). If T is bounded on A (H), then (CCP n ) holds. Proof. Since (n + ) > + 2 and T is bounded on A (H) by assumtion, we see from Lemma 4.4 that the function c n;k,l m f (l) (γ ν ) λ j b l k l! jw j z k+l where b j = φ j( ) (k,l) E n ν= j J ν is L -integrable near against the measure da for any f A (H). For n = 0, note that this function is constant and thus must be trivial. Also, for n, note that the functions with (k, l) E z k+l n are not L -integrable near against the measure da, because n + 2 by assumtion. So, the function above again must be trivial. Accordingly, we see that all the coefficients must vanish. Recall c n;k,l 0 for (k, l) E n. It follows that m f (l) (γ ν ) λ j b l jw k j = 0, j J ν ν= (k, l) E n

31 LINEAR FRACTIONAL COMPOSITION OPERATORS 3 for any f A (H). Now, aly this equality to the function f := τw s where w H and s > +2. Then the resulting equality simlifies to m (γ ν= ν w) s+l λ j b l jw k j = 0, (k, l) E n. j J ν This being true for all w H, we conclude that (CCP n ) holds, as required. The roof is comlete. For n = 0, note that the ranges for in Theorem 4.5 and Theorem 4.6 coincide. So, one may combine those two theorems with Theorem 4.2 to obtain the next result. Note that this result, when > + 2, together with Theorem 4.3(b), rovides a comlete characterization for bounded/comact linear combinations. Corollary 4.7. Let >, 0 < < and assume > +2. For distinct mas φ,..., φ N Sf, let T be as in (4.25). Then the following statements are equivalent: (a) T is bounded on A (H); (b) (CCP 0 ) holds and φ j ( ) H for all j. If φ j Sc Sh for all j, then the above statements are also equivalent to (c) T is comact on A (H). We rovide a concrete examle for linear combinations of three comosition oerators. One may also construct various examles of general linear combinations. Furthermore, the next examle reveals a new half-lane henomenon whose disk analogue is not available. In fact it shows (for +2 2 < + 2) the so-called double cancelation henomenon on the half-lane, meaning that a difference of two noncomact differences involving exactly three distinct comosition oerators is ossibly comact. Such double cancelation henomenon is not ossible in the disk setting; see [5]. Examle 4.8. Consider functions φ S c and φ 2, φ 3 S h φ (z) = i 2 z + i, Then the following statements hold: φ 2(z) = i z given by and φ 3 (z) = i 3 z. (a) C φ C φ2 and C φ C φ3 are bounded/comact on A (H) if and only if > + 2. (b) 2C φ C φ2 C φ3 is comact on A (H) for > Proof. Note φ j ( ) = i for j =, 2, 3. Meanwhile, we have φ ( ) = 2, φ 2 ( ) =, and φ 3 ( ) = 3. Thus (a) holds by Theorem 3.9 and (b) holds by Theorem 4.5 with n =. We now close the aer with the next remark.