NEW ZEALAN JOURNAL OF MATHEMATICS Volume 44 (204), 93 0 INTEGRATION OPERATORS FROM CAUCHY INTEGRAL TRANSFORMS TO WEIGHTE IRICHLET SPACES Ajay K. Sharma and Anshu Sharma (Received 6 April, 203) Abstract. The boundedness and compactness of integration operators between Cauchy integral transforms and weighted irichlet spaces are characterized.. Introduction Let be the open unit disk in the complex plane C, its boundary, H() the class of all holomorphic functions on, da(z) = π dxdy = rdrdθ the normalized π area measure on, H () the space of all bounded analytic function on with the norm f = z f(z) and M the space of all complex Borel measures on. For 0 < p <, the Hardy space H p is the space of all f H() such that f p admits a harmonic majorant. If we take as the norm of f H p the p-th root of the value at some fixed point z of the least harmonic majorant of f p, then H p is a Banach space for p < (p-banach space for 0 < p < ). Moreover, regardless of z all these norms are equivalent. It is, however, more customary to work with another definition of H p and with another equivalent norm. Recall that f H() belongs to H p if and only if the integrals M p (r, f) := f(rζ) p dm(ζ), 0 < r <, are bounded. In this case f p := M p (r, f) /p 0<r< is a norm (p-norm if 0 < p < ) on H p which is equivalent to norms described above. Also it is well known that if 0 < p < q <, then H q H p. Let ω be a positive integrable function. If we extend it on by ω(z) = ω( z ), z, we call it a weight or a weight function. By ω we denote the weighted irichlet space consisting of all f H() such that f 2 ω = f(0) 2 + f (z) 2 ω(z)da(z) <. The Bergman space A 2 ω is a Hilbert space of holomorphic functions on with the norm f 2 A = f(z) 2 ω(z)da(z) <. 2 ω 200 Mathematics Subject Classification Primary 47B33, 46E0; Secondary 3055. Key words and phrases: Integration operator, Cauchy integral transforms, weighted irichlet space.
94 AJAY K. SHARMA AN ANSHU SHARMA A simple computation shows that a function f(z) = only if where ω 0 = and a n 2 ω n <, n=0 a n z n belongs to ω if and n=0 ω n = ω(n) = 2n 2 r 2n ω(r)dr, n N. 0 The sequence (ω n ) n N0 is called the weight sequence of the weighted irichlet space ω. The properties of the weighted irichlet space with the weight sequence (ω n ) n N0, clearly depends upon ω n. For more about Hardy and weighted irichlet spaces and some related topics, see [7], [9] and [22]. We may assume that the weight ω satisfies the following three properties: (W ) ω is non-increasing; ω(r) (W 2 ) is non-decreasing for some δ > 0; ( r) +δ (W 3 ) lim r ω(r) = 0. If ω also satisfies one of the following properties: (W 4 ) ω is convex and lim r ω(r) = 0; or (W 5 ) ω is concave, then such a weight function is called admissible (see [9]). If ω satisfies conditions (W ), (W 2 ), (W 3 ) and (W 4 ), then it is said that ω is I-admissible. If ω satisfies conditions (W ), (W 2 ), (W 3 ) and (W 5 ), then it is said that ω is II-admissible. I- admissibility corresponds to the case H 2 H ω A 2 α for some α >, whereas II-admissibility corresponds to the case H ω H 2. If we say that a weight is admissible it means that it is I-admissible or II-admissible. A function f in H() is in the space of Cauchy integral transforms K if dµ(ζ) f(z) = () ζz for some µ M. The space K becomes a Banach space under the norm { } dµ(ζ) f K = inf µ : f(z) =, (2) ζz where µ denotes the total variation of measure µ. It is well known known that H K 0<p< H p. Let g H(), n N {0} and ϕ be a holomorphic self-map of. We define the generalized integration operator as follows I (n) g,ϕf(z) = z 0 f (n) (ϕ(ζ))g(ζ)dζ, z. (3) Operator (3) is an extension of many operators appearing in the literature. For example, if n = 0, then is obtained an operator, which is a natural extension of the integral operator by Pommerenke [0]. If n = then is obtained, so called generalized composition operator, which is a natural extension of the integral operator by Yoneda [29]. Recently, several authors have studied these operators along
INTEGRATION OPERATORS 95 with composition and weighted composition operators on different spaces of analytic functions. For example, one can refer to ([]-[9], []-[29] and the related references therein for the study of these operators on different spaces of analytic functions. Here we provide complete characterizations of when g and ϕ induce bounded or compact integration operator I g,ϕ (n) from the space K of Cauchy integral transforms into weighted irichlet spaces. Throughout this paper constants are denoted by C, they are positive and not necessarily the same at each occurrence. We write A B if there is a positive constant C such that CA B A/C. 2. Boundedness and Compactness of I (n) g,ϕ : K ω Theorem. Let g H(), n N {0} and ϕ be a holomorphic self-map of. Then I (n) g,ϕ : K ω is bounded if and only if the family {g/( ζϕ) n+ : ζ } is a norm bounded subset of A 2 ω, that is there exists a constant M > 0 such that ω(z)da(z) M <. (4) ζϕ(z) 2(n+) ζ Proof. First pose that (4) holds. If f K, then there is µ M with µ = f K such that f(z) = ζz dµ(ζ). Thus we have f (n) (ζ) n (z) = n! dµ(ζ), n N. (5) ( ζz) n+ Replacing z in (5) by ϕ(z), using Jensen s inequality and multiplying such obtained inequality by ω(z), we obtain (ϕ(z)) 2 ω(z) (n!) 2 µ 2 ω(z)d µ (ζ) ζϕ(z) 2(n+) µ, (6) Integrating (6) with respect to da(z) and applying Fubini s theorem yield (ϕ(z)) 2 ω(z)da(z) [ (n!) 2 ] µ ω(z)da(z) d µ (ζ), (7) ζϕ(z) 2(n+) Since I g,ϕf(0) (n) = 0 and (I g,ϕf) (n) (z) = g(z)f (n) (ϕ(z)), so by (4), the inequality in (7) reduces to I g,ϕf (n) 2 ω (n!) 2 M µ d µ (ζ) = (n!) 2 M µ 2 = (n!) 2 M f 2 K. Thus I (n) g,ϕ : K ω is bounded. Conversely, pose that I (n) g,ϕ : K ω is bounded. Let f ζ (z) = / ζz.then the fact that f ζ K = for each ζ and the boundedness of I (n) g,ϕ : K
96 AJAY K. SHARMA AN ANSHU SHARMA implies that I g,ϕf (n) ζ ω for every ζ. In particular, the fact that I g,ϕf (n) ζ (0) = 0 asserts that g ( ζϕ) n+ A2 ω for every ζ. Moreover, ζ ω(z)da(z) ζϕ(z) 2(n+) ( ) = I(n) g,ϕ M ζz ζz = M K ζ ω ζ and so (4) holds, as desired. To prove the next theorem, we need the following lemma. Lemma. Let g H(), n N {0} and ϕ be a holomorphic self-map of. Then I g,ϕ (n) : K ω is compact if and only if for any sequence {f m } in K with K and which converges to zero locally uniformly, we have lim m I g,ϕf (n) m ω 0. Sine the unit ball of K is a normal family of holomorphic functions. A standard normal family argument then yields the proof. See Proposition 3. of [7]. Theorem 2. Let g H(), n N {0} and ϕ be a holomorphic self-map of. Suppose that I g,ϕ (n) : K α ω is bounded. Then the following statements are equivalent: () I g,ϕ (n) : K ω is compact; (2) The integral ω(z)da(z) ζϕ(z) 2(n+) is a continuous function of ζ. (3) The family of measures {ν ζ : ζ } defined by ν ζ (E) = ω(z)da(z) ζϕ(z) 2(n+) E is equi-absolutely continuous. That is, given ε > 0, there exists a δ > 0 such that ν ζ (E) < ε for all ζ, whenever A(E) < δ. (4) g A 2 ω and lim ω(z)da(z) = 0. (8) r ζ ζϕ(z) 2n+2 Proof. () (2). Let ζ m with ζ m ζ as m, and let f ζm (z) = ζ m z. Then f ζm K = and f ζm f ζ uniformly on compact subsets of. Since I g,ϕ (n) : K ω is compact, by Lemma, we have I (n) g,ϕf ζm I (n) g,ϕf ζ ω 0
INTEGRATION OPERATORS 97 as m. Since I g,ϕ (n) : K ω is bounded, there is a constant L > 0 such that I g,ϕf (n) ζ ω L f ζ K = L for all ζ. Let dλ g,ω (z) = ω(z)da(z). Thus by Cauchy-Schwarz inequality, we have 2 2 dλg,ω (z) ( ζ m ϕ(z)) n+ ( ζϕ(z)) n+ C ζ m (ϕ(z)) 2 ζ (ϕ(z)) 2 dλg,ω (z) ( ) /2 C ζ m (ϕ(z)) f (n) ζ (ϕ(z)) 2 dλ g,ω (z) = C I g,ϕf (n) ζm I g,ϕf (n) ζ ω 0 as m. Thus ω(z)da(z) ω(z)da(z), ζ m ϕ(z) 2(n+) ζϕ(z) 2(n+) which shows the continuity of the integral in (2). (2) (3). Suppose that (3) does not holds. Then there exists a sequence {ζ k } in with ζ k ζ and a sequence of Borel sets {E k } in such that A(E k ) 0 as k, but ν ζk (E k ) C > 0 for all k N. Note that 2 2 dλg,ω ν ζk (E k ) ν ζ (E k ) (z). ( ζ k ϕ(z)) n+ ( ζϕ(z)) n+ Thus ν ζk (E k ) E k E k 2 2 dλg,ω (z) + ν ( ζ k ϕ(z)) n+ ( ζϕ(z)) n+ ζ (E k ). (9) Since I g,ϕ (n) : K ω is bounded, so equation (4) holds. Therefore, ν ζ (E k ) 0 as k. Moreover, as in first part first term in (9) is dominated by E k ( ζ k ϕ(z)) 2 dλg,ω (z). n+ ( ζϕ(z)) n+ Therefore, ν ζk (E k ) 0 as k. This contradiction shows that (2) (3). (3) (). Let {f m } be a sequence in K such that m K and f m 0 uniformly on compact subsets of. We have to show that I g,ϕf (n) m ω 0 as m. For each m, we can find µ m M with µ m = K such that f m (z) = ζz dµ m(ζ). Composing with ϕ and applying Jensen s inequality, we have (n) (ϕ(w)) 2 µ m 2 d µ m (ζ) ζϕ(w) 2(n+) µ m. Integrating with respect to dλ g,ω (w) and then applying Fubini s theorem, we have (n) (ϕ(w)) 2 dλ g,α (w) µ m ζϕ(w) dλ g,ω(w)d µ 2(n+) m (ζ).
98 AJAY K. SHARMA AN ANSHU SHARMA Let ɛ > 0 be given. Now choose a compact set F such that A( \ F ) < δ. Thus (n) (ϕ(w)) 2 dλ g,ω (w) \F µ m \F ζϕ(w) dλ g,ω(w)d µ 2(n+) m (ζ) ɛ µ m d µ m (ζ) = ɛ 2 K < ɛ. (0) On F, (n) (ϕ(w)) 2 < ɛ as m m 0. Moreover, by taking f(z) = z n /n! K, the boundedness of I g,ϕ (n) : K gives Thus \F a dλ g,ω (w) C. m (ϕ(w)) 2 dλ g,ω (w) < ɛc () as m m 0. Therefore, by (0) and (), I g,ϕf (n) m ω 0 as m. () (4). Since I ϕ,g (n) : K ω is bounded, for f(z) = z n /n! K, we have that g A 2 ω. Let f m (z) = z m, m N. It is a norm bounded sequence in K converging to zero uniformly on compacts of. Hence by Lemma, it follows that I ϕ,gf (n) m ω 0 as m. Thus for every ε > 0, there is an m 0 N such that for m m 0, we have ( n ) 2 (m j) ϕ(z) 2(m n) ω(z)da(z) < ε. (2) j=0 From (2), we have that for each r (0, ) r 2(m n)( n ) 2 (m j) j=0 ω(z)da(z) < ε. (3) Hence for r [ n m j=0 (m j)) 0 n, ), we have ω(z)da(z) < ε. (4) Let f B K and f t (z) = f(tz), 0 < t <. Then 0<t< f t K f K, f t K, t (0, ) and f t f uniformly on compacts of as t. The compactness of I (n) g,ϕ : K ω implies that lim t I(n) g,ϕf t I ϕ,gf (n) ω = 0. Hence for every ε > 0, there is a t (0, ) such that a t (ϕ(z)) f (n) (ϕ(z)) 2 ω(z)da(z) < ε. (5)
INTEGRATION OPERATORS 99 Inequalities (4) and (5), give (ϕ(z)) 2 ω(z)da(z) a 2 + 2 t 2ε( + f (n) t 2 ). (ϕ(z)) f (n) (ϕ(z)) 2 ω(z)da(z) t (ϕ(z)) 2 ω(z)da(z) Hence for every f B K, there is a δ 0 (0, ), δ 0 = δ 0 (f, ε), such that for r (δ 0, ) (ϕ(z)) 2 ω(z)da(z) < ε. (6) From the compactness of I g,ϕ (n) : K, we have that for every ε > 0 there is a finite collection of functions f, f 2,..., f k B K such that for each f B K, there is a j {, 2,..., k} such that a (ϕ(z)) f (n) j (ϕ(z)) 2 ω(z)da(z) < ε. (7) On the other hand, from (6) it follows that if δ := max j k δ j (f j, ε), then for r (δ, ) and all j {, 2,..., k} we have j (ϕ(z)) 2 ω(z)da(z) < ε. (8) From (7) and (8) we have that for r (δ, ) and every f B K (ϕ(z)) 2 ω(z)da(z) < 4ε. (9) ζ Applying (9) to the functions f ζ (z) = /( ζz), ζ, we obtain ζϕ(z) ω(z)da(z) < 2(n+) 4ε/(n!)2, from which (8) follows. (4) (). Assume that (f m ) m N is a bounded sequence in K, say by L, converging to 0 uniformly on compacts of as m. Then by the Weierstrass theorem, also converges to 0 uniformly on compacts of, for each k N. We show that ϕ,gf m ω 0 as m, and then apply Lemma. For each m N, we can find a µ m M with µ m = f m K such that dµ m (ζ) f m (z) = ζz. (20) f m (k) I (n) ifferentiating (20) n times, composing such obtained equation by ϕ, applying Jensen s inequality, as well as the boundedness of sequence (f m ) m N, we obtain (n) (ϕ(w)) 2 L(n!) 2 d µ m (ζ). (2) ζϕ(w) 2(n+)
00 AJAY K. SHARMA AN ANSHU SHARMA By the second condition in (4), we have that for every ε > 0, there is an r (0, ) such that for r (r, ), we have ω(z)da(z) < ε. (22) ζϕ(z) 2n+2 Now ζ ( ) I g,ϕf (n) m 2 ω = + ϕ(z) r m (ϕ(z)) 2 ω(z)da(z). Using (2), (22), Fubini s theorem and the fact that w r (n) (w) 2 < ε, for sufficiently large m, say m m 0, we have that for m m 0 I (n) ϕ,gf m 2 ω C + C ϕ(z) r ( C M + (n) (ϕ(z)) 2 ω(z)da(z) ϕ(z) r ζϕ(w) ω(z)da(z)d µ m (ζ) 2n+2 ) d µ m (ζ) ε Cε. From this and the fact that ε is arbitrary, the result follows. References [] A. Aleman and J. A. Cima, An integral operator on H p and Hardy inequality, J. Anal. Math. 85 (200), 57-76. [2] A. Aleman and A. G. Siskakis, An integral operator on H p, Complex variables. 28 (995), 49-58. [3] A. Aleman and A. G. Siskakis, Integration operators on Bergman spaces, Indiana Univ. Math. J. 46 (997), 337-356. [4] P. Bourdon and J. A. Cima, On integrals of Cauchy-Stieltjes type, Houston J. Math., 4 (988), 465 474. [5] J. S. Choa and H. O. Kim, Composition operators from the space of Cauchy transforms into its Hardy-type subspaces, Rocky Mountain J. Math. 3 (200), 95-3. [6] J. A. Cima and A. L. Matheson, Cauchy transforms and composition operators, Illinois J. Math. 4 (998), 58 69. [7] C. C. Cowen and B.. MacCluer, Composition Operators on Spaces of Analytic Functions, CRC Press Boca Raton, New York, 995. [8]. Girela and J. A. Pelaez, Carleson measures, multipliers and integration operators for spaces of irichlet type, J. Funct. Anal. 24 (2006), 334-358. [9] K. Kellay and P. Lefevre, Compact composition operators on weighted Hilbert spaces of analytic functions, J. Math. Anal. Appl. 386 (202), 78-727. [0] C. Pommerenke, Schlichte funktionen und analytische funktionen von beschränkter mittlerer oszillation, Comment. Math. Helv. 52 (977), 59-602. [] J. Rättyä, Integration operator acting on Hardy and weighted Bergman spaces, Bull. Austral. Math. Soc. 75 (2007), 43-446. [2] J. H. Shapiro, Composition Operators and Classical Function Theory, Springer- Verlag, New York, 993.
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