CUBO A Mathematical Journal Vol.16, No¯ 01, (81–93). March 2014 Characterizations for certain analytic functions by series expansions with Hadamard gaps A. El-Sayed Ahmed Mathematics Department, Faculty of Science, Sohag University, 82524 Sohag, Egypt ahsayed80@hotmail.com A. Kamal Department of Mathematics, Faculty of Science, Port Said University, Port Said, Egypt alaa mohamed1@yahoo.com T.I. Yassen Department of Mathematics, Faculty of Science,, El Azhar University at Assiut, Assiut, Egypt taha hmour@yahoo.com ABSTRACT In this paper we characterize QK,ω(p,q) functions by lacunary series under mild con- ditions posed on the weight functions K and ω, where QK,ω(p,q) is a space of analytic functions defined in the unit disk generalizing the well known analytic Besov-type space. RESUMEN En este art́ıculo caracterizamos las funciones QK,ω(p,q) por series lacunarias bajo condiciones medianas impuestas en las funciones de peso K y ω, donde QK,ω(p,q) es un espacio de funciones anaĺıticas definidas en el disco unitario generalizando el conocido espacio anaĺıtico del tipo Besov. Keywords and Phrases: QK,ω(p,q)-type spaces, Lacunary series. 2010 AMS Mathematics Subject Classification: 30B10, 30B50, 46E15. 82 A. El-Sayed Ahmed, A. Kamal & T.I. Yassen CUBO 16, 1 (2014) 1 Introduction Let D = {z ∈ C : |z| < 1} be the open unit disk in the complex plane C, let H(D) denote the class of functions analytic in the unit disc D, while dA(z) denotes the Lebesgue area measure on the plane, normalized so that A(D) = 1. Let the Green’s function of D be defined as g(z,a) = log 1 |ϕa(z)| , where ϕa(z) = a−z 1−āz is the Möbius transformation related to the point a ∈ D. For 0 < r < 1, let D(a,r) = {r ∈ D : |ϕa(z)| < r} be the pseudo-hyperbolic disk with center a ∈ D and radius r. Definition 1.1. [17] Let K : [0,∞) → [0,∞) be right-continuous and nondecreasing function, 0 < p < ∞,−2 < q < ∞ and for given reasonable function ω : (0,1] → (0,∞), an analytic function f in D is said to belong to the space QK,ω(p,q) if ||f||QK,ω(p,q) = sup a∈D ∫ D |f′(z)|p (1 − |z|)q ωp(1 − |z|) K(g(z,a))dA(z) < ∞. In the past few decades both Taylor and Fourier series expansions for various classes of analytic function spaces where the studies are done by the help of Hadamard gap class (see [1, 3, 10, 16] and others). It is well known that a lacunary series belongs to BMOA if and only if it is in the Hardy space H2, (see [2]) for example. Very recently, in [6, 7, 14] there are some characterizations for some classes of meromorphic functions by the coefficients of certain lacunary series expansions in the unit disk. On the other hand there are some studies of the same problem in Clifford analysis (see [4, 5, 12, 13]). We assume throughout the paper that ∫1 0 (1 − r)q K ( log 1 r ) ωp ( log 1 r ) rdr < ∞. An important tool in the study of QK,ω(p,q) space is the auxiliary functions φK and ψω defined by φK(s) = sup 0 0, independent of t, such that ω1(t) ≤ Cω2(t), K1(t) ≤ CK2(t) for all t. The notation ω1 & ω2, K1 & K2 is used in a similar fashion. When ω1 . ω2 . ω1, we write ω1 ≈ ω2. Also for K1 . K2 . K1, we write K1 ≈ K2. 2 Auxiliary Lemmas In what follows we say f . g (for two functions f and g ) if there is a constant C such that f ≤ Cg. We say f ≈ g (that is, f is comparable with g ) whenever g . f . g. In this section we prove several result about the weight function that are needed for subsequent sections and are of some independent interest. Lemma 2.1. [20] If K satisfies condition (1), then the function K∗(t) = t ∫ ∞ t K(s) s2 ds (where, 0 < t < ∞), has the following properties : (A) K∗ is nondecreasing on (0,∞). (B) K∗(t)/t is nondecreasing on (0,∞). (C) K∗(t) ≥ K(t) for all t ∈ (0,∞). (D) K∗ . K on (0,1]. If K(t) = K(1) for t ≥ 1, then we also have (E) K∗(t) = K∗(1) = K(1) for t ≥ 1, so K∗ ≈ K on (0,∞). Lemma 2.2. [20] If K satisfies condition (1), then we can find another non-negative weight func- tion given by K∗(t) = t ∫ ∞ t K(s) s2 ds (where, 0 < t < ∞), such that QK,ω(p,q) = QK∗,ω(p,q) and that the new function K ∗ has the following properties : (A) K∗ is nondecreasing on (0,∞). (B) K∗(t)/t is nondecreasing on (0,∞). 84 A. El-Sayed Ahmed, A. Kamal & T.I. Yassen CUBO 16, 1 (2014) (c) K∗(t) satisfies condition (1) (d) K∗(2t) ≈ K∗(t) on (0,∞). (e) K∗(t) ≈ K(t) on (0,1]. (f) K∗ is differentiable on (0,∞). (g) K∗ is concave on (0,∞). (h) K∗(t) = K∗(1) for t ≥ 1, Lemma 2.3. [8] If ω satisfies condition (2), then the function ω∗(t) = t ∫1 t ω(s) s2 ds (where, 0 < t < 1), has the following properties : (A) ω∗ is nondecreasing on (0,1). (B) ω∗(t)/t is nondecreasing on (0,1). (C) ω∗(t) ≥ ω(t) for all t ∈ (0,1). (D) ω∗ . ω on (0,1). If ω(t) = ω(1) for t ≥ 1, then we also have (E) ω∗(t) = ω∗(1) = ω(1) for t ≥ 1, so ω∗ ≈ ω on (0,1). Lemma 2.4. Let α ≥ 1 and β > 0. If K satisfies (1) and ω satisfies (2), then ∫1 0 rα−1(log 1 r )β−1 K(log 1 r ) ωp(log 1 r ) dr ≈ ( 1 α )β K( 1 α ) ωp( 1 α ) Proof. Let I = ∫1 0 rα−1(log 1 r )β−1 K(log 1 r ) ωp(log 1 r ) dr. By the change variables we have I = ∫1 0 e−αttβ−1 K(t) ωp(t) dt. CUBO 16, 1 (2014) Characterizations for certain analytic functions by series . . . 85 We write I = I1 + I2 where I1 = ∫1 1 α e−αt tβ−1 K(t) ωp(t) dt, and I2 = ∫ 1 α 0 e−αt tβ−1 K(t) ωp(t) dt. By Lemma 2.1 and Lemma 2.2, we have I1 ≤ K( 1 α ) ( 1 α ) × ( 1 α ) ωp( 1 α ) ∫1 1 α e−αttβ−1 dt, = K( 1 α ) ωp( 1 α ) ∫1 1 α e−αttβ−1 dt. Making the change of variables s = αt, we have I1 ≤ K( 1 α ) ωp( 1 α ) ∫α 1 e−ssβ−1 ( 1 α )β−1( 1 α ) ds = K( 1 α ) ωp( 1 α ) ( 1 α )β∫α 1 e−ssβ−1 ds. Then I1 ≤ C(β) K( 1 α ) ωp( 1 α ) ( 1 α )β . Since K(t) and ω(t) are non-decreasing, then by making the change of variables s = αt, we obtain I2 ≤ K( 1 α ) ωp( 1 α ) ∫ 1 α 0 e−αttβ−1 dt, and I2 ≤ K( 1 α ) ωp( 1 α ) ( 1 α )β∫1 0 e−ssβ−1 ds. Then I2 ≤ C(β) K( 1 α ) ωp( 1 α ) ( 1 α )β . Combining this with what was proved in the previous paragraph, we have ∫1 0 rα−1(log 1 r )β−1 K(log 1 r ) ωp(log 1 r ) dr ≤ C(β) K( 1 α ) ωp( 1 α ) ( 1 α )β , where C(β) is constant which only depends on (β). On the other hand, recall that K(t) and ω(t) are non-decreasing. Then, I ≥ ∫1 1 α e−αttβ−1 K(t) ωp(t) dt ≥ K( 1 α ) ωp( 1 α ) ∫1 1 α e−αttβ−1 dt. 86 A. El-Sayed Ahmed, A. Kamal & T.I. Yassen CUBO 16, 1 (2014) Making the change of variables s = αt, we have I ≥ C(β) K( 1 α ) ωp( 1 α ) ( 1 α )β . Then ∫1 0 rα−1(log 1 r ) β−1 K(log 1 r ) ωp(log 1 r ) dr ≈ C(β) K( 1 α ) ωp( 1 α ) ( 1 α )β . This completes the proof of the lemma. Lemma 2.5. Let 0 < γ ≤ 1 and η(r) = ∞∑ n=0 2nγr2 n , 0 ≤ r < 1, then η(r) ≤ 2Γ(γ)(log 1 r )−γ. Then above result can be found in [15]. Lemma 2.6. [19] 1 2π ∫2π 0 g(reiθ,a)dθ =    log 1 |a| , 0 < r ≤ |a|, log 1 r , |a| < r ≤ 1. By Jensen’s formula(see [11, 18]), we can directly obtain the above result. Theorem 2.7. Let K satisfy condition (1) and ω satisfy condition (2). Suppose that f(x) = ∞∑ n=1 anx n, with an ≥ 0. For α > 0, p > 0, we have that ∫1 0 (1 − x)α−1(f(x))p K(log 1 x ) ωp(log 1 x ) dx ≈ ∞∑ n=0 2−nα tpn K( 1 2n ) ωp( 1 2n ) (3) where tn = ∑ k∈In ak, n ∈ N, In = {k : 2 n ≤ k < 2n+1; k ∈ N}. Proof. Let rn = 1 − 2 −n, n = 1,2, ..., then r2 n −1 n ≥ 1 e . By Lemma 2.3, we have that ∫1 0 ( ∞∑ n=1 anr n)p (1 − r)α−1 K(log 1 r ) ωp(log 1 r ) dr ≥ ∫1 1 2 ( ∞∑ n=0 tnr 2 n+1 −1)p (1 − r)α−1 K(log 1 r ) ωp(log 1 r ) dr & ∞∑ n=0 tpn (r 2 n+1 −1 n+1 ) p ∫rn+2 rn+1 (1 − r)α−1 K(log 1 rn+1 ) ωp(log 1 rn+1 ) dr & ∞∑ n=0 2−nα tpn K( 1 2n ) ωp( 1 2n ) . CUBO 16, 1 (2014) Characterizations for certain analytic functions by series . . . 87 The last inequality holds because of (log 1 r ) ≥ 1 − r. Conversely, we first suppose that p > 1. Let γ = min{1,α/p} and η(r) = ∞∑ k=0 2kγ r2k, 0 ≤ r < 1. Then by Jensen’s inequality (see [11, 18]), we have ( ∞∑ n=0 anr n )p ≤ ( ∞∑ n=0 tnr 2 n )p . (η(r))p−1 ∞∑ n=0 2nγ(1−p) r2n|tn| p. From Lemma 2.2, Lemma 2.3, Lemma 2.4 and Lemma 2.5, it follows that ∫1 0 ( ∞∑ n=1 anr n )p (1 − r)α−1 K(log 1 r ) ωp(log 1 r ) dr ≤ ∫1 0 (2Γ(γ))p−1(log 1 r )−γ(p−1) ( ∞∑ n=0 2nγ(1−p) r2n |tn| p(1 − r)α−1 ) K(log 1 r ) ωp(log 1 r ) dr ≤ 2nγ(1−p) ∞∑ n=0 |tn| p ∫1 0 r2n−1(log 1 r ) −γ(p−1)+α−1 K(log 1 r ) ωp(log 1 r ) dr ≤ 2nγ(1−p) ∞∑ n=0 |tn| p( 1 2n )−γ(p−1)+α K(log 1 r ) ωp(log 1 r ) ≤ ∞∑ n=0 2−nα tpn K( 1 2n ) ωp( 1 2n ) . (4) Denote r(1 − r)α−1 ≤ (log 1 r )α−1, 0 ≤ r < 1. Secondly suppose that p = 1, by Lemma 2.4, we obtain that ∫1 0 ( ∞∑ n=1 anr n ) (1 − r)α−1 K(log 1 r ) ωp(log 1 r ) dr ∫1 0 ( ∞∑ n=0 tnr 2 n ) (1 − r)α−1 K(log 1 r ) ωp(log 1 r ) dr ≤ ∞∑ n=0 tn ∫1 0 ( r2 n −1 ) r(1 − r)α−1 K(log 1 r ) ωp(log 1 r ) dr ≤ ∞∑ n=0 tn ∫1 0 ( r2 n −1 ) (log 1 r )α−1 K(log 1 r ) ωp(log 1 r ) dr ≤ ∞∑ n=0 2−nα tn K( 1 2n ) ωp( 1 2n ) . (5) 88 A. El-Sayed Ahmed, A. Kamal & T.I. Yassen CUBO 16, 1 (2014) Finally, if p < 1, we have ∫1 0 ( ∞∑ n=1 anr n )p (1 − r)α−1 K(log 1 r ) ωp(log 1 r ) dr ∫1 0 ( ∞∑ n=0 tnr 2 n )p (1 − r)α−1 K(log 1 r ) ωp(log 1 r ) dr ≤ ∞∑ n=0 tpn ∫1 0 ( rp2 n −1 ) r(1 − r)α−1 K(log 1 r ) ωp(log 1 r ) dr ≤ ∞∑ n=0 tpn ∫1 0 ( rp2 n −1 ) (log 1 r )α−1 K(log 1 r ) ωp(log 1 r ) dr ≤ ∞∑ n=0 2−nα tpn K( 1 2n ) ωp( 1 2n ) (6) From (4), (5) and (6), we obtain the desired result and the proof is therefore established. 3 Main results We prove that an analytic function f on the unit disk D with Hadamard gaps, that is, f(z) = ∞∑ n=1 anz n satisfying nk+1 nk ≥ λ > 1 for all k ∈ N, belongs to the space QK,ω(p,q) with mild conditions on the weight functions K and ω if and only if ∞∑ k=0 n p−q−1 k |ak| p K( 1 nk ) ωp( 1 nk ) < ∞, where 0 < p < ∞ and −1 < q < ∞. Now, we give the following result: Theorem 3.1. Let K satisfy condition (1) and ω satisfy condition (2). Suppose that f(z) = ∞∑ n=1 anz n, with an ≥ 0. For α > 0, p > 0, we have ∫ D |f′(z)|p (1 − |z|)q ωp(1 − |z|) K(log 1 |z| )dA(z) ≈ ∞∑ n=1 np−q−1 |an| p K( 1 n ) ωp( 1 n ) , (7) where tn = ∑ k∈In ak, n ∈ N, In = {k : 2 n ≤ k < 2n+1; k ∈∈ N}. Proof. We write I(f) = ∫ D |f′(z)|p (1 − |z|)q ωp(1 − |z|) K(log 1 |z| )dA(z). CUBO 16, 1 (2014) Characterizations for certain analytic functions by series . . . 89 Integrating in polar coordinates we gets I(f) = ∫1 0 ∫2π 0 ( ∞∑ n=1 n |an|r n−1 )p (1 − r)q K(log 1 r ) ωp(log 1 r ) rdrdθ . ∫1 0 ( ∞∑ n=1 n |an|r n−1 )p (1 − r)q K(log 1 r ) ωp(log 1 r ) dr, by theorem 2.7, we obtain I(f) . ∞∑ n=0 2−n(q+1) tpn K( 1 2n ) ωp( 1 2n ) ≈ ∞∑ n=1 np−q−1 |an| p K( 1 n ) ωp( 1 n ) . The proof is therefore established. Theorem 3.2. Let 0 < p < ∞, −1 < q < ∞. Let K satisfy condition (1) and ω satisfy condition (2). Suppose that f(z) = ∞∑ k=1 ak z nk, has Hadamard gaps, then f belongs to QK,ω(p,q) if and only if ∞∑ k=1 n p−q−1 k |ak| p K( 1 nk ) ωp( 1 nk ) < ∞. (8) Proof. Suppose that f(z) = ∑ ∞ k=1 ak z nk is a lacunary series in QK,ω(p,q). Without loss general- ity, we assume that nk ≥ 1. If f has Hadamard gaps, then Mp(r,f ′) ≈ M2(r,f ′) (see [21]), where Mpp(r,f ′) = 1 2π ∫2π 0 |f′(reiθ)|pdθ. since f ∈ QK,ω(p,q), by theorem 2.7, we obtain ∞ > ∫ D |f′(z)|p (1 − |z|)q ωp(1 − |z|) K(log 1 |z| )dA(z) = ∫1 0 Mpp(r,f ′) (1 − r)q K(log 1 r ) ωp(log 1 r ) rdr, & ∫1 0 ( ∞∑ k=1 n2k |ak| 2r2(nk−1) ) p 2 (1 − r)q K(log 1 r ) ωp(log 1 r ) rdr, & ∞∑ k=1 2−k(q+1) t p 2 k K( 1 2k ) ωp( 1 2k ) , where tk = ∑ nj∈Ik n2j |aj| 2. The Taylor series of f has at most [logλ 2] + 1 terms aj z nj when nj ∈ Ik for k ≥ 1. By Hölder’s inequality, we note that t p 2 k & ∑ nj∈Ik n p j |aj| p. Then 90 A. El-Sayed Ahmed, A. Kamal & T.I. Yassen CUBO 16, 1 (2014) ∞∑ k=0 n p−q−1 k |ak| p K( 1 nk ) ωp( 1 nk ) . ∞∑ k=0 2−k(q+1) ( ∑ nj∈Ik n p j |aj| p ) K( 1 2k ) ωp( 1 2k ) < ∞. Next suppose that condition (8) holds, we write z = reiθ in polar form and observe that |f(z)| = ∞∑ k=1 |ak| r nk. Since K and ω satisfies conditions (1) and (2) respectively, K is concave. Then, by Jensen’s inequality, Lemma 2.2 and Lemma 2.6, we deduce that ∫2π 0 K(g(reiθ,a))dθ ≤ K (∫2π 0 (g(reiθ,a))dθ ) ≤ K ( log 1 r ) . Hence, ||f|| p K,ω = sup a∈D ∫ D |f′(z)|p (1 − |z|)q ωp(1 − |z|) K(log 1 |z| )dA(z) ≤ sup a∈D ∫1 0 ( ∞∑ k=1 nk |ak|r (nk−1) )p (1 − r)q ωp(1 − r) × { 1 2π ∫2π 0 K(g(reiθ,a))dθ } rdr ≤ sup a∈D ∫1 0 ( ∞∑ k=1 nk |ak|r (nk−1) )p (1 − r)q ωp(1 − r) K ( log 1 r ) rdr by Theorem 2.7,we obtain ||f||QK,ω(p,q) . ∞∑ n=0 2−n(q+1) tpn K( 1 2n ) ωp( 1 2n ) . ∞∑ K=1 n p−q−1 k |ak| p K( 1 nk ) ωp( 1 nk ) < ∞. The proof is therefore established. Theorem 3.3. Let 0 < p < ∞, −1 < q < ∞. Let K satisfy condition (1) and ω satisfy condition (2). Suppose that f(z) = ∞∑ k=1 ak z nk, has Hadamard gaps, ω > 0, then f belongs to QK,ω(p,q) if and only if f ∈ QK,ω,0(p,q). Proof. Since sufficiency is obvious because of QK,ω,0(p,q) ⊂ QK,ω(p,q). Now we will prove necessity of Theorem 3.3. Suppose that the lacunary series f belongs to QK,ω(p,q). We must show that I(a) → 0 as |a| → 1 −, where I(a) = ∫ D |f′(z)|p (1 − |z|)q ωp(1 − |z|) K(log 1 |z| )dA(z). CUBO 16, 1 (2014) Characterizations for certain analytic functions by series . . . 91 From the proof of Theorem 3.2, we know that f ∈ QK,ω(p,q) implies that ∫1 0 ( ∞∑ k=1 nk |ak|r (nk−1) )p (1 − r)q ωp(log 1 r ) K ( log 1 r ) rdr < ∞. For any � < 0, there is a δ ∈ (0,1) such that ∫1 δ ( ∞∑ k=1 nk |ak|r (nk−1) )p (1 − r)q K ( log 1 r ) ωp(log 1 r ) dr < �. We may as well assume that lim |a|→1− K(log 1 |a| ) = 0. If K satisfies the condition (1). Then we choose a such that 1 > |a| > δ. By Lemma 2.6, Theorem 2.7 and Theorem 3.2, we have ∫δ 0 ( ∞∑ k=1 nk |ak|r (nk−1) )p (1 − r)q ωp(1 − r) ∫2π 0 K(g(z,a))rdrdθ ≤ ∫δ 0 ( ∞∑ k=1 nk |ak|r (nk−1) )p (1 − r)q ωp(1 − r) K ( log 1 |a| ) dr = K ( log 1 |a| ) ∫δ 0 ( ∞∑ k=1 nk |ak|r (nk−1) )p (1 − r)q ωp(1 − r) dr ≤ K ( log 1 |a| ) K ( log 1 δ ) ∫δ 0 ( ∞∑ k=1 nk |ak|r (nk−1) )p (1 − r)q K ( log 1 r ) ωp(log 1 r ) dr ≤ K ( log 1 |a| ) K ( log 1 δ ) ∞∑ k=1 n p−q−1 k |ak| p K( 1 nk ) ωp( 1 nk ) . Since � is arbitrary, we conclude that I(a) → 0 as |a| → 1−. So f ∈ QK,ω,0(p,q) and the proof is complete. Carefully checking the proof of Theorems 3.2 and 3.3, we also obtain the following sufficient condition for f ∈ QK,ω,0(p,q) and hence in f ∈ QK,ω,0(p,q) in terms of Taylor coefficients. Theorem 3.4. Let 0 < p < ∞ and −1 < q < ∞. Let K satisfy condition (1) and ω satisfy condition (2). Suppose that f(z) = ∞∑ k=0 ak z nk, has Hadamard gaps, then f belongs to QK,ω,0(p,q) if and only if ∞∑ k=0 n p−q−1 k |ak| p K( 1 nk ) ωp( 1 nk ) < ∞. (9) 4 Conclusion From Theorems 3.2, 3.3 and3.4, we can give the following theorem; 92 A. El-Sayed Ahmed, A. Kamal & T.I. Yassen CUBO 16, 1 (2014) Theorem 4.1. Let 0 < p < ∞, −1 < q < ∞. Let K satisfy condition (1) and ω satisfy condition (2). Suppose that f(z) = ∞∑ k=0 ak z nk ∈ H(D) has Hadamard gaps, then the following statements are equivalent: (i) f ∈ QK,ω(p,q); (ii) f ∈ QK,ω,0(p,q); (iii) ∑ ∞ k=0 n p−q−1 k |ak| p K( 1 nk ) ωp( 1 nk ) < ∞. Received: April 2013. Accepted: October 2013. References [1] K. L. Avetisyan, Sharp inclusions and lacunary series in mixed-norm spaces on the polydisc, Complex variables and elliptic equations, 58(2)(2013), 185195. [2] A. Baernstein II, Analytic functions of bounded mean oscillation, in Aspects of Contemporary Complex Analysis, Academic Press, London, (1980), 3-36. [3] J. S. Choa, A property of series of holomorphic homogeneous polynomials with Hadamard gaps, Bull. Aust. Math. Soc. 53(3)(1996), 479-484. [4] A. 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