Guanlong Bao, Zengjian Lou, Ruishen Qian, and Hasi Wulan

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1 Bull. Korean Math. Soc. 53 (2016, No. 2, pp ON ABSOLUTE VALUES OF Q K FUNCTIONS Guanlong Bao, Zengjian Lou, Ruishen Qian, and Hasi Wulan Abstract. In this paper, the effect of absolute values on the behavior of functions f in the spaces Q K is investigated. It is clear that g Q K ( g Q K (, but the converse is not always true. For f in the Hardy space H 2, we give a condition involving the modulus of the function only, such that the condition together with f Q K ( is equivalent to f Q K. As an application, a new criterion for inner-outer factorisation of Q K spaces is given. These results are also new for Q p spaces. 1. Introduction enote by the boundary of the unit disk in the complex plane C. Let H( be the space of functions analytic in. Throughout this paper, we assume that K : [0, [0, is a right-continuous and increasing function. A function f H( belongs to the space Q K if f 2 Q K = f (z 2 K(g(a,zdA(z <, where da is the area measure on and g(a,z is the Green function in with singularity at a. By [5, Theorem 2.1], we know that f 2 Q K is equivalent to f (z 2 K ( 1 σ a (z 2 da(z, where σ a (z = a z 1 az is a Möbius transformation of. If K(t = tp, 0 p <, then the space Q K gives the space Q p (cf. [11, 13]. In particular, Q 0 is the irichlet space; Q 1 = BMOA, the space of functions with bounded mean oscillation on ; Q p is the Bloch space for all p > 1. See [5] and [6] for more Received March 17, 2015; Revised July 13, Mathematics Subject Classification. Primary 3050, 30H25, 46E15. Key words and phrases. Q K spaces, absolute values, inner-outer factorisation. The work is ported by NSF of China (No , No and No , NSF of Guangdong Province (No. 2014A and Project of International Science and Technology Cooperation Innovation Platform in Universities in Guangdong Province (No. 2014KGJHZ c 2016 Korean Mathematical Society

2 562 G. BAO, Z. LOU, R. QIAN, AN H. WULAN results on Q K spaces. Let Q K ( be the space of f L 2 ( with f 2 Q = f(ζ f(η 2 ( ζ η K( I ζ η 2 K dζ dη <. I I I Clearly, if K(t = t 2, then Q K ( is equal to BMO(, the space of functions having bounded mean oscillation on (see [7]. To study Q K and Q K (, we usually need two constraints on K as follows. (1.1 and ( ϕ K (s ds < s ϕ K (s s 2 ds <, where ϕ K (s = K(st/K(t, 0 < s <. 0<t 1 If K satisfies (1.2, then Q K BMOA H 2, where H 2 denotes the Hardy space in (see [3, 7]. Thus, if K satisfies (1.2, then the function f Q K has its non-tangential limit f almost everywhere on. We also know that for f H 2 if K satisfies (1.1 and (1.2, then f Q K if and only if f Q K (. Using the triangle inequality, one gets that if g Q K (, then g alsobelongs to Q K (. In general, the converse is not true. Consider { g(e it logt, 0 < t < π, = log t, π < t < 0. By [8, p. 66], g BMO(, but g BMO(. For g H 2, it is natural to seek a condition which together with g Q K ( is equivalent to g Q K (. Our main result, Theorem 1.1, is even new for Q p spaces. Theorem 1.1. Suppose that K satisfies (1.1 and (1.2. Let f H 2. Set dµ z (ζ = 1 z 2 2π ζ z 2 dζ, z, ζ. Then the following conditions are equivalent. (i f Q K. (ii f Q K (. (iii f Q K ( and ( (1.3 f(ζ dµ z (ζ f(z K(1 σ a (z 2 (1 z 2 da(z <. Applying Theorem 1.1, in Section 4, we will show a new criterion for innerouter factorisation of Q K spaces. In this article, the symbol A B means that A B A. We say that A B if there exists a constant C such that A CB.

3 ON ABSOLUTE VALUES OF Q K FUNCTIONS Preliminaries Given f L 2 (, let f be the Poisson extension of f. Namely, f(z = f(ζdµ z (ζ, z. We first give the following characterization of Q K ( spaces. In particular, if K(t = t p, 0 < p < 1, the corresponding result was proved in [12]. Theorem 2.1. Suppose that K satisfies (1.1 and (1.2. Let f L 2 (. Then f Q K ( if and only if ( (2.1 f(ζ 2 dµ z (ζ f(z 2 K(1 σa (z 2 (1 z 2 da(z <. To prove Theorem 2.1, we need the following estimate. Lemma 2.2. Let (1.1 and (1.2 hold for K. If s < 1+c and 2s+r 4 0, then K ( 1 σ a (w 2 (1 w 2 s 1 wz rda(w K ( 1 σa (z 2 (1 z 2 s+r 2 for all a,z. Here c is a small enough positive constant which depends only on (1.1 and (1.2. Proof. We point out that K ( 1 σ a (w 2 (1 w 2 s 1 wz ( K 1 σa (z 2 rda(w (1 z 2 s+r 2 was proved in [1]. So we need only to prove the reverse. For any z, let E(z,1/2 = {w : σ z (w < 1/2} be the pseudo-hyperbolic disk. It is well known that 1 z 1 w 1 wz for all w E(z,1/2. Furthermore, by [14, Lemma 4.30], we have that 1 aw 1 az for all a and w E(z,1/2. Since K satisfies (1.2, K(2t K(t for all t (0,1. We obtain K ( 1 σ a (w 2 (1 w 2 s 1 wz rda(w which gives the desired result. E(z,1/2 K( 1 σ a (z 2 (1 z 2 s+r 2, K ( 1 σ a (w 2 (1 w 2 s 1 wz rda(w Proof of Theorem 2.1. For any f L 2 (, the Littlewood-Paley identity ([7, p. 228] shows that (2.2 f(w 2 log 1 w da(w = 1 f(ζ 2π f(0 2 dζ.

4 564 G. BAO, Z. LOU, R. QIAN, AN H. WULAN Replacing f by f σ z in (2.2 for z, one obtains f(ζ 2 dµ z (ζ f(z 2 f(w 2 (1 σ z (w 2 da(w. Using Fubini s theorem and Lemma 2.2, we obtain, for all a, that ( f(ζ 2 dµ z (ζ f(z 2 K(1 σa (z 2 (1 z 2 da(z ( f(w 2 (1 σ z (w 2 K(1 σa (z 2 da(w (1 z 2 da(z f(w 2 (1 σ z (w 2 K(1 σ a (z 2 da(w (1 z 2 da(z f(w 2 K(1 σ a (w 2 da(w. By [9], we know that f Q K ( if and only if f(z 2 K(1 σ a (z 2 da(z <. Therefore, f Q K ( if and only if ( f(ζ 2 dµ z (ζ f(z 2 K(1 σa (z 2 (1 z 2 da(z <. By [6], for f H 2, if (1.1 and (1.2 hold for K, then f Q K if and only if f Q K (. This, together with Theorem 2.1, gives the following result immediately which was also obtained in [10] by a different method. Corollary 2.3. Suppose that K satisfies (1.1 and (1.2. Let f H 2. Then f Q K if and only if ( f(ζ 2 dµ z (ζ f(z 2 K(1 σa (z 2 (1 z 2 da(z <. 3. Proof of Theorem 1.1 Recall that B H( is called an inner function if B is bounded in and B(ζ = 1 for almost every ζ. An outer function for the Hardy space H 2 is the function of the form ( ζ +z O(z = ηexp ζ z logψ(ζ dζ, η, 2π where ψ > 0 a.e. on, logψ L 1 ( and ψ L 2 (. See [3] for more results on inner and outer functions. Using a technique in [2], we give the proof of Theorem 1.1 as follows.

5 ON ABSOLUTE VALUES OF Q K FUNCTIONS 565 Proof of Theorem 1.1. Note that (i (ii was proved in [6]. (i (iii. For f Q K, we have that f Q K (. The triangle inequality gives that f Q K (. For any z, it follows by Hölder s inequality that ( ( 2 f(ζ dµ z (ζ f(z = f(ζ f(z dµ z (ζ f(ζ f(z 2 dµ z (ζ f(ζ 2 dµ z (ζ f(z 2. Since f Q K, the above estimate, together with Corollary 2.3, gives (1.3. (iii (i. If f 0, the result is true. Note that f H 2. If f 0, then f must be of the form BO, where B is an inner function and O is an outer function of H 2 (see [3]. By the estimates of B and O respectively, Böe [2, p. 237] gave that for any z, f (z 4 1 z ( f(ζ f (z dµ z(ζ+ f (z f(z. Here we remind that f (z = f(ζ dµ z (ζ. Thus, for any a, by Hölder s inequality, we deduce that f (z 2 K(1 σ a (z 2 da(z ( f(ζ f (z dµ K(1 σ a (z 2 z(ζ (1 z 2 da(z K(1 σ a (z + ( f (z f(z 2 (1 z 2 da(z ( ( f (z K(1 σ a (z 2 f(ζ dµ z (ζ (1 z 2 da(z K(1 σ a (z + ( f (z f(z 2 (1 z 2 2 da(z ( f(ζ 2 K(1 σ a (z dµ z (ζ ( f (z (1 z 2 da(z K(1 σ a (z + ( f (z f(z 2 (1 z 2 da(z. By Theorem 2.1 and (1.3, we have that f Q K. The proof is complete.

6 566 G. BAO, Z. LOU, R. QIAN, AN H. WULAN Remark. J. Xiao [12] gave an interesting characterization of Q p spaces in terms of functions with absolute values. Namely, for f H 2, if 0 < p < 1, then f Q p if and only if f Q p ( and ( ( f(ζ dµ z (ζ f(z 2 (1 σ a (z 2 p (1 z 2 da(z <. We show that our Theorem 1.1 implies Xiao s result above. In fact, set K(t = t p, 0 < p < 1, in our Theorem 1.1 and Corollary 2.3. Note that ( 2 ( f(ζ dµ z (ζ f(z 2 f(ζ dµ z (ζ f(z and ( 2 f(ζ dµ z (ζ Thus, one can obtain Xiao s result directly. f(ζ 2 dµ z (ζ. 4. An application to inner-outer factorisation of Q K spaces In this section, we will show a new criterion for inner-outer decomposition of Q K spaces. In fact, an inner-outer factorisation characterization of Q K spaces has been obtained in [6] as follows. Theorem A. Let K satisfy (1.1 and (1.2 with K( z 2 = 2 K(1 z 2, z. z z Let f H 2 with f 0. Then f Q K if and only if f = BO, where B is an inner function and O is an outer function in Q K for which (4.1 O(z 2 (1 B(z 2 K ( σ a (z 2 σ a (z 2 da(z <. As an application of Theorem 1.1, we obtain the following result. Theorem 4.1. Let K satisfy (1.1 and (1.2 with K( z 2 = 2 K(1 z 2, z. z z Let f H 2 with f 0. Then f Q K if and only if f = BO, where B is an inner function and O is an outer function in Q K for which (4.2 O(z 2 (1 B(z 2 K( σa (z 2 σ a (z 2 da(z <. Remark. Theorem 4.1 shows that formula (4.1 in Theorem A can be replaced by the weaker condition (4.2, and this result is also new for Q p spaces.

7 ON ABSOLUTE VALUES OF Q K FUNCTIONS 567 Proof. Necessity. This is a direct result from Theorem A. Sufficiency. Let f = BO and O Q K. Note that O Q K is equivalent to Õ Q K(. By the triangle inequality, one gets Õ Q K(. Hence f Q K (. Observe that f(ζ dµ z (ζ f(z (4.3 = Õ(ζ dµ z(ζ O(z + O(z B(zO(z. Wulan and Ye [10] gave that if K satisfies (1.1 and (1.2, then for all z (4.4 K( z 2 K(1 z 2 (1 z 2. By Hölder s inequality, Õ Q K ( and Corollary 2.3, we show that for any a, ( Õ(ζ dµ K(1 σ a (z 2 z(ζ O(z (1 z 2 da(z ( Õ(ζ O(z K(1 σa (z 2 dµz (ζ (1 z 2 da(z ( = Õ(ζ 2 dµ z (ζ O(z 2 K(1 σa (z 2 (1 z 2 da(z <. Combining the above inequality, (4.2, (4.3 and (4.4, we get ( K(1 σ a (z 2 f(ζ dµ z (ζ f(z (1 z 2 da(z <. Applying Theorem 1.1, we get f Q K. The proof is complete. For f Q K H 2, if we ignore the choice of a constant with modulus one, then f has a unique decomposition with the form f(z = B(zO(z, where B is an inner function and O is an outer function. Combining this with Theorem A and Theorem 4.1, we obtain an interesting result as follows. Corollary 4.2. Suppose that K satisfies (1.1 and (1.2. Let B be an inner function and let O be an outer function in Q K. Then the following conditions are equivalent. (i For some p [1,2], O(z 2 (1 B(z 2 p K( σa (z 2 σ a (z 2 da(z <. (ii For all p [1,2], O(z 2 (1 B(z 2 p K( σa (z 2 σ a (z 2 da(z <.

8 568 G. BAO, Z. LOU, R. QIAN, AN H. WULAN References [1] G. Bao, Z. Lou, R. Qian, and H. Wulan, Improving multipliers and zero sets in Q K spaces, Collect. Math. 66 (2015, no. 3, [2] B. Böe, A norm on the holomorphic Besov space, Proc. Amer. Math. Soc. 131 (2003, no. 1, [3] P. uren, Theory of H p Spaces, Academic Press, New York, [4] K. yakonov, Besov spaces and outer functions, Michigan Math. J. 45 (1998, no. 1, [5] M. Essén and H. Wulan, On analytic and meromorphic function and spaces of Q K -type, Illionis J. Math. 46 (2002, no. 4, [6] M. Essén, H. Wulan, and J. Xiao, Several function-theoretic characterizations of Möbius invariant Q K spaces, J. Funct. Anal. 230 (2006, no. 1, [7] J. Garnett, Bounded Analytic Functions, Springer, New York, [8]. Girela, Analytic functions of bounded mean oscillation, In: Complex Function Spaces, Mekrijärvi 1999, , Editor: R. Aulaskari. Univ. Joensuu ept. Math. Rep. Ser. 4, Univ. Joensuu, Joensuu, [9] J. Pau, Bounded Möbius invariant Q K spaces, J. Math. Anal. Appl. 338 (2008, no. 2, [10] H.Wulan and F. Ye, Some results in Möbius invariant Q K spaces, Complex Var.Elliptic Equ. 60 (2015, no. 11, [11] J. Xiao, Holomorphic Q Classes, Springer, LNM 1767, Berlin, [12], Some results on Q p spaces, 0 < p < 1, continued, Forum Math. 17 (2005, no. 4, [13], Geometric Q p Functions, Birkhäuser Verlag, Basel-Boston-Berlin, [14] K. Zhu, Operator Theory in Function Spaces, American Mathematical Society, Providence, RI, Guanlong Bao epartment of Mathematics Shantou University Shantou, Guangdong , P. R. China address: glbaoah@163.com Zengjian Lou epartment of Mathematics Shantou University Shantou, Guangdong , P. R. China address: zjlou@stu.edu.cn Ruishen Qian School of Mathematics and Computation Science Lingnan Normal University Zhanjiang, Guangdong , P. R. China address: qianruishen@sina.cn Hasi Wulan epartment of Mathematics Shantou University Shantou, Guangdong , P. R. China address: wulan@stu.edu.cn