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Composition operators from Zygmund spaces into Q K spaces

Abstract

The boundedness and compactness of composition operators from Zygmund and little Zygmund spaces into Q K and Q K , 0 spaces are characterized in this paper.

MSC:47B33, 30H99.

1 Introduction

Let D={z:|z|<1} be the open unit disk of complex plane . Denote by H(D) the class of functions analytic in D. Let g(z,a) denote the Green’s function with pole at aD, i.e., g(z,a)=log 1 | φ a ( z ) | , where φ a (z)= a z 1 a ¯ z is a Möbius transformation of D. An fH(D) is said to belong to the Zygmund space, denoted by Z, if

sup | f ( e i ( θ + h ) ) + f ( e i ( θ h ) ) 2 f ( e i θ ) | h <,

where the supremum is taken over all e i θ D and h>0. By Theorem 5.3 in [1], we see that fZ if and only if

f Z = | f ( 0 ) | + | f ( 0 ) | + sup z D ( 1 | z | 2 ) | f ( z ) | <.
(1)

It is easy to check that Z is a Banach space under the above norm. Let Z 0 denote the subspace of Z consisting of those fZ for which

lim | z | 1 ( 1 | z | 2 ) | f ( z ) | =0.

The space Z 0 is called the little Zygmund space. Throughout this paper, the closed unit ball in Z and Z 0 will be denoted by B Z and B Z 0 , respectively.

Let K:[0,)[0,) be a nondecreasing continuous function. We say that an fH(D) belongs to the space Q K if (see, e.g., [24])

f Q K 2 = sup a D D | f ( z ) | 2 K ( g ( z , a ) ) dA(z)<.
(2)

Here, dA is the normalized Lebesgue area measure in D. Modulo constants, Q K is a Banach space under the norm |f(0)|+ f Q K and Q K is Möbius invariant. When (see [2])

0 1 K(logr) ( 1 r 2 ) 2 rdr<,

Q K =B. Here, is the Bloch space defined as follows:

B= { f : f H ( D ) , f B = | f ( 0 ) | + sup a D | f ( z ) | ( 1 | z | 2 ) < } .

If K(t)= t p , then Q K = Q p (see [5, 6]). If fH(D) such that

lim | a | 1 D | f ( z ) | 2 K ( g ( z , a ) ) dA(z)=0,
(3)

we say that f belongs to the space Q K , 0 . If Q K consists of just constant functions, we say that it is trivial. Q K is nontrivial if and only if (see [2])

0 1 / e K(logr)rdr<.
(4)

To avoid that Q K is trivial, we assume from now that (4) is satisfied. See [24, 715] for the study of the space Q K .

Let φ be an analytic self-map of D. The composition operator C φ is defined by

C φ (f)(z)=f ( φ ( z ) ) ,fH(D).

It is interesting to provide a function theoretic characterization of when φ induces a bounded or compact composition operator on various spaces. For a study of the composition operators, see [16] and [17].

The composition operator from Bloch spaces to Q K and Q K , 0 was studied in [9, 10, 18]. Some characterizations of the boundedness and compactness of the composition operator, as well as Volterra type operator, on the Zygmund space can be found in [1923].

The purpose of this paper is to study the boundedness and compactness of the operator C φ from the Zygmund space and little Zygmund space into Q K and Q K , 0 .

Throughout this paper, constants are denoted by C, they are positive and may differ from one occurrence to the other.

2 Main results and proofs

In this section, we state and prove our main results. In order to formulate our main results, we need some auxiliary results which are incorporated in the following lemmas. The following lemma, can be proved in a standard way (see, e.g., Theorem 3.11 in [16]).

Lemma 1 Let K be a nonnegative nondecreasing function on [0,). Assume that φ is an analytic self-map of D. Then C φ :Z Q K is compact if and only if C φ :Z Q K is bounded and for every bounded sequence { f n } in Z which converges to 0 uniformly on compact subsets of D as n, lim n C φ f n Q K =0.

By using the methods of [10] (see also [24]), we can obtain the following lemma. Since the proof is similar, we omit the details.

Lemma 2 Let K be a nonnegative nondecreasing function on [0,). Assume that φ is an analytic self-map of D. If C φ :Z( Z 0 ) Q K is compact, then for any ε>0 there exists a δ, 0<δ<1, such that for all f in Z( Z 0 ),

sup a D | φ ( z ) | > r | f ( φ ( z ) ) | 2 | φ ( z ) | 2 K ( g ( z , a ) ) dA(z)<ε
(5)

holds whenever δ<r<1.

By modifying the proof of Theorem 3.1 of [7] (or see [25]), we can prove the following lemma. We omit the details.

Lemma 3 Let K be a nonnegative nondecreasing function on [0,). Assume that φ is an analytic self-map of D. Then C φ :Z Q K , 0 is compact if and only if C φ :Z Q K , 0 is bounded and

lim | a | 1 sup f Z 1 D | ( C φ f ) ( z ) | 2 K ( g ( z , a ) ) dA(z)=0.

Lemma 4 [20]

Suppose that f Z 0 , then

lim | z | 1 | f ( z ) | /ln e 1 | z | 2 =0.

Lemma 5 [26]

Suppose that { n k } is an increasing sequence of positive integers satisfying n k + 1 n k λ>1 for all kN. Let 0<p<. Then there are two positive constants C 1 and C 2 , depending only on p and λ such that

C 1 ( k = 1 | a k | 2 ) 1 2 ( 1 2 π 0 2 π | k = 1 a k e i n k θ | p d θ ) 1 p C 2 ( k = 1 | a k | 2 ) 1 2 .

Now we are in a position to state and prove our main results in this paper.

Theorem 1 Let K be a nonnegative nondecreasing function on [0,). Assume that φ is an analytic self-map of D. Then the following statements hold:

  1. (i)

    If

    sup a D D | φ ( z ) | 2 ( ln e 1 | φ ( z ) | 2 ) 2 K ( g ( z , a ) ) dA(z)<,
    (6)

then C φ :Z( Z 0 ) Q K is bounded.

  1. (ii)

    If C φ :Z( Z 0 ) Q K is bounded, then

    sup a D D | φ ( z ) | 2 ln 1 1 | φ ( z ) | 2 K ( g ( z , a ) ) dA(z)<.
    (7)

Proof (i) Let fZ. Then by the following result (see [20]):

| f ( z ) | C f Z ln e 1 | z | 2 ,
(8)

we have

In addition, by the well-known fact that f C f Z , we obtain

| f ( φ ( 0 ) ) | C f Z .

Therefore, C φ :Z Q K is bounded, and hence C φ : Z 0 Q K is bounded.

  1. (ii)

    First, we suppose that C φ :Z Q K is bounded. Let g(z)=zZ. By the boundedness of C φ :Z Q K we have that φ= C φ g Q K . Hence, we have

    (9)

For zD, such that |z|=r 1 e . Let

f(z)= k = 0 1 2 k + 1 z 2 k + 1 .

Then by the fact that p(z)= k = 0 z 2 k belongs to Bloch space (see [[27], Theorem 1]) and the relationship of Bloch function and Zygmund function, we see that fZ. Let

h θ (z)=f ( e i θ z ) = k = 0 1 2 k + 1 ( e i θ z ) 2 k + 1 .

Then h θ Z and h θ Z = f Z . We have

> C φ 2 h θ Z 2 C φ h θ Q K 2 sup a D D | ( C φ h θ ) ( z ) | 2 K ( g ( z , a ) ) d A ( z ) sup a D | φ ( z ) | > 1 e | k = 0 e i ( 2 k + 1 ) θ φ 2 k ( z ) | 2 | φ ( z ) | 2 K ( g ( z , a ) ) d A ( z ) .
(10)

Since

1 2 π 0 2 π C φ 2 h θ Z 2 dθ= 1 2 π 0 2 π C φ 2 f Z 2 dθ= C φ 2 f Z 2 = C φ 2 h θ Z 2 ,

by (10), Lemma 5 and Fubini’s theorem we have

> 1 2 π 0 2 π C φ 2 h θ Z 2 d θ 1 2 π 0 2 π sup a D | φ ( z ) | > 1 e | k = 0 e i ( 2 k + 1 ) θ φ 2 k ( z ) | 2 | φ ( z ) | 2 K ( g ( z , a ) ) d A ( z ) d θ = sup a D | φ ( z ) | > 1 e { 1 2 π 0 2 π | k = 0 e i ( 2 k + 1 ) θ φ 2 k ( z ) | 2 d θ } | φ ( z ) | 2 K ( g ( z , a ) ) d A ( z ) sup a D | φ ( z ) | > 1 e k = 0 | φ ( z ) | 2 k + 1 | φ ( z ) | 2 K ( g ( z , a ) ) d A ( z ) .

For any r(0,1), a calculation shows that

ln 1 1 r 2 = ln ( 1 + r ) ln ( 1 r ) = 0 r ( n = 0 ( 1 ) n t n + n = 0 t n ) d t = n = 0 ( 1 ( 1 ) n ) r n + 1 n + 1 = 2 k = 1 r 2 k 2 k = k = 1 r 2 k k = k = 0 j = 2 k 2 k + 1 1 r 2 j j k = 0 ( 1 2 k + + 1 2 k ) r 2 2 k = k = 0 r 2 k + 1 ,
(11)

since the number of terms in the sum from 2 k to 2 k + 1 1 is 2 k . Therefore,

> 1 2 π 0 2 π C φ 2 h θ Z 2 d θ sup a D | φ ( z ) | > 1 e | φ ( z ) | 2 ln 1 1 | φ ( z ) | 2 K ( g ( z , a ) ) d A ( z ) ,
(12)

which together with (9) implies that (7) holds.

Now suppose that C φ : Z 0 Q K is bounded. Take the function f(z) given by the above. Set

f r (z)=f(rz)= k = 1 1 2 k + 1 ( r z ) 2 k + 1 ,r(0,1).

Then f r Z 0 . Then, as argued the same with the case of C φ :Z Q K and let r1, we get the desired result. The proof of the theorem is finished. □

Theorem 2 Let K be a nonnegative nondecreasing function on [0,). Assume that φ is an analytic self-map of D. Then the following statements holds:

  1. (i)

    If φ Q K and

    lim r 1 sup a D | φ ( z ) | > r | φ ( z ) | 2 ( ln e 1 | φ ( z ) | 2 ) 2 K ( g ( z , a ) ) dA(z)=0,
    (13)

    then C φ :Z( Z 0 ) Q K is compact.

  2. (ii)

    If C φ :Z( Z 0 ) Q K is compact, then φ Q K and

    lim r 1 sup a D | φ ( z ) | > r | φ ( z ) | 2 ln 1 1 | φ ( z ) | 2 K ( g ( z , a ) ) dA(z)=0.
    (14)

Proof (i) Assume that φ Q K and (13) holds. Let { f k } be a bounded sequence in Z which converges to 0 uniformly on compact subsets of D. We need to show that { C φ f k } converges to 0 in Q K norm. By (13), for given ε>0, there is an r(0,1), such that

sup a D | φ ( z ) | > r | φ ( z ) | 2 ( ln e 1 | φ ( z ) | 2 ) 2 K ( g ( z , a ) ) dA(z)<ε.

Therefore, by (8), we have

(15)

From the assumption, we see that { f k } also converges to 0 uniformly on compact subsets of D by Cauchy’s estimates. It follows that C φ f k Q K 0 since | f k (φ(0))|0 and sup | w | r | f k (w)|0 as k. By Lemma 1, C φ :Z Q K is compact, and hence C φ : Z 0 Q K is also compact.

(ii) We only need to prove the case of C φ : Z 0 Q K . Assume that C φ : Z 0 Q K is compact. By taking g(z)=z Z 0 we get φ Q K . Now we choose the function f(z) given in the proof of Theorem 1. Then fZ. Choose a sequence { λ j } in D which converges to 1 as j, and let f j (z)=f( λ j z) for jN. Then, f j Z 0 for all jN and f j Z C. Let f j , θ (z)= f j ( e i θ z). Then f j , θ Z 0 . Replace f by f j , θ in (5) and then integrate both sides with respect to θ. By Fubini’s theorem, we obtain

ε > sup a D 1 2 π | φ ( z ) | > r ( 0 2 π | f j ( e i θ φ ( z ) ) | 2 d θ ) | φ ( z ) | 2 K ( g ( z , a ) ) d A ( z ) = sup a D 1 2 π | φ ( z ) | > r 0 2 π | k = 1 ( λ j φ ( z ) e i θ ) 2 k | 2 d θ | λ j | 2 | φ ( z ) | 2 K ( g ( z , a ) ) d A ( z ) = sup a D | φ ( z ) | > r ( k = 1 | λ j φ ( z ) | 2 k + 1 ) | λ j | 2 | φ ( z ) | 2 K ( g ( z , a ) ) d A ( z ) .
(16)

From the proof of Theorem 1, for 1/ e <r<1 and for sufficiently large j, (16) gives

sup a D | φ ( z ) | > r | λ j | 2 | φ ( z ) | 2 ln 1 1 | λ j φ ( z ) | 2 K ( g ( z , a ) ) dA(z)<ε.

By Fatou’s lemma, we get (14). □

Theorem 3 Let K be a nonnegative nondecreasing function on [0,). Assume that φ is an analytic self-map of D. Then the following statements hold:

  1. (i)

    If C φ : Z 0 Q K , 0 is bounded, then φ Q K , 0 and

    sup a D D | φ ( z ) | 2 ln 1 1 | φ ( z ) | 2 K ( g ( z , a ) ) dA(z)<.
    (17)
  2. (ii)

    If φ Q K , 0 and

    sup a D D | φ ( z ) | 2 ( ln e 1 | φ ( z ) | 2 ) 2 K ( g ( z , a ) ) dA(z)<,
    (18)

then C φ : Z 0 Q K , 0 is bounded.

Proof (i) Assume that C φ : Z 0 Q K , 0 is bounded. Then it is obvious that C φ : Z 0 Q K is bounded. By Theorem 1, (17) holds. Taking g(z)=z and using the boundedness of C φ : Z 0 Q K , 0 , we get φ Q K , 0 .

(ii) Suppose that φ Q K , 0 and (18) holds. From Theorem 1, we see that C φ : Z 0 Q K is bounded. To prove that C φ : Z 0 Q K , 0 is bounded, it suffices to prove that C φ f Q K , 0 for any f Z 0 . Let f Z 0 . By Lemma 4, for every ε>0, we can choose ρ(0,1) such that | f (w)|<εln e 1 | w | 2 for all wDρ D ¯ . Then by (8), we have

which together with the assumed conditions imply the desired result. □

Theorem 4 Let K be a nonnegative nondecreasing function on [0,). Assume that φ is an analytic self-map of D. Then the following statements holds:

  1. (i)

    If

    lim | a | 1 D | φ ( z ) | 2 ( ln e 1 | φ ( z ) | 2 ) 2 K ( g ( z , a ) ) dA(z)=0,
    (19)

then C φ :Z( Z 0 ) Q K , 0 is compact.

  1. (ii)

    If C φ :Z( Z 0 ) Q K , 0 is compact, then

    lim | a | 1 D | φ ( z ) | 2 ln 1 1 | φ ( z ) | 2 K ( g ( z , a ) ) dA(z)=0.
    (20)

Proof (i) Assume that (19) holds. Set

h φ , K (a)= D | φ ( z ) | 2 ( ln e 1 | φ ( z ) | 2 ) 2 K ( g ( z , a ) ) dA(z).

From the assumption, we have that for every ε>0, there is a s(0,1) such that for |a|>s, h φ , K (a)<ε. Similarly to the proof of Lemma 2.3 of [25], we see that h φ , K is continuous on |a|s, hence is bounded on |a|s. Therefore, h φ , K is bounded on D. From Theorem 1, we see that C φ :Z Q K is bounded.

For any fZ, by (8), we have

(21)

which together with (19) imply that C φ :Z Q K , 0 is bounded. Fix f B Z . The right-hand side of (21) tends to 0, as |a|1 by (19). From Lemma 3, we see that C φ :Z Q K , 0 is compact, and hence C φ : Z 0 Q K is compact.

(ii) From the assumption, we see that C φ : Z 0 Q K , 0 is bounded and C φ : Z 0 Q K is compact. From Theorems 2 and 3, we have φ Q K , 0 and

lim r 1 sup a D | φ ( z ) | > r | φ ( z ) | 2 ln 1 1 | φ ( z ) | 2 K ( g ( z , a ) ) dA(z)=0.
(22)

Hence, for any given ε>0, there exists a s(0,1) such that

sup a D | φ ( z ) | > s | φ ( z ) | 2 ln 1 1 | φ ( z ) | 2 K ( g ( z , a ) ) dA(z)<ε.
(23)

Therefore, by (23) and the fact that φ Q K , 0 , we have

By the arbitrary of ε, we get the desired result. The proof of the theorem is completed. □

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Acknowledgements

The authors are supported by the project of Department of Education of Guangdong Province (No. 2012KJCX0096), NNSF of China (No. 11001107).

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Fu, X., Li, S. Composition operators from Zygmund spaces into Q K spaces. J Inequal Appl 2013, 175 (2013). https://doi.org/10.1186/1029-242X-2013-175

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Keywords

  • Zygmund space
  • Q K space
  • composition operator