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Product of differentiation and composition operators on the logarithmic Bloch space

Abstract

We obtain a criterion for the boundedness and compactness of the products of differentiation and composition operators C φ D m on the logarithmic Bloch space in terms of the sequence { z n }. An estimate for the essential norm of C φ D m is given.

MSC:47B38, 30H30.

1 Introduction

Denote by H(D) the space of all analytic functions on the unit disk D={z:|z|<1} in the complex plane. Let H = H (D) denote the space of bounded analytic functions on A. An fH(D) is said to belong to the Bloch space if

f B = sup z D | f ( z ) | ( 1 | z | 2 ) <.

The logarithmic-Bloch space, denoted by LB, consists of all fH(D) satisfying

f log = sup z D ( 1 | z | ) | f ( z ) | log e 1 | z | <.

LB is a Banach space with the norm f LB =|f(0)|+ f log . It is well known that LB H is the space of multipliers of the Bloch space (see [1, 2]). For some results on logarithmic-type spaces and operators on them, see, for example, [310].

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

C φ (f)=fφ,fH(D).

The differentiation operator D is defined by Df= f , fH(D). For a nonnegative integer mN, we define

D m f= f ( m ) ,fH(D).

The product of differentiation and composition operators C φ D m is defined as follows:

C φ D m f= f ( m ) φ,fH(D).

A basic problem concerning concrete operators on various Banach spaces is to relate the operator theoretic properties of the operators to the function theoretic properties of their symbols, which attracted a lot of attention recently, the reader can refer to [437].

It is a well-known consequence of the Schwarz-Pick lemma that the composition operator is bounded on . See [2124, 27, 3335, 37] for the study of composition operators and weighted composition operators on the Bloch space. The product-type operators on or into Bloch type spaces have been studied in many papers recently; see [1220, 26, 2832, 34, 36] for example.

Let X and Y be two Banach spaces. Recall that a linear operator T:XY is said to be compact if it takes bounded sets in X to sets in Y which have compact closure. The essential norm of an operator T between X and Y is the distance to the compact operators K, that is, T e X Y =inf{TK:K is compact}, where is the operator norm. It is easy to see that T e X Y =0 if and only if T is compact. For some results in the topic, see, for example, [11, 20, 22, 24, 26, 28, 37].

In [34], Wu and Wulan obtained a characterization for the compactness of the product of differentiation and composition operators acting on the Bloch space as follows:

Theorem A Let φ be an analytic self-map of D, mN. Then C φ D m :BB is compact if and only if

lim n C φ D m ( z n ) B =0.

The purpose of the paper is to extend Theorem A to the case of LB. We will characterize the boundedness and compactness of C φ D m in terms of the sequence { z n }. Moreover, an estimate for the essential norm of C φ D m will be given. The main results are given in Sections 3 and 4.

In the paper, we say that a real sequence { a n } n N is asymptotic to another real sequence of { b n } n N and write ‘ a n b n ’ if and only if

lim n a n b n =1.

In addition, we say that AB if there exists a constant C such that ACB. The symbol AB means that ABA.

2 Auxiliary lemmas

In this section, we state and prove some auxiliary results which will be used to prove the main results in this paper.

Lemma 2.1 For m,nN, define the function H m , n :[0,1)[0,) by

H m , n (x)= n ! ( n m 1 ) ! x n m 1 ( 1 x ) m + 1 log e 1 x .
(2.1)

Then the following statements hold:

  1. (i)

    For n,mN and nm+1, there is a unique x m , n [0,1) such that H m , n ( x m , n ) is the absolute maximum of H m , n .

  2. (ii)
    lim n x m , n =1
    (2.2)

and

lim n [ n ( 1 x m , n ) ] =m+1.
(2.3)
  1. (iii)
    lim n max 0 < t < 1 H m , n ( t ) log ( n + 1 ) = ( m + 1 e ) m + 1 .
    (2.4)

Proof Directly computing we have

H m , n (x)= n ! ( n m 1 ) ! x n m 2 ( 1 x ) m ( ( n m 1 n x ) log e 1 x + x ) .

Define

g m , n (x)=(nm1nx)log e 1 x +x,x[0,1).
(2.5)

It is easy to see that g m , n is continuous on [0,1) and g m , n (0)=nm10, lim x 1 g m , n (x)=. Furthermore,

g m , n (x)=nlog e 1 x +n m + 1 1 x +1<0,x[0,1).

Then g m , n is decreasing on [0,1). When n=m+1, we get max 0 x < 1 H m , n (x)= H m , n (0). When n>m+1, the intermediate value theorem of continuous function gives the result that there exists a unique x m , n (0,1) such that g m , n ( x m , n )=0. So we have

max 0 < t < 1 H m , n (x)= H m , n ( x m , n ).

(i) has been proved. By (2.5), we have g m , n ( x m , n )=0. Thus

( n m 1 n x m , n ) log e 1 x m , n = x m , n n .

It follows from lim n x m , n n =0 and log e 1 x m , n 1 that (2.2) holds. Also, g m , n ( x m , n )=0 gives the result that

n m 1 n x m , n = x m , n n log e 1 x m , n .

So we have

n(1 x m , n )m1= x m , n log e 1 x m , n .

This gives the result (2.3). The proof of (ii) is complete.

Note that

nlog x m , n nlog [ 1 + ( x m , n 1 ) ] n( x m , n 1)m1as n.

This and (2.2) give

lim n x m , n n m 1 = e m 1 .
(2.6)

By (2.3) and (2.6) we obtain

lim n H m , n ( x m , n ) log ( n + 1 ) = lim n n ! x m , n n m 1 ( 1 x m , n ) m + 1 log e 1 x m , n ( n m 1 ) ! log ( n + 1 ) = e m 1 lim n n ! ( ( m + 1 ) / n ) m + 1 log e n m + 1 ( n m 1 ) ! log ( n + 1 ) = ( m + 1 e ) m + 1 ,

which shows that (iii) hold. The proof is complete. □

Lemma 2.2 Let m,nN and nm1>0. Let r m , n =(nm1)/n. Then H m , n is increasing on [ r m , n m , r m , n ] and

min r m , n m x r m , n H m , n (x)= H m , n ( r m , n m ) ( m + 1 e ) m + 1 log(n+1)as n.
(2.7)

Consequently,

min r m , n m x r m , n H m , n ( x ) z n LB = H m , n ( r m , n m ) z n LB ( m + 1 ) m + 1 e m as n.
(2.8)

Proof Since nm1>0, we have

H m , n ( r m , n )= n ! ( n m 1 ) ! ( n m 1 n ) n m 2 ( m + 1 n ) m ( n m 1 n ) >0.

By Lemma 2.1, we have r m , n < x m , n , where x m , n is given as in Lemma 2.1. Since H m , n (x)>0 for x(0, x m , n ), we see that H m , n is increasing on [ r m , n m , r m , n ]. Thus

min r m , n m x r m , n H m , n ( x ) = H m , n ( r m , n m ) = n ! ( n m 1 ) ! ( n 2 m 1 n m ) n m 1 ( m + 1 n m ) m + 1 log e ( n m ) m + 1 .

Applying the important limit lim n ( 1 + 1 n ) n =e we obtain the result that (2.7) holds.

By Lemma 2.1 we have

z n LB = sup | z | < 1 n | z | n 1 ( 1 | z | ) log e 1 | z | = H 0 , n ( x 0 , n ),
(2.9)

where x 0 , n is given in Lemma 2.1. By Lemma 2.1 we have

lim n H m , n ( r m , n m ) z n LB = lim n H m , n ( r m , n m ) log ( n + 1 ) log ( n + 1 ) z n LB = lim n H m , n ( r m , n m ) log ( n + 1 ) lim n log ( n + 1 ) z n LB = ( m + 1 ) m + 1 e m .

This gives (2.8). The proof is complete. □

Lemma 2.3 [3]

For mN. Then fLB if and only if

sup z D ( 1 | z | ) m | f ( m ) ( z ) | log e 1 | z | <.

Moreover,

f LB j = 0 m 1 | f ( j ) ( 0 ) | + sup z D ( 1 | z | ) m | f ( m ) ( z ) | log e 1 | z | .

3 The boundedness of C φ D m on LB

In this section, we will state the boundedness criterion for the operator C φ D m on LB. Since the boundedness of C φ D m on LB gives φLB, we may always assume that φLB. The main result of this section is stated as follows.

Theorem 3.1 Let mN and φ be an analytic self-map of D such that φLB. Then C φ D m is bounded on LB if and only if

sup n N C φ D m ( z n ) LB z n LB <.
(3.1)

Proof ) Assume that C φ D m is bounded on LB, that is, C φ D m LB LB <. Since the sequence { z n / z n LB } is bounded in the logarithmic Bloch space LB, we have

C φ D m ( z n ) LB z n LB C φ D m LB LB z n z n LB LB C φ D m LB LB <,

for any nN, from which the implication follows.

) We now assume that the condition (3.1) holds. On the one hand, for the case sup z D |φ(z)|<1, there is an r(0,1) such that |φ(z)|<r. By (3.1), for any given fLB, we have

C φ D m f LB = sup z D ( 1 | z | ) log e 1 | z | | f ( m + 1 ) ( φ ( z ) ) φ ( z ) | sup z D φ LB | f ( m + 1 ) ( φ ( z ) ) | ( 1 | φ ( z ) | ) m + 1 log e 1 | φ ( z ) | ( 1 | φ ( z ) | ) m + 1 log e 1 | φ ( z ) | sup z D φ LB f LB ( 1 r ) m + 1 ln e 1 r < .

The last estimate shows that the operator C φ is bounded on LB.

On the other hand, for the case sup z D |φ(z)|=1. Let N be the smallest positive integer such that D N is not empty, where

D n = { z D : r m , n m | φ ( z ) | r m , n }

and r m , n is given in Lemma 2.2. Note that H m , n (|φ(z)|)>0, when z D n , nN, by (2.8) we obtain

ϵ:= inf z D n H m , n ( | φ ( z ) | ) z n LB >0.

For any given fLB, by Lemma 2.3 we have

C φ D m f LB = sup z D ( 1 | z | ) log e 1 | z | | f ( m + 1 ) ( φ ( z ) ) φ ( z ) | = sup n N sup z D n ( 1 | z | ) log e 1 | z | | f ( m + 1 ) ( φ ( z ) ) φ ( z ) | = sup n N sup z D n ( 1 | z | ) log e 1 | z | | f ( m + 1 ) ( φ ( z ) ) φ ( z ) | z n LB H m , n ( | φ ( z ) | ) H m , n ( | φ ( z ) | ) z n LB f LB ϵ sup n N sup z D n n ! ( n m 1 ) ! ( 1 | z | ) log e 1 | z | | φ ( z ) | | φ ( z ) | n m 1 z n LB f LB ϵ sup n N C φ D m ( z n ) LB z n LB .

The proof is complete. □

4 The essential norm of C φ D m on LB

Denote K r f(z)=f(rz) for r(0,1). Then K r is a compact operator on the space LB. It is easy to see that K r 1. We denote by I the identity operator.

In order to give the lower and upper estimate for the essential norm of C φ D m on LB, we need the following result.

Lemma 4.1 There is a sequence { r k }, with 0< r k <1 tending to 1, such that the compact operator

L n = 1 n k = 1 n K r k

on LB satisfies:

  1. (i)

    for any t(0,1), lim n sup f LB t sup | z | t | ( ( I L n ) f ) (z)|=0,

(iia) lim n sup f LB 1 sup | z | < 1 |(I L n )f(z)|1,

(iib) lim n sup f LB 1 sup | z | < s |(I L n )f(z)|=0, for any s(0,1),

  1. (iii)

    lim sup n I L n 1.

Proof (i) follows from (iib) by Cauchy’s formula. The proof of (iii) is similar to the proof of Proposition 8 in [25]. Hence we omit it. Next we prove (iia) and (iib). The argument is much like that given in the proof of Proposition 2.1 of [25] or Lemmas 1 and 2 in [22]. For any 0<s<1, we choose an increasing sequence r k tending to 1 such that r k 1 1 s k 2 . For any given zD and r k , k=1,2,3, , there exists an s k ( r k ,1) such that

| f ( z ) f r k ( z ) | =z f ( s k z)(z r k z).
(4.1)

For any fLB with f LB 1, we have

| ( I L n ) f ( z ) | 1 n k = 1 n | f ( z ) f r k ( z ) | 1 n k = 1 n | f ( s k z ) | ( 1 r k ) 1 n k = 1 n 1 r k ( 1 | r k z | ) log e 1 | r k z | 1 n k = 1 n 1 = 1 .

Thus

lim sup n sup f LB 1 sup | z | < 1 | ( I L n ) f ( z ) | 1.

This shows that (iia) holds.

If |z|s, by the equality (4.1), we have

| ( I L n ) f ( z ) | 1 n k = 1 n 1 r k ( 1 | s z | ) log e 1 | s z | 1 n k = 1 n 1 r k ( 1 s ) = 1 n k = 1 n 1 k 2 π 2 6 n .

The above estimate gives (iib). The proof is complete. □

The following lemma can be proved in a standard way; see, for example Proposition 3.11 in [11].

Lemma 4.2 Let mN and φ be an analytic self-map of D. Then C φ D m is compact on LB if and only if C φ D m is bounded on LB and for any bounded sequence { f n } in LB which converges to zero uniformly on compact subsets of D, then C φ D m f n LB 0 as n.

Theorem 4.3 Let mN and φ be an analytic self-map of D. Suppose that C φ D m is bounded on LB. Then the estimate for the essential norm of C φ D m on LB is

C φ D m e LB LB lim sup n C φ D m ( z n ) LB z n LB .
(4.2)

Proof We first give the lower estimate for the essential norm. Without loss of generality, we assume that nm+1. Choose the sequence of function f n (z)= z n / z n LB , nN. Then f n LB =1, and { f n } converges to zero weakly on LB as n. Thus we have

lim n K f n LB =0

for any given compact operator K on LB. The basic inequality gives

C φ D m K LB LB ( C φ D m K ) f n LB C φ D m f n LB K f n LB .

Thus we obtain

C φ D m K LB LB lim sup n C φ D m f n LB lim sup n C φ D m f n LB .

So we have

C φ D m e LB LB = inf K C φ D m K lim sup n C φ D m ( z n ) LB z n LB .

Now we give the upper estimate for the essential norm. For the case of sup z D |φ(z)|<1, there is a number δ(0,1) such that sup z D |φ(z)|<δ. In this case, the operator C φ D m is compact on LB. In fact, choose a bounded sequence { f n } n N in LB which converges to zero uniformly on compact subset of D. From Cauchy’s integral formula, { f n ( m + 1 ) } converges to zero on compact subsets of D as n. It follows that

lim n C φ D m f n LB = lim n ( | f n ( m ) ( φ ( 0 ) ) | + C φ D m f n log ) = lim n sup z D ( 1 | z | ) log e 1 | z | | f n ( m + 1 ) ( φ ( z ) ) φ ( z ) | φ LB lim n sup z D | f n ( m + 1 ) ( φ ( z ) ) | = φ LB lim n sup | w | δ | f n ( m + 1 ) ( w ) | = 0 .

Then the operator C φ D m is compact on LB by Lemma 4.2. This gives

C φ D m e LB LB =0.
(4.3)

On the other hand, by Lemma 2.1 and (2.9) we obtain

z n LB = H 0 , n ( x 0 , n ) H 0 , n ( r 0 , n ) 1 2 log(en),

which implies that

lim sup n C φ D m ( z n ) LB z n LB e lim sup n sup z D ( 1 | z | ) log e 1 | z | n ! ( n m 1 ) ! | φ ( z ) | n m 1 | φ ( z ) | e φ LB lim n n m δ n m 1 = 0 .

Combining the last inequality with (4.3), we get the desired result.

Next, we assume that sup z D |φ(z)|=1. Let L n be the sequence of operators given in Lemma 4.1. Since L n is compact on LB and C φ D m is bounded on LB, then C φ D m L n is also compact on LB. Hence

C φ D m e LB LB lim sup n C φ D m C φ D m L n LB LB = lim sup n C φ D m ( I L n ) LB LB = lim sup n sup f LB 1 C φ D m ( I L n ) f LB = lim sup n sup f LB 1 ( ( I L n ) f ) ( m ) φ LB I 1 + I 2 ,

where

I 1 = lim sup n sup f LB 1 | ( ( I L n ) f ) ( m ) ( φ ( 0 ) ) |

and

I 2 = lim sup n sup f LB 1 sup z D | ( ( I L n ) f ) ( m + 1 ) ( φ ( z ) ) φ ( z ) | ( 1 | z | ) log e 1 | z | .

It follows from Lemma 4.1(iib) and Cauchy’s integral formula that I 1 =0.

For each positive integer nm+1, we define

D n = { z D : r m , n m | φ ( z ) | < r m , n } ,

where r m , n is given in Lemma 2.1. Let k be the smallest positive integer such that D k 0. Since sup z D |φ(z)|=1, D n is not empty for every integer nk and D= n = k D n , we have

sup f LB 1 sup z D | ( ( I L n ) f ) ( m + 1 ) ( φ ( z ) ) φ ( z ) | ( 1 | z | ) log e 1 | z | = I 21 + I 22 ,

where

I 21 = sup f LB 1 sup k i N 1 sup z D i | ( ( I L n ) f ) ( m + 1 ) ( φ ( z ) ) φ ( z ) | ( 1 | z | ) log e 1 | z |

and

I 22 = sup f LB 1 sup N i sup z D i | ( ( I L n ) f ) ( m + 1 ) ( φ ( z ) ) φ ( z ) | ( 1 | z | ) log e 1 | z | .

Here N is a positive integer determined as follows.

By (2.8),

lim i z i LB H m , i ( r m , i m ) = e m ( m + 1 ) m + 1 .

Hence, for any given ε>0, there exists an N such that

z i LB H m , i ( r m , i m ) e m ( m + 1 ) m + 1 +ε

when iN. For such N it follows that

I 22 = sup f LB 1 sup N i sup z D i | ( ( I L n ) f ) ( m + 1 ) ( φ ( z ) ) φ ( z ) | ( 1 | z | ) log e 1 | z | = sup f LB 1 sup N i sup z D i | ( ( I L n ) f ) ( m + 1 ) ( φ ( z ) ) φ ( z ) | ( 1 | z | ) log e 1 | z | H m , i ( | φ ( z ) | ) z i LB z i LB H m , i ( | φ ( z ) | ) ( e m ( m + 1 ) m + 1 + ε ) sup f LB 1 ( I L n ) f LB sup N i sup z D i | φ ( z ) | ( 1 | z | ) log e 1 | z | i ! ( i m 1 ) ! | φ ( z ) | i m 1 z i LB ( e m ( m + 1 ) m + 1 + ε ) I L n sup N i C φ D m ( z i ) LB z i LB .

Thus

lim sup n I 22 ( e m ( m + 1 ) m + 1 + ε ) sup N i C φ D m ( z i ) LB z i LB .
(4.4)

By (ii) of Lemma 4.1 and Cauchy’s integral formula, we have

lim sup n I 21 = lim sup n sup f LB 1 sup k i < N 1 sup z D i | ( ( I L n ) f ) ( m + 1 ) ( φ ( z ) ) | | φ ( z ) | ( 1 | z | ) log e 1 | z | φ LB lim sup n sup f LB 1 sup | φ ( z ) | < r m , N 1 | ( ( I L n ) f ) ( m + 1 ) ( φ ( z ) ) | = 0 ,

which together with (4.4) implies that

I 2 ( e m ( m + 1 ) m + 1 + ε ) sup N i C φ D m ( z i ) LB z i LB .
(4.5)

From (4.5) we obtain

C φ D m e LB LB I 1 + I 2 ( e m ( m + 1 ) m + 1 + ε ) sup N i C φ D m ( z i ) LB z i LB .

By the arbitrariness of ε, we get

C φ D m e LB LB e m ( m + 1 ) m + 1 lim sup n C φ D m ( z n ) LB z n LB .

The proof is complete. □

From Theorem 4.3, we obtain the following result.

Corollary 4.4 Let mN and φ be an analytic self-map of D such that C φ D m is bounded on LB. Then C φ D m is compact on LB if and only if

lim sup n C φ D m ( z n ) LB z n LB =0.

Especially, when m=0, from the proof of Theorem 4.3, we get the exact formula for essential norm of composition operator on LB.

Corollary 4.5 Let φ be an analytic self-map of D. Suppose that C φ is bounded on LB; then

C φ e LB LB = lim sup n φ n LB z n LB .

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Acknowledgements

The project is supported by Department of education of Anhui Province of China (No. KJ2013A101), the Natural Science Foundation of Guangdong (No. S2013010011978) and NSF of China (No. 11471143).

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Zhou, J., Zhu, X. Product of differentiation and composition operators on the logarithmic Bloch space. J Inequal Appl 2014, 453 (2014). https://doi.org/10.1186/1029-242X-2014-453

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