# Isometries of a Bergman-Privalov-Type Space on the Unit Ball

Authors: Stevo Stević; Sei-Ichiro Ueki
Journal: Discrete Dynamics in Nature and Society
 (2009)

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## Abstract

We introduce a new spaceANlog,α(𝔹) consisting of all holomorphic functions on the unit ball 𝔹⊂ℂn such that ∥f∥ANlog,α:=∫𝔹φe(ln(1+|f(z)|))dVα(z)<∞, where α>−1, dVα(z)=cα,n(1−|z|2)αdV(z) (dV(z) is the normalized Lebesgue volume measure on 𝔹, and cα,n is a normalization constant, that is, Vα(𝔹)=1), and φe(t)=tln(e+t) for t∈[0,∞). Some basic properties of this space are presented. Among other results we proved that ANlog,α(𝔹) with the metric d(f,g)=∥f−g∥ANlog,α is an F-algebra with respect to pointwise addition and multiplication. We also prove that every linear isometry T of ANlog,α(𝔹) into itself has the form Tf=c(f∘ψ) for some c∈ℂ such that |c|=1 and some ψ which is a holomorphic self-map of 𝔹 satisfying a measure-preserving property with respect to the measure dVα. As a consequence of this result we obtain a complete characterization of all linear bijective isometries of ANlog,α(𝔹).


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## Body

## 1. Introduction

Let𝔹 denote the open unit ball in the n-dimensional complex vector space ℂn, H(𝔹) the space of all holomorphic functions on 𝔹, dV(z) the normalized Lebesgue measure on 𝔹, dσ the normalized surface measure on the boundary ∂𝔹 of the unit ball, and dVα(z)=cα,n(1-|z|2)αdV(z), where α>-1 and cα,n is a normalization constant, that is, Vα(𝔹)=1. For each α>-1 we define the holomorphic function space ANlog,α(𝔹)=ANlog,α as follows:(1.1)ANlog,α(𝔹)={f∈H(𝔹)∣∥f∥ANlog,α:=∫𝔹φe(ln(1+|f(z)|))dVα(z)<∞},
where φe(t)=tln(e+t) for t∈[0,∞).Since(1.2)φe′(t)=ln(e+t)+te+t≥1,φe′′(t)=2e+t(e+t)2>0,t∈[0,∞),
it follows that φe(t) is an increasing convex function which is obviously nonnegative.Lets(t)=φe(ln(1+t)), then(1.3)s′′(t)=φe′′(ln(1+t))-φs′(ln(1+t))(1+t)2.
Let F(u)=φs′(u)-φe′′(u), then F(0)=1-(2/e), F′(u)>0, so that φe′′(u)≤φs′(u) for u≥0. Thus s(t) is a nonnegative concave function on the interval [0,∞), so that(1.4)s(x+y)-s(x)y≤s(y)-s(0)y
for x≥0 and y>0, and consequently s(x+y)≤s(x)+s(y),x,y≥0, which implies the following inequality:(1.5)φe(ln(1+x+y))≤φe(ln(1+x))+φe(ln(1+y))
for all x,y∈[0,∞). It is easy to see that ∥cf∥ANlog,α need not be equal to |c|∥f∥ANlog,α, for every c∈ℂ and f∈ANlog,α(𝔹). These facts imply that ∥·∥ANlog,α is not a norm on ANlog,α(𝔹) but satisfies the triangle inequality(1.6)∥f+g∥ANlog,α≤∥f∥ANlog,α+∥g∥ANlog,α.
Furthermore if we define d(f,g)=∥f-g∥ANlog,α for any f,g∈ANlog,α(𝔹), then we see that ANlog,α(𝔹) is a metric space with respect to d(f,g).LetX be a space of all holomorphic functions on some domain and T a linear isometry of X into X [1]. When X is the Hardy space Hp(𝔹)(0<p<∞,p≠2), Forelli [2, 3] and Rudin [4] have determined the injective and/or surjective isometries of Hp(𝔹). For the case when X is the weighted Bergman spaces Aαp(𝔹)(0<p<∞,p≠2), the isometries were completely characterized in a sequence of papers by Kolaski [5–7]. By these works we see that the isometries on these holomorphic function spaces are described as weighted composition operators, which is one of the reasons why these operators have been investigated so much recently in the settings of the unit ball or the unit polydisk (see, e.g., monograph [8], recent papers [9–19], and references therein). See also paper [20] for integral-type operators closely related with weighted composition operators. The case when X is not a Banach space has also been studied by many authors. The Smirnov classN* and the Privalov spaceNp(1<p<∞) which are contained in the Nevanlinna classN are examples of such spaces. These type of spaces are F-spaces with respect to a suitable metric on them. For properties of these space, we can refer [21]. Stephenson [22], Iida and Mochizuki [23] and Subbotin [24–26] have studied linear isometries on these spaces. Their works showed that the injective isometries are weighted composition operators induced by some inner functions and inner maps of 𝔹 whose radial limit maps satisfy a measure-preserving property. Recently Matsugu and the second author of the present paper have studied the weighted Bergman-Privalov spaceANαp(𝔹)(1≤p<∞) and characterized the isometries of this space in [27]. They showed that in this case, T has the form of a constant multiple composition operator which satisfies a measure-preserving property with respect to the measure dVα.Motivated by paper [28], in this paper, we investigate the space ANlog,α(𝔹). Some basic properties of the space ANlog,α(𝔹) are presented in Section 2; among other results it was proved that ANlog,α(𝔹) with the metric d(f,g)=∥f-g∥ANlog,α is an F-algebra with respect to pointwise addition and multiplication. Also an estimate for the point evaluation functional on ANlog,α(𝔹) is given. In Section 3, we will prove that every linear isometry T of ANlog,α(𝔹) has the form Tf=c(f∘ψ), where c∈ℂ such that |c|=1 and ψ is a holomorphic self-map of 𝔹 satisfying a measure-preserving property with respect to the measure dVα. As a consequence of this result we show that every surjective isometry T of ANlog,α(𝔹) is of the form Tf=c(f∘U) for any f∈ANlog,α(𝔹), where c∈ℂ with |c|=1 and U is a unitary operator on ℂn.


## 2. Basic Properties ofANlog,α(𝔹)

This section is devoted to collecting fundamental results onANlog,α(𝔹) which will be used in the proofs of the main results.Recall that theweighted Bergman spaceAα1(𝔹) is defined as follows:(2.1)Aα1(𝔹)={f∈H(𝔹)∣∥f∥Aα1:=∫𝔹|f(z)|dVα(z)<∞}.First we prove an elementary inequality, which has the main role in determining the relationship betweenANlog,α(𝔹) space and the weighted Bergman space Aα1(𝔹).Lemma 2.1.
The following inequality holds:(2.2)φe(ln(1+x))≤x,x∈[0,∞).Proof.
Note that inequality (2.2) is transformed into
(2.3)et-1-φe(t)≥0,t∈[0,∞)
by the change x=et-1.
LetF(t)=et-1-φe(t). Then F(0)=0 and
(2.4)F′(t)=et-ln(e+t)-te+t,
which implies F′(0)=0 and
(2.5)F′′(t)=et-1e+t-e(e+t)2≥1-2e>0.
From all these relations it follows that inequality (2.3) holds and consequently inequality (2.2).From Lemma2.1 we obtain the following corollary.Corollary 2.2.
For eachα>-1, Aα1(𝔹)⊂ANlog,α(𝔹).Proof.
From Lemma2.1 it follows that
(2.6)φe(ln(1+|f(z)|))≤|f(z)|
for every f∈Aα1(𝔹) and z∈𝔹. Multiplying this inequality by dVα(z), then integrating such obtained inequality over 𝔹 we obtain
(2.7)∥f∥ANlog,α≤∥f∥Aα1,
from which the result follows.Remark 2.3.
Note that from inequality (2.7) it follows that the inclusion i:Aα1(𝔹)→ANlog,α(𝔹) has “norm" less than one, if we define operator norm as usual by
(2.8)∥i∥Aα1→ANlog,α=sup∥f∥Aα1≤1∥i(f)∥ANlog,α.Lemmas2.5 and 2.7 are based on the following lemma.Lemma 2.4.
LetΨ(x)=ln(1+ex)ln{e+ln(1+ex)}(x∈(-∞,∞)). Then Ψ is a positive continuous, increasing, and convex function on (-∞,∞).Proof.
It is clear thatΨ is a positive and continuous function on the interval (-∞,∞). Now we prove that it is increasing and convex on (-∞,∞). Note that Ψ(x)=φe(ln(1+ex)). Hence
(2.9)Ψ′(x)=φe′(ln(1+ex))ex1+ex,Ψ′′(x)=φe′′(ln(1+ex))(ex1+ex)2+φ′e(ln(1+ex))ex(1+ex)2.
From (1.2) and (2.9), and since ln(1+ex)∈(0,∞), when x∈(-∞,∞), the lemma follows.Lemma 2.5.
Letα>-1. If f∈ANlog,α(𝔹), then it holds that
(2.10)limr→1-∥fr-f∥ANlog,α=0,
where fr(z)=f(rz) for each r∈(0,1) and z∈𝔹.Proof.
Take anf∈ANlog,α(𝔹) and ε>0. Then we can choose an r0∈(0,1) such that
(2.11)∫𝔹∖r0𝔹¯φe(ln(1+|f(z)|))dVα(z)<ε3,
where r0𝔹¯={z∈𝔹:|z|≤r0}. Since φe(ln(1+|f(z)|))=Ψ(ln|f(z)|), from Lemma 2.4 it follows that φe(ln(1+|f(z)|)) is a positive plurisubharmonic function in 𝔹. Hence we have
(2.12)∫∂𝔹φe(ln(1+|f(rtζ)|))dσ(ζ)≤∫∂𝔹φe(ln(1+|f(tζ)|))dσ(ζ)
for any r,t∈(0,1). This inequality implies that
(2.13)∫𝔹∖r0𝔹¯φe(ln(1+|fr(z)|))dVα(z)≤∫𝔹∖r0𝔹¯φe(ln(1+|f(z)|))dVα(z)<ε3
for any r∈(0,1).
Now we choose anε0>0 such that φe(ln(1+ε0))=ε/3. By the continuity of f on the compact subset r0𝔹¯, we see that there exists a δ∈(0,1) such that if z,w∈r0𝔹¯ with |z-w|<δ, then
(2.14)|f(z)-f(w)|<ε0.
Setr1=1-δ. If r1<r<1, then
(2.15)∫r0𝔹¯φe(ln(1+|fr(z)-f(z)|))dVα(z)≤φe(ln(1+ε0))=ε3.
By (2.13), (2.15), and (1.5), we obtain
(2.16)∥fr-f∥ANlog,α=(∫r0𝔹¯+∫𝔹∖r0𝔹¯)φe(ln(1+|fr(z)-f(z)|))dVα(z)≤∫r0𝔹¯φe(ln(1+|fr(z)-f(z)|))dVα(z)+∫𝔹∖r0𝔹¯{φe(ln(1+|fr(z)|))+φe(ln(1+|f(z)|))}dVα(z)<ε3+ε3+ε3=ε.
This completes the proof.Corollary 2.6.
For eachα>-1 the space ANlog,α(𝔹) is separable.Proof.
Since the dilated functionfr is approximated by the nth partial sum of its Taylor expansion uniformly on 𝔹, Lemma 2.5 implies that polynomials are dense in ANlog,α(𝔹). Since the polynomials with rational coefficients approximate any polynomial on the close unit ball 𝔹¯, the corollary follows.Lemma 2.7.
Letα>-1, f∈H(𝔹) and z∈𝔹. Then it holds that
(2.17)ln(1+|f(z)|)ln{e+ln(1+|f(z)|)}≤(1+|z|1-|z|)n+α+1∥f∥ANlog,α.
Moreover, if f∈ANlog,α(𝔹), then f satisfies
(2.18)ln(1+|f(z)|)ln{e+ln(1+|f(z)|)}=o((1-|z|)-(n+α+1))(as|z|→1-).Proof.
Fixf∈ANlog,α(𝔹) and z∈𝔹. Let φz be the biholomorphic involution of 𝔹 described in [29, page 25]. Since φe(ln(1+|f∘φz|)) is a positive plurisubharmonic function in 𝔹 by Lemma 2.4, we have
(2.19)φe(ln(1+|f(z)|))=φe(ln(1+|f∘φz(0)|))=2ncα,n∫01r2n-1(1-r2)αφe(ln(1+|f(φz(0))|))dr≤2ncα,n∫01r2n-1(1-r2)αdr∫∂𝔹φe(ln(1+|f(φz(rζ))|))dσ(ζ)=cα,n∫𝔹φe(ln(1+|f(φz(w))|))(1-|w|2)αdV(w)=cα,n∫𝔹φe(ln(1+|f(w)|))(Jℝφz)(w)(1-|φz(w)|2)αdV(w),
where (Jℝφz)(w) denotes the real Jacobian of φz at w. By [29, Theorem 2.2.2 and 2.2.6], for w∈𝔹 we have
(2.20)(Jℝφz)(w)(1-|φz(w)|2)α≤(1+|z|1-|z|)n+α+1(1-|w|2)α.
By (2.19) and (2.20), we obtain
(2.21)φe(ln(1+|f(z)|))≤(1+|z|1-|z|)n+α+1∥f∥ANlog,α,
which completes the proof of the first claim.
Next we prove the second claim. Fix anε>0. By Lemma 2.5, there exists an r∈(0,1) such that ∥fr-f∥ANlog,α<ε. From, (1.5) and (2.21) applied to the function fr-f we have that
(2.22)φe(ln(1+|f(z)|))(1-|z|)n+α+1≤2n+α+1∥fr-f∥ANlog,α+(1-|z|)n+α+1Mr<2n+α+1ε+(1-|z|)n+α+1Mr,
where Mr=maxw∈r𝔹¯φe(ln(1+|f(w)|))<∞. Letting |z|→1- in (12) and using the fact that ε is an arbitrary positive number we obtain
(2.23)φe(ln(1+|f(z)|))(1-|z|)n+α+1→0,as|z|→1-.
This completes the proof of this lemma.Lemma 2.8.
ANlog,α(𝔹) is a complete metric space.Proof.
In the introduction we have seen thatANlog,α(𝔹) is a metric space. Since the convergence in ANlog,α(𝔹) implies the uniform convergence on compact subsets of 𝔹 by Lemma 2.7, a usual normal family argument along with Fatou's lemma shows that every Cauchy sequence in ANlog,α(𝔹) converges to an element of ANlog,α(𝔹). Hence the space ANlog,α(𝔹) equipped with the metric d(f,g)=∥f-g∥ANlog,α is a complete metric space.Recall that a metric space(X,d) is called an F-space if it is a complete and(i)
d(x,y)=d(x-y,0) for every x,y∈X;(ii)
for each sequence{xk}⊂X such that d(xk,0)→0 as k→∞ it follows that d(cxk,0)→0 for every c∈ℂ;(iii)
for each sequence{ck}⊂ℂ such that ck→0 as k→∞, d(ckx,0)→0 for each x∈X.If in anF-space (X,d) is introduced an operation of pointwise multiplication such that X becomes an algebra and the operation of multiplication is continuous in the metric d, then the F-space (X,d) is called an  F-algebra.Before we prove our next result we will prove two technical lemmas.Lemma 2.9.
The following inequality holds:(2.24)φe(ln(1+cx))≤max{1,c2}φe(ln(1+x))
for x,c≥0.Proof.
It is enough to prove the casec>1. Since φe is an increasing function and ln(1+cx)≤cln(1+x), we have
(2.25)φe(ln(1+cx))≤φe(cln(1+x))=cln(1+x)ln(e+cln(1+x)).
Let
(2.26)η(t)=(e+t)c-(e+ct),fort∈[0,∞).
Then from
(2.27)η′(t)=c((e+t)c-1-1),
we see that η is an increasing function on the interval [0,∞), so that η(t)≥η(0)=ec-e>0. Hence (e+ct)<(e+t)c. This implies that
(2.28)ln(e+ct)<cln(e+t).
So we obtain
(2.29)φe(ln(1+cx))≤c2ln(1+x)ln(e+ln(1+x))=c2φe(ln(1+x)),
completing the proof of the lemma.The next lemma improves inequality (19) in [28].Lemma 2.10.
The following inequality holds:(2.30)φe(x+y)≤2(φe(x)+φe(y))
for x,y≥0.Proof.
First note that the function(2.31)f(x,y)=2(φe(x)+φe(y))-φe(x+y)
satisfies the following condition: f(x,y)=f(y,x). Hence it is enough to prove inequality (2.30) for the case x≥y≥0. Note that f(0,y)=φe(y)≥0 and
(2.32)∂f∂x(x,y)=2(ln(e+x)+xe+x)-ln(e+x+y)-x+ye+x+y≥2(ln(e+x)+xe+x)-ln(e+2x)-2xe+2x≥ln(e+x)2-ln(e+2x)>0.
Hence, for each fixed y we have that f(x,y)≥f(0,y)≥0, from which inequality (2.30) follows.Theorem 2.11.
ANlog,α(𝔹) is an F-algebra with respect to pointwise addition and multiplication.Proof.
By Lemma2.8  ANlog,α(𝔹) is a complete metric space satisfying the condition d(f,g)=d(f-g,0). From Lemma 2.9 we have the inequality
(2.33)φe(ln(1+st))≤max{1,s2}φe(ln(1+t)),t,s≥0,
which implies that
(2.34)∥cf∥ANlog,α≤max{1,|c|2}∥f∥ANlog,α
for each c∈ℂ, so that the operation of the multiplication by complex numbers is closed in ANlog,α(𝔹). On the other hand, by inequality (1.5) it follows that the pointwise addition is also closed in ANlog,α(𝔹). From (2.34) it follows that the condition d(fk,0)→0 implies d(cfk,0)→0 for every c∈ℂ, since
(2.35)d(cfk,0)=∥cfk∥ANlog,α≤max{1,|c|2}∥fk∥ANlog,α=max{1,|c|2}d(fk,0).
Now assume that {ck}k∈ℕ⊂ℂ is a sequence such that ck→0 as k→∞. Note that for each such sequence there is k0∈ℕ such that |ck|≤1 for k≥k0. Hence we have that for each f∈ANlog,α(𝔹) and z∈𝔹(2.36)φe(ln(1+|ckf(z)|))≤φe(ln(1+|f(z)|)),k≥k0.
From (2.36), since
(2.37)φe(ln(1+|ckf(z)|))→0ask→∞
for each z∈𝔹, and by the Lebesgue dominated convergence theorem, we get d(ckf,0)=∥ckf∥ANlog,α→0 as k→∞, which proves (iii) and that ANlog,α(𝔹) is an F-space.
Now we prove thatANlog,α(𝔹) is closed with respect to the pointwise multiplication. From Lemma 2.10 we have that
(2.38)φe(t+s)≤2(φe(t)+φe(s)),t,s≥0,
from which it follows that
(2.39)∥fg∥ANlog,α≤2(∥f∥ANlog,α+∥g∥ANlog,α),
so that ANlog,α(𝔹) is closed with respect to the pointwise multiplication.
Finally, we prove that the pointwise multiplication is continuous with respect to the metricd, that is, that from d(fk,f)→0 as k→∞ and d(gl,g)→0 as l→∞ it follows that d(fkgl,fg)→0 as k,l→∞. Note that from triangle inequality and by (2.39) it follows that
(2.40)∥fkgl-fg∥ANlog,α≤∥(fk-f)(gl-g)∥ANlog,α+∥(fk-f)g∥ANlog,α+∥f(gl-g)∥ANlog,α≤2(∥fk-f∥ANlog,α+∥gl-g∥ANlog,α)+∥(fk-f)g∥ANlog,α+∥f(gl-g)∥ANlog,α.
From (2.40) and the symmetry ∥fg∥ANlog,α=∥gf∥ANlog,α we see that it is enough to prove that from d(fk,f)→0 as k→∞, for each g∈ANlog,α(𝔹), it follows that d(fkg,fg)→0 as k→∞. Fix a g∈ANlog,α(𝔹) and put Es={z∈𝔹:|g(z)|≥s} for each s∈[0,∞). Hence we see that for every δ>0 there is an s0>0 such that for every s≥s0(2.41)Vα(Es)≤1φe(ln(1+s))∫𝔹φe(ln(1+|g(z)|))dVα(z)<δ.
Fix an ε>0. Since φe(ln(1+|g|))∈L1(dVα), we have that for every ε>0 there is a δ1>0 such that
(2.42)∫Eφe(ln(1+|g(z)|))dVα(z)<ε
for every set E such that Vα(E)<δ1. From (2.41) with δ=δ1 we have that Vα(Es)<δ for sufficiently large s. From this and inequalities (2.34) and (2.39) we have that
(2.43)∥g(fk-f)∥ANlog,α=∫𝔹φe(ln(1+|g(z)(fk(z)-f(z))|))dVα(z)=(∫𝔹∖Es+∫Es)φe(ln(1+|g(z)(fk(z)-f(z))|))dVα(z)≤2∫Esφe(ln(1+|g(z)|))dVα(z)+2∫Esφe(ln(1+|fk(z)-f(z)|))dVα(z)+∫𝔹∖Esφe(ln(1+s|fk(z)-f(z)|))dVα(z)≤2ε+max{3,2+s2}∥fk-f∥ANlog,α.
Letting k→∞ in (2.43) and since ε is an arbitrary positive number, it follows that ∥g(fk-f)∥ANlog,α→0 as k→∞, finishing the proof of the theorem.


## 3. Linear Isometries ofANlog,α(𝔹)

In this section, we investigate linear isometries ofANlog,α(𝔹) into ANlog,α(𝔹), by modifying ideas from some earlier papers in this area (see, e.g., [23, 27, 28]).The following two lemmas play an important role in the proofs of the main results in this section.Lemma 3.1.
IfT is a linear isometry of ANlog,α(𝔹) into itself, then the restriction of T to Aα1(𝔹) is also a linear isometry of Aα1(𝔹) into itself.Proof.
Take anf∈Aα1(𝔹) and put g=Tf. For each m∈ℕ we have g/m=T(f/m), and so the assumption T which is an isometry of ANlog,α(𝔹) gives
(3.1)∫𝔹φe(ln(1+|f(z)|m))dVα(z)=∫𝔹φe(ln(1+|g(z)|m))dVα(z).
By inequality (2.2), we have
(3.2)mφe(ln(1+|f(z)|m))≤|f(z)|
for any m∈ℕ and z∈𝔹. Also, it is easy to see that
(3.3)limm→∞mφe(ln(1+|f(z)|m))=|f(z)|
for each z∈𝔹. Hence the Lebesgue-dominated convergence theorem gives
(3.4)limm→∞∫𝔹mφe(ln(1+|f(z)|m))dVα(z)=∫𝔹|f(z)|dVα(z).
On the other hand, Fatou's lemma (3.1) and (3.4) show that
(3.5)∫𝔹|g(z)|dVα(z)≤lim infm→∞∫𝔹mφe(ln(1+|g(z)|m))dVα(z)=lim infm→∞∫𝔹mφe(ln(1+|f(z)|m))dVα(z)=∫𝔹|f(z)|dVα(z),
and so g∈Aα1(𝔹). By applying the Lebesgue-dominated convergence theorem once more again, we have
(3.6)limm→∞∫𝔹mφe(ln(1+|g(z)|m))dVα(z)=∫𝔹|g(z)|dVα(z).
By (3.1), (3.4), and (3.6), we see that T is a linear isometry of Aα1(𝔹) into Aα1(𝔹).Lemma 3.2.
There exists a bounded continuous functionθ(x) on [0,∞) such that
(3.7)(12-1e)x2-x3θ(x)=x-φe(ln(1+x))≥0x∈[0,∞).Proof.
Set(3.8)θ(x):=φe(ln(1+x))-x-(-1/2+1/e)x2x3
for x∈(0,∞). Since θ(x) is a continuous function on (0,∞) such that θ(x)→0 as x→+∞, it is enough to prove that θ(x) has a finite limit as x→0+. By the application of Taylor's theorem to φe(ln(1+x)), we have
(3.9)φe(ln(1+x))=x+(-12+1e)x2+(13-1e-12e2)x3+R4(x),
where R4(x) denotes the remainder term of order 4. Since R4(x)/x3→0 as x→0+, we obtain
(3.10)limx→0+θ(x)=13-1e-12e2,
which completes the proof.Theorem 3.3.
Every linear isometryT of ANlog,α(𝔹) into itself has the form Tf=c(f∘ψ) for any f∈ANlog,α(𝔹), where c∈ℂ with |c|=1 and ψ:𝔹→𝔹 is a holomorphic map which satisfies the condition
(3.11)∫𝔹h(ψ(z))dVα(z)=∫𝔹h(z)dVα(z)
for every bounded or positive Borel function h in 𝔹.Proof.
First suppose thatT:ANlog,α(𝔹)→ANlog,α(𝔹) is a linear isometry. By Lemma 3.1, the restriction of T to Aα1(𝔹) is a linear isometry of Aα1(𝔹) into Aα1(𝔹). Hence Kolaski's theorem [6, page 911, Theorem 2.11] implies that T has the following form:
(3.12)Tf(z)=g(z)f(ψ(z))(f∈Aα1(𝔹),z∈𝔹),
where g=T1 and ψ is a holomorphic self-map of 𝔹 such that
(3.13)∫𝔹h(ψ(z))|g(z)|dVα(z)=∫𝔹h(z)dVα(z)
for every bounded or positive Borel function h in 𝔹. Fix an f∈ANlog,α(𝔹). Since {fr}0<r<1⊂Aα1(𝔹), the representation (3.12) and Lemmas 2.5 and 2.7 give
(3.14)Tf(z)=limr→1-Tfr(z)=limr→1-g(z)fr(ψ(z))=g(z)f(ψ(z))
for all z∈𝔹. Since g=T1∈Aα1(𝔹) and Vα(𝔹)=1, Hölder's inequality gives
(3.15)1=∥1∥Aα1=∥g∥Aα1≤∥g∥Aα2,
where ∥g∥Aα22=∫𝔹|g|2dVα. On the other hand, from Lemma 3.2 it follows that
(3.16)∫𝔹{(12-1e)|tg(z)|2-|tg(z)|3θ(|tg(z)|)}dVα(z)=∫𝔹{|tg(z)|-φe(ln(1+|tg(z)|))}dVα(z)=∫𝔹|T(t)(z)|dVα(z)-∫𝔹φe(ln(1+|T(t)(z)|))dVα(z)=∫𝔹tdVα(z)-∫𝔹φe(ln(1+|t|))dVα(z)=(12-1e)t2-t3θ(t).
By Lemma 3.2 and Fatou's lemma we have
(3.17)∫𝔹(12-1e)|g(z)|2dVα(z)≤lim inft→0+∫𝔹{(12-1e)|g(z)|2-t|g(z)|3θ(|tg(z)|)}dVα(z)=lim inft→0+{(12-1e)-tθ(t)}=12-1e,
and so ∥g∥Aα2≤1. Combining this with (3.15), we obtain
(3.18)∥g∥Aα1=∥g∥Aα2=1.
This implies that g≡c in 𝔹 for some c∈ℂ with |c|=1. Hence (3.13) and (3.14) show that Tf=c(f∘ψ) for any f∈ANlog,α(𝔹) and ψ satisfies the condition in (3.11).
Conversely, for some complex numberc with |c|=1 and holomorphic self-map ψ of 𝔹 which satisfies (3.11) we define an operator T on ANlog,α(𝔹) by Tf=c(f∘ψ) for f∈ANlog,α(𝔹). Since φe(ln(1+|f|)) is a positive Borel function in 𝔹, condition (3.11) implies that T is a linear isometry of ANlog,α(𝔹) into ANlog,α(𝔹). This completes the proof.Corollary 3.4.
Every surjective isometryT of ANlog,α(𝔹) is of the form Tf=c(f∘U) for any f∈ANlog,α(𝔹), where c∈ℂ with |c|=1 and U is a unitary operator on ℂn.Proof.
Assume thatT:ANlog,α(𝔹)→ANlog,α(𝔹) is a surjective isometry. Then Theorem 3.3 implies that T is written in the form Tf=cT(f∘ψT) where cT∈ℂ with |cT|=1 and ψT is a holomorphic self-map of 𝔹 which satisfies (3.11). Since T-1 is also a surjective isometry of ANlog,α(𝔹), we see that
(3.19)f=T-1(T(f))=T-1(cT(f∘ψT))=cT-1cT(f∘ψT∘ψT-1),
so that cT-1=cT-1 and ψT-1=ψT-1, that is, ψT is an automorphism of 𝔹.
We prove thatψ:=ψT fixes the origin. Let ψj(1≤j≤n) be the components of ψ. For each j∈{1,…,n} and an r∈(0,1) we have
(3.20)∫∂𝔹ψj(rζ)dσ(ζ)=∫∂𝔹dσ(ζ)12π∫-ππψj(reiθζ)dθ.
Since the slice function (ψj)ζ of ψj is holomorphic in the open unit disc of ℂ, the mean value theorem gives (1/2π)∫-ππψj(reiθζ)dθ=ψj(0) for each ζ∈∂𝔹 and j∈{1,…,n}. Using this fact in formula (3.20), multiplying such obtained equality by 2ncα,n(1-r2)αr2n-1dr, then integrating it from 0 to 1 and using the polar coordinates on the unit ball we get
(3.21)∫𝔹ψj(z)dVα(z)=ψj(0)
for each j∈{1,…,n}. Now we fix a j∈{1,…,n}. By applying the condition (3.11) to a bounded Borel function h(w)=〈w,ej〉, j∈{1,…,n}, where ej is the standard orthonormal base vector in ℂn, we have
(3.22)∫𝔹ψj(z)dVα(z)=∫𝔹〈ψ(z),ej〉dVα(z)=∫𝔹〈z,ej〉dVα(z)=2ncα,n∫01r2n(1-r2)αdr∫∂𝔹〈ζ,ej〉dσ(ζ).
From [29, page 15, § 1.4.5(2)] we have that
(3.23)∫∂𝔹〈ζ,ej〉dσ(ζ)=n-1π∫-ππ∫01(1-t2)n-2t2eiθdtdθ=0.
Hence (3.21), (3.22), and (3.23) show that ψj(0)=0 for each j∈{1,…,n}, and so ψ fixes the origin. By a well-known theorem [29] we see that ψ is a unitary operator on ℂn. This completes the proof.Remark 3.5.
LetIsomANlog,α(𝔹) denote the set of all surjective isometries of ANlog,α(𝔹). Since the transformations ψ in Corollary 3.4 form a group of unitary transformations the corollary implies the following isomorphism:
(3.24)IsomANlog,α(𝔹)≃𝕋×U(n,ℂ).
Here 𝕋 denotes the set of all complex numbers with |c|=1 and U(n,ℂ) being the group of all unitary operators on ℂn. For the Smirnov class N* and the Privalov space Np(p>1), analogous results are obtained by Stephenson and Subbotin (see [22, 24–26]). So we see that every surjective isometry of ANlog,α(𝔹) has the same form as one of the Smirnov class or the Privalov space.



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