Isometries of a Bergman-Privalov-Type Space on the Unit Ball
We introduce a new space consisting of all holomorphic functions on the unit ball such that , where , ( is the normalized Lebesgue volume measure on , and is a normalization constant, that is, ), and for . Some basic properties of this space are presented. Among other results we proved that with the metric is an -algebra with respect to pointwise addition and multiplication. We also prove that every linear isometry of into itself has the form for some such that and some which is a holomorphic self-map of satisfying a measure-preserving property with respect to the measure . As a consequence of this result we obtain a complete characterization of all linear bijective isometries of .
Let denote the open unit ball in the -dimensional complex vector space , the space of all holomorphic functions on , the normalized Lebesgue measure on , the normalized surface measure on the boundary of the unit ball, and , where and is a normalization constant, that is, . For each we define the holomorphic function space as follows:
where for .
it follows that is an increasing convex function which is obviously nonnegative.
Let , then , , so that for . Thus is a nonnegative concave function on the interval , so that
for and , and consequently , which implies the following inequality:
for all . It is easy to see that need not be equal to for every and These facts imply that is not a norm on but satisfies the triangle inequality
Furthermore if we define for any , then we see that is a metric space with respect to .
Let be a space of all holomorphic functions on some domain and a linear isometry of into . When is the Hardy space , Forelli [2, 3] and Rudin  have determined the injective and/or surjective isometries of . For the case when is the weighted Bergman spaces , 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 , recent papers [9–19], and references therein). See also paper  for integral-type operators closely related with weighted composition operators. The case when is not a Banach space has also been studied by many authors. The Smirnov class and the Privalov space which are contained in the Nevanlinna class are examples of such spaces. These type of spaces are -spaces with respect to a suitable metric on them. For properties of these space, we can refer . Stephenson , Iida and Mochizuki  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 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 space and characterized the isometries of this space in . They showed that in this case, has the form of a constant multiple composition operator which satisfies a measure-preserving property with respect to the measure .
Motivated by paper , in this paper, we investigate the space . Some basic properties of the space are presented in Section 2; among other results it was proved that with the metric is an -algebra with respect to pointwise addition and multiplication. Also an estimate for the point evaluation functional on is given. In Section 3, we will prove that every linear isometry of has the form , where such that and is a holomorphic self-map of satisfying a measure-preserving property with respect to the measure As a consequence of this result we show that every surjective isometry of is of the form for any , where with and is a unitary operator on .
2. Basic Properties of
This section is devoted to collecting fundamental results on which will be used in the proofs of the main results.
Recall that the weighted Bergman space is defined as follows:
First we prove an elementary inequality, which has the main role in determining the relationship between space and the weighted Bergman space
Lemma 2.1. The following inequality holds:
From Lemma 2.1 we obtain the following corollary.
Corollary 2.2. For each , .
Proof. From Lemma 2.1 it follows that for every and Multiplying this inequality by , then integrating such obtained inequality over we obtain from which the result follows.
Remark 2.3. Note that from inequality (2.7) it follows that the inclusion has “norm" less than one, if we define operator norm as usual by
Lemma 2.4. Let . Then is a positive continuous, increasing, and convex function on .
Proof. It is clear that is a positive and continuous function on the interval . Now we prove that it is increasing and convex on . Note that . Hence From (1.2) and (2.9), and since , when , the lemma follows.
Lemma 2.5. Let . If , then it holds that where for each and .
Proof. Take an and . Then we can choose an such that
where . Since , from Lemma 2.4 it follows that is a positive plurisubharmonic function in . Hence we have
for any . This inequality implies that
for any .
Now we choose an such that . By the continuity of on the compact subset , we see that there exists a such that if with , then
Set . If , then By (2.13), (2.15), and (1.5), we obtain This completes the proof.
Corollary 2.6. For each the space is separable.
Proof. Since the dilated function is approximated by the th partial sum of its Taylor expansion uniformly on , Lemma 2.5 implies that polynomials are dense in . Since the polynomials with rational coefficients approximate any polynomial on the close unit ball , the corollary follows.
Lemma 2.7. Let , and . Then it holds that Moreover, if , then satisfies
Proof. Fix and . Let be the biholomorphic involution of described in [29, page 25]. Since is a positive plurisubharmonic function in by Lemma 2.4, we have
where denotes the real Jacobian of at . By [29, Theorem and ], for we have
By (2.19) and (2.20), we obtain
which completes the proof of the first claim.
Next we prove the second claim. Fix an . By Lemma 2.5, there exists an such that . From, (1.5) and (2.21) applied to the function we have that where . Letting in (12) and using the fact that is an arbitrary positive number we obtain This completes the proof of this lemma.
Lemma 2.8. is a complete metric space.
Proof. In the introduction we have seen that is a metric space. Since the convergence in implies the uniform convergence on compact subsets of by Lemma 2.7, a usual normal family argument along with Fatou's lemma shows that every Cauchy sequence in converges to an element of . Hence the space equipped with the metric is a complete metric space.
Recall that a metric space is called an -space if it is a complete and(i) for every ; (ii)for each sequence such that as it follows that for every ; (iii)for each sequence such that as , for each .
If in an -space is introduced an operation of pointwise multiplication such that becomes an algebra and the operation of multiplication is continuous in the metric , then the -space is called an -algebra.
Before we prove our next result we will prove two technical lemmas.
Lemma 2.9. The following inequality holds: for .
Proof. It is enough to prove the case . Since is an increasing function and , we have Let Then from we see that is an increasing function on the interval , so that . Hence . This implies that So we obtain completing the proof of the lemma.
The next lemma improves inequality () in .
Lemma 2.10. The following inequality holds: for .
Proof. First note that the function satisfies the following condition: Hence it is enough to prove inequality (2.30) for the case . Note that and Hence, for each fixed we have that , from which inequality (2.30) follows.
Theorem 2.11. is an -algebra with respect to pointwise addition and multiplication.
Proof. By Lemma 2.8 is a complete metric space satisfying the condition . From Lemma 2.9 we have the inequality
which implies that
for each , so that the operation of the multiplication by complex numbers is closed in . On the other hand, by inequality (1.5) it follows that the pointwise addition is also closed in . From (2.34) it follows that the condition implies for every , since
Now assume that is a sequence such that as . Note that for each such sequence there is such that for . Hence we have that for each and
From (2.36), since
for each , and by the Lebesgue dominated convergence theorem, we get as , which proves (iii) and that is an -space.
Now we prove that is closed with respect to the pointwise multiplication. From Lemma 2.10 we have that from which it follows that so that is closed with respect to the pointwise multiplication.
Finally, we prove that the pointwise multiplication is continuous with respect to the metric , that is, that from as and as it follows that as . Note that from triangle inequality and by (2.39) it follows that From (2.40) and the symmetry we see that it is enough to prove that from as , for each , it follows that as . Fix a and put for each . Hence we see that for every there is an such that for every Fix an . Since , we have that for every there is a such that for every set such that . From (2.41) with we have that for sufficiently large . From this and inequalities (2.34) and (2.39) we have that Letting in (2.43) and since is an arbitrary positive number, it follows that as , finishing the proof of the theorem.
3. Linear Isometries of
The following two lemmas play an important role in the proofs of the main results in this section.
Lemma 3.1. If is a linear isometry of into itself, then the restriction of to is also a linear isometry of into itself.
Proof. Take an and put . For each we have , and so the assumption which is an isometry of gives By inequality (2.2), we have for any and . Also, it is easy to see that for each . Hence the Lebesgue-dominated convergence theorem gives On the other hand, Fatou's lemma (3.1) and (3.4) show that and so . By applying the Lebesgue-dominated convergence theorem once more again, we have By (3.1), (3.4), and (3.6), we see that is a linear isometry of into .
Lemma 3.2. There exists a bounded continuous function on such that
Proof. Set for . Since is a continuous function on such that as , it is enough to prove that has a finite limit as . By the application of Taylor's theorem to , we have where denotes the remainder term of order . Since as , we obtain which completes the proof.
Theorem 3.3. Every linear isometry of into itself has the form for any , where with and is a holomorphic map which satisfies the condition for every bounded or positive Borel function in .
Proof. First suppose that is a linear isometry. By Lemma 3.1, the restriction of to is a linear isometry of into . Hence Kolaski's theorem [6, page 911, Theorem 2.11] implies that has the following form:
where and is a holomorphic self-map of such that
for every bounded or positive Borel function in . Fix an . Since , the representation (3.12) and Lemmas 2.5 and 2.7 give
for all . Since and , Hölder's inequality gives
where . On the other hand, from Lemma 3.2 it follows that
By Lemma 3.2 and Fatou's lemma we have
and so . Combining this with (3.15), we obtain
This implies that in for some with . Hence (3.13) and (3.14) show that for any and satisfies the condition in (3.11).
Conversely, for some complex number with and holomorphic self-map of which satisfies (3.11) we define an operator on by for . Since is a positive Borel function in , condition (3.11) implies that is a linear isometry of into . This completes the proof.
Corollary 3.4. Every surjective isometry of is of the form for any , where with and is a unitary operator on .
Proof. Assume that is a surjective isometry. Then Theorem 3.3 implies that is written in the form where with and is a holomorphic self-map of which satisfies (3.11). Since is also a surjective isometry of , we see that
so that and , that is, is an automorphism of .
We prove that fixes the origin. Let be the components of . For each and an we have Since the slice function of is holomorphic in the open unit disc of , the mean value theorem gives for each and . Using this fact in formula (3.20), multiplying such obtained equality by , then integrating it from 0 to 1 and using the polar coordinates on the unit ball we get for each . Now we fix a . By applying the condition (3.11) to a bounded Borel function , , where is the standard orthonormal base vector in , we have