Abstract
Let \({\mathbb {D}}\) be the unit disc in \({\mathbb {C}}\). If \(\mu \) is a finite positive Borel measure on the interval [0, 1) and f is an analytic function in \({\mathbb {D}}\), \(f(z)=\sum _{n=0}^\infty a_nz^n\) (\(z\in {\mathbb {D}}\)), we define
where, for \(n\ge 0\), \(\mu _n\) denotes the n-th moment of the measure \(\mu \), that is, \(\mu _n=\int _{[0, 1)}t^nd\mu (t).\) In this way, \({\mathcal {C}}_\mu \) becomes a linear operator defined on the space \({\mathrm{Hol}}({\mathbb {D}})\) of all analytic functions in \({\mathbb {D}}\). We study the action of the operators \({\mathcal {C}}_\mu \) on distinct spaces of analytic functions in \({\mathbb {D}}\), such as the Hardy spaces \(H^p\), the weighted Bergman spaces \(A^p_\alpha \), BMOA, and the Bloch space \({\mathcal {B}}\).
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1 Introduction and main results
Let \({{\mathbb {D}}}=\{z\in {{\mathbb {C}}}: |z|<1\}\) denote the open unit disc in the complex plane \({{\mathbb {C}}}\) and let \({\mathrm{Hol}}({{\mathbb {D}}})\) be the space of all analytic functions in \({{\mathbb {D}}}\) endowed with the topology of uniform convergence in compact subsets.
If \(\,0<r<1\,\) and \(\,f\in {\mathrm{Hol}}({{\mathbb {D}}})\), we set
For \(\,0<p\le \infty \), the Hardy space \(H^p\) consists of those \(f\in {\mathrm{Hol}}({\mathbb {D}})\) such that
We refer to [13] for the notation and results regarding Hardy spaces.
Let dA denote the area measure on \({\mathbb {D}}\), normalized so that the area of \({\mathbb {D}}\) is 1. Thus \(dA(z)\,=\,\frac{1}{\pi }\,dx\,dy\,=\,\frac{1}{\pi }\,r\,dr\,d\theta \). For \(0<p<\infty \) and \(\alpha >-1\) the weighted Bergman space \(A^p_\alpha \) consists of those \(f\in {\mathrm{Hol}}({\mathbb {D}})\) such that
where \(dA_\alpha (z)=(\alpha +1)(1-\vert z\vert ^2)^{\alpha }dA(z)\). We refer to [14, 25, 39] for the notation and results about Bergman spaces.
The space BMOA consists of those functions \(f\in H^1\) whose boundary values have bounded mean oscillation on \(\partial \mathbb D\). We refer to [16] for the theory of BMOA-functions.
Finally, we recall that a function \(f\in {\mathrm{Hol}}({\mathbb {D}})\) is said to be a Bloch function if
The space of all Bloch functions is denoted by \({\mathcal {B}}\). A classical reference for the theory of Bloch functions is [2]. Let us recall that
The Cesàro operator \({\mathcal {C}}\) is defined over the space of all complex sequences as follows: If \((a)=\{ a_k\} _{k=0}^\infty \) is a sequence of complex numbers then
The operator \({\mathcal {C}}\) is known to be bounded from \(\ell ^p\) to \(\ell ^p\) for \(1<p<\infty \). In fact, the sharp inequalities
were proved by Hardy [21] and Landau [29] (see also [24, Theorem 326, p.239 ]).
Identifying any given function \(f\in {\mathrm{Hol}}({\mathbb {D}})\) with the sequence \(\{ a_k\}_{k=0}^\infty \) of its Taylor coefficients, the Cesàro operator \({\mathcal {C}}\) becomes a linear operator from \({\mathrm{Hol}}({\mathbb {D}})\) into itself as follows:
If \(f\in {\mathrm{Hol}}({\mathbb {D}})\), \(f(z)=\sum _{k=0}^\infty a_kz^k\) (\(z\in {\mathbb {D}}\)), then
The Cesàro operator is bounded on \(H^p\) for \(0<p<\infty \). For \(1<p<\infty \), this follows from a result of Hardy on Fourier series [22] together with the M. Riesz’s theorem on the conjugate function [13, Theorem 4.1]. Siskakis [33] used semigroups of composition operators to give an alternative proof of this result and to extend it to \(p=1\). A direct proof of the boundedness on \(H^1\) was given by Siskakis in [34]. Miao [31] dealt with the case \(0<p<1\). Stempak [36] gave a proof valid for \(0<p\le 2\) and Andersen [1] provided another proof valid for all \(p<\infty \).
In this paper we associate to every positive finite Borel measure on [0, 1) a certain operator \({\mathcal {C}}_\mu \) acting on \({\mathrm{Hol}}({\mathbb {D}})\) which is a natural generalization of the classical Cesàro operator \({\mathcal {C}}\).
If \(\mu \) is a positive finite Borel measure on [0, 1) and n is a non-negative integer, we let \(\mu _n\) denote the moment of order n of \(\mu \), that is,
If \(f\in {\mathrm{Hol}}({\mathbb {D}})\), \(f(z)=\sum _{n=0}^\infty a_nz^n\) (\(z\in \mathbb D\)), we define \({\mathcal {C}}_\mu (f)\) as follows
It is clear that \(\mathcal C_\mu \) is a well defined linear operator \({\mathcal {C}}_\mu :{\mathrm{Hol}}({\mathbb {D}})\rightarrow {\mathrm{Hol}}({\mathbb {D}})\). When \(\mu \) is the Lebesgue measure on [0, 1), the operator \({\mathcal {C}}_\mu \) reduces to the classical Cesàro operator \({\mathcal {C}}\).
Our main objective in this work is to characterize those positive finite Borel measures \(\mu \) on [0, 1) such that the operator \({\mathcal {C}}_\mu \) is bounded or compact on classical subspaces of \({\mathrm{Hol}}({\mathbb {D}})\) such as the Hardy spaces \(H^p\), the weighted Bergman spaces \(A^p_\alpha \), and the spaces BMOA and \({\mathcal {B}}\).
Measures of Carleson type will play a basic role in the sequel. If \(I\subset \partial {{\mathbb {D}}}\) is an interval, \(\vert I\vert \) will denote the length of I. The Carleson square S(I) is defined as
If \(\, s>0\) and \(\mu \) is a positive Borel measure on \({{\mathbb {D}}}\), we shall say that \(\mu \) is an s-Carleson measure if there exists a positive constant C such that
If \(\mu \) satisfies \(\mu \left( S(I)\right) \,=\,{{\,\mathrm{o}\,}}\left( |I|^s\right) \), as \(\vert I\vert \rightarrow 0\), then we say that \(\mu \) is a vanishing s-Carleson measure.
A 1-Carleson measure, respectively, a vanishing 1-Carleson measure, will be simply called a Carleson measure, respectively, a vanishing Carleson measure.
We recall that Carleson [7] proved that \(H^p\,\subset \,L^p(d\mu )\) (\(0<p<\infty \)), if and only if \(\mu \) is a Carleson measure (see [13, Chapter 9]).
Following [38], if \(\mu \) is a positive Borel measure on \({{\mathbb {D}}}\), \(0\le \alpha <\infty \), and \(0<s<\infty \), we say that \(\mu \) is an \(\alpha \)-logarithmic s-Carleson measure if there exists a positive constant C such that
If \(\mu \left( S(I)\right) \left( \log \frac{2}{\vert I\vert }\right) ^\alpha \,=\,{{\,\mathrm{o}\,}}\left( |I|^s\right) \), as \(\vert I\vert \rightarrow 0\), we say that \(\mu \) is a vanishing \(\alpha \)-logarithmic s-Carleson measure.
A measure \(\mu \) on [0, 1) can be seen as a measure on \({\mathbb {D}}\) with support contained in the radius [0, 1). In this way, a positive Borel measure \(\mu \) on [0, 1) is an s-Carleson measure if and only if there exists a positive constant C such that
and we have similar statements for vanishing s-Carleson measures, for \(\alpha \)-logarithmic s-Carleson measures, and for vanishing \(\alpha \)-logarithmic s-Carleson measures.
Among other, we shall prove the following results.
Theorem 1
Suppose that \(1\le p<\infty \) and let \(\mu \) be a positive finite Borel measure on [0, 1). Then the following conditions are equivalent.
-
(i)
The measure \(\mu \) is a Carleson measure.
-
(ii)
The operator \({\mathcal {C}}_\mu \) is bounded from \(H^p\) into itself.
Theorem 2
Suppose that \(1\le p<\infty \) and let \(\mu \) be a positive finite Borel measure on [0, 1). Then the following conditions are equivalent.
-
(i)
The measure \(\mu \) is a vanishing Carleson measure.
-
(ii)
The operator \({\mathcal {C}}_\mu \) is compact from \(H^p\) into itself.
Danikas and Siskakis [12] observed that \({\mathcal {C}}(H^\infty )\not \subset H^\infty \) and \({\mathcal {C}}(BMOA)\not \subset BMOA\) and studied the action of the the Cesàro operator on these spaces. We will devote Sects. 3.3 and 5 to study the Cesàro-like operators \({\mathcal {C}}_\mu \) acting on these spaces. Let us just mention here the following result.
Theorem 3
Let \(\mu \) be a positive finite Borel measure on [0, 1). Then the following conditions are equivalent.
-
(i)
The measure \(\mu \) is a 1-logarithmic 1-Carleson measure.
-
(ii)
The operator \({\mathcal {C}}_\mu \) is bounded from BMOA into itself.
-
(iii)
The operator \({\mathcal {C}}_\mu \) is bounded from the Bloch space \({\mathcal {B}}\) into itself.
Section 3 will be devoted to present the proofs of Theorem 1 and Theorem 2 as well as some further results concerning the action of the operators \(\mathcal C_\mu \) on Hardy spaces. Section 4 will deal with the action of the operators \({\mathcal {C}}_\mu \) on Bergman spaces and, as we have already mentioned, Sect. 5 will be devoted to study the operators \({\mathcal {C}}_\mu \) acting on BMOA, the Bloch space, and some related spaces. In particular, Sect. 5 will include a proof of Theorem 3 and the substitute of this result concerning compactness.
In Sect. 2 we shall give two alternative representations of the operator \({\mathcal {C}}_\mu \), one of them is an integral representation and the other one involves the convolution with a fixed analytic function in \({\mathbb {D}}\). We shall also introduce a related operator which will be denoted \(T_\mu \) and which will play a basic role in the proofs of some of our results.
Throughout the paper, if \(\mu \) is a finite positive Borel measure on [0, 1), for \(n\ge 0\), \(\mu _n\) will denote the moment of order n of \(\mu \). Also, we shall be using the convention that \(C=C(p, \alpha ,q,\beta , \dots )\) will denote a positive constant which depends only upon the displayed parameters \(p, \alpha , q, \beta \dots \) (which sometimes will be omitted) but not necessarily the same at different occurrences. Furthermore, for two real-valued functions \(K_1, K_2\) we write \(K_1\lesssim K_2\), or \(K_1\gtrsim K_2\), if there exists a positive constant C independent of the arguments such that \(K_1\le C K_2\), respectively \(K_1\ge C K_2\). If we have \(K_1\lesssim K_2\) and \(K_1\gtrsim K_2\) simultaneously, then we say that \(K_1\) and \(K_2\) are equivalent and we write \(K_1\asymp K_2\).
Let us close this section noticing that, since the subspaces X of \({\mathrm{Hol}}({\mathbb {D}})\) we shall be dealing with are Banach spaces continuously embedded in \({\mathrm{Hol}}({\mathbb {D}})\), to prove that the operator \({\mathcal {C}}_\mu \) (or \(T_\mu \), to be defined below) is bounded on X it suffices to show that it maps X into X by appealing to the closed graph theorem.
2 Alternative representations of \({\mathcal {C}}_\mu \) and a related operator
A simple calculation with power series gives the following integral representation of the operators \({\mathcal {C}}_\mu \).
Proposition 1
If \(\mu \) is a positive finite Borel measure on [0, 1) and \(f\in {\mathrm{Hol}}({\mathbb {D}})\) then
Next we shall give another expression for \(\mathcal C_\mu (f)\) involving the convolution of analytic functions. If f and g are two analytic functions in the unit disc,
the convolution \(f\star g\) of f and g is defined by
Lemma 1
Let \(\mu \) be a positive finite Borel measure on [0, 1) and set
If \(f\in {\mathrm{Hol}}({\mathbb {D}})\) and
then \({\mathcal {C}}_\mu (f)=F\star g\).
The proof is elementary and will be omitted.
The following result regarding the radial measures \(\mu \) we are considering will be used in our work.
Lemma 2
Let \(\mu \) be a finite positive Borel measure on the interval [0, 1) and, for \(n\ge 0\), let \(\mu _n\) denote the moment of order n of \(\mu \).
-
(i)
\(\mu \) is a Carleson measure if and only if \(\mu _n={{\,\mathrm{O}\,}}(\frac{1}{n})\).
-
(ii)
\(\mu \) is a vanishing Carleson measure if and only if \(\mu _n={{\,\mathrm{o}\,}}(\frac{1}{n}\)).
-
(iii)
\(\mu \) is a 1-logarithmic 1-Carleson measure if and only if \(\mu _n={{\,\mathrm{O}\,}}(\frac{1}{n\log n})\).
-
(iv)
\(\mu \) is a vanishing 1 logarithmic 1-Carleson measure if and only if \(\mu _n={{\,\mathrm{o}\,}}(\frac{1}{n\log n})\).
Proof
(i) is Proposition 8 of [8] and (ii) follows with a similar argument. Lemma 2. 7 of [19] gives one implication of (iii) and the other one follows from the from the simple inequality
Finally, (iv) can be proved with an argument similar the the one used to prove (iii). \(\square \)
Now we define a new operator operator \(T_\mu \) associated to \(\mu \) which will be important in our work because it will become the adjoint of \({\mathcal {C}}_\mu \) in distinct instances.
If \(\mu \) is a finite positive Borel measure on [0, 1) and \(f\in {\mathrm{Hol}}({\mathbb {D}})\), \(f(z)=\sum _{n=0}^\infty a_nz^n\) (\(z\in {\mathbb {D}}\)) we set
whenever the right hand side makes sense and defines an analytic function in \({\mathbb {D}}\).
Clearly, the operator \(T_\mu \) is not defined over the whole space \({\mathrm{Hol}}({\mathbb {D}})\). We have the following result.
Proposition 2
Let \(\mu \) is a finite positive Borel measure on [0, 1).
-
(a)
If P is a polynomial then \(T_\mu (P)\) is well defined and it also a polynomial.
-
(b)
If \(\mu \) is a Carleson measure then \(T_\mu \) is well defined on \(H^1\).
Proof
(a) is clear. To prove (b) we use the fact that if \(\mu \) is a Carleson measure then \(\mu _n={{\,\mathrm{O}\,}}\left( n^{-1}\right) \) (see Lemma 2). This and Hardy’s inequality [13, p. 48] shows that if \(f\in H^1\), \(f(z)=\sum _{k=1}^\infty a_kz^k\), then there exists \(C>0\) such that
for all n. Clearly, this implies (b). \(\square \)
It is well known that, for \(1<p<\infty \), the dual of \(H^p\) is identifiable with \(H^q\), \(\frac{1}{p}+\frac{1}{q}=1\), with the pairing
where \(f(z)=\sum _{n=0}^\infty a_nz^n\in H^p\) and \(g(z)=\sum _{n=0}^\infty b_nz^n\in H^q\) (see [13, Theorem 7.3]).
Similarly, if \(1<p<\infty \) and \(\alpha >1\), the dual of \(A^p_\alpha \) is identifiable with \(A^q_{\alpha }\) with the pairing
where
and \(f(z)=\sum _{n=0}^\infty a_nz^n\in A^p_\alpha \), \(g(z)=\sum _{n=0}^\infty b_nz^n\in A^q_\alpha \) (see [25, Theorem 1.16 and p. 5]). A simple calculation gives the following result.
Proposition 3
Let \(\mu \) be a positive finite Borel measure on [0, 1).
-
(i)
If \(1<p<\infty \), \(f\in H^p\), and g is a polynomial then
$$\begin{aligned}<{\mathcal {C}}_\mu (f), g>_{H^p}\,=\,<f, T_\mu (g)>_{H^p}. \end{aligned}$$ -
(ii)
If \(1<p<\infty \), \(\alpha >-1\), \(f\in A^p_\alpha \), and g is a polynomial then
$$\begin{aligned}<{\mathcal {C}}_\mu (f), g>_{p,\alpha }\,=\,<f, T_\mu (g)>_{p,\alpha }.\end{aligned}$$
Proposition 3, together with the fact that the polynomials are dense in all the spaces \(H^p\) (\(p<\infty \)) and \(A^p_\alpha \) (\(p<\infty ,\,\alpha >-1\)), readily implies the following result.
Proposition 4
Suppose that \(1<p<\infty \) and let \(\mu \) be a positive finite Borel measure on [0, 1). Let q be the conjugate exponent of p, that is, \(\frac{1}{p}+\frac{1}{q}=1\).
-
(i)
If \({\mathcal {C}}_\mu \) is a bounded operator from \(H^p\) into itself, then there exists a positive constant C such that
$$\begin{aligned} \Vert T_\mu (P)\Vert _{H^q}\le C\Vert P\Vert _{H^q} \end{aligned}$$for every polynomial P. Consequently, \(T_\mu \) extends to a bounded linear operator from \(H^q\) into itself. This extension, which will be also denoted by \(T_\mu \), is the adjoint of \({\mathcal {C}}_\mu \).
-
(ii)
Suppose that \(\alpha >-1\). If \({\mathcal {C}}_\mu \) is a bounded operator from \(A^p_\alpha \) into itself, then there exists a positive constant C such that
$$\begin{aligned} \Vert T_\mu (P)\Vert _{A^q_\alpha }\le C\Vert P\Vert _{A^q_\alpha } \end{aligned}$$for every polynomial P. Consequently, \(T_\mu \) extends to a bounded linear operator from \(A^q_\alpha \) into itself. This extension, which will be also denoted by \(T_\mu \), is the adjoint of \({\mathcal {C}}_\mu \).
3 The operators \({\mathcal {C}}_\mu \) acting on Hardy spaces
In this section we shall study the action of the operators \(\mathcal C_\mu \) on Hardy spaces.
We shall use complex interpolation to prove some of our results. Let us refer to [39, Chapter 2] for the terminology and basic results concerning complex interpolation.
If \(X_0\) and \(X_1\) are two compatible Banach spaces then, for \(0<\theta <1\), \((X_0, X_1)_{\theta }\) stands for the space obtained by the complex method of interpolation of Calderón [5]. It is well known (see [6, 26, 32]) that if \(1\le p_0, p_1\le \infty \), \(0<\theta <1\), and \(1/p=(1-\theta )/p_0+\theta /p_1\), then
In particular,
3.1 Proof of Theorem 1
We shall split it in several cases.
Proof of the implication (i) \(\Rightarrow \) (ii) when \(p=1\). Assume that \(\mu \) is a Carleson measure and take \(f\in H^1\). Set
and
Using the integral representation on \({\mathcal {C}}_\mu \), we see that, for \(0<r<1\),
Since \(\mu \) is a Carleson measure, \(\mu \left( [t_{k-1},t_k]\right) \lesssim \frac{1}{2^k}\). Using this, the Hardy-Littlewood maximal theorem [13, Theorem 1.9], the fact that integral means \(M_1(s,g)\) increase with s, and the Cauchy-Schwarz inequality, we obtain
Making the change of variables \(t=rs\) in the last integral and setting \(f_r(z)=f(rz)\) (\(z\in {\mathbb {D}}\)), it follows that
Using a result of Hardy and Littlewood [23] (see also [34]) we see that
Then it follows that
This implies that \({\mathcal {C}}_\mu (f)\in H^1\) and that \(\Vert \mathcal C_\mu (f)\Vert _{H^1}\lesssim \Vert f\Vert _{H^1}\). \(\square \)
Proof of the implication (i) \(\Rightarrow \) (ii) when \(p=2\). Assume that \(\mu \) is a Carleson measure and take \(f\in H^2\), \(f(z)=\sum _{n=0}^\infty a_nz^n\) (\(z\in {\mathbb {D}}\)). Using [8, Proposition 1] we see that \(\vert \mu _n\vert \lesssim \frac{1}{n+1}\). Using this, the definition of \(\mathcal C_\mu (f)\), and the fact that the Cesàro operator is bounded on \(H^2\), it follows that
\(\square \)
Proof of the implication (i) \(\Rightarrow \) (ii) for \(1<p<2\). Since (i) \(\Rightarrow \) (ii) when \(p=1\) and \(p=2\), the fact that (i) \(\Rightarrow \) (ii) when \(1<p<2\) follows using (3) and Theorem 2. 4 of [39]. \(\square \)
To prove the remaining case, that is, the implication (i) \(\Rightarrow \) (ii) for \(2<p<\infty \) we shall use ideas of Andersen [1]. Actually, our next argument works for \(1<p<\infty \).
Proof of the implication (i) \(\Rightarrow \) (ii) for \(1<p<\infty \). Assume that \(\mu \) is a Carleson measure, \(1<p<\infty \), and \(f\in H^p\).
For \(0<r<1\), set
Arguing just as in [1, p. 621], using Fubini’s theorem, we have that
Now, letting \(\{ t_k\} _{k=0}^\infty \) be as above, using the fact that \(\mu ([t_k,t_{k+1}))\lesssim \frac{1}{2^k}\), and simple estimates, we obatin
Then, using Lemma 2.1 of [1], we see that for all \(\theta , \varphi \in (-\pi ,\pi )\setminus \{ 0\} \) and \(r\in (0,1)\), we have
where
Using this and (5) it follows that
Then the argument in p. 622 of [1] yields that
and, hence \(\Vert {\mathcal {C}}_\mu (f)\Vert _{H^p}\lesssim \Vert f\Vert _{H^p}.\) \(\square \)
Proof of the implication (ii) \(\Rightarrow \) (i) for \(1\le p\le 2\). Suppose that \(1\le p\le 2\) and that \({\mathcal {C}}_\mu \) is bounded on \(H^p\). Recall that, for \(\alpha >0\),
where
For \(0<a<1\), set
We have that
Since \({\mathcal {C}}_\mu \) is bounded on \(H^p\), we have
Now
Using the fact that \(1\le p\le 2\), [13, Theorem 6.2], (7), and the fact that the sequence \(\{ \mu _n\} \) is decreasing, we obtain
for every positive integer N and every \(a\in (0,1)\). Taking \(a=1-\frac{1}{N}\) and using the fact that \({\mathcal {C}}_\mu \) is bounded on \(H^p\), we obtain
This and (8) imply that \(\mu _N\lesssim \frac{1}{N}\). Using again Lemma 2, this yields that \(\mu \) is a Carleson measure. \(\square \)
Proof of the implication (ii) \(\Rightarrow \) (i) for \(2\le p<\infty \). Suppose that \(2<p<\infty \) and that \({\mathcal {C}}_\mu \) is a bounded operator on \(H^p\). Let q be the conjugate exponent of p, that is, \(\frac{1}{p}+\frac{1}{q}=1\). Bearing in mind Proposition 2 and Proposition 3, we see that the operator \(T_\mu \), initially defined over polynomials, extends to a bounded operator on \(H^q\).
For \(0<a<1\) and \(N\in {\mathbb {N}}\), set
We have that for all \(a\in (0,1)\), \(f_a\in H^q\) and \(\Vert f_a\Vert _{H^q}=1\). Since \(T_\mu \) is bounded on \(H^q\), it follows that
Also, for every a, \(\,f_{a,N}\, \rightarrow f_a\), as \(N\rightarrow \infty \) in \(H^q\) and uniformly on compact subsets of \({\mathbb {D}}\). Now, \(T_\mu \left( f_{a,N}\right) (z)=(1-a^2)^{1/q}\sum _{n=0}^N\left( \sum _{k=n}^N\mu _ka_k(2/q)a^k\right) z^n\) (\(z\in {\mathbb {D}}\)) and then, using that \(1<q<2\) and [13, Theorem 6.2], we have that
Letting N tend to \(\infty \), we obtain
Taking \(a=1-\frac{1}{N}\) and letting [N/2] denote the largest integer less than or equal to N/2, we obtain
Using (10), it follows that \(\mu _N\lesssim \frac{1}{N}\) and then Lemma 2 implies that \(\mu \) is a Carleson measure. \(\square \)
3.2 Proof of Theorem 2
Proof
Let us start with the implication (ii) \(\Rightarrow \) (i). We shall consider the cases \(1\le p\le 2\) and \(2<p<\infty \) separately.
Suppose first that \(1\le p\le 2\) and \({\mathcal {C}}_\mu \) is compact from \(H^p\) into itself. As in the proof of Theorem 1, for \(0<a<1\), set
We have that \(\Vert f_a\Vert _{H^p}=1\) for all a and, also, \(f_a\,\rightarrow \, 0,\) as \(a\rightarrow 1\), uniformly on compact subsets of \({\mathbb {D}}\). Hence, \(\Vert {\mathcal {C}}_\mu (f_a)\Vert _{H^p}\rightarrow 0\), as \(a\rightarrow 1\). But in the course of the proof of the implication (ii) \(\Rightarrow \) (i) of Theorem 1, we obtained that \(\mu _NN\lesssim \Vert {\mathcal {C}}_\mu (f_a)\Vert _{H^p}\) for \(a=1-\frac{1}{N}\) (see (9)). Then it follows that \(\mu _N={{\,\mathrm{o}\,}}\left( \frac{1}{N}\right) \) and this implies that \(\mu \) is a vanishing Carleson measure.
Suppose now that \(2<p<\infty \) and \({\mathcal {C}}_\mu \) is compact from \(H^p\) into itself. By Theorem 1, \(\mu \) is a Carleson measure and then it follows that the operator \(T_\mu \) is well defined on \(H^q\) (\(\frac{1}{p}+\frac{1}{q}=1\)) and it is the adjoint of \(\mathcal C_\mu \). For \(0<a<1\), set \(f_a(z)=\left( \frac{1-a^2}{(1-az)^2}\right) ^{1/q}\) , (\(z\in {\mathbb {D}}\)). We have that \(\Vert f_a\Vert _{H^q}=1\) for all a and, also, \(f_a\,\rightarrow \, 0,\) as \(a\rightarrow 1\), uniformly on compact subsets of \({\mathbb {D}}\). By Schauder’s theorem [10, p. 174], \(T_\mu \) is a compact operator from \(H^q\) into itself and, hence, \(\Vert T_\mu (f_a)\Vert _{H^q}\rightarrow 0\). In the course of the proof of the implication (ii) \(\Rightarrow \) (i) of Theorem 1, we obtained that \(\mu _NN\lesssim \Vert T_\mu (f_a)\Vert _{H^q}\) for \(a=1-\frac{1}{N}\) (see (11)). Then it follows that \(\mu _N={{\,\mathrm{o}\,}}\left( \frac{1}{N}\right) \) and, hence, \(\mu \) is a vanishing Carleson measure.
To prove the other implication we shall consider the cases \(p=2\), \(p=1\), \(1<p<2\), and \(2<p<\infty \) separately.
Let us start with the case \(p=2\). So assume that \(\mu \) is a vanishing Carleson measure and let \(\{ f_n\} \) be a sequence of functions in \(H^2\) with \(\Vert f_n\Vert _{H^2}\le 1\), for all n, and such that \(\, f_n\, \rightarrow 0\), uniformly on compact subsets of \({\mathbb {D}}\).
Since \(\mu \) is a vanishing Carleson measure \(\mu _k={{\,\mathrm{o}\,}}\left( \frac{1}{k}\right) \), as \(k\rightarrow \infty \). Say
Then \(\{ \varepsilon _k\} \rightarrow 0\). Say that, for every n,
Since the Cesàro operator \({\mathcal {C}}\) is bounded on \(H^2\), there exists \(M>0\) such that
Take \(\varepsilon >0\) and next take a natural number N such that
We have
Now, since \(\, f_n\, \rightarrow 0\), uniformly on compact subsets of \({\mathbb {D}}\), it follows that
Then it follows that that there exist \(n_0\in {\mathbb {N}}\) such that \(\Vert {\mathcal {C}}_\mu (f_n)\Vert _{H^2}^2<\varepsilon \) for all \(n\ge n_0\). So, we have proved that \(\Vert {\mathcal {C}}_\mu (f_n)\Vert _{H^2}^2\rightarrow 0\). The compactness of \({\mathcal {C}}_\mu \) on \(H^2\) follows.
Let us move to the case \(p=1\). Assume that \(\mu \) is a vanishing Carleson measure and let \(\{ f_n\} \) be a sequence of functions in \(H^1\) with \(\Vert f_n\Vert _{H^1}\le 1\), for all n, and such that \(\, f_n\, \rightarrow 0\), uniformly on compact subsets of \({\mathbb {D}}\).
Set
and
As in the proof of the implication (i) \(\Rightarrow \) (ii) in Theorem 1 when \(p=1\) we see that, for \(0<r<1\) and \(n\in {\mathbb {N}}\),
and, hence,
Since \(\mu \) is a vanishing Carleson measure \(\mu \left( [t_{k-1},t_k]\right) ={{\,\mathrm{o}\,}}(2^{-k})\) and, hence, we have
On the other hand, looking at the proof of Theorem 1, we see that there exists \(C>0\) such that
Take \(\varepsilon >0\) and then take \(N\in {\mathbb {N}}\) so that \(\varepsilon _k\le \frac{\varepsilon }{2CK}\), for all \(k\ge N\), where K is the constant in the Hardy-Littlewood maximal estimate
Using (13) we see that
where
Using (14), we obtain
Since \(\, f_n\, \rightarrow 0\), uniformly on compact subsets of \({\mathbb {D}}\), it is clear that \(I(n)\rightarrow 0\), as \(n\rightarrow \infty \). Then it follows that there exists \(n_0\in {\mathbb {N}}\) such that \(\Vert {\mathcal {C}}_\mu (f_n)\Vert _{H^1}<\varepsilon \) whenever \(n\ge n_0\). Thus,we have shown that \(\Vert {\mathcal {C}}_\mu (f_n)\Vert _{H^1}\rightarrow 0\), as \(n\rightarrow \infty \) and the compactness of \({\mathcal {C}}_\mu \) on \(H^1\) follows.
To deal with the cases \(1<p<2\) and \(2<p<\infty \), we use again complex interpolation.
Suppose first that \(1<p<2\) and \(\mu \) is a vanishing Carleson measure. Recall that
We have also that if \(2<s<\infty \) then
for a certain \(\alpha \in (0,1)\), namely, \(\alpha =\left( \frac{1}{2}-\frac{1}{s}\right) /\left( 1-\frac{1}{s}\right) \). Since \(H^2\) is reflexive, and \({\mathcal {C}}_\mu \) is compact from \(H^2\) into \(H^2\) and from \(H^1\) into \(H^1\), Theorem 10 of [11] gives that and \({\mathcal {C}}_\mu \) is compact from \(H^p\) into \(H^p\).
Suppose now that \(2<p<\infty \) and \(\mu \) is a vanishing Carleson measure. Let q be conjugate exponent of p. Take \(q_1\) with \(1<q_1<q<2\). We have that \(T_\mu \) is compact from \(H^2\) into itself and continuous from \(H^{q_1}\) into \(H^{q_1}\). Also, \(H^q=(H^2, H^{q_1})_{\theta }\) for a certain \(\theta \in (0,1)\). Then, Theorem 10 of [11] gives that and \(T_\mu \) is compact from \(H^q\) into \(H^q\) and, hence, \({\mathcal {C}}_\mu \) is compact from \(H^p\) into itself. \(\square \)
3.3 The operators \({\mathcal {C}}_\mu \) acting on \(H^\infty \)
For the constant function 1 we have
Consequently, \({\mathcal {C}} (H^\infty )\not \subset H^\infty \).
If \(\mu \) is positive finite Borel measure on [0, 1) then
So, it follows that
This easily implies the following result.
Theorem 4
Let \(\mu \) be positive finite Borel measure on [0, 1). Then the following conditions are equivalent.
-
(i)
\({\mathcal {C}}_\mu \) is a bounded operator from \(H^\infty \) into itself.
-
(ii)
\(\int _{[0,1]}\frac{d\mu (t)}{1-t}<\infty \).
-
(iii)
\(\sum _{n=0}^\infty \mu _n<\infty .\)
Danikas and Siskakis [12] proved that
We extend this result obtaining a characterization of those positive finite Borel measure \(\mu \) on [0, 1) for which \({\mathcal {C}}_\mu (H^\infty )\subset {\mathcal {B}}\).
Theorem 5
Let \(\mu \) be positive finite Borel measure on [0, 1). Then the following conditions are equivalent
-
(i)
\({\mathcal {C}}_\mu \) is a bounded operator from \(H^\infty \) into the Bloch space \({\mathcal {B}}\).
-
(ii)
\(\mu \) is a Carleson measure.
Proof
Let us start with the implication (i) \(\Rightarrow \) (ii). So, assume that \({\mathcal {C}}_\mu (H^\infty )\subset {\mathcal {B}}\). Then \({\mathcal {C}}_\mu (1)\in {\mathcal {B}}\), but, as we have seen above
and then, using the fact that the sequence \(\{ \mu _n\} \) is a decreasing sequence of nonnegative numbers and Lemma B, we see that \(\mu _n={{\,\mathrm{O}\,}}\left( \frac{1}{n}\right) \) which is equivalent to saying that \(\mu \) is a Carleson measure.
Let us turn now to prove the other implication. So, assume that \(\mu \) is a Carleson measure and take \(f\in H^\infty \). Using the integral representation of \({\mathcal {C}}_\mu \) we see that
Hence, using that \(f\in H^\infty \subset {\mathcal {B}}\), we obtain
Take \(z\in {\mathbb {D}}\) and set \(r=\vert z\vert \). Set also
Integrating by parts and using the fact that \(\mu \) is a Carleson measure, we obtain
This and (15) yield that \({\mathcal {C}}_\mu (f)\in {\mathcal {B}}\). \(\square \)
It is natural to ask whether or not \(\mu \) being a Carleson measure implies that \({\mathcal {C}}_\mu (H^\infty )\subset BMOA\). We do not know the answer to this question.
4 The operators \({\mathcal {C}}_\mu \) acting on Bergman spaces
The boundedness of the Cesàro operator on Bergman spaces was studied in [1] and [35] where the following result was proved.
Theorem A
If \(p>0\) and \(\alpha >-1\), then the Cesàro operator is bounded from \(A^p_\alpha \) into itself.
In the course of our proof of Theorem 1, we proved that if \(\mu \) is a Carleson measure, \(1\le p<\infty \), and \(f\in H^p\), then \(M_p(r,{\mathcal {C}}_\mu (f)\lesssim M_p(r,f)\) (see (4) and (6)). This readily yields that that if \(\mu \) is a Carleson measure, \(1\le p<\infty \), and \(\alpha >-1\), then \({\mathcal {C}}_\mu \) is bounded from \(A^p_\alpha \) into itself.
For \(p>1\) we shall give a different proof of this result and we shall also prove that the converse is true. Hence, our work in particular will lead to a new proof of the boundedness of the classical Cesàro operator on the spaces \(A^p_\alpha \) (\(1<p<\infty \), \(\alpha >-1\)).
Theorem 6
Suppose that \(1<p<\infty \) and \(\alpha >-1\). Let \(\mu \) be a positive finite Borel measure on [0, 1).Then the following conditions are equivalent.
-
(i)
The measure \(\mu \) is a Carleson measure.
-
(ii)
The operator \({\mathcal {C}}_\mu \) is bounded from \(A^p_\alpha \) into itself.
Let us collect several results which will be needed in the proof of Theorem 6.
Let us start recalling the given \(1\le p\le \infty \) and \(0<\alpha \le 1\), the mean Lipschitz space \(\varLambda _ \alpha ^p\) consists of those functions f analytic in \({{\mathbb {D}}}\) having a non-tangential limit almost everywhere for which \(\omega _ p(\delta , f)=O(\delta ^\alpha )\), as \(\delta \rightarrow 0\), where \(\omega _ p(., f)\) is the integral modulus of continuity of order p of the boundary values \(f(e^{i\theta })\) of f. A classical result of Hardy and Littlewood [23] (see also Chapter 5 of [13]) asserts that for \(1\le p\le \infty \) and \(0<\alpha \le 1\), we have that \(\varLambda _ \alpha ^p\subset H^p\) and
The space \(\varLambda _ \alpha ^p\) is a Banach space with the norm \(\Vert \cdot \Vert _{p, \alpha }\) given by
Of special interest are the spaces \(\varLambda _ {1/p}^ p\) since they lie in the border of continuity. Indeed, if \(1<p<\infty \) and \(\alpha >\frac{1}{p}\) then each \(f\in \varLambda _\alpha ^p\) has a continuous extension to the closed unit disc. This is not true for \(\alpha =\frac{1}{p}\). This follows easily noticing that the function \(f(z)=\log (1-z)\) belongs to \(\varLambda _ {1/p}^p\) for all \(p\in (1, \infty )\). Cima and Petersen proved in [9] that \(\varLambda _{1/2}^2\subset BMOA\) and this result was generalized by Bourdon, Shapiro and Sledd who proved in [4] that
This was shown to be sharp in a very strong sense in [3].
The following result of Merchán [30, Lemma 1] (see also [18, Theorem 2] and [17, Theorem 2]) will be needed in our work.
Lemma B
Let \(f\in {\mathrm{Hol}}({\mathbb {D}})\), \(f(z)=\sum _{n=0}^\infty a_nz^n\) (\(z\in {\mathbb {D}}\)). Suppose that \(1<p<\infty \) and that the sequence \(\{ a_n\} \) is a decreasing sequence of nonnegative numbers. If \(1<p<\infty \) and X is a subspace of \({\mathrm{Hol}}({\mathbb {D}})\) with \(\varLambda ^p_{1/p}\subset X\subset {\mathcal {B}}\), then
We shall also use some results on pointwise multipliers and coefficient multipliers of Bergman spaces and Hardy spaces.
Let us start recalling that for \(g\in {\mathrm{Hol}}({{\mathbb {D}}})\), the multiplication operator \(M_g\) is defined by
If X and Y are two spaces of analytic functions in \({\mathbb {D}}\) (which will always be assumed to be Banach or F-spaces continuously embedded in \({\mathrm{Hol}}({\mathbb {D}} )\)) and \(g\in {\mathrm{Hol}}(\mathbb D )\) then g is said to be a pointwise multiplier from X to Y if \(M_g(X)\subset Y\). The space of all multipliers from X to Y will be denoted by M(X, Y). Using the closed graph theorem we see that if \(g\in M(X, Y)\) then \(M_g\) is a bounded operator from X into Y. The following result is a particular case of Theorem C of [37].
Theorem C
Suppose that \(1<p<\infty \) and \(\alpha >-1\). Then
If X and Y are two spaces of analytic functions in \(\mathbb D\), a function \(F\in {\mathrm{Hol}}({\mathbb {D}})\) is said to be a coefficient multiplier (or a convolution multiplier) from X to Y if
The following result is due to Duren and Shields, it is a particular case of [15, Theorem 4].
Theorem D
Suppose that \(1<p<\infty \) and \(F\in {\mathrm{Hol}}({\mathbb {D}})\). Let m be a positive integer such that \((m +1)^{-1}\le \frac{p}{p+1}<m^{-1}\). Then F is a coefficient multiplier from \(H^{p/(p+1)}\) to \(H^p\) if and only if the \((m+1)\)-th derivative \(F^{(m+1)}\) of F satisfies
We can now proceed to prove Theorem 6.
Proof of the implication (i) \(\Rightarrow \) (ii) in Theorem 6. Assume that \(\mu \) is a Carleson measure and set
Since \(\mu \) is a Carleson measure \(\mu _n={{\,\mathrm{O}\,}}\left( \frac{1}{n}\right) \). This, the simple fact that \(\{ \mu _n\} \) is a deceasing sequence of nonnegative numbers, and Lemma B imply that \(F\in \varLambda ^p_{1/p}\) and, hence
Using [13, Theorem 5.5], we see that this implies
and then Theorem D gives that F is a coefficient multiplier from \(H^{p/(p+1)}\) into \(H^p\). Trivially, this implies that
Take \(f\in A^p_\alpha \). We have to prove that \({\mathcal {C}}_\mu (f)\in A^p_\alpha \). Set \(g(z)=\frac{f(z)}{1-z}\) (\(z\in \mathbb D\)). A simple computation shows that \(\frac{1}{1-z}\in A^1_\alpha \). Then, using Theorem C we deduce that \(g\in A^{p/(p+1)}_\alpha \). This and (16) imply that \(F\star g\in A^p_\alpha \). By Lemma 1 this is equivalent to saying that \({\mathcal {C}}_\mu (f)\in A^p_\alpha \). \(\square \)
Proof of the implication (ii) \(\Rightarrow \) (i) in Theorem 6. Suppose that \({\mathcal {C}}_\mu \) is a bounded operator on \(A^p_\alpha \). Let q be the exponent conjugate to p, that is, \(\frac{1}{p}+\frac{1}{q}=1\). Let \(T_\mu \) be the adjoint of \({\mathcal {C}}_\mu \), it is a bounded operator on \(A^q_\alpha \).
For \(0<b<1\), set
Using [39, Lemma 3.10], we see that
Also,
For \(N\in {\mathbb {N}}\), set
Bearing in mind Proposition 2 and Proposition 3, we see that
Since the coefficients \(a_{k,b}\) are nonnegative, it follows that the sequence of the Taylor coefficients of \(T_\mu (f_{b,N})\) is a decreasing sequence of nonnegative numbers, then (see, e. g., [20, Proposition 1])
Since \(\, f_{b,N}\, \rightarrow f_b\) in \(A^q_\alpha \) as \(N\rightarrow \infty \), using the fact that \(T_\mu \) is bounded on \(A^q_\alpha \), (17), and simple estimations, we deduce that
Hence, \(\mu \) is a Carleson measure. \(\square \)
5 The operators \({\mathcal {C}}_\mu \) acting on BMOA and on the Bloch space
Let \(\lambda \) be defined by \(\lambda (z)=\log \frac{1}{1-z}\) (\(z\in {\mathbb {D}}\)). Then \(\lambda \in BMOA\). In fact, it is true that \(\lambda \in \varLambda ^p_{1/p}\) for all \(p>1\). Danikas and Siskakis [12] observed that \({\mathcal {C}} (\lambda )\notin BMOA\). This implies that the Cesàro operator does not map BMOA into itself. Our Theorem 3 includes a characterization of those \(\mu \) so that \({\mathcal {C}}_\mu \) maps BMOA into itself.
Since \(\varLambda ^2_{1/2}\subset BMOA\subset {\mathcal {B}}\), Theorem 3 follows from the following result.
Theorem 7
Let \(\mu \) be a positive finite Borel measure on [0, 1) and let X and Y be two Banach subspaces of \({\mathrm{Hol}}({\mathbb {D}})\) with \(\varLambda ^2_{1/2}\subset X \subset {\mathcal {B}}\) and \(\varLambda ^2_{1/2}\subset Y \subset {\mathcal {B}}\). Then the following conditions are equivalent.
-
(i)
The measure \(\mu \) is a 1-logarithmic 1-Carleson measure.
-
(ii)
The operator \({\mathcal {C}}_\mu \) is bounded from X into Y.
Proof
Let us start showing that (i) \(\Rightarrow \) (ii). So assume that \(\mu \) is a 1-logarithmic 1-Carleson measure and take \(f\in X\). We recall that \(\mu \) being a 1-logarithmic 1-Carleson measure is equivalent to
Take \(f\in X\), \(f(z)=\sum _{n=0}^\infty a_nz^n\) (\(z\in {\mathbb {D}}\)). Since \(X\subset {\mathcal {B}}\), we have that \(f\in {\mathcal {B}}\). Then, using a result of Kayumov and Wirths (see [27, Corollary 4] or [28, Corollary D]), we have
The estimates (18) and (19) yield
Hence \({\mathcal {C}}_\mu (f)\in \varLambda ^2_{1/2}\subset Y\).
Suppose now that \({\mathcal {C}}_\mu (X)\subset Y\). As above, set \(\lambda (z)=\log \frac{1}{1-z}=\sum \limits _{n=1}^\infty \frac{z^n}{n}\) (\(z\in {\mathbb {D}}\)). We have that \(\lambda \in X\) and then \({\mathcal {C}}_\mu (\lambda )\in Y\subset {\mathcal {B}}\). Now, \({\mathcal {C}}_\mu (\lambda )(z)=\sum \limits _{n=1}^\infty \mu _n\left( \sum \limits _{k=1}^n\frac{1}{k}\right) z^n\) and then it follows that
For \(N\ge 2\) take \(r_N=1-\frac{1}{N}\). Bearing in mind that the sequence \(\{ \mu _n\} \) is decreasing, simple estimations lead us to the following
Hence \(\mu _N\lesssim \frac{1}{N\log N}\) which implies that \(\mu \) is a 1-logarithmic 1-Carleson measure. \(\square \)
We have the following result concerning compactness.
Theorem 8
Let \(\mu \) be a positive finite Borel measure on [0, 1) and let X and Y be two Banach subspaces of \({\mathrm{Hol}}({\mathbb {D}})\) with \(\varLambda ^2_{1/2}\subset X \subset {\mathcal {B}}\) and \(\varLambda ^2_{1/2}\subset Y \subset {\mathcal {B}}\). Then the following four conditions are equivalent.
-
(i)
\(\mu \) is a vanishing 1-logarithmic 1-Carleson measure.
-
(ii)
The operator \({\mathcal {C}}_\mu \) is a compact operator from X into Y.
-
(iii)
The operator \({\mathcal {C}}_\mu \) is a compact operator from the Bloch space \({\mathcal {B}}\) into itself.
-
(iv)
The operator \({\mathcal {C}}_\mu \) is a compact operator from the BMOA into itself.
Proof
Clearly, it suffices to prove that (i) and (ii) are equivalent. Let us prove first that (i) implies (ii). So, assume that \(\mu \) is a vanishing 1-logarithmic 1-Carleson measure and \(\varLambda ^2_{1/2}\subset X, Y \subset {\mathcal {B}}\).
Take \(\{ f_j\} \subset X\) with \(\Vert f_j\Vert _{X}\le 1\), for all j, and \(\, f_j\, \rightarrow 0\), as \(j\rightarrow \infty \), uniformly on compact subsets of \({\mathbb {D}}\). Since X is continuously embedded in \({\mathcal {B}}\), \(\{ f_j\} \subset {\mathcal {B}}\) and there exists \(K_1>0\) such that \(\Vert f\Vert _{{\mathcal {B}}}\le K_1\), for all j.
Say \(f_j(z)=\sum _{k=0}^\infty a_k^{(j)}z^k\) (\(z\in {\mathbb {D}}\)). Using the result of Kayumov and Wirths that we have mentioned above, we see that there exists \(K_2>0\) such that
Set \(K=K_1K_2\).
Since \(\mu \) is a vanishing 1-logarithmic 1-Carleson measure, \(\mu _n={{\,\mathrm{o}\,}}\left( \frac{1}{n\log (n+1)}\right) \). Say \(\mu _n=\frac{\varepsilon _n}{n\log (n+1)}\), with \(\{ \varepsilon _n\} \rightarrow 0\). Take \(\varepsilon >0\). Take \(N\in {\mathbb {N}}\) such that \(\varepsilon _n^2K^2<\frac{\varepsilon }{2}\) if \(n\ge N\). We have, for all \(j\in {\mathbb {N}}\) and \(0<r<1\),
Thus,
Now, since \(\sum _{n=1}^Nn^2\mu _n^2\left| \sum _{k=0}^na_k^{(j)}\right| ^2\rightarrow 0\) and \(f_j(0)\rightarrow 0\), as \(j\rightarrow \infty \), it follows that there exists \(j_0\in {\mathbb {N}}\) such that
for all \(j\ge j_0\). With this we have proved that \({\mathcal {C}}_\mu (f_j)\rightarrow 0\) in \(\varLambda ^2_{1/2}\). Since \(\varLambda ^2_{1/2}\) is continuously embedded in Y, it follows that \({\mathcal {C}}_\mu (f_j)\rightarrow 0\) in Y.
Let us prove now that (ii) implies (i). Assume that \(\varLambda ^2_{1/2}\subset X, Y \subset {\mathcal {B}}\) and that \({\mathcal {C}}_\mu \) is compact from X into Y. For \(0<a<1\), set
We have that
Then it is clear that \(f_a\in \varLambda ^2_{1/2}\) for all \(a\in [0, 1)\) and that there exists a constant \(M_1>0\) such that \(\Vert f_a\Vert _{2, 1/2}\le M_1,\) for all \(a\in (0, 1)\). Since \(\varLambda ^2_{1/2}\) is continuously embedded in X, it follows that \(f_a\in X\) for all \(a\in [0, 1)\) and that there exists \(M>0\) such that \(\Vert f_a\Vert _{X}\le M,\) for all \(a\in (0, 1)\). Also, \(\, f_a\, \rightarrow 0\), as \(a\rightarrow 1\), uniformly on compact subsets of \(\mathbb D\). Since \({\mathcal {C}}_\mu \) is compact from X into Y, we have that \(\Vert {\mathcal {C}}_\mu (f_a)\Vert _{Y}\rightarrow 0\), as \(a\rightarrow 1\). This, together with the fact that Y is continuously embedded in \({\mathcal {B}}\), implies that
A simple calculation gives that for \(0<a<1\) and \(z\in {\mathbb {D}}\),
Then it follows that, for \(0<a<1\),
This gives that
This and (20) imply that \(\mu \) is a vanishing 1-logarithmic 1-Carleson measure. \(\square \)
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Galanopoulos, P., Girela, D. & Merchán, N. Cesàro-like operators acting on spaces of analytic functions. Anal.Math.Phys. 12, 51 (2022). https://doi.org/10.1007/s13324-022-00649-x
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DOI: https://doi.org/10.1007/s13324-022-00649-x