# New approach on interpolating sequences for the Bloch space

Alejandro Miralles  and  Mario P. Maletzki  Departament de Matemàtiques and
IMAC, Universitat Jaume I, Castelló (Spain). _e_.mail<EMAIL_ADDRESS>and
<EMAIL_ADDRESS>
###### Abstract.

When we deal with $H^{\infty}$, it is known that $c_{0}-$interpolating
sequences are interpolating and it is sufficient to interpolate idempotents of
$\ell_{\infty}$ in order to interpolate the whole $\ell_{\infty}$. We will
extend these results to the frame of interpolating sequences in the classical
Bloch space $\mathcal{B}$ and we will provide new characterizations of
interpolating sequences for $\mathcal{B}$. For that, bearing in mind that
$\mathcal{B}$ is isomorphic but not isometric to the bidual of the little
Bloch space $\mathcal{B}_{0}$, we will prove that given an interpolating
sequence $(z_{n})$ for $\mathcal{B}$, the second adjoint of the interpolating
operator which maps $f\in\mathcal{B}_{0}$ to the sequence
$(f^{\prime}(z_{n}))(1-|z_{n}|^{2}))$ can be perfectly identified with the
corresponding interpolating operator from $\mathcal{B}$ onto $\ell_{\infty}$.
Furthermore, we will prove that interpolating sequences for $H^{\infty}$ are
also interpolating for $\mathcal{B}$. This yields us to provide examples of
interpolating sequences which are $\varepsilon-$separated for the
pseudohyperbolic distance for $\varepsilon>0$ as small as we want such that
$\varepsilon$ cannot be increased.

###### Key words and phrases:

Bloch function, interpolating sequence, idempotents

###### Key words and phrases:

Bloch space, interpolating sequence, idempotent

###### 1991 Mathematics Subject Classification:

Primary 30H30, Secondary 30H05.

###### 2020 Mathematics Subject Classification:

Primary 30H30, Secondary 30H05

Supported by Project PGC2018-094431-B-I00 (MICINN. Spain)

##  1\. Introduction and Background

###  1.1. The spaces

Let $\mathbf{D}$ be the open unit disk of the complex plane $\mathbb{C}$.
Recall that $H^{\infty}$ is the classical space of bounded analytic functions
$f:\mathbf{D}\to\mathbb{C}$ endowed with the supremum norm
$\|\cdot\|_{\infty}$. The classical Bloch space $\mathcal{B}$ is given by the
space of analytic functions $f:\mathbf{D}\to\mathbb{C}$ such that
$\|f\|_{B}=\sup_{z\in\mathbf{D}}(1-|z|^{2})|f^{\prime}(z)|<\infty$ and the
little Bloch space $\mathcal{B}_{0}$ is the closed subspace of $\mathcal{B}$
of analytic functions $f:\mathbf{D}\to\mathbb{C}$ that satisfy
$\lim\limits_{|z|\rightarrow 1}(1-|z|^{2})|f^{\prime}(z)|=0.$ It is well known
that $\mathcal{B}$ becomes a Banach space under the norm:

$||f||_{Bloch}:=|f(0)|+\|f\|_{B}.$

It is well-known that $H^{\infty}$ is properly contained in $\mathcal{B}$.

###  1.2. Interpolating sequences

Recall that a sequence $(z_{n})\subset\mathbf{D}$ is said to be interpolating
for $H^{\infty}$ if for any $(a_{n})\in\ell_{\infty}$ there exists $f\in
H^{\infty}$ such that $f(z_{n})=a_{n}$ for any $n\in\mathbb{N}$. However, if
we deal with $\mathcal{B}$ or $\mathcal{B}_{0}$, the natural way to
interpolate is by taking the derivatives multiplied by the weight $1-|z|^{2}$.
A sequence $(z_{n})\subset\mathbb{D}$ is said to be an interpolating sequence
for $\mathcal{B}$ if for any $(a_{n})\in\ell_{\infty}$ there exists
$f\in\mathcal{B}$ such that $(1-|z_{n}|^{2})f^{\prime}(z_{n})=a_{n}\mbox{ for
any }n\in\mathbb{N}.$ Interpolating sequences for $\mathcal{B}_{0}$ are
defined in the same way just considering sequences $(a_{n})\in c_{0}$ instead
of $\ell_{\infty}$.

Given $(z_{n})\subset\mathbb{D}$ we define the interpolating operator
$L:H^{\infty}\to\ell_{\infty}$ by $L(f)=(f(z_{n}))$ for $H^{\infty}$ and
$L:\mathcal{B}\rightarrow\ell^{\infty}$ given by
$L(f)=((1-|z_{n}|^{2})f^{\prime}(z_{n}))$ for $\mathcal{B}$. Notice that $L$
is clearly linear and $(z_{n})$ is interpolating for $H^{\infty}$ or
$\mathcal{B}$ if and only if $L$ is surjective.

For $H^{\infty}$ and $\mathcal{B}$, we can also consider $c_{0}-$interpolating
sequences just by considering $(a_{n})\in c_{0}$ instead of $\ell_{\infty}$.
The connection between $c_{0}-$interpolating sequences and interpolating
sequences has been studied in the context of uniform algebras (see [5]). In
particular, the authors proved that $c_{0}-$interpolating sequences for
$H^{\infty}$ are indeed interpolating for $H^{\infty}$. In the sequel, we will
prove that this result remains true if we deal with $\mathcal{B}$.

The most important result on interpolating sequences for $H^{\infty}$ is the
classical Carleson’s Theorem [4], which states that $(z_{n})\subset\mathbf{D}$
is interpolating for $H^{\infty}$ if and only if $(z_{n})$ is uniformly
separated, that is, if there exists $\delta>0$ such that:

$\inf_{k\in\mathbb{N}}\prod_{n\neq k}\rho(z_{n},z_{k})\geq\delta$

where the pseudohyperbolic distance $\rho(z,w)$ for $z,w\in\mathbf{D}$ is
given by:

$\rho(z,w)=\left|\frac{z-w}{1-\bar{z}w}\right|.$

A sequence $(z_{n})\subset\mathbf{D}$ is said to be separated if there exists
$\delta>0$ such that:

(1.1) $\displaystyle\rho(z_{n},z_{k})\geq\delta\ \mbox{ for any }n\neq k.$

With respect to $\mathcal{B}$, Attele studied necessary and sufficient
conditions for a sequence to be interpolating in [1]. He proved that such
sequences must be separated and, conversely, if they are sufficiently
separated, that is, $\delta$ in (1.1) is sufficiently close to $1$, then the
sequence is interpolating. Madigan and Matheson also studied such conditions
and used them to study composition operators on $\mathcal{B}$ in [8].

###  1.3. Interpolating with idempotents

An idempotent $(a_{n})$ of $\ell_{\infty}$ is a sequence that satisfies
$a_{n}^{2}=a_{n}$ for any $n\in\mathbb{N}$, that is, $a_{n}=0$ or $a_{n}=1$
for any $n\in\mathbb{N}$. Hayman proved that it is sufficient to interpolate
idempotent elements $(a_{n})\in\ell_{\infty}$ to assure that the sequence
$(z_{n})\subset\mathbf{D}$ is interpolating for $H^{\infty}$ (see [7]). We
will prove that this result also remains true when we deal with interpolating
sequences for the Bloch space $\mathcal{B}$.

##  2\. Interpolating operators on $\mathcal{B}_{0}$ and $\mathcal{B}$

###  2.1. Anti-linear mappings

Let $X$ and $Y$ be complex Banach spaces. A mapping $T:X\to Y$ is said to be
anti-linear if $T(x_{1}+x_{2})=T(x_{1})+T(x_{2})$ and $T(\lambda
x)=\bar{\lambda}T(x)$ for any $\lambda\in\mathbb{C}$. It is easy that $T$ is
continuous if and only if $T$ is bounded. We define the anti-adjoint
$T^{*}:Y^{*}\to X^{*}$ of $T$ by the mapping which maps $y^{*}$ to
$T^{*}(y^{*})\in X^{*}$ such that $T^{*}(y^{*})(x)=(\overline{y^{*}}\circ
T)(x)$ and $\overline{y^{*}}(u):=\overline{y^{*}(u)}$. Notice that $T^{*}$ is
also anti-linear. In addition, if $T:X\to Y$ and $S:Y\to Z$ are anti-linear,
then $S\circ T:X\to Z$ is a linear operator and thus $T^{*}$ is well-defined
and it is easy that if $(y_{\alpha})\subset Y^{*}$ is a net then
$T^{*}(y_{\alpha}^{*})\stackrel{{\scriptstyle
w^{*}}}{{\rightarrow}}T^{*}(y^{*})$.

###  2.2. Isomorphism of $\mathcal{B}_{0}$ and $\mathcal{B}$

It is well-known that the bidual of $\mathcal{B}_{0}$ is isomorphic to
$\mathcal{B}$ (see for instance [9]). However, this isomorphism is not an
isometry. Denote by $i$ the natural inclusion of $\mathcal{B}_{0}$ into its
bidual $\mathcal{B}_{0}^{**}$, that is,
$i:\mathcal{B}_{0}\to\mathcal{B}_{0}^{**}$ given by $i(f)(u)=u(f)$ for
$f\in\mathcal{B}_{0}$ and $u\in\mathcal{B}_{0}^{*}$.

Let $(z_{n})\subset\mathbb{D}$ such that $|z_{n}|\rightarrow 1$ when
$n\rightarrow\infty$. Define $T:\mathcal{B}_{0}\rightarrow c_{0}$ by:

$T(f):=\\{(1-|z_{n}|^{2})f^{\prime}(z_{n})\\}.$

Notice that $T$ is well-defined since $f\in\mathcal{B}_{0}$ and
$(1-|z_{n}|^{2})f^{\prime}(z_{n})\rightarrow 0$ because $|z_{n}|\rightarrow
1$. On the one hand, it is clear that the interpolating operator
$T:\mathcal{B}_{0}\to c_{0}$ can be extended by considering
$T^{**}:\mathcal{B}_{0}^{**}\to\ell_{\infty}$ since $T^{**}|_{i(B_{0})}=T$. On
the other hand, we can also consider the interpolating operator
$L:\mathcal{B}\to\ell_{\infty}$ given by
$L(f)=((1-|z_{n}|^{2})f^{\prime}(z_{n}))$ which is also a extension of $T$
since $L|_{\mathcal{B}_{0}}=T$. We will prove that $T^{**}$ and $L$ can be
perfectly identified since there exists an isomorphism
$R:\mathcal{B}_{0}^{**}\to\mathcal{B}$ such that $T^{**}=L\circ R$. In
particular, $R\circ i|_{\mathcal{B}_{0}}=Id|_{\mathcal{B}_{0}}$ where
$Id|_{\mathcal{B}_{0}}$ denotes the identity operator on $\mathcal{B}_{0}$.

Recall that the Bergman space $L_{a}^{1}$ is defined to be the set of analytic
functions $f:\mathbf{D}\rightarrow\mathbb{C}$ such that:

$||f||_{1}:=\left(\int_{\mathbb{D}}|f(z)|dm(z)\right)<\infty,$

where $m$ denotes the usual normalized Lebesgue measure on $\mathbf{D}$. It is
well-known that $L_{a}^{1}$ endowed with the norm $||\cdot||_{1}$ becomes a
Banach space. Given $g\in L^{1}_{a}$, define the mapping
$L_{g}:\mathcal{B}_{0}\rightarrow\mathbb{C}$ given by:

$L_{g}(f):=\int_{\mathbb{D}}f(z)\overline{g(z)}(1-|z|^{2})\ dm(z)\quad\mbox{
for }f\in\mathcal{B}_{0}$

which is clearly linear and continuous, so $L_{g}\in\mathcal{B}_{0}^{*}$.
Hence the mapping $R_{1}:L^{1}_{a}\rightarrow(\mathcal{B}_{0})^{*}$ given by
$R_{1}(g):=L_{g}$ is well-defined and it becomes an anti-linear bijection
correspondence (see Theorem 5.15 in [9]).

In the same way, given $h\in\mathcal{B}$ we have that
$S_{h}:L^{1}_{a}\rightarrow\mathbb{C}$ defined by:

$S_{h}(f):=\int_{\mathbb{D}}f(z)\overline{h(z)}(1-|z|^{2})\ dm(z)\quad\mbox{
for }f\in L^{1}_{a}$

satisfies that $S_{h}\in(L_{a}^{1})^{*}$ and the mapping
$R_{2}:\mathcal{B}\rightarrow(L^{1}_{a})^{*}$ given by $R_{2}(h):=S_{h}$ is
also well-defined and it becomes an anti-linear bijection correspondence (see
Theorem 5.3 in [9]).

######  Proposition 2.1.

The mapping $R=R_{2}^{-1}\circ R_{1}^{*}:\mathcal{B}_{0}^{**}\to\mathcal{B}$
defines a linear isomorphism and $R(i(f))=f$ for any $f\in\mathcal{B}_{0}$.

Proof. Notice that $R$ is a bijection since it is a composition of bijections.
It is also clear that $R$ is linear since it is the composition of two anti-
linear mappings. Notice that for any $f\in\mathcal{B}_{0}$ and $g\in
L_{a}^{1}$ we have:

$\displaystyle
R_{1}^{*}(i(f))(g)=\overline{i(f)}(R_{1}(g))=\overline{i(f)}(L_{g})=\overline{i(f)(L_{g})}=\overline{L_{g}(f)}=$
$\displaystyle\overline{\int_{\mathbb{D}}f(z)\overline{g(z)}(1-|z|^{2})\
dm(z)}=\int_{\mathbb{D}}\overline{f(z)}g(z)(1-|z|^{2})\ dm(z)=S_{f}(g)$

so:

$R_{2}^{-1}(R_{1}^{*}(i(f))=R_{2}^{-1}(S_{f})=f$

and we are done. ∎

Notice that from Proposition 2.1 we have that for any $f\in\mathcal{B}_{0}$:

(2.1) $\displaystyle R_{1}^{*}(i(f))(g)=S_{f}(g)\ \mbox{ for any }\ g\in
L_{a}^{1}.$

The following remark is clear since $L_{a}^{1}$ is separable, so bounded sets
of $\mathcal{B}$ are metrizable and convergence in $\mathcal{B}$ can be
characterized by sequences instead of nets:

######  Remark 2.2.

Consider the isomorphism $R_{2}:\mathcal{B}\to(L_{a}^{1})^{*}$ defined above.
The topology $\tau$ inherited by the weak-star topology from $(L_{a}^{1})^{*}$
to $\mathcal{B}$ is given by the following convergence:
$h_{n}\stackrel{{\scriptstyle\tau}}{{\to}}h$ in $\mathcal{B}$ if and only if
$S_{h_{n}}(f)\rightarrow S_{h}(f)$ for any $f\in L_{a}^{1}$.

The following result proves that primitives of Bloch functions are also Bloch
functions and will be used in Theorem 2.4.

######  Proposition 2.3.

For any $f\in\mathcal{B}$ we have that
$\sup_{z\in\mathbf{D}}|f(z)|(1-|z|^{2})\leq||f||_{Bloch}$.

Proof. Let $f\in\mathcal{B}$. It is clear that:

$f(z)-f(0)=\int_{[0,z]}f^{\prime}(w)\ dw=\int_{0}^{1}f^{\prime}(zt)zdt$

and thus:

$\displaystyle(1-|z|^{2})|f(z)-f(0)|=(1-|z|^{2})\left|\int_{0}^{1}f^{\prime}(zt)z\
dt\right|\leq$
$\displaystyle(1-|z|^{2})|z|\sup_{t\in[0,1]}|f^{\prime}(zt)|=(1-|z|^{2})|z||f^{\prime}(zt_{0})|$

for some $t_{0}\in[0,1]$ such that
$|f^{\prime}(zt_{0})|=\max_{t\in[0,1]}|f^{\prime}(zt)|$ since $f^{\prime}(zt)$
is continuous in $[0,1]$. Hence:

$(1-|z|^{2})|f(z)-f(0)|\leq(1-|z|^{2})|z||f^{\prime}(zt_{0})|\leq(1-|zt_{0}|^{2})|z||f^{\prime}(zt_{0})|\leq\|f\|_{\mathcal{B}}$

so:

$(1-|z|^{2})|f(z)|\leq(1-|z|^{2})|f(0)|+\|f\|_{\mathcal{B}}\leq|f(0)|+\|f\|_{\mathcal{B}}=\|f\|_{Bloch}$

and we are done. ∎

Hence:

######  Theorem 2.4.

The isomorphism $R:(\mathcal{B}_{0})^{**}\rightarrow\mathcal{B}$ defined as
above satisfies $T^{**}=L\circ R$.

Proof. If $h\in(\mathcal{B}_{0})^{**}$ satisfies $h=i(f)$ for some
$f\in\mathcal{B}_{0}$, then $R(i(f))=f$ by Propositon 2.1 and just making
calculations so $(L\circ R)(i(f))=L(f)=(f^{\prime}(z_{n})(1-|z_{n}|^{2}))$ and
$T^{**}(i(f))=T(f)=(f^{\prime}(z_{n})(1-|z_{n}|^{2}))$ so both sequences are
the same. Otherwise $h\in\mathcal{B}_{0}^{**}\setminus\mathcal{B}_{0}$ and
there exists $(f_{n})\subset\mathcal{B}_{0}$ such that $i(f_{n})\rightarrow h$
for the weak-star topology of $\mathcal{B}_{0}^{**}$. Notice that $R_{1}^{*}$
is an adjoint map so it is $w^{*}-w^{*}-$continuous and we obtain by (2.1):

$R_{1}^{*}(h)=R_{1}^{*}(\lim_{n\rightarrow\infty}i(f_{n}))=\lim
R_{1}^{*}(i(f_{n}))=\lim_{n\rightarrow\infty}S_{f_{n}}$

and since $i(f_{n})$ is weak-star convergent and $R_{1}^{*}$ is
$w^{*}-w^{*}-$continuous, then $S_{f_{n}}$ is weak-star convergent to
$S_{f_{0}}$ for some $f_{0}\in\mathcal{B}$. Hence
$R(h)=R_{2}^{-1}(h)=R_{2}^{-1}(S_{f_{0}})=f_{0}$ and:

(2.2) $\displaystyle\hskip
22.76228ptT^{**}(h)=w^{*}-\lim_{n\rightarrow\infty}T^{**}(i(f_{n}))=w^{*}-\lim_{n\rightarrow\infty}T(f_{n})=w^{*}-\lim_{n\rightarrow\infty}(f_{n}^{\prime}(z_{k})(1-|z_{k}|^{2}))$

and:

(2.3) $\displaystyle L(R(h))=L(f_{0})=(f_{0}^{\prime}(z_{k})(1-|z_{k}|^{2}))$

so we will show that sequences given in (2.2) and (2.3) are the same. For it,
we will prove that $f_{n}^{\prime}(z)$ converges to $f_{0}^{\prime}(z)$ for
any $z\in\mathbf{D}$. Since $i(f_{n})$ is weak-star convergent, then it is
bounded, so $(f_{n})\subset\mathcal{B}_{0}$ is bounded, that is, there exists
$M>0$ such that $\|f_{n}\|_{Bloch}\leq M$, so
$|f_{n}(0)|+\sup_{z\in\mathbf{D}}(1-|z|^{2})|f_{n}^{\prime}(z)|\leq M$ for any
$n\in\mathbb{N}$. Hence the sequence $(f_{n}(0))$ is bounded and
$(f_{n}^{\prime})$ is clearly bounded on any compact subset of $\mathbf{D}$.

Since $(f_{n}(0))$ is bounded, there exists a subsequence $(f_{n_{k}}(0))$
which is convergent to some $a\in\mathbb{C}$. Consider the sequence
$(f_{n_{k}}^{\prime})$. By the Montel’s Theorem, taking subsequences if
necessary, there exists an analytic function which can be supposed to be of
the form $g^{\prime}$ defined on $\mathbf{D}$ such that $f_{n_{k}}^{\prime}$
converges to $g^{\prime}$ on compact sets (recall that any analytic function
can be shown to be the derivative of another analytic function). Take a
primitive $g$ of $g^{\prime}$ such that $g(0)=a$. Notice that $g$ belongs to
$\mathcal{B}$. Indeed,
$\sup_{z\in\mathbf{D}}(1-|z|^{2})|g^{\prime}(z)|=\lim_{n\rightarrow\infty}\sup_{z\in\mathbf{D}}(1-|z|^{2})|f_{n_{k}}^{\prime}(z)|\leq
M$. It is easy that $f_{n_{k}}$ converges to $g$ for the compact-open
topology.

For any $h\in L_{a}^{1}$ we have $f_{n_{k}}(z)h(z)(1-|z|^{2})\rightarrow
g(z)h(z)(1-|z|^{2})$ pointwise. By Proposition 2.3, we have that
$f_{n_{k}}(z)h(z)(1-|z|^{2})$ is bounded by $Kh(z)$ which is integrable since
$h\in L_{a}^{1}$. We conclude that:

$\int_{\mathbf{D}}f_{n_{k}}(z)h(z)(1-|z|^{2})dz\rightarrow\int_{\mathbf{D}}g(z)h(z)(1-|z|^{2})dz$

by the Dominated Convergence Theorem, so $S_{f_{n_{k}}}\rightarrow S_{g}$ and
applying the isomorphism $R_{2}^{-1}$ we have that $f_{n_{k}}\rightarrow g$.
Bearing in mind that $S_{f_{n}}\rightarrow S_{f_{0}}$, then $g=f_{0}$ and we
are done since $f_{n}$ converges to $g=f_{0}$ on compact sets. ∎

###  2.3. Interpolating sequences for $\mathcal{B}$ and $c_{0}-$interpolating
sequences for $\mathcal{B}_{0}$

In the following we gather some well-known about linear operators
$S:X\rightarrow Y$ for normed spaces $X$ and $Y$.

######  Proposition 2.5.

Let $X$ and $Y$ be normed spaces and let $S:X\rightarrow Y$ be a linear
operator. Then the following statements hold:

  1. a)

$S(X)$ is dense in $Y$ if and only if $S^{*}$ es injective.

  2. b)

If $S^{*}(Y^{*})$ is dense in $X^{*}$ then $S$ is injective.

Furthermore, if $X$ and $Y$ are both Banach spaces, then:

  1. c)

If $S$ is injective and $S(X)$ is closed in $Y$, then $S^{*}$ es surjective.

  2. d)

If $S^{*}$ is injective and $S^{*}(Y^{*})$ is closed in $X^{*}$, then $S$ es
surjective.

Dealing with the interpolating operator $T:\mathcal{B}_{0}\to c_{0}$ we have:

######  Theorem 2.1.

The operator $T:\mathcal{B}_{0}\to c_{0}$ is surjective if and only if
$T^{**}:\mathcal{B}_{0}^{**}\to\ell_{\infty}$ is surjective.

Proof. Since $\mathcal{B}_{0}$ and $c_{0}$ are both Banach spaces, applying
twice the Closed Range Theorem to $T$ and $T^{*}$, we conclude that $T$ has
closed range if and only if $T^{*}$ has a closed range if and only if $T^{**}$
has closed range.

Suppose that $T$ is surjective, so $T$ has a closed range and $T^{*}$ also has
a closed range. Since $T$ is surjective, by Proposition 2.5 a) we have that
$T^{*}$ is also injective. In addition, $T^{*}$ has a closed range, so
$T^{**}$ is surjective by Proposition 2.5 c).

Conversely, if $T^{**}$ is surjective, we have that $T^{*}$ has closed range
and that by Proposition 2.5 b) it is also injective, and by Proposition 2.5 d)
we conclude that $T$ has a closed dense range, and is thus surjective.

######  Corollary 2.6.

A sequence $(z_{n})\subset\mathbb{D}$ is an interpolating sequence for
$\mathcal{B}$ if and only if it is a $c_{0}$-interpolating sequence for
$\mathcal{B}_{0}$.

Proof. The result is clear by Proposition 2.1 and Theorem 2.4. ∎

As we have mentioned in the Introduction, we prove that $c_{0}-$interpolating
sequences for $\mathcal{B}$ are interpolating for $\mathcal{B}$.

######  Theorem 2.2.

Let $(z_{n})\subset\mathbb{D}$ and suppose that for any $(a_{n})\in c_{0}$
there exists $f\in\mathcal{B}$ such that
$(1-|z_{n}|^{2})f^{\prime}(z_{n})=a_{n}$ for any $n\in\mathbb{N}$. Then
$(z_{n})$ is an interpolating sequence for $\mathcal{B}$.

Proof. Let $a\in\ell_{\infty}$. Since $c_{0}^{**}=\ell_{1}^{*}=\ell_{\infty}$
and $\ell_{1}$ is separable, then bounded sets of $\ell_{\infty}$ are
$w^{*}-$metrizable. By Goldstein’s Theorem, we have that there exists
$\\{b_{k}\\}\subset c_{0}$ such that $b_{k}\stackrel{{\scriptstyle
w^{*}}}{{\rightarrow}}a$ when $k\rightarrow\infty$, so in particular
$\\{b_{k}\\}$ is a norm bounded sequence. Notice that by hypothesis, for any
$k\in\mathbb{N}$ there exists $g_{k}\in\mathcal{B}$ such that
$R(g_{k})=b_{k}$. Consider the interpolating operator
$R:\mathcal{B}\to\ell_{\infty}$ given by
$R(f)=((1-|z_{n}|^{2})f^{\prime}(z_{n}))$ and take $A=R^{-1}(c_{0})$. Since
$c_{0}$ is a closed linear subspace of $\ell_{\infty}$, then $A$ is also a
closed linear subspace of $\mathcal{B}$, hence a Banach space and we can
consider the linear operator $R|_{A}:A\to c_{0}$ which is surjective by
hypothesis. By the Open Mapping Theorem, since $\\{b_{k}\\}$ is a bounded
sequence in $c_{0}$, there exists $M>0$ such that $\|g_{k}\|_{\mathcal{B}}\leq
M$ for any $k\in\mathbb{N}$. Since $\mathcal{B}=(L_{a}^{1})^{*}$ and
$\|g_{k}\|_{\mathcal{B}}\leq M$, and we know by Alaoglu’s theorem that the
closed unit ball centered at $0$ and radius $M$ is $w^{*}-$compact, there
exists a $w^{*}-$convergent subsequence $g_{k_{m}}$ which $w^{*}-$converges to
$g\in\mathcal{B}$. Notice that $R:\mathcal{B}\to\ell_{\infty}$ is
$w^{*}-w^{*}-$continuous since $R=\left(R|_{\mathcal{B}_{0}}\right)^{**}$ and
in particular, $R$ is an adjoint operator. Hence $R(g_{k_{m}})=b_{k_{m}}$
converges to $R(g)$. However, it is clear that $b_{k_{m}}$ converges to $a$,
so $R(g)=a$ because the limit is unique and we are done. ∎

Hence the following corollary is clear:

######  Corollary 2.7.

Let $(z_{n})\subset\mathbb{D}$. Then the following assertions are equivalent:

  * •

$(z_{n})$ is $c_{0}-$interpolating for $\mathcal{B}$.

  * •

$(z_{n})$ is $c_{0}-$interpolating for $\mathcal{B}_{0}$.

  * •

$(z_{n})$ is interpolating for $\mathcal{B}$.

  * •

$(z_{n})$ is interpolating for $\mathcal{B}$ only considering idempotents of
$\ell_{\infty}$.

###  2.4. Examples of interpolating sequences for $\mathcal{B}$

As we have mentioned in the Introduction, sufficiently separated sequences in
$\mathbf{D}$ for the pseudohyperbolic distance are interpolating for the Bloch
space $\mathcal{B}$. However, we will provide examples of interpolating
sequences for $\mathcal{B}$ which are so close as we want, that is, for any
$\varepsilon>0$, these sequences will be $\varepsilon-$separated and
$\varepsilon$ will not be able to be increased. First we prove that
interpolating sequences for $H^{\infty}$ are also interpolating for
$\mathcal{B}$. For this, we adapt and modify Proposition 5.1 in [3].

######  Proposition 2.8.

If $(z_{n})$ is interpolating for $H^{\infty}$ then it is interpolating for
$\mathcal{B}$.

Proof. Let $(z_{n})$ an interpolating sequence for $H^{\infty}$. Then there
exists a sequence of Beuring functions $(f_{k})\subset H^{\infty}$, that is,
for any $k\in\mathbb{N}$ it is satisfied that $f_{k}(z_{k})=1$,
$f_{k}(z_{j})=0$ if $j\neq k$ and there exists $M>0$ such that
$\sum_{k=1}^{\infty}|f_{k}(z)|\leq M$ for any $z\in\mathbf{D}$ (see [5]). For
any $w\in\mathbf{D}\setminus\\{0\\}$ consider $g_{w}:\mathbf{D}\to\mathbb{C}$
given by:

$g_{w}(z)=\frac{1-|w|^{2}}{(1-\bar{w}z)^{2}}$

which is clearly analytic on $z\in\mathbf{D}$. For any
$(a_{n})\in\ell_{\infty}$ define $f:\mathbf{D}\to\mathbb{C}$ such that:

$f^{\prime}(z)=\sum_{k=1}^{\infty}a_{k}g_{z_{k}}(z)f_{k}(z).$

It is clear that $f$ is analytic and $f\in\mathcal{B}$ since:

$|f^{\prime}(z)|(1-|z|^{2})|\leq
M\sup_{k\in\mathbb{N}}|g_{z_{k}}(z)|(1-|z|^{2})\leq
M\sup_{k\in\mathbb{N}}\frac{(1-|z_{k}|^{2})(1-|z|^{2})}{|1-\bar{z_{k}}z|^{2}}\leq
4M$

and satisfies $g_{z_{k}}(z_{k})=\frac{1}{1-|z_{k}|^{2}}$ so
$f^{\prime}(z_{k})(1-|z_{k}|^{2})=a_{k}$ for any $k\in\mathbb{N}$ so we are
done. ∎

Recall that a sequence $(z_{n})\subset\mathbf{D}$ is said to be a Blaschke
sequence if $\sum_{n=1}^{\infty}(1-|z_{n}|)<\infty$. It is well-known that the
sequence of zeros of a non-null bounded analytic function on $\mathbf{D}$
satisfies the Blaschke condition (see [6]). It is also well-known that
interpolating sequences for $H^{\infty}$ satisfy the Blaschke condition.
However, there exist interpolating sequences for $\mathcal{B}$ which does not
satisfy this condition (see for instance [3]).

######  Proposition 2.9.

If $(z_{n})\subset\mathbf{D}$ is interpolating for $H^{\infty}$ and
$z_{0}\in\mathbf{D}$ is such that $z_{0}\neq z_{n}$ for every
$n\in\mathbb{N}$, then the sequence $(w_{n})$ defined by $w_{1}=z_{0}$ and
$w_{n+1}=z_{n}$ for $n\geq 1$ is also interpolating for $H^{\infty}$.

Proof. Let $(z_{n})\subset\mathbf{D}$ an interpolating sequence for
$H^{\infty}$. As interpolating sequences in $H^{\infty}$ satisfy the Blaschke
condition, we can consider the Blaschke product $B:\mathbf{D}\to\mathbb{C}$ of
$(z_{n})$, which is a bounded analytic function which has the property that
$B(z)=0$ if and only if $z=z_{n}$ for some positive integer $n$.

Consider a bounded sequence $(a_{n})$. We know that there exists a function
$f\in H^{\infty}$ satisfying $f(z_{n})=a_{n+1}$ for every $n\in\mathbb{N}$, so
if we let $\alpha:=B(z_{0})\neq 0$ and define $g:\mathbf{D}\to\mathbb{C}$ by:

$g(z):=\frac{(a_{1}-f(z_{0}))}{\alpha}B(z)+f(z)$

we will have that $g\in H^{\infty}$ and $g(w_{n})=a_{n}$ for all $n$, so we
are done. ∎

It is well-known that interpolating sequences for $H^{\infty}$ are separated
for the pseudohyperbolic distance. With respect to necessary conditions,
Attele proved that this is also true for interpolating sequences for the Bloch
space $\mathcal{B}$ (see [1]). If we deal with sufficient conditions, as we
have mentioned above, he also proved that if a sequence is sufficiently
separated for the pseudohyperbolic distance, then it is interpolating for
$\mathcal{B}$. We prove that this condition is not necessary since we can find
interpolating sequences for $\mathcal{B}$ as close as we want:

######  Example 2.10.

For any $\varepsilon>0$ there exist interpolating sequences for $\mathcal{B}$
which are $\varepsilon-$separated and $\varepsilon$ cannot be increased.

Proof. Let $\varepsilon>0$. Consider an interpolating sequence $(z_{n})$ for
$H^{\infty}$, for example, $z_{n}:=1-\frac{1}{2^{n}}$, and add a point $z_{0}$
not in $(z_{n})$ such that $\rho(z_{0},z_{1})<\varepsilon$. By Proposition 2.9
the sequence given by $\\{z_{0}\\}\cup(z_{n})$ is interpolating for
$H^{\infty}$ hence interpolating for $\mathcal{B}$ by Proposition 2.8 and we
are done. ∎

##  3\. Interpolating idempotents of $\mathcal{B}$

###  3.1. Results on interpolating idempotents of $\mathcal{B}$

The following theorem is essentially an unpublished result of Beurling
mentioned in [2]. We wil use it to prove Theorem 3.2 and we include the proof
for the shake of completeness. Recall that a Banach space $E$ is said to be a
Schur space if for any $(x_{n})\subset E$ weakly convergent to $0$ satisfies
that $x_{n}\rightarrow 0$ in the norm of $E$.

######  Theorem 3.1 (Beurling).

If $X$ is a Banach space and $S:\ell_{1}\rightarrow X$ is a linear operator
such that $S^{*}(X^{*})$ is dense in $\ell_{\infty}=\ell_{1}^{*}$, then
$S^{*}(X^{*})=\ell_{\infty}$.

Proof. It is enough to prove that $S^{*}$ has closed range. Notice that this
is true if and only if $S$ has closed range because of the Closed Range
Theorem. We will show that $S:\ell^{1}\to S(X)$ is an isomorphism. By
hypothesis we have that $S^{*}(X^{*})$ is dense in $\ell^{\infty}$, which
implies that $S$ is injective by Proposition 2.5 a), so we only have to prove
that $S^{-1}$ is bounded. Otherwise, there would be a sequence $(x_{n})\subset
X$ such that $||x_{n}||_{X}=1$ but $||S^{-1}(x_{n})||_{1}\geq n$ for every
$n\in\mathbb{N}$. Thus the sequence $(\xi_{n})$ defined by:

$\xi_{n}:=\frac{S^{-1}(a_{n})}{||S^{-1}(a_{n})||_{1}}\in\ell^{1}$

satisfies $||(\xi_{n})||_{1}=1$ and $S(\xi_{n})\rightarrow 0$ in $X$ when
$n\rightarrow\infty$. We claim that $\\{\xi_{n}\\}$ converges weakly to $0$,
since for every $x^{*}\in X^{*}$ we have that $x^{*}(S(\xi_{n}))\rightarrow
0$, and thereby:

$\sum_{m=1}^{\infty}\xi_{n}(m)S^{*}(x^{*})(m)=S^{*}(x^{*})(\xi_{n})=x^{*}(S(\xi_{n}))\rightarrow
0,$

which due to the density of $S^{*}(X^{*})$ in $\ell^{\infty}$ implies that
$\sum_{m=1}^{\infty}\xi_{n}(m)y(m)\rightarrow 0$ for every
$(y(m))\in\ell^{\infty}$. So $(\xi_{n})$ converges weakly to $0$ but it
doesn’t converge in norm since $||(\xi_{n})||_{1}=1$ for every $n$, which is a
contradiction since $\ell_{1}$ is a Schur space. ∎

######  Theorem 3.2.

Let $(z_{n})\subset\mathbb{D}$ such that for any idempotent
$(a_{n})\in\ell_{\infty}$, there exists $f\in\mathcal{B}$ such that
$(1-|z_{n}|^{2})f^{\prime}(z_{n})=a_{n}$ for any $n\in\mathbb{N}$. Then
$(z_{n})$ is an interpolating sequence for the $\mathcal{B}$.

Proof. We will prove that the interpolation operator
$L:\mathcal{B}\rightarrow\ell^{\infty}$ has dense range. Then, by Proposition
2.4 we will obtain that $T^{**}$ has also dense range and thus it is
surjective by Theorem 3.1. Then we will apply Proposition 2.4 to prove that
$L$ is surjective and we will be done. So first we prove that $L$ has dense
range. Indeed, consider $E\subseteq\mathbb{N}$ and denote by $\chi_{E}$ the
sequence in $\ell^{\infty}$ given by:

$\chi_{E}(n):=\left\\{\begin{array}[]{ll}1&\mbox{ si }n\in E\\\ 0&\mbox{ si
}n\in\mathbb{N}\setminus E.\end{array}\right.$

Define $S\subset\ell^{\infty}$ to be the set of functions on $\mathbb{N}$ of
the form $\sum_{i=1}^{m}a_{i}\chi_{A_{i}}$ such that $m\in\mathbb{N}$,
$a_{i}\in\mathbb{C}$ for any $i=1,2,\ldots,m$ and $(A_{i})$ are pairwise
disjoint sets. It is clear that $S$ is dense in $\ell^{\infty}$. Indeed, it
follows from the fact that $S$ is the set of simple functions in
$\ell_{\infty}=L^{\infty}(\mathbb{N},c)$ where $c$ is the cardinal measure.

Let $x\in S$, that is, $x:=\sum_{i=1}^{m}a_{i}\chi_{A_{i}}\in\ell^{\infty}$
where $(A_{i})$ are pairwise disjoint sets and $a_{i}$ are complex numbers. By
hypothesis, for any $i=1,2,\ldots,m$ there are functions $f_{i}\in\mathcal{B}$
such that:

$(1-|z_{n}|^{2})f_{i}^{\prime}(z_{n})=\chi_{A_{i}}(n)\quad\mbox{ for every
}n\in\mathbb{N}$

and defining $f:=\sum_{i=1}^{m}a_{i}f_{i}\in\mathcal{B}$ we have that:

$L(f)=((1-|z_{n}|^{2})f^{\prime}(z_{n}))=\left(\sum_{i=1}^{m}a_{i}[(1-|z_{n}|^{2})f_{i}^{\prime}(z_{n})]\right)=\left(\sum_{i=1}^{m}a_{i}\chi_{A_{i}}(n)\right)=x$

and we are done. ∎

###  3.2. Interpolating sequences for $H^{\infty}$

As we have mentioned at the beginning of Section 1.3, Hayman proved in [7]
that Theorem 3.2 is true in the context of interpolating sequences in
$H^{\infty}$. Indeed he proved that if $(z_{n})\subset\mathbf{D}$ interpolates
idempotents of $\ell_{\infty}$, then it is uniformly separated, which clearly
implies that $(z_{n})$ is interpolating for $H^{\infty}$ to classical
Carleson’s Theorem stated in Section 1.2. We will modify Hayman’s Theorem to
provide a new proof that $c_{0}-$interpolating sequences are interpolating for
$H^{\infty}$.

######  Theorem 3.3.

Suppose that $(z_{n})\subset\mathbb{D}$ is a $c_{0}-$interpolating sequence
for $H^{\infty}$. Then $(z_{n})$ is interpolating for $H^{\infty}$.

Proof. First observe that with our assumptions, there exists a non-zero
function $f\in H^{\infty}$ such that $f(z_{1})=1$ and $f(z_{n})=0$ for any
$n\geq 2$. Since the zeros of a bounded analytic function in $\mathbb{D}$
satisfy the Blaschke condition it follows that
$\sum_{n=1}^{\infty}(1-|z_{n}|)<\infty$.

We prove that $(z_{n})$ is interpolating. Otherwise, by Carleson’s Theorem
$(z_{n})$ would not be uniformly separated, and we could find an increasing
sequence of positive integers $(n_{p})$ such that
$|B_{n_{p}}(z_{n_{p}})|<\frac{1}{p^{2}}$ for any $p\in\mathbb{N}$. For every
$n\in\mathbb{N}$ consider the Möbius transformation
$b_{n}:\mathbf{D}\to\mathbf{D}$ given by:

$b_{n}(z):=\frac{z-z_{n}}{1-\overline{z_{n}}z}$

which satisfies that $\lim\limits_{m\rightarrow\infty}|b_{n}(z_{m})|=1$ and
thus, constructing a subsequence of $(n_{p})$ if necessary, we can suppose
that for every $p\in\mathbb{N}$ we have:

$|b_{n_{k}}(z_{n_{p}})|>\frac{1}{e^{1/2^{p-k}}}\quad\mbox{ for }k=1,...,p-1.$

Define $Q_{k}:\mathbf{D}\to\mathbb{C}$ given by:

$Q_{k}(z)=\prod_{p=1,p\neq k}^{\infty}|b_{n_{k}}(z_{n_{p}})|$

which satisfies:

$Q_{k}(z)>\prod_{p=1,p\neq
k}^{\infty}\frac{1}{e^{(1/2^{|p-k|})}}=\left(e^{\sum_{p=1,p\neq
k}^{\infty}(1/2^{|p-k|})}\right)^{-1}>\left(e^{2\sum_{n=1}^{\infty}(1/2^{n})}\right)^{-1}=\frac{1}{e^{2}},$

where we have used that $|b_{n_{k}}(z_{n_{p}})|=|b_{n_{p}}(z_{n_{k}})|$ to
prove the first inequality. By hypothesis, there is a function $f\in
H^{\infty}$ such that:

$\left\\{\begin{array}[]{cl}f(z_{n})=\frac{1}{p},&\quad\mbox{if there is some
}p\mbox{ such that }n=n_{p},\\\ f(z_{n})=0,&\quad\mbox{in other
case.}\end{array}\right.$

We write $M:=||f||_{\infty}<\infty$. For any fixed number $N$, we define:

$\varphi_{N}(z):=f(z)\cdot\prod_{n=1,n\notin\\{n_{p}\\}_{p}}^{N}\frac{1-\overline{z_{n}}z}{z-z_{n}}.$

Since $z_{n}$ is a zero of $f$ whenever $n$ is not in $(n_{p})$, the
singularities of $\varphi_{N}$ are all removable, and if we still denote
$\varphi_{N}$ its holomorphic extension we have that $\varphi_{N}\in
H^{\infty}$. So:

$\lim\limits_{|z|\rightarrow 1}|\varphi_{N}(z)|\leq
M\cdot\lim\limits_{|z|\rightarrow
1}\prod_{n=1,n\notin\\{n_{p}\\}_{p}}^{N}\left|\frac{1-\overline{z_{n}}z}{z-z_{n}}\right|=M,$

and by the Maximum Modulus Principle we conclude that $|\varphi_{N}(z)|\leq M$
for every $z\in\mathbb{D}$, and in particular:

$|f(z)|\leq
M\cdot\prod_{n=1,n\notin(n_{p})}^{N}\left|\frac{z-z_{n}}{1-\overline{z_{n}}z}\right|\mbox{
for every }z\in\mathbb{D}.$

Taking limits when $N\rightarrow\infty$ in the last inequality, we deduce
that:

$|f(z)|\leq
M\cdot\prod_{n=1,n\notin\\{n_{p}\\}_{p}}^{\infty}\left|\frac{z-z_{n}}{1-\overline{z_{n}}z}\right|$

for every $z\in\mathbb{D}$, and in particular in $z_{n_{k}}$ where $k$ is
chosen to be $k\geq Me^{2}$. Finally, we conclude that:

$\frac{1}{k}=|f(z_{n_{k}})|\leq
M\prod_{n=1,n\notin\\{n_{p}\\}_{p}}^{\infty}\left|\frac{z_{n_{k}}-z_{n}}{1-\overline{z_{n}}z_{n_{k}}}\right|=M\frac{|B_{n_{k}}(z_{n_{k}})|}{Q_{k}}<M\frac{e^{2}}{k^{2}},$

which is a contradiction, and the proof is completed. ∎

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