# Linearization of holomorphic Lipschitz functions

Richard Aron  Department of Mathematical Sciences, Kent State University,
Kent, OH 44242 USA<EMAIL_ADDRESS>,  Verónica Dimant  Departamento de
Matemática y Ciencias, Universidad de San Andrés, Vito Dumas 284, (B1644BID)
Victoria, Buenos Aires, Argentina and CONICET<EMAIL_ADDRESS>,  Luis C.
García-Lirola  Departamento de Matemáticas, Universidad de Zaragoza, 50009,
Zaragoza, Spain<EMAIL_ADDRESS>and  Manuel Maestre  Departamento de
Análisis Matemático, Universidad de Valencia, Doctor Moliner 50, 46100
Burjasot, Valencia, Spain<EMAIL_ADDRESS>
###### Abstract.

Let $X$ and $Y$ be complex Banach spaces with $B_{X}$ denoting the open unit
ball of $X.$ This paper studies various aspects of the holomorphic Lipschitz
space $\mathcal{H}L_{0}(B_{X},Y)$, endowed with the Lipschitz norm. This space
is the intersection of the spaces, $\operatorname{Lip}_{0}(B_{X},Y)$ of
Lipschitz mappings and $\mathcal{H}^{\infty}(B_{X},Y)$ of bounded holomorphic
mappings, from $B_{X}$ to $Y$. Thanks to the Dixmier-Ng theorem,
$\mathcal{H}L_{0}(B_{X},\mathbb{C})$ is indeed a dual space, whose predual
$\mathcal{G}_{0}(B_{X})$ shares linearization properties with both the
Lipschitz-free space and Dineen-Mujica predual of
$\mathcal{H}^{\infty}(B_{X})$. We explore the similarities and differences
between these spaces, and combine techniques to study the properties of the
space of holomorphic Lipschitz functions. In particular, we get that
$\mathcal{G}_{0}(B_{X})$ contains a 1-complemented subspace isometric to $X$
and that $\mathcal{G}_{0}(X)$ has the (metric) approximation property whenever
$X$ has it. We also analyze when $\mathcal{G}_{0}(B_{X})$ is a subspace of
$\mathcal{G}_{0}(B_{Y})$, and we obtain an analogous to Godefroy’s
characterization of functionals with a unique norm preserving extension to the
holomorphic Lipschitz context.

###### Key words and phrases:

Banach space; holomorphic function; Lipschitz function; Linearization;
Symmetric regularity

###### 2020 Mathematics Subject Classification:

Primary 46E15, 46E50; Secondary 46B20, 46B28

##  1\. Introduction

Linearizing non-linear functions is a typical procedure in infinite
dimensional analysis. Originating nearly 70 years ago with Grothendieck [31]
(and his research about linearization of bilinear mappings through the
projective tensor product), the practice of identifying spaces of continuous
_non-linear_ functions with spaces of continuous _linear_ mappings defined on
Banach spaces has proved to be a useful technique. Accordingly, the study of
geometric and topological properties of these _linearizing Banach spaces_ has
increasingly attracted interest.

Lipschitz functions (defined on pointed metric spaces) and holomorphic bounded
functions (defined on the open unit ball of a Banach space) are really
different both as sets and as function spaces. However, when looking at their
linearization processes several similarities emerge. The purpose of this
article is to study, in light of these resemblances, the new set of functions
consisting of the intersection of the previous sets. Lipschitz holomorphic
functions defined on the open unit ball of a Banach space taking the value 0
at 0 will be our focus of attention. In the exploration of this set we take
advantage of a result of Ng [39] concerning the existence of preduals and all
the background about related linearization processes.

We begin with a brief review of important terms and concepts. General
references for Lipschitz functions include [30] and [44] and a standard
reference for holomorphic functions on finite or infinite dimensional domains
is [38]. The linearization process for bounded holomorphic functions is
originally developed in [36]. A review about linearization procedures both for
Lipschitz functions and for bounded holomorphic functions appeared in the
recent survey [26] while a general approach to linearizing non-linear sets of
functions was settled in [17].

For a metric space $(M,d)$ and a Banach space $Y,$ let
$\operatorname{Lip}(M,Y)$ be the vector space of all $f\colon M\to Y$ such
that $\left\|{f(x_{1})-f(x_{2})}\right\|\leq Cd(x_{1},x_{2})$ for some $C>0$
and for all $x_{1}\neq x_{2}\in M$. The smallest $C$ in the above definition
is the Lipschitz constant of $f,L(f)$. Let $0\in M$ denote an arbitrary fixed
point. In order to get a normed space, we will be particularly interested in
the subspace $\operatorname{Lip}_{0}(M,Y)$ consisting of those
$f\in\operatorname{Lip}(M,Y)$ such that $f(0)=0.$ In this way, $L(f)=0$ if and
only if $f=0$, and so $\left\|{\cdot}\right\|=L(\cdot)$ defines a norm on
$\operatorname{Lip}_{0}(M,Y)$.

For complex Banach spaces $X$ and $Y$ and open set $U\subset X$, denote by
$\mathcal{H}^{\infty}(U,Y)$ the vector space of all $f\colon U\to Y$ such that
$f$ is holomorphic (i.e. complex Fréchet differentiable) and bounded on $U$,
endowed with the supremum norm. In both the Lipschitz and
$\mathcal{H}^{\infty}$ situations, if the range $Y=\mathbb{K},$ then the
notation is shortened to $\operatorname{Lip}_{0}(M)$ and
$\mathcal{H}^{\infty}(U)$.

It is known that $\operatorname{Lip}_{0}(M)$ and $\mathcal{H}^{\infty}(U)$ are
dual spaces and that in some special situations, the predual is unique. The
construction of a (or, in some cases, the) predual follows the same lines for
both the Lipschitz and $\mathcal{H}^{\infty}$ situations: Calling $X$ one of
$\operatorname{Lip}_{0}$ or $\mathcal{H}^{\infty},$ we consider those
functionals $\varphi\in X^{*}$ such that $\varphi|_{\overline{B}_{X}}$ is
continuous when $\overline{B}_{X}$ is endowed with the compact-open topology.
Among such functionals are the evaluations $f\leadsto\delta(x)(f)\equiv f(x)$
where $x$ ranges over the domain of $f\in X.$ In the case of
$\operatorname{Lip}_{0}(M),$ the closed span of the set of such $\varphi$ will
be denoted $\mathcal{F}(M)$ while the analogous closed span for
$\mathcal{H}^{\infty}(U)$ is $\mathcal{G}^{\infty}(U).$ Each of these is a
Banach space, being a closed subspace of $\operatorname{Lip}_{0}(M)^{*},$ and
$\mathcal{H}^{\infty}(U)^{*},$ respectively. Using a standard technique
developed by Ng [39], it follows that
$\mathcal{F}(M)^{*}\equiv\operatorname{Lip}_{0}(M)$ and
$\mathcal{G}^{\infty}(U)^{*}\equiv\mathcal{H}^{\infty}(U).$

Among the most important common features of $\operatorname{Lip}_{0}$ and
$\mathcal{H}^{\infty}$ is linearization. In each of the two cases below,
$\delta$ is the evaluation inclusion taking $x\leadsto\delta(x).$ Also, for
$f$ in either $\operatorname{Lip}_{0}(M,Y)$ or $\mathcal{H}^{\infty}(U,Y),$
$T_{f}$ is the unique linear mapping making the diagram commute. Moreover,
$\|f\|=\|T_{f}\|.$

$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{\delta}$$\textstyle{Y}$$\textstyle{\mathcal{F}(M)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{T_{f}}$
$\textstyle{U\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{\delta}$$\textstyle{Y}$$\textstyle{\mathcal{G}^{\infty}(U)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{T_{f}}$

#### Notation

$X,Y$ will stand for complex Banach spaces. We denote by $B_{X}$ (respectively
$S_{X}$) its open unit ball (respectively unit sphere). $\mathcal{L}(X,Y)$
denotes the space of continuous linear maps from $X$ to $Y$, and
$X^{*}=\mathcal{L}(X,\mathbb{C})$. $\mathcal{P}(^{m}X,Y)$ stands for the space
of continuous $m$-homogeneous polynomials, that is, those $P\colon X\to Y$ so
that there exists a continuous $m$-linear symmetric map $\widecheck{P}\colon
X\times\cdots\times X\to Y$ with $P(x)=\widecheck{P}(x,\ldots,x)$. We also
write $\mathcal{P}(^{m}X)=\mathcal{P}(^{m}X,\mathbb{C})$. We say that
$P\in\mathcal{P}(^{m}X,Y)$ is of _finite type_ if
$P(x)=\sum_{j=1}^{n}[x_{j}^{*}(x)]^{m}y_{j}$ for certain $x_{j}^{*}\in X^{*}$
and $y_{j}\in Y$. $\mathcal{P}_{f}(^{m}X,Y)$ stands for the space of finite
type $m$-homogeneous polynomials. Moreover, we set $\mathcal{P}(X,Y)$ (resp.
$\mathcal{P}_{f}(X,Y)$) to be the space of finite sums of continuous
homogeneous polynomials (resp. homogeneous polynomials of finite type) from
$X$ to $Y$. Also, $\mathbb{D}(z,r)$ (resp. $C(z,r)$) denotes the open disc
(resp. the circumference) in $\mathbb{C}$ centered at $z$ with radius $r$, in
particular $\mathbb{D}=\mathbb{D}(0,1)$.

Recall that $X$ is said to have the Bounded Approximation Property (BAP) if
there is $\lambda>0$ such that the identity $I\colon X\to X$ can be
approximated by finite-rank operators in $\lambda B_{\mathcal{L}(X,X)}$
uniformly on compact sets (equivalently, pointwise). If $\lambda=1$, then $X$
is said to have the Metric Approximation Property (MAP). If $X$ has
$\lambda$-BAP and $Y$ is $\lambda^{\prime}$-complemented in $X$, then $Y$ has
$\lambda\lambda^{\prime}$-BAP. Recall, also, the version of this notion
without control of the norms: $X$ has the Approximation Property (AP) if the
identity $I\colon X\to X$ can be approximated by finite-rank operators in
$\mathcal{L}(X,X)$ uniformly on compact sets. We refer the reader to [18] for
examples and applications.

#### Organization of the paper

Section 2 introduces the main space of interest, $\mathcal{H}L_{0}(B_{X},Y),$
consisting of those functions that are in both
$\operatorname{Lip}_{0}(B_{X},Y)$ and $\mathcal{H}^{\infty}(B_{X},Y).$ A
number of properties of $\mathcal{H}L_{0}(B_{X},Y)$ are discussed and it is
proved that this space really differs from $\operatorname{Lip}_{0}(B_{X},Y)$
and $\mathcal{H}^{\infty}(B_{X},Y)$ (in the sense that a nonseparable space
can be injected in between). Then we focus on the predual
$\mathcal{G}_{0}(B_{X})$ of $\mathcal{H}L_{0}(B_{X})$ (where $Y=\mathbb{C}$).
Specifically, we will see that $\mathcal{H}L_{0}(B_{X})$ has a canonical
predual whose properties echo those of $\mathcal{H}^{\infty}(B_{X})$ and
$\operatorname{Lip}_{0}(B_{X}).$ When $X=\mathbb{C}$ with open unit disc
$\mathbb{D},$ one consequence of our work is a characterization of the extreme
points of the closed ball of $\mathcal{H}L_{0}(\mathbb{D})$ and of the norm
attaining elements of $\mathcal{H}L_{0}(\mathbb{D})$ considered as the dual of
$\mathcal{G}_{0}(\mathbb{D})$. In Section 3 we deal with the (metric)
approximation property for $\mathcal{G}_{0}(B_{X})$, again inspired by the
results for $\mathcal{G}^{\infty}(B_{X})$. The final two sections involve a
closer inspection of $\mathcal{G}_{0}(B_{X})$ and its relationship with
$\mathcal{G}_{0}(B_{X^{**}})$. Section 4 begins by considering the interaction
between $\mathcal{G}_{0}(B_{X})$ and $\mathcal{G}_{0}(B_{Y})$ when $X\subset
Y$ and then focuses on the case of $X\subset X^{\ast\ast}$. The final section
studies a natural connection between $\mathcal{G}_{0}(B_{X^{**}})$ and
$\mathcal{G}_{0}(B_{X})^{**}$ under the hypothesis of $X^{**}$ having the MAP.
Among other things, this enables us to characterize, under natural conditions
on $X$ and $X^{\ast\ast},$ when a function $f\in\mathcal{H}L_{0}(B_{X})$ has a
unique norm preserving extension to $\mathcal{H}L_{0}(B_{X^{\ast\ast}}).$ Both
sections make use of the concept of _(Arens) symmetric regularity,_ which is
reviewed in Section 4.

##  2\. The space of holomorphic Lipschitz functions and its predual

In the case that the metric space $M$ is $B_{X}$, the open unit ball of a
complex Banach space $X$, and $Y$ is another complex Banach space,
$\operatorname{Lip}_{0}(B_{X},Y)$ is the space of Lipschitz functions $f\colon
B_{X}\to Y$ with $f(0)=0$ and:

$L(f)=\sup\left\\{\frac{\|f(x)-f(y)\|}{\|x-y\|}:\ x\neq y\in B_{X}\right\\}.$

It is well known that $L(\cdot)$ defines a norm on
$\operatorname{Lip}_{0}(B_{X},Y)$ and
$(\operatorname{Lip}_{0}(B_{X},Y),L(\cdot))$ is a Banach space. Indeed,
$\operatorname{Lip}_{0}(B_{X},Y)$ is isometrically isomorphic to the space of
operators $\mathcal{L}(\mathcal{F}(B_{X}),Y)$, where $\mathcal{F}(B_{X})$
denotes the _Lipschitz-free space_ over $B_{X}$ (see e.g. [28, 44]).

Next, $\mathcal{H}^{\infty}(B_{X},Y)$ stands for the space of bounded
holomorphic functions from $B_{X}$ to $Y$, which is a Banach space when
endowed with the supremum norm. Analogous to the Lipschitz case above, we have
that $\mathcal{H}^{\infty}(B_{X},Y)$ is isometrically isomorphic to
$\mathcal{L}(\mathcal{G}^{\infty}(B_{X}),Y)$, where
$\mathcal{G}^{\infty}(B_{X})$ is Mujica’s canonical predual of
$\mathcal{H}^{\infty}(B_{X})$ [36] (we will review the space
$\mathcal{G}^{\infty}(B_{X})$ later in this section).

The parallel behavior of these Lipschitz and $\mathcal{H}^{\infty}$ spaces was
the authors’ motivation to introduce and study the following space and its
canonical predual:

$\displaystyle\mathcal{H}L_{0}(B_{X},Y)$
$\displaystyle=\operatorname{Lip}_{0}(B_{X},Y)\cap\mathcal{H}^{\infty}(B_{X},Y).$

We will also denote
$\mathcal{H}L_{0}(B_{X})=\mathcal{H}L_{0}(B_{X},\mathbb{C})$. Sometimes we
will deal with holomorphic Lipschitz functions without assuming $f(0)=0$, and
then we use the notation $\mathcal{H}L(B_{X},Y)$ and $\mathcal{H}L(B_{X})$.

Since both normed spaces $\mathcal{H}^{\infty}(B_{X},Y)$ and
$\operatorname{Lip}_{0}(B_{X},Y)$ are complete (with their respective norms)
and each $f\in\mathcal{H}L_{0}(B_{X},Y)$ satisfies $\|f\|_{\infty}\leq L(f)$
we easily derive that $\mathcal{H}L_{0}(B_{X},Y)$ is a Banach space with norm
$L(\cdot)$.

Given $f\in\mathcal{H}^{\infty}(B_{X},Y)$ such that
$df\in\mathcal{H}^{\infty}(B_{X},\mathcal{L}(X,Y))$ and $f(0)=0$, by the Mean
Value Theorem, we have that $\|f(x)-f(y)\|\leq\|df\|\|x-y\|$ for any $x,y\in
B_{X}.$ Then, $f\in\operatorname{Lip}_{0}(B_{X},Y)$ and $L(f)\leq\|df\|$.
Conversely, if $f\in\mathcal{H}L_{0}(B_{X},Y)$ we know that
$df\in\mathcal{H}(B_{X},\mathcal{L}(X,Y))$. Also, for $x,y\in B_{X}$,

$\|df(x)(y)\|=\lim_{h\to 0}\left\|\frac{f(x+hy)-f(x)}{h}\right\|\leq L(f).$

This means that $df$ belongs to $\mathcal{H}^{\infty}(B_{X},\mathcal{L}(X,Y))$
and $\|df\|\leq L(f)$.

This shows that there is another useful representation of our primary space of
interest.

######  Proposition 2.1.

$\mathcal{H}L_{0}(B_{X},Y)=\\{f\in\mathcal{H}^{\infty}(B_{X},Y):\
df\in\mathcal{H}^{\infty}(B_{X},\mathcal{L}(X,Y));\ f(0)=0\\}$. Moreover, for
every $f\in\mathcal{H}L_{0}(B_{X},Y),\ L(f)=\left\|{df}\right\|;$ that is,
$L(f)=\sup_{x\in B_{X}}\left\|{df(x)}\right\|.$

######  Remark 2.2.

If $f\colon B_{X}\to Y$ is a holomorphic function and $x\in B_{X}$ then
$f(x+h)=\sum_{m=1}^{\infty}P_{m}(x)(h)$ for $h$ in a suitable neighborhood of
0, where $P_{m}$ is an $m$-homogeneous polynomial. Recall that
$df(x)(h)=P_{1}(x)(h)$ for every $h\in X$.

Note that $P|_{B_{X}}\in\mathcal{H}L_{0}(B_{X},Y)$ for every
$P\in\mathcal{P}(X,Y)$ such that $P(0)=0$, a fact that will be useful later.

When $Y=\mathbb{C},$ we can define a mapping

$\displaystyle\Phi\colon\mathcal{H}L_{0}(B_{X})$
$\displaystyle\to\mathcal{H}^{\infty}(B_{X},X^{*})$ $\displaystyle f$
$\displaystyle\mapsto df$

In general, $\Phi$ is an isometry into $\mathcal{H}^{\infty}(B_{X},X^{*}),$
although if $X$ also equals $\mathbb{C}$, then $\Phi$ is onto. Indeed, in the
one-dimensional case, $\Phi$ is surjective since every holomorphic function
$f$ on $\mathbb{D}$ has a primitive that is Lipschitz whenever $f$ is bounded.
However, $\Phi$ is not surjective for $X\neq\mathbb{C}$. Indeed, given
$P\in\mathcal{P}(^{2}X)$, we have that $P|_{B_{X}}\in\mathcal{H}L_{0}(B_{X})$
and $dP\in\mathcal{L}(X,X^{*})$ is symmetric (i.e. $dP(x)(y)=dP(y)(x)$ for
every $x,y\in X$). Note that $df$ is linear only when $f$ is a 2-homogeneous
polynomial. Hence, a non-symmetric element of $\mathcal{L}(X,X^{*})$ (which
always exists whenever the dimension of $X$ is strictly bigger than one)
cannot be in the range of $\Phi$.

In particular, we see that

$\mathcal{H}L_{0}(\mathbb{D})=\\{f\in\mathcal{H}^{\infty}(\mathbb{D}):\
f(0)=0\ {\rm and\ }f^{\prime}\in\mathcal{H}^{\infty}(\mathbb{D})\\}.$

A lot of research has been done on $\mathcal{H}L_{0}(\mathbb{D})$ and on
$\mathcal{H}L_{0}(U)$ for certain domains $U\subset\mathbb{C}^{n}$ such as the
Euclidean ball. See, e.g., [1, 10, 11, 12, 14, 25, 40, 42] where this topic is
approached from different viewpoints than what is done here.

For the case of $\mathcal{H}L(B_{X},Y)$ we consider the norm
$\|f\|_{\mathcal{H}L}=\max\\{\|f(0)\|,L(f)\\}$. The fact that this is a norm
and that $(\mathcal{H}L(B_{X},Y),\|\cdot\|_{\mathcal{H}L})$ is a Banach space
follows easily. Note that $\|f\|_{\infty}\leq 2\|f\|_{\mathcal{H}L}$ for any
$f\in\mathcal{H}L(B_{X},Y)$. Also, it is plain to see that
$\mathcal{H}L_{0}(B_{X},Y)$ is a 1-complemented subspace of
$\mathcal{H}L(B_{X},Y)$. Moreover, motivated by a similar result for
$\operatorname{Lip}_{0}$-spaces (see [44, Th. 1.7.2]) we get:

######  Proposition 2.3.

Let $X,Y$ be complex Banach spaces. Then $\mathcal{H}L(B_{X},Y)$ is isometric
to a $1$-complemented subspace of
$\mathcal{H}L_{0}(B_{X\oplus_{1}\mathbb{C}},Y)$.

###### Proof.

Consider
$\Phi\colon\mathcal{H}L(B_{X},Y)\to\mathcal{H}L_{0}(B_{X\oplus_{1}\mathbb{C}},Y)$
given by $\Phi f(x,\lambda)=f(x)+(\lambda-1)f(0)$. It is easy to check that
$\Phi f$ is Lipschitz with $L(\Phi f)\leq\left\|{f}\right\|_{\mathcal{H}L}$
for every $f\in\mathcal{H}L(B_{X},Y)$. Note that

$L(\Phi f)\geq\sup\left\\{\frac{\left\|{\Phi f(x,0)-\Phi
f(y,0)}\right\|}{\left\|{x-y}\right\|}:x\neq y\in B_{X}\right\\}=L(f)$

and also

$L(\Phi f)\geq\frac{\left\|{\Phi f(0,1)-\Phi
f(0,0)}\right\|}{\left\|{(0,1)-(0,0)}\right\|_{1}}=\left\|{f(0)}\right\|,$

so we actually have $L(\Phi f)=\left\|{f}\right\|_{\mathcal{H}L}$. Thus $\Phi$
is an into isometry.

Now consider
$T\colon\mathcal{H}L_{0}(B_{X\oplus_{1}\mathbb{C}},Y)\to\mathcal{H}L(B_{X},Y)$
given by $Tg(x)=g(x,0)+g(0,1)$. One can easily check that
$\left\|{T}\right\|\leq 1$ and $T\circ\Phi=I_{\mathcal{H}L(B_{X},Y)}$.
Therefore $P=\Phi\circ T$ is a norm-one projection from
$\mathcal{H}L_{0}(B_{X\oplus_{1}\mathbb{C}},Y)$ onto
$\Phi(\mathcal{H}L(B_{X},Y))$.

∎

Note that there are plenty of examples of non-Lipschitz functions in
$\mathcal{H}^{\infty}(\mathbb{D})$. For instance, given a sequence
$(b_{n})\subset\mathbb{C}\setminus\\{1\\}$ with $|b_{n}|=1$ and $b_{n}\to 1$,
define $f\colon\\{b_{n}\\}\cup\\{1\\}\to\mathbb{C}$ by $f(1)=0$ and
$f(b_{n})=\sqrt{|b_{n}-1|}$. Then the Rudin-Carleson theorem provides an
extension of $f$ which lies in the disc algebra $\mathcal{A}(\mathbb{D})$
(that is, the space of uniformly continuous functions in
$\mathcal{H}^{\infty}(\mathbb{D})$) and has the same supremum norm, but it is
not Lipschitz.

Our next goal is to show that $\mathcal{H}L_{0}(B_{X})$ is indeed much smaller
than both $\mathcal{H}^{\infty}(B_{X})$ and $\operatorname{Lip}_{0}(B_{X})$.
More precisely, we will prove the following result, where we denote
$\mathcal{H}^{\infty}_{0}(B_{X})=\\{f\in\mathcal{H}^{\infty}(B_{X}):\
f(0)=0\\}$.

######  Theorem 2.4.

Let $X$ be a non-null complex Banach space. Then

  1. (a)

$\ell_{\infty}$ is isomorphic to a subspace of
$\mathcal{H}^{\infty}_{0}(B_{X})\setminus\mathcal{H}L_{0}(B_{X})\cup\\{0\\}$.

  2. (b)

$\ell_{\infty}$ is isomorphic to a subspace of
$\operatorname{Lip}_{0}(B_{X})\setminus\mathcal{H}L_{0}(B_{X})\cup\\{0\\}$.

We will provide a different proof of Theorem 2.4. $(a)$ in the Appendix.
Indeed, there we build an isomorphism into its image
$F\colon\ell_{\infty}\longrightarrow\mathcal{H}^{\infty}(B_{X})$ such that,
additionally, its restriction to $c_{0}$ satisfies that $F|_{c_{0}}\colon
c_{0}\longrightarrow\mathcal{A}(B_{X})$.

###### Proof.

$(a)$ For the case $X=\mathbb{C}$, it has been observed to the authors that a
classical result is that given $(z_{j})$ an interpolating sequence on
$\mathbb{D}$ there exists an topological isomorphism
$S\colon\ell_{\infty}\longrightarrow\mathcal{H}^{\infty}(\mathbb{D})$ such
that $S(c)(z_{j})=c_{j}$ for every $j$ and every $c=(c_{n})\in\ell_{\infty}$
(see e.g. [27, Theorem VII.2.1 and applications, p. 285] where it is made for
the upper half plane), we can also get that $S(c)(0)=0$. Hence, if
$\mathbb{N}$ is partitioned into infinitely many infinite sequences $n_{ik}$
with $i,k\in\mathbb{N}$ and for $c\in\ell_{\infty}$ it is defined
$x_{c}\in\ell_{\infty}$, $x_{c}(n_{ik})=(-1)^{i}c_{k}$, and
$Y=\\{S(x_{c}):c\in\ell_{\infty}\\}$, then $Y$ is an subspace of
$\mathcal{H}^{\infty}(\mathbb{D})$ isomorphic to $\ell_{\infty}$. And if
$c\neq 0$, then $c_{k}\neq 0$ for some $k$, and $S(x_{c})$ takes values $\pm
c_{k}$ along a sequence tending to 1. So it cannot be uniformly continuous,
hence it is not Lipschitz.

Now, to get the general case, we fix $x_{0}\in S_{X}$ and consider
$x^{\ast}\in X^{\ast}$ such that $x^{\ast}(x_{0})=1=\|x^{\ast}\|$. We define

$\Psi\colon\mathcal{H}^{\infty}(\mathbb{D})\longrightarrow\mathcal{H}^{\infty}(B_{X})$

by $\Psi(f)=f\circ x^{\ast}$. Clearly $\Psi$ is a well-defined linear mapping
and since $x^{\ast}(B_{X})=\mathbb{D}$ we have that $\Psi$ is an isometry onto
its image. Moreover, considering its restriction we are going to have

$\Psi\colon\mathcal{H}L(\mathbb{D})\longrightarrow\mathcal{H}L(B_{X}),$

that again is an isometry, now with the Lipschitz norms. Indeed, if
$f\in\mathcal{H}L(\mathbb{D})$ then

$L(\Psi(f))=L(f\circ x^{*})\leq L(f)L(x^{*})=L(f).$

But if $\lambda,\mu\in\mathbb{D}$, then

$\displaystyle|f(\lambda)-f(\mu)|=$ $\displaystyle|f\circ x^{\ast}(\lambda
x_{0})-f\circ x^{\ast}(\mu x_{0})|=|\Psi(f)(\lambda x_{0})-\Psi(f)(\mu
x_{0})|$ $\displaystyle\leq$ $\displaystyle L(\Psi(f))\|\lambda x_{0}-\mu
x_{0}\|=L(\Psi(f))|\lambda-\mu|,$

and we get $L(f)\leq L(\Psi(f))$. Finally, due to the injectivity of $\Psi$ (a
direct proof is also elementary) we have that

$\Psi(\mathcal{H}^{\infty}_{0}(\mathbb{D})\setminus\mathcal{H}L(\mathbb{D}))\subset\mathcal{H}^{\infty}_{0}(B_{X})\setminus\mathcal{H}L(B_{X}).$

Now the claim follows.

$(b)$ First we consider the 1-dimensional case $X=\mathbb{C}$. Let
$l\colon\mathbb{R}\to[0,1]$ be a $C^{\infty}$ function such that $l$ is
strictly increasing on $(1/2,1)$, $0<l(x)<1$ if $1/2<x<1$, $l(x)=0$ for $x\leq
1/2$, $l(x)=1$ for $x\geq 1$, $l^{(k)}(1/2)=0$ if $k\geq 0$ and $l^{(k)}(1)=0$
if $k\geq 1$. Define $f\colon\mathbb{C}\to[0,1]$ as $f(z)=l(|z|)$. Considered
it as being defined on $\mathbb{R}^{2}$, $f$ is $C^{\infty}$ and
$df\colon\mathbb{R}^{2}\to\mathbb{R}$ is a continuous function. Hence, by the
Mean Value Theorem, $f\in\operatorname{Lip}_{0}(\mathbb{D})$. Now we define
$T\colon\mathcal{H}L_{0}(\mathbb{D})\to\operatorname{Lip}_{0}(\mathbb{D})$ as
$T(g)=f\cdot g$. We claim that $T$ is an isomorphism onto its image. Indeed,
given $g\in\mathcal{H}L_{0}(\mathbb{D})$ and $z,u\in\mathbb{D}$,

$|f(z)g(z)-f(u)g(u)|\leq|f(z)-f(u)||g(z)|+|f(u)||g(z)-g(u)|\leq
2L(f)L(g)|z-u|.$

Thus $T$ is a continuous linear mapping with $\|T\|\leq 2L(f)$. Now we check
that $T$ is bounded below. As $f(x,y)=l(\sqrt{x^{2}+y^{2}})$ we have that
$df(x,y)=0$ if $z=x+iy$ satisfies $|z|\geq 1$. By continuity on a compact set,
given $\varepsilon>0$ there exists $0<r<1$ such that if $|z|\geq r$, then both
$\|df(x,y)\|<\varepsilon$ and $f(z)>1-\varepsilon$. Thus, for
$g\in\mathcal{H}L_{0}(\mathbb{D})$,

$L(fg)=\|d(fg)\|_{\mathbb{D}}=\|gdf+fg^{\prime}\|_{\mathbb{D}}\geq\|fg^{\prime}\|_{\mathbb{D}\setminus
r\mathbb{D}}-\|g\|_{\mathbb{D}}\|df\|_{\mathbb{D}\setminus
r\mathbb{D}}\geq\|fg^{\prime}\|_{\mathbb{D}\setminus
r\mathbb{D}}-L(g)\varepsilon.$

But, by the maximum modulus theorem

$\|fg^{\prime}\|_{\mathbb{D}\setminus
r\mathbb{D}}\geq(1-\varepsilon)\|g^{\prime}\|_{\mathbb{D}\setminus
r\mathbb{D}}=(1-\varepsilon)\|g^{\prime}\|_{\mathbb{D}}=(1-\varepsilon)L(g).$

and we get $L(fg)\geq(1-2\varepsilon)L(g)$, for every $\varepsilon>0$. As a
consequence

$L(Tg)=L(fg)\geq L(g),$

and $T$ is bounded below. Moreover, $T(g)=f\cdot g$ is never holomorphic on
$\mathbb{D}$ for any $g\in\mathcal{H}L_{0}(\mathbb{D})\setminus\\{0\\}$, and
$T(\mathcal{H}L_{0})(\mathbb{D})$ is isomorphic to
$\mathcal{H}L_{0}(\mathbb{D})$ which in turn is isometric to
$\mathcal{H}^{\infty}(\mathbb{D})$ that has a subspace isomorphic to
$\ell_{\infty}$.

The general case is a straightforward consequence of the above argument in the
following natural way. Let $X$ a non-null complex Banach space and take
$x^{\ast}\in S_{X^{\ast}}$. Defining
$R\colon\operatorname{Lip}_{0}(\mathbb{D})\to\operatorname{Lip}_{0}(B_{X})$ by
$R(h)=h\circ x^{\ast}$, we are going to have that $R$ is an isometry into.
Hence, $R\circ
T\colon\mathcal{H}L_{0}(\mathbb{D})\to\operatorname{Lip}_{0}(B_{X})$ is an
isomorphism into its image and we get that $\ell_{\infty}$ is isomorphic to a
subspace of $\mathcal{H}L_{0}(B_{X})$. But if
$g\in\mathcal{H}L_{0}(\mathbb{D})\setminus\\{0\\}$, then $R\circ T(g)=(f\cdot
g)x^{\ast}$ is not a Gateaux holomorphic function since its restriction to
$\\{zx\,:z\in\mathbb{D}\\}$ is not holomorphic. We conclude that
$\ell_{\infty}\setminus\\{0\\}\subset\operatorname{Lip}_{0}(B_{X})\setminus\mathcal{H}L_{0}(B_{X})$.
∎

In this rest of the section, we will focus the attention on the canonical
predual of space $\mathcal{H}L_{0}(B_{X})$ and show that it shares many
properties with the canonical preduals of $\mathcal{H}^{\infty}(B_{X})$ and
$\operatorname{Lip}_{0}(B_{X})$.

Let us denote by $\tau_{0}$ the compact-open topology on
$\mathcal{H}L_{0}(B_{X})$. An easy argument using Montel’s theorem [21, Th.
15.50] and Remark 2.2 shows that $\overline{B}_{\mathcal{H}L_{0}(B_{X})}$ is
$\tau_{0}$-compact. In fact, on this ball, convergence in the topology
$\tau_{0}$ coincides with pointwise convergence. Thus, the Dixmier-Ng theorem
[39] says that $\mathcal{H}L_{0}(B_{X})$ is a dual space with predual given by

$\mathcal{G}_{0}(B_{X}):=\\{\varphi\in\mathcal{H}L_{0}(B_{X})^{*}:\
\varphi|_{\overline{B}_{\mathcal{H}L_{0}(B_{X})}}\textrm{ is
}\tau_{0}-\textrm{continuous}\\}.$

For $x\in B_{X}$ and $f\in\mathcal{H}L_{0}(B_{X})$, denote
$\delta(x)(f)=f(x)$. Clearly
$\delta(x)\colon\mathcal{H}L_{0}(B_{X})\to\mathbb{C}$ is linear and continuous
meaning that $\delta(x)\in\mathcal{H}L_{0}(B_{X})^{*}$. Also,
$\delta(x)|_{\overline{B}_{\mathcal{H}L_{0}(B)}}$ is $\tau_{0}$-continuous so
$\delta(x)\in\mathcal{G}_{0}(B_{X})$.

######  Proposition 2.5.

Let $X$ be a complex Banach space.

  1. (a)

The mapping

$\displaystyle\delta\colon B_{X}$ $\displaystyle\to\mathcal{G}_{0}(B_{X})$
$\displaystyle x$ $\displaystyle\mapsto\delta(x)$

is holomorphic and $\left\|{\delta(x)-\delta(y)}\right\|=\left\|{x-y}\right\|$
for every $x,y\in B_{X}$. In particular,
$\delta\in\mathcal{H}L_{0}(B_{X},\mathcal{G}_{0}(B_{X}))$ with $L(\delta)=1$.

  2. (b)

$\mathcal{G}_{0}(B_{X})=\overline{\operatorname{span}}\\{\delta(x):\ x\in
B_{X}\\}$.

  3. (c)

For any complex Banach space $Y$ and any $f\in\mathcal{H}L_{0}(B_{X},Y)$,
there is a unique operator $T_{f}\in\mathcal{L}(\mathcal{G}_{0}(B_{X}),Y)$
such that the following diagram commutes:

$\textstyle{B_{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{\delta}$$\textstyle{Y}$$\textstyle{\mathcal{G}_{0}(B_{X})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{T_{f}}$

The map $f\mapsto T_{f}$ defines an isometric isomorphism from
$\mathcal{H}L_{0}(B_{X},Y)$ onto $\mathcal{L}(\mathcal{G}_{0}(B_{X}),Y)$.
These properties characterize $\mathcal{G}_{0}(B_{X})$ uniquely up to an
isometric isomorphism.

  4. (d)

A bounded net $(f_{\alpha})\subset\mathcal{H}L_{0}(B_{X})$ is weak-star
convergent to a function $f\in\mathcal{H}L_{0}(B_{X})$ if and only if
$f_{\alpha}(x)\to f(x)$ for every $x\in B_{X}$.

###### Proof.

$(a)$ The map $\delta$ is weakly holomorphic since for any
$f\in\mathcal{G}_{0}(B_{X})^{*}=\mathcal{H}L_{0}(B_{X})$ we have that
$f\circ\delta=f$ is holomorphic. Thus, $\delta$ is holomorphic (see [38, Th.
8.12]). Also, given $x,y\in B_{X}$, we have

$\left\|{\delta(x)-\delta(y)}\right\|=\sup_{f\in
B_{\mathcal{H}L_{0}(B_{X})}}|\langle f,\delta(x)-\delta(y)\rangle|=\sup_{f\in
B_{\mathcal{H}L_{0}(B_{X})}}|f(x)-f(y)|\leq\left\|{x-y}\right\|,$

and equality holds since we may take $f=x^{*}|_{B_{X}}$ where
$\left\|{x^{*}}\right\|=1$ and $x^{*}(x-y)=\left\|{x-y}\right\|$.

$(b)$ Just observe that for every
$f\in\mathcal{H}L_{0}(B_{X})=\mathcal{G}_{0}(B_{X})^{*}$ we have that $f=0$
whenever $f|_{\\{\delta(x):x\in B_{X}\\}}=0$.

$(c)$ First, note that an interpolation argument shows that the set
$\\{\delta(x):x\in B_{X}\setminus\\{0\\}\\}$ is linearly independent in
$\mathcal{G}_{0}(B_{X})$. Indeed, assume that
$\sum_{j=1}^{n}\lambda_{j}\delta(x_{j})=0$ for different points $x_{j}\in
B_{X}\setminus\\{0\\}$ and $\lambda_{j}\in\mathbb{C}$. Let $x_{0}=0$ and
$\lambda_{0}=0$. Take $x_{ij}^{*}\in S_{X^{*}}$ with
$x_{ij}^{*}(x_{i}-x_{j})=\left\|{x_{i}-x_{j}}\right\|$ and define
$f(x)=\sum_{j=0}^{n}\overline{\lambda_{j}}\prod_{i\neq
j}\frac{x_{ij}^{*}(x_{i}-x)}{\left\|{x_{i}-x_{j}}\right\|}$. Then
$f\in\mathcal{H}L_{0}(B_{X})$ and $0=\langle
f,\sum_{j=1}^{n}\lambda_{j}\delta(x_{j})\rangle=\sum_{j=1}^{n}|\lambda_{j}|^{2}$.

Now, given $f\in\mathcal{H}L_{0}(B_{X},Y)$, we define $T_{f}(\delta(x)):=f(x)$
for every $x\in B_{X}$ (this is the only possibility to get a commutative
diagram) and extend it linearly to $\operatorname{span}\\{\delta(x):x\in
B_{X}\\}$. Note that, given $u=\sum_{j=1}^{n}\lambda_{j}\delta(x_{j})$,

$\displaystyle\left\|{T_{f}u}\right\|$
$\displaystyle=\left\|{\sum_{j=1}^{n}\lambda_{j}f(x_{j})}\right\|=\sup_{y^{*}\in
B_{Y^{*}}}\left|\sum_{j=1}^{n}\lambda_{j}(y^{*}\circ
f)(x_{j})\right|=\sup_{y^{*}\in B_{Y^{*}}}\left|\langle u,y^{*}\circ
f\rangle\right|$ $\displaystyle\leq\sup\\{L(y^{*}\circ f):y^{*}\in
B_{Y}\\}\left\|{u}\right\|=L(f)\left\|{u}\right\|.$

Thus, $T_{f}$ extends uniquely to an operator
$T_{f}\in\mathcal{L}(\mathcal{G}_{0}(B),Y)$ with $\left\|{T_{f}}\right\|\leq
L(f)$. Since $L(\delta)=1$ and $f=T_{f}\circ\delta$, indeed we get that
$\left\|{T_{f}}\right\|=L(f)$.

Moreover, the map $f\mapsto T_{f}$ is onto since, given any $T\in
L(\mathcal{G}_{0}(B_{X}),Y)$, we have that $f:=T\circ\delta$ is a holomorphic
Lipschitz map with $f(0)=0$ and $T=T_{f}$.

The uniqueness of $\mathcal{G}_{0}(B_{X})$ follows from the diagram property
and the fact that $\left\|{T_{f}}\right\|=L(f)$.

$(d)$ The ball $\overline{B}_{\mathcal{H}L_{0}(B_{X})}$ is $\tau_{0}$-compact
and the weak-star topology is coarser than $\tau_{0}$, so they coincide on
$\overline{B}_{\mathcal{H}L_{0}(B_{X})}$. ∎

######  Proposition 2.6.

For every complex Banach space $X$ we have that $X$ is isometric to a
1-complemented subspace of $\mathcal{G}_{0}(B_{X})$.

###### Proof.

In the particular case of $f=Id\colon B_{X}\to X$, differentiating the diagram
in Proposition 2.5 and using that $d(Id)(x)=Id$ for all $x\in B_{X}$, we
obtain another commutative diagram where all the arrows are linear:

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{Id}$$\scriptstyle{d\delta(0)}$$\textstyle{X}$$\textstyle{\mathcal{G}_{0}(B_{X})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{T_{Id}}$

Moreover, $d\delta(0)$ is an isometry. Indeed, given $x\in X$ and
$f\in\mathcal{H}L_{0}(B_{X})$ we have

$\langle f,d\delta(0)(x)\rangle=\lim_{t\to 0}\langle
f,\frac{\delta(tx)-\delta(0)}{t}\rangle=\lim_{t\to
0}\frac{f(tx)-f(0)}{t}=df(0)(x)$

and so

$\left\|{d\delta(0)(x)}\right\|=\sup\\{|df(0)(x)|:f\in
B_{\mathcal{H}L_{0}(B_{X})}\\}\leq\left\|{x}\right\|.$

The other inequality is clear due to the commutative diagram:

$\|x\|=\|T_{Id}\circ d\delta(0)(x)\|\leq\|d\delta(0)(x)\|.$

Finally, let $P=d\delta(0)\circ T_{Id}$. Then, using that $T_{Id}\circ
d\delta(0)=Id$, we have

$P^{2}=d\delta(0)\circ T_{Id}\circ d\delta(0)\circ T_{Id}=d\delta(0)\circ
T_{Id}=P,$

so $P$ is a norm-one projection from $\mathcal{G}_{0}(B_{X})$ onto
$d\delta(0)(X)$. ∎

Note that this result also holds for $\mathcal{G}^{\infty}(B_{X})$ [36] but
not in general for $\mathcal{F}(B_{X})$. In [30] it is proved that this is
true for $X$ separable although for nonseparable $X$ it could even occur that
$\mathcal{F}(B_{X})$ does not contain a subspace isomorphic to $X$. Another
useful property of Lipschitz-free spaces is the fact that they contain a
complemented copy of $\ell_{1}$ [19], the same holds for
$\mathcal{G}_{0}(B_{X})$.

######  Proposition 2.7.

Let $X$ be a complex Banach space. Then there is a complemented subspace of
$\mathcal{G}_{0}(B_{X})$ isomorphic to $\ell_{1}$.

###### Proof.

$\ell_{\infty}$ is isomorphic to a subspace of
$\mathcal{H}^{\infty}(\mathbb{D})$. Since $\mathcal{H}^{\infty}(\mathbb{D})$
is isometric to $\mathcal{H}L_{0}(\mathbb{D})$, which is a complemented
subspace in $\mathcal{H}L_{0}(B_{X})$, the same holds for
$\mathcal{H}L_{0}(B_{X})$. It is a classical result (see [13, Th. 4]) that
this implies its predual $\mathcal{G}_{0}(B_{X})$ contains a complemented copy
of $\ell_{1}$. ∎

Next, we want to describe the closed unit ball of $\mathcal{G}_{0}(B_{X})$.
For that, we introduce some more notation. We denote by $\operatorname{conv}$
the convex hull of a set and by $\Gamma$ the absolute convex hull of a set. As
usual in the Lipschitz world, for every $x,y\in B_{X}$ with $x\neq y$,
$m_{x,y}$ stands for the _elementary molecule_
$\frac{\delta(x)-\delta(y)}{\|x-y\|}$. Also, for every $x\in B_{X}$, $y\in X$
and $f\in\mathcal{H}L_{0}(B_{X})$, we denote $e_{x,y}(f):=df(x)(y)$. Then
$e_{x,y}\in\mathcal{G}_{0}(B_{X})$ with
$\left\|{e_{x,y}}\right\|=\left\|{y}\right\|$. Indeed, it is clear that

$\left\|{e_{x,y}}\right\|=\sup\\{|df(x)(y)|:f\in
B_{\mathcal{H}L_{0}(B_{X})}\\}\leq\sup\\{\left\|{df(x)}\right\|:f\in
B_{\mathcal{H}L_{0}(B_{X})}\\}\left\|{y}\right\|\leq\left\|{y}\right\|.$

Conversely, take $x^{*}\in X^{*}$ with $x^{*}(y)=\left\|{y}\right\|$ and
$\left\|{x^{*}}\right\|=1$. Then $x^{*}|_{B_{X}}\in\mathcal{H}L_{0}(B_{X})$
and $e_{x,y}(x^{*}|_{B_{X}})=x^{*}(y)=\left\|{y}\right\|$. This shows that
$e_{x,y}$ belongs to $\mathcal{H}L_{0}(B_{X})^{*}$ and the equality of norms.
Finally, by a simple application of a Cauchy’s integral formula we derive that
the restriction of $e_{x,y}$ to $\overline{B}_{\mathcal{H}L_{0}(B_{X})}$ is
$\tau_{0}$-continuous and so it belongs to $\mathcal{G}_{0}(B_{X})$.

######  Proposition 2.8.

Let $X$ be a complex Banach space. Then,

$\overline{B}_{\mathcal{G}_{0}(B_{X})}=\overline{\Gamma}\\{m_{x,y}:\ x,y\in
B_{X},\ x\neq y\\}=\overline{\operatorname{conv}}\\{e_{x,y}:\ x\in B_{X},y\in
S_{X}\\}$

###### Proof.

By Proposition 2.5, we have that $\left\|{m_{x,y}}\right\|=1$ for every
$x,y\in B_{X}$ with $x\neq y$. Also,

$L(f)=\sup\\{|\langle f,m_{x,y}\rangle|:x,y\in B_{X},x\neq y\\}\,\text{for all
}f\in\mathcal{H}L_{0}(B_{X}).$

Thus, $\\{m_{x,y}:x,y\in B_{X},x\neq y\\}$ is $1$-norming for
$\mathcal{H}L_{0}(B_{X})$. Equivalently,
$\overline{B}_{\mathcal{G}_{0}(B_{X})}=\overline{\Gamma}\\{m_{x,y}:\ x,y\in
B_{X},\ x\neq y\\}$. Analogously, we have that

$L(f)=\left\|{df}\right\|=\sup\\{\left\|{df(x)}\right\|:x\in
B_{X}\\}=\sup\\{|\langle f,e_{x,y}\rangle|:x\in B_{X},y\in S_{X}\\}$

and so $\overline{B}_{\mathcal{G}_{0}(B_{X})}=\overline{\Gamma\\{e_{x,y}:x\in
B_{X},y\in S_{X}\\}}$. But $e_{x,\lambda y_{1}+\eta y_{2}}=\lambda
e_{x,y_{1}}+\eta e_{x,y_{2}}$ for every $\lambda,\eta\in\mathbb{C}$ so
actually
$\overline{B}_{\mathcal{G}_{0}(B_{X})}=\overline{\operatorname{conv}}\\{e_{x,y}:x\in
B_{X},y\in S_{X}\\}$. ∎

As a consequence, the density characters of $X$ and $\mathcal{G}_{0}(B_{X})$
coincide. In particular $X$ is separable if and only if
$\mathcal{G}_{0}(B_{X})$ is separable.

We will now relate $\mathcal{G}_{0}(B_{X})$ with the Lipschitz-free space
$\mathcal{F}(B_{X})$ and Mujica’s predual $\mathcal{G}^{\infty}(B_{X})$ of
$\mathcal{H}^{\infty}(B_{X})$. Note that each element of $\mathcal{F}(B_{X})$
can be seen also as an element of $\mathcal{G}_{0}(B_{X})$, but maybe with a
different behavior. For instance, consider $z\in B_{X}\setminus\\{0\\}$ and
$\mu$ given by $\langle\mu,f\rangle=\int_{C(0,1)}f(\lambda z)d\lambda$ for
$f\in\operatorname{Lip}_{0}(B_{X})$. Then $\mu\neq 0$ in $\mathcal{F}(B_{X})$
but $\langle\mu,f\rangle=0$ for all $f\in\mathcal{H}L_{0}(B_{X})$, so $\mu=0$
when considered as an element of $\mathcal{G}_{0}(B_{X})$. The next
proposition formalizes this situation. We say that an operator $T\colon X\to
Y$ is a _quotient operator_ if $T$ is surjective and
$\left\|{y}\right\|=\inf\\{\left\|{x}\right\|:Tx=y\\}$ for every $y\in Y$;
this implies that $X/\ker T$ is isometrically isomorphic to $Y$.

######  Proposition 2.9.

Let $X$ be a complex Banach space.

  1. (a)

The operator

$\displaystyle\pi\colon\mathcal{F}(B_{X})$
$\displaystyle\to\mathcal{G}_{0}(B_{X})$ $\displaystyle\delta(x)$
$\displaystyle\mapsto\delta(x)$

is a quotient operator with kernel
$\mathcal{H}L_{0}(B_{X})_{\bot}=\\{\mu\in\mathcal{F}(B_{X}):\langle
f,\mu\rangle=0\,\forall f\in\mathcal{H}L_{0}(B_{X})\\}$. Thus
$\mathcal{G}_{0}(B_{X})\equiv\mathcal{F}(B_{X})/\mathcal{H}L_{0}(B_{X})_{\bot}$
isometrically.

  2. (b)

The operator

$\displaystyle\Psi\colon\mathcal{G}^{\infty}(B_{X})\widehat{\otimes}_{\pi}X$
$\displaystyle\to\mathcal{G}_{0}(B_{X})$ $\displaystyle\delta(x)\otimes y$
$\displaystyle\mapsto e_{x,y}$

is a quotient map with $\left\|{\Psi}\right\|=1$. In addition, the operator
$\Psi$ is injective if and only if $X=\mathbb{C}$.

###### Proof.

$(a)$ First note that the existence of such an operator $\pi$ follows from the
linearization property of Lipschitz-free spaces applied to the 1-Lipschitz map
$B_{X}\to\mathcal{G}_{0}(B_{X})$ given by $x\mapsto\delta(x)$. Also,
$\pi^{*}\colon\mathcal{H}L_{0}(B_{X})\to\operatorname{Lip}_{0}(B_{X})$ is just
the inclusion map since

$\pi^{*}f(x)=\langle\pi^{*}f,\delta(x)\rangle=\langle
f,\pi(\delta(x))\rangle=\langle f,\delta(x)\rangle=f(x)\quad\forall
f\in\mathcal{H}L_{0}(B_{X}),\forall x\in B_{X}.$

Thus, $\pi^{*}$ is an isometry into. It is a standard fact that this implies
that $\pi$ is a quotient operator. Moreover,
$\ker\pi=\pi^{*}(\mathcal{H}L_{0}(B_{X}))_{\bot}=\mathcal{H}L_{0}(B_{X})_{\bot}$.

$(b)$ Consider the into isometry

$\displaystyle\Phi\colon\mathcal{H}L_{0}(B_{X})$
$\displaystyle\to\mathcal{H}^{\infty}(B_{X},X^{*})$ $\displaystyle f$
$\displaystyle\mapsto df$

defined after Proposition 2.1. Recall that
$\mathcal{G}^{\infty}(B_{X})\widehat{\otimes}_{\pi}X$ is a predual of
$\mathcal{L}(\mathcal{G}^{\infty}(B_{X}),X^{*})\simeq\mathcal{H}^{\infty}(B_{X},X^{*})$
(see e.g. [43]). Thus, if we restrict $\Phi^{*}$ to this predual we obtain
$\Psi=\Phi^{*}|_{\mathcal{G}^{\infty}(B_{X})\widehat{\otimes}_{\pi}X}$, note
that $\Psi(\delta(x)\otimes y)=e_{x,y}\in\mathcal{G}_{0}(B_{X})$ for all $x,y$
and so
$\Psi(\mathcal{G}^{\infty}(B_{X})\widehat{\otimes}_{\pi}X)\subset\mathcal{G}_{0}(B_{X})$.
Then $\left\|{\Psi}\right\|=1$ and $\Psi$ is a quotient operator since
$\Psi^{*}=\Phi$ is an into isometry. In the case $X=\mathbb{C}$, we have
indeed that
$\Phi\colon\mathcal{H}L_{0}(\mathbb{D})\to\mathcal{H}^{\infty}(\mathbb{D})$ is
an onto isometry, and thus $\Psi$ is also an isometry from
$\mathcal{G}^{\infty}(\mathbb{D})$ onto $\mathcal{G}_{0}(\mathbb{D})$.
However, $\Psi$ is not injective for $X\neq\mathbb{C}$ since $\Phi$ is not
surjective. ∎

Thus, $\mathcal{G}_{0}(\mathbb{D})$ is isometric to
$\mathcal{G}^{\infty}(\mathbb{D})$ (which is the unique predual of
$\mathcal{H}^{\infty}(\mathbb{D})$ [3]). We have some immediate consequences.

######  Corollary 2.10.

A function $f$ is an extreme point of
$\overline{B}_{\mathcal{H}L_{0}(\mathbb{D})}$ if and only if $f^{\prime}$ is
an extreme point of $\overline{B}_{\mathcal{H}^{\infty}(\mathbb{D})}$.

######  Corollary 2.11.

A function $f\in\mathcal{H}L_{0}(\mathbb{D})$ attains its norm as a functional
on $\mathcal{G}_{0}(\mathbb{D})$ if and only if
$f^{\prime}\in\mathcal{H}^{\infty}(\mathbb{D})$ attains its norm as a
functional on $\mathcal{G}^{\infty}(\mathbb{D})$.

Let us state one more consequence of Proposition 2.9.

######  Corollary 2.12.

Let $X$ be a complex Banach space and $\varphi\in\mathcal{G}_{0}(B_{X})$.

  1. (a)

There are sequences $(x_{n}),(y_{n})\subset B_{X}$ with $x_{n}\neq y_{n}$ and
$(a_{n})\subset\ell_{1}$ such that

$\varphi=\sum_{n=1}^{\infty}a_{n}m_{x_{n},y_{n}}.$

Moreover, $\left\|{\varphi}\right\|=\inf\sum_{n=1}^{\infty}|a_{n}|$ where the
infimum is taken over all such representations of $\varphi$.

  2. (b)

There are sequences $(x_{n})\subset B_{X}$, $(y_{n})\subset S_{X}$ and
$(a_{n})\subset\ell_{1}$ such that

$\varphi=\sum_{n=1}^{\infty}a_{n}e_{x_{n},y_{n}}.$

Moreover, $\left\|{\varphi}\right\|=\inf\sum_{n=1}^{\infty}|a_{n}|$ where the
infimum is taken over all such representations of $\varphi$.

###### Proof.

Given $\varepsilon>0$, Proposition 2.9 $(a)$ provides an element
$\mu\in\mathcal{F}(B_{X})$ with $\pi(\mu)=\varphi$ and
$\left\|{\pi(\mu)}\right\|\leq\left\|{\varphi}\right\|+\varepsilon$. It is
known (see e.g. [2, Lem. 3.3]) that there are points $x_{n},y_{n}\in B_{X}$
and $(a_{n})\subset\ell_{1}$ with
$\mu=\sum_{n=1}^{\infty}a_{n}\frac{\delta(x_{n})-\delta(y_{n})}{\left\|{x_{n}-y_{n}}\right\|}$
and
$\sum_{n=1}^{\infty}|a_{n}|\leq\left\|{\mu}\right\|+\varepsilon\leq\left\|{\varphi}\right\|+2\varepsilon$
(here $\delta$ denotes the canonical embedding $\delta\colon
B_{X}\to\mathcal{F}(B_{X})$). Then
$\varphi=\sum_{n=1}^{\infty}a_{n}\pi(\frac{\delta(x_{n})-\delta(y_{n})}{\left\|{x_{n}-y_{n}}\right\|})=\sum_{n=1}^{\infty}a_{n}m_{x_{n},y_{n}}$.

Item $(b)$ follows similarly using the corresponding property for projective
tensor products (see e.g. [43, Prop. 2.8]) and $\mathcal{G}^{\infty}(B_{X})$
[37, Th. 5.1]. ∎

Another consequence of the linearization process shows that functions in
$\mathcal{H}L_{0}$ behave similarly to functions in
$\operatorname{Lip}_{0}(B_{X},B_{Y})$ (that can be isometrically factorized
through the free-Lipschitz spaces $\mathcal{F}(B_{X})$ and
$\mathcal{F}(B_{Y})$). Given $f\in\mathcal{H}L_{0}(B_{X},Y)$ with
$f(B_{X})\subset B_{Y}$ we can easily obtain a commutative diagram:

(1)
$\textstyle{B_{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{\delta_{X}}$$\textstyle{B_{Y}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\delta_{Y}}$$\textstyle{\mathcal{G}_{0}(B_{X})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{T_{\delta_{Y}\circ
f}}$$\textstyle{\mathcal{G}_{0}(B_{Y}),}$

where $T_{\delta_{Y}\circ f}$ is linear and $\|T_{\delta_{Y}\circ f}\|=L(f)$.

##  3\. Approximation properties on $\mathcal{G}_{0}(B_{X})$

Following Mujica’s ideas [36] we devote this section to study the metric
approximation property (MAP) and the approximation property (AP) for
$\mathcal{G}_{0}(B_{X})$ whenever $X$ has the same property. Beginning with
the MAP, we prove the following result about approximation of elements in the
closed unit ball of the dual space. We first introduce the notation:

  * •

$\mathcal{P}_{0}(X,Y)$: The vector space of polynomials $P\colon X\to Y$ such
that $P(0)=0$ endowed with the norm $\|dP\|=L(P|_{B_{X}})$.

  * •

$\mathcal{P}_{f,0}(X,Y)$: The subspace of $\mathcal{P}_{0}(X,Y)$ consisting of
finite type polynomials.

######  Proposition 3.1.

Let $X$ and $Y$ be complex Banach spaces. Then

  1. (a)

$\overline{B}_{\mathcal{H}L_{0}(B_{X},Y)}=\overline{B_{\mathcal{P}_{0}(X,Y)}}^{\tau_{0}}$.

  2. (b)

If $X$ has the MAP then
$\overline{B}_{\mathcal{H}L_{0}(B_{X},Y)}=\overline{B_{\mathcal{P}_{f,0}(X,Y)}}^{\tau_{0}}$.

###### Proof.

$(a)$ If $f\in\overline{B}_{\mathcal{H}L_{0}(B_{X},Y)}$ then
$f\in\mathcal{H}^{\infty}(B_{X},Y)$ and $f(0)=0$. Consider the Taylor series
expansion of $f$ at 0: $f(x)=\sum_{k=0}^{\infty}P^{k}f(0)(x)$. As in [36], for
each $m\in\mathbb{N}\cup\\{0\\}$, we denote

$S_{m}f(x)=\sum_{k=0}^{m}P^{k}f(0)(x)\qquad\textrm{and}\qquad\sigma_{m}f(x)=\frac{1}{m+1}\sum_{k=0}^{m}S_{k}f(x).$

Since
$df=\sum_{k=0}^{\infty}dP^{k}f(0)\in\mathcal{H}^{\infty}(B_{X},\mathcal{L}(X,Y))$
it follows from [36, Prop. 5.2] that $\sigma_{m}f(x)\to f(x)$ for all $x\in
B_{X}$ and

$\|d\sigma_{m}f\|=\|\sigma_{m}(df)\|\leq\|df\|\leq 1.$

This implies that $f\in\overline{B_{\mathcal{P}_{0}(X,Y)}}^{\tau_{0}}$.

For the reverse inclusion, let $f\in\mathcal{H}L_{0}(B_{X},Y)$ and
$(P_{\alpha})\subset B_{\mathcal{P}_{0}(X,Y)}$ such that $P_{\alpha}(x)\to
f(x)$ for all $x\in B_{X}$. Then $L(f)\leq 1$ and so
$f\in\overline{B}_{\mathcal{H}L_{0}(B_{X},Y)}$.

$(b)$ If $X$ has the MAP there is a net of finite rank operators
$(T_{\alpha})\subset\mathcal{L}(X,X)$ such that $T_{\alpha}(x)\to x$ for all
$x\in X$ and $\|T_{\alpha}\|\leq 1$ for every $\alpha$. Given $P\in
B_{\mathcal{P}_{0}(X,Y)}$ we have that $P\circ T_{\alpha}$ belongs to
$B_{\mathcal{P}_{f,0}(X,Y)}$ (since $L(P\circ T_{\alpha}|_{B_{X}})<1$) and
$P(T_{\alpha}x)\to P(x)$ for every $x$. This means that
$P\in\overline{B_{\mathcal{P}_{f,0}(X,Y)}}^{\tau_{0}}$. Finally, an appeal to
$(a)$ yields the result. ∎

######  Theorem 3.2.

$X$ has the MAP if and only if $\mathcal{G}_{0}(B_{X})$ has the MAP.

###### Proof.

$X$ being isometric to a 1-complemented subspace of $\mathcal{G}_{0}(B_{X})$
it is clear that $X$ has the MAP when $\mathcal{G}_{0}(B_{X})$ has it.

Now, suppose that $X$ has the MAP and consider the mapping
$\delta\in\overline{B}_{\mathcal{H}L_{0}(B_{X},\mathcal{G}_{0}(B_{X}))}$. By
Proposition 3.1 there exist a net $(P_{\alpha})\subset
B_{\mathcal{P}_{f,0}(X,\mathcal{G}_{0}(B_{X}))}$ such that
$P_{\alpha}(x)\to\delta(x)$ for all $x\in B_{X}$. Applying a linearization as
in Proposition 2.5 we obtain finite rank linear mappings $T_{P_{\alpha}}$ with
norm bounded by 1, such that the following diagram commutes:

$\textstyle{B_{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{P_{\alpha}}$$\scriptstyle{\delta}$$\textstyle{\mathcal{G}_{0}(B_{X})}$$\textstyle{\mathcal{G}_{0}(B_{X})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{T_{P_{\alpha}}}$

Note that $T_{P_{\alpha}}(\delta(x))=P_{\alpha}(x)\to\delta(x)=Id(\delta(x))$.
Then, we have that $T_{P_{\alpha}}\to Id$ on
$\operatorname{span}\\{\delta(x):x\in B_{X}\\}$. Since the net
$(T_{P_{\alpha}})$ is bounded the same holds for the closure. Hence,
$\mathcal{G}_{0}(B_{X})$ has the MAP. ∎

Note that our arguments cannot be adapted to the case in which $X$ has the BAP
since the approximations of the identity could send the unit ball $B_{X}$ to a
bigger ball (and, hence, we cannot control the Lipschitz norm of $P\circ
T_{\alpha}|_{B_{X}}$ as in Proposition 3.1 $(b)$).

######  Question 1.

Does $\mathcal{G}_{0}(B_{X})$ have the BAP whenever $X$ has the BAP?

The same question for $\mathcal{G}^{\infty}(B_{X})$ was posed by Mujica in
[36]. As far we know, this question is still open.

In contrast to this unknown case about the BAP, the analogous statement for
the AP (Approximation Property -without bounds-) was successfully solved by
Mujica [36] for $\mathcal{G}^{\infty}(B_{X})$. We now turn to this goal for
our space $\mathcal{G}_{0}(B_{X})$, following Mujica’s scheme but somewhat
simplifying the arguments.

Note that in the results about the MAP we used several times that a bounded
net of linear operators converges uniformly on compact sets if and only if it
converges pointwise on a dense set. For the AP we cannot make use of this kind
of argument so our first step will be to describe a locally convex topology
$\tau_{\gamma}$ such that the following topological isomorphism holds:

(2)
$(\mathcal{H}L_{0}(B_{X},Y),\tau_{\gamma})\cong(\mathcal{L}(\mathcal{G}_{0}(B_{X}),Y),\tau_{0}).$

######  Remark 3.3.

Note that for a topology $\tau_{\gamma}$ satisfying (2), if $(f_{\alpha})$ is
a bounded net in $\mathcal{H}L_{0}(B_{X},Y)$ which converges pointwise to
$f\in\mathcal{H}L_{0}(B_{X},Y)$ then
$f_{\alpha}\overset{\tau_{\gamma}}{\to}f$. Indeed, linearizating we obtain a
bounded net $(T_{f_{\alpha}})\subset\mathcal{L}(\mathcal{G}_{0}(B_{X}),Y)$
which converges pointwise to $T_{f}$. Then,
$T_{f_{\alpha}}\overset{\tau_{0}}{\to}T_{f}$ implying that
$f_{\alpha}\overset{\tau_{\gamma}}{\to}f$.

As a consequence, we derive from Proposition 3.1 $(a)$ the following identity:

(3)
$\overline{B}_{\mathcal{H}L_{0}(B_{X},Y)}=\overline{B_{\mathcal{P}_{0}(X,Y)}}^{\tau_{\gamma}}.$

In order to work with the $\tau_{0}$-topology in
$\mathcal{L}(\mathcal{G}_{0}(B_{X}),Y)$ it would be good to have a useful
description of the compact sets of the space $\mathcal{G}_{0}(B_{X})$. For
that, we appeal to the following variation of the classical Grothendieck
description of compact sets (which can be proved, for instance, by slightly
modifying the proof of [41, Prop. 9, pg 134]):

######  Lemma 3.4.

Let $X$ be a Banach space and $V\subset S_{X}$ such that
$\overline{B}_{X}=\overline{\Gamma}(V)$. For each compact set $K\subset X$
there exist sequences $(\alpha_{j})\in c_{0}$ (with $\alpha_{j}>0$ for all
$j$) and $(v_{j})\subset V$ such that
$K\subset\overline{\Gamma}(\\{\alpha_{j}v_{j}\\})$.

A direct consequence of this lemma, along with Proposition 2.8 is the
following:

######  Corollary 3.5.

Let $K\subset\mathcal{G}_{0}(B_{X})$ be a compact set. Then there exist
sequences $(\alpha_{j})\in c_{0}$ and $(x_{j},y_{j})\subset B_{X}\times B_{X}$
(with $\alpha_{j}>0$ and $x_{j}\not=y_{j}$ for all $j$) such that
$K\subset\overline{\Gamma}(\\{\alpha_{j}m_{x_{j}y_{j}}\\})$.

Now we can introduce, as in [36, Th. 4.8], a topology $\tau_{\gamma}$
satisfying (2).

######  Theorem 3.6.

Let $\tau_{\gamma}$ be the locally convex topology on
$\mathcal{H}L_{0}(B_{X},Y)$ generated by the seminorms

$p(f)=\sup_{j}\alpha_{j}\frac{\|f(x_{j})-f(y_{j})\|}{\|x_{j}-y_{j}\|}$

where $(\alpha_{j})\in c_{0}$, $(x_{j},y_{j})\subset B_{X}\times B_{X}$ and
$\alpha_{j}>0$, $x_{j}\not=y_{j}$ for all $j$. Then, the mapping

$\displaystyle(\mathcal{H}L_{0}(B_{X},Y),\tau_{\gamma})$ $\displaystyle\to$
$\displaystyle(\mathcal{L}(\mathcal{G}_{0}(B_{X}),Y),\tau_{0})$ $\displaystyle
f$ $\displaystyle\mapsto$ $\displaystyle T_{f}$

is a topological isomorphism.

###### Proof.

If $K\subset\mathcal{G}_{0}(B_{X})$ is a compact set, by the previous
corollary there are sequences $(\alpha_{j})\in c_{0}$, $(x_{j},y_{j})\subset
B_{X}\times B_{X}$ with $\alpha_{j}>0$, $x_{j}\not=y_{j}$ for all $j$, such
that $K\subset\overline{\Gamma}(\\{\alpha_{j}m_{x_{j}y_{j}}\\})$. Then, for
all $f\in\mathcal{H}L_{0}(B_{X},Y)$,

$\sup_{u\in
K}\|T_{f}u\|\leq\sup_{j}\|T_{f}(\alpha_{j}m_{x_{j}y_{j}})\|=\sup_{j}\alpha_{j}\frac{\|f(x_{j})-f(y_{j})\|}{\|x_{j}-y_{j}\|},$

showing that the mapping $f\mapsto T_{f}$ is $\tau_{\gamma}-\tau_{0}$
continuous.

To prove the continuity of the inverse mapping note that for a seminorm $p$ of
$\tau_{\gamma}$, the associated sequence $\alpha_{j}m_{x_{j}y_{j}}$ converges
to 0 in $\mathcal{G}_{0}(B_{X})$. Thus, the set
$K=\\{\alpha_{j}m_{x_{j}y_{j}}\\}\cup\\{0\\}$ is a compact set in
$\mathcal{G}_{0}(B_{X})$ and
$p(f)=\sup_{j}\|T_{f}(\alpha_{j}m_{x_{j}y_{j}})\|=\sup_{u\in K}\|T_{f}u\|$. ∎

Let us state separately the corresponding result for Lipschitz-free spaces,
that we will not use but it might be of independent interest.

######  Theorem 3.7.

Let $M$ be a complete pointed metric space. Then

  1. (i)

For each compact subset $K$ of $\mathcal{F}(M)$ there exists sequences
$(\alpha_{j})\in c_{0}$ and $(x_{j},y_{j})\subset M\times M$ (with
$\alpha_{j}>0$ and $x_{j}\not=y_{j}$ for all $j$) such that
$K\subset\overline{\Gamma}(\\{\alpha_{j}m_{x_{j}y_{j}}\\})$.

  2. (ii)

Given a Banach space $Y$, let $\tau_{\gamma}$ be the locally convex topology
on $\operatorname{Lip}_{0}(M,Y)$ generated by the seminorms

$p(f)=\sup_{j}\alpha_{j}\frac{\|f(x_{j})-f(y_{j})\|}{d(x_{j},y_{j})}$

where $(\alpha_{j})\in c_{0}$, $(x_{j},y_{j})\subset M\times M$ and
$\alpha_{j}>0$, $x_{j}\not=y_{j}$ for all $j$. Then, the mapping

$\displaystyle(\mathcal{\operatorname{Lip}}_{0}(M,Y),\tau_{\gamma})$
$\displaystyle\to$ $\displaystyle(\mathcal{L}(\mathcal{F}(M),Y),\tau_{0})$
$\displaystyle f$ $\displaystyle\mapsto$ $\displaystyle T_{f}$

is a topological isomorphism.

Now we examine the relationship between the topologies $\tau_{\gamma}$ and
$\tau_{0}$ in $\mathcal{H}L_{0}(B_{X},Y)$.

######  Proposition 3.8.

Let $X$ and $Y$ be complex Banach spaces. Then, $\tau_{\gamma}$ is finer than
$\tau_{0}$ in $\mathcal{H}L_{0}(B_{X},Y)$, and these topologies are equivalent
in $\mathcal{P}(^{m}X,Y)$ for each $m\in\mathbb{N}$.

###### Proof.

If $K\subset B_{X}$ is a compact set, then
$\delta(K)\subset\mathcal{G}_{0}(B_{X})$ is compact. By Corollary 3.5, there
exist sequences $(\alpha_{j})\in c_{0}$ and $(x_{j},y_{j})\subset B_{X}\times
B_{X}$ (with $\alpha_{j}>0$ and $x_{j}\not=y_{j}$ for all $j$) such that
$\delta(K)\subset\overline{\Gamma}(\\{\alpha_{j}m_{x_{j}y_{j}}\\})$. Hence,
for all $f\in\mathcal{H}L_{0}(B_{X},Y)$,

$\sup_{x\in
K}\|f(x)\|\leq\sup_{j}\alpha_{j}\frac{\|f(x_{j})-f(y_{j})\|}{\|x_{j}-y_{j}\|},$

proving the first assertion.

For the second statement, take a seminorm $p$ that generates $\tau_{\gamma}$:
$p(f)=\sup_{j}\alpha_{j}\frac{\|f(x_{j})-f(y_{j})\|}{\|x_{j}-y_{j}\|}$, with
$(\alpha_{j})\in c_{0}$, $(x_{j},y_{j})\subset B_{X}\times B_{X}$,
$\alpha_{j}>0$ and $x_{j}\not=y_{j}$ for all $j$. For a homogeneous polynomial
$P\in\mathcal{P}(^{m}X,Y)$ we have:

$\displaystyle p(P)$ $\displaystyle=$
$\displaystyle\sup_{j}\alpha_{j}\frac{\|P(x_{j})-P(y_{j})\|}{\|x_{j}-y_{j}\|}=\sup_{j}\frac{\|P(\alpha_{j}^{1/m}x_{j})-P(\alpha_{j}^{1/m}y_{j})\|}{\|x_{j}-y_{j}\|}$
$\displaystyle=$
$\displaystyle\sup_{j}\frac{\left\|\sum_{k=1}^{m}\binom{m}{k}\widecheck{P}\left((\alpha_{j}^{1/m}(x_{j}-y_{j}))^{k},(\alpha_{j}^{1/m}y_{j})^{m-k}\right)\right\|}{\|x_{j}-y_{j}\|}$
$\displaystyle=$
$\displaystyle\sup_{j}\left\|\sum_{k=1}^{m}\binom{m}{k}\widecheck{P}\left(\left(\frac{\alpha_{j}^{1/m}(x_{j}-y_{j})}{\|x_{j}-y_{j}\|^{1/k}}\right)^{k},(\alpha_{j}^{1/m}y_{j})^{m-k}\right)\right\|.$

Note that there exist compact sets $K_{1}$ and $K_{2}$ in $X$ such that
$\left\\{\alpha_{j}^{1/m}\frac{(x_{j}-y_{j})}{\|x_{j}-y_{j}\|^{1/k}}\right\\}\subset
K_{1}$ and $\left\\{\alpha_{j}^{1/m}y_{j}\right\\}\subset K_{2}$ (since both
sequences go to 0). Then,

$p(P)\leq\sum_{k=1}^{m}\binom{m}{k}\sup_{a\in K_{1},b\in
K_{2}}\|\widecheck{P}(a^{k},b^{m-k})\|.$

Using the polarization formula, for each $k\in\\{1,\dots,m\\}$,

$\widecheck{P}(a^{k},b^{m-k})=\frac{1}{2^{m}m!}\sum_{\varepsilon_{i}=\pm
1}\varepsilon_{1}\cdots\varepsilon_{m}P\left(\left(\sum_{i=1}^{k}\varepsilon_{i}\right)a+\left(\sum_{i=k+1}^{m}\varepsilon_{i}\right)b\right).$

Taking into account that the following set is compact

$C(K_{1},K_{2})=\left\\{\left(\sum_{i=1}^{k}\varepsilon_{i}\right)a+\left(\sum_{i=k+1}^{m}\varepsilon_{i}\right)b:\
a\in K_{1},b\in K_{2},k\in\\{1,\dots,m\\},\varepsilon_{i}=\pm 1\right\\},$

and that

$\sup_{a\in K_{1},b\in
K_{2}}\|\widecheck{P}(a^{k},b^{m-k})\|\leq\frac{1}{m!}\sup_{u\in
C(K_{1},K_{2})}\|P(u)\|$

we derive the intended inequality:

$p(P)\leq\frac{2^{m}-1}{m!}\sup_{u\in C(K_{1},K_{2})}\|P(u)\|.$

∎

We can now combine all the pieces of our study of the topology $\tau_{\gamma}$
to obtain the following:

######  Proposition 3.9.

If $X$ has the AP, for a given $f\in\mathcal{H}L_{0}(B_{X},Y)$ there exists a
net $(P_{\alpha})\subset\mathcal{P}_{f,0}(X,Y)$ such that
$P_{\alpha}\overset{\tau_{\gamma}}{\to}f$.

###### Proof.

It is enough to consider $f\in\overline{B}_{\mathcal{H}L_{0}(B_{X},Y)}$.
Moreover, taking into account the equality (3) we just need to prove the
result for each homogeneous polynomial $P\in\mathcal{P}(^{m}X,Y)$ (for any
$m$). Applying [36, Lem. 5.3] (or composing the polynomial with the
approximations of the identity supplied by the AP of $X$) we obtain a net
$(P_{\alpha})\subset\mathcal{P}_{f,0}(X,Y)$ such that
$P_{\alpha}\overset{\tau_{0}}{\to}P$. Now, Proposition 3.8 implies that
$P_{\alpha}\overset{\tau_{\gamma}}{\to}P$, which finishes the proof. ∎

Finally, we are in the position of proving the announced result:

######  Theorem 3.10.

$X$ has the AP if and only if $\mathcal{G}_{0}(B_{X})$ has the AP.

###### Proof.

One implication is clear because $X$ is isometric to a complemented subspace
of $\mathcal{G}_{0}(B_{X})$.

For the other, take $\delta\in\mathcal{H}L_{0}(B_{X},\mathcal{G}_{0}(B_{X}))$.
By Proposition 3.9 there exists a net
$(P_{\alpha})\subset\mathcal{P}_{f,0}(X,\mathcal{G}_{0}(B_{X}))$ such that
$P_{\alpha}\overset{\tau_{\gamma}}{\to}\delta$. By the linearization process,
appealing to the isomorphism (2), we obtain that
$(T_{P_{\alpha}})\subset\mathcal{L}(\mathcal{G}_{0}(B_{X}),\mathcal{G}_{0}(B_{X}))$
is a net of finite rank linear mappings satisfying
$T_{P_{\alpha}}\overset{\tau_{0}}{\to}Id$. ∎

######  Remark 3.11.

With the same procedure as at the beginning of the previous section we can
produce a canonical predual $\mathcal{G}(B_{X})$ of $\mathcal{H}L(B_{X})$ made
up of elements of $\mathcal{H}L(B_{X})^{*}$ which are $\tau_{0}$-continuous
when restricted to the closed unit ball. The fact that
$\mathcal{H}L_{0}(B_{X})$ is a 1-complemented subspace of
$\mathcal{H}L(B_{X})$ and that the projection from $\mathcal{H}L(B_{X})$ onto
$\mathcal{H}L_{0}(B_{X})$ is $\tau_{0}-\tau_{0}$ continuous allow us to derive
that $\mathcal{G}_{0}(B_{X})$ is isometric to a 1-complemented subspace of
$\mathcal{G}(B_{X})$.

With standard adaptations most of the results of this and the previous
sections can be stated for $\mathcal{G}(B_{X})$ instead of
$\mathcal{G}_{0}(B_{X})$. That is the case of Propositions 2.5, 2.6, 3.1 and
Theorem 3.2. The version of Proposition 2.8 for $\mathcal{G}(B_{X})$ requires
the addition of $\delta(0)$ to both considered sets. This addition has impact
in Corollary 3.5 and Theorem 3.6, which in turn affects the proofs of
Propositions 3.8 and 3.9 and Theorem 3.10. All these results are valid for
$\mathcal{G}(B_{X})$ after the mentioned modifications. Alternatively, this
also follows from the fact that $\mathcal{G}(B_{X})$ is isometric to a
$1$-complemented subspace of $\mathcal{G}_{0}(B_{X\oplus_{1}\mathbb{C}})$
(just note that the map $\Phi$ in Proposition 2.3 is the adjoint of the
linearization $T_{F}$ of the map
$F(x,\lambda)=\delta(x)+(\lambda-1)\delta(0)$). Also note that the square
diagram (1) can be made for $\mathcal{G}(B_{X})$ but there is no equality
between the norms of $T_{\delta_{Y}\circ f}$ and $f$.

##  4\. Relation between $\mathcal{G}_{0}(B_{X})$ and $\mathcal{G}_{0}(B_{Y})$
when $X\subset Y$

Recall that, given metric spaces $M,N$ with $0\in M\subset N$, the (real)
Lipschitz-free space $\mathcal{F}(M)$ canonically identifies with a subspace
of $\mathcal{F}(N)$. This follows from the McShane extension theorem asserting
that for every $f\in\operatorname{Lip}_{0}(M,\mathbb{R})$ there is
$\tilde{f}\in\operatorname{Lip}_{0}(N,\mathbb{R})$ with $\tilde{f}|_{M}=f$ and
$L(f)=L(\tilde{f})$, see e.g. [44, Th. 1.33]. Note in passing that in the
complex-valued case all extensions can have a larger Lipschitz constant. This
is why our next goal is to analyze the corresponding relation between
$\mathcal{G}_{0}(B_{X})$ and $\mathcal{G}_{0}(B_{Y})$ when $X\subset Y$. Then
$B_{X}\subset B_{Y}$ and the restriction mapping has norm one:

$\displaystyle\mathcal{H}L_{0}(B_{Y})$
$\displaystyle\to\mathcal{H}L_{0}(B_{X})$ $\displaystyle f$
$\displaystyle\mapsto f|_{B_{X}}.$

Then, the following mapping also has norm one:

$\displaystyle\rho:\mathcal{G}_{0}(B_{X})$
$\displaystyle\to\mathcal{G}_{0}(B_{Y})$ $\displaystyle\varphi$
$\displaystyle\mapsto\widehat{\varphi},$

where $\widehat{\varphi}(f)=\varphi(f|_{B_{X}})$.

Whenever $\rho$ is an isometry, we write
$\mathcal{G}_{0}(B_{X})\subset\mathcal{G}_{0}(B_{Y})$. Then, by the Hahn-
Banach theorem, every element of $\mathcal{H}L_{0}(B_{X})$ would have a norm
preserving extension to $\mathcal{H}L_{0}(B_{Y})$. Since there exist
polynomials which cannot be extended to a larger space it is not always true
that $\mathcal{G}_{0}(B_{X})\subset\mathcal{G}_{0}(B_{Y})$. Moreover, the
previous argument can be clearly reversed, so:
$\mathcal{G}_{0}(B_{X})\subset\mathcal{G}_{0}(B_{Y})$ if and only if every
$f\in\mathcal{H}L_{0}(B_{X})$ has a norm preserving extension to
$\mathcal{H}L_{0}(B_{Y})$.

We study some cases where this norm preserving extension occurs. All are cases
where we have an extension morphism. The simplest occurs when $X$ is
1-complemented in $Y$. Here, the complementation also spreads to
$\mathcal{G}_{0}(B_{X})$.

######  Proposition 4.1.

If $X$ is 1-complemented in $Y$ then $\rho$ is an isometry and
$\mathcal{G}_{0}(B_{X})$ is a 1-complemented subspace of
$\mathcal{G}_{0}(B_{Y})$.

###### Proof.

Let $\pi\colon Y\to X$ be a norm-one projection. Given
$f\in\mathcal{H}L_{0}(B_{X})$ the mapping $f\circ\pi$ belongs to
$\mathcal{H}L_{0}(B_{Y})$ with $L(f\circ\pi)\leq L(f)$ and
$(f\circ\pi)|_{B_{X}}=f$. Now, for each $\varphi\in\mathcal{G}_{0}(B_{X})$,

$\|\varphi\|=\sup_{f\in B_{\mathcal{H}L_{0}(B_{X})}}|\varphi(f)|=\sup_{f\in
B_{\mathcal{H}L_{0}(B_{X})}}|\widehat{\varphi}(f\circ\pi)|\leq\|\widehat{\varphi}\|.$

Thus, $\|\varphi\|=\|\widehat{\varphi}\|$, meaning that $\rho$ is an isometry.
Finally, we derive that $\mathcal{G}_{0}(B_{X})$ is 1-complemented in
$\mathcal{G}_{0}(B_{Y})$ through the following projection:

$\displaystyle\mathcal{G}_{0}(B_{Y})$ $\displaystyle\to\mathcal{G}_{0}(B_{X})$
$\displaystyle\psi$ $\displaystyle\mapsto[f\mapsto\psi(f\circ\pi)].$

∎

M. Jung has proved recently that $\mathcal{G}^{\infty}(B_{X})$ does not have
the Radon-Nikodym property (RNP) for any $X$ [33]. Here we obtain the same
result for $\mathcal{G}_{0}(B_{X})$.

######  Corollary 4.2.

The space $\mathcal{G}_{0}(B_{X})$ fails to have the Radon-Nikodym Property
for every complex Banach space $X$.

###### Proof.

The space $\mathcal{G}^{\infty}(\mathbb{D})$ fails to have the RNP since its
the unit ball does not have extreme points [3]. Thus, by the isometry
presented in Proposition 2.9, the same holds for
$\mathcal{G}_{0}(\mathbb{D})$. Since $\mathbb{C}$ is $1$-complemented in $X$,
Proposition 4.1 yields that $\mathcal{G}_{0}(\mathbb{D})$ is a subspace of
$\mathcal{G}_{0}(B_{X})$ and we are done. ∎

Another situation when we have an extension morphism is when $Y=X^{**}$.
Recall that, given $f\in\mathcal{H}^{\infty}(B_{X})$, we can consider its AB
extension $\tilde{f}\in\mathcal{H}^{\infty}(B_{X^{**}})$ [6]. The AB
extension, which defines an isometry from $\mathcal{H}^{\infty}(B_{X})$ to
$\mathcal{H}^{\infty}(B_{X^{**}})$ [20], is a topic widely developed in the
literature. For instance, it is essential in the description of the spectrum
(or maximal ideal space) of the Banach algebra $\mathcal{H}^{\infty}(B_{X})$.
Another ingredient that usually appears associated with the AB extension and
its properties is the notion of _symmetrically regular space_. Both these
concepts have their origin in the study initiated by Arens [4, 5] about
extending the product of a Banach algebra to its bidual.

For an $n$-linear mapping $A:X\times\cdots\times X\to Y$ the canonical
extension $\widetilde{A}:X^{**}\times\cdots\times X^{**}\to Y^{**}$ is given
by consecutive weak-star convergence in the following way:

$\widetilde{A}(x^{**}_{1},\dots,x^{**}_{n})(y^{*})=\lim_{\alpha_{1}}\dots\lim_{\alpha_{n}}y^{*}(A(x_{\alpha_{1}},\dots,x_{\alpha_{n}}))$

where each $(x_{\alpha_{i}})\subset X$ is a net which is weak-star convergent
to $x_{i}^{**}$ and $y^{*}\in Y^{*}$. Now, the AB extension of a homogeneous
polynomial $P\in\mathcal{P}(^{n}X,Y)$ is given by
$\widetilde{P}\in\mathcal{P}(^{n}X^{**},Y^{**})$ which is defined, for
$x^{**}\in X^{**}$, in the expected way:

$\widetilde{P}(x^{**})=\widetilde{\widecheck{P}}(x^{**},\dots,x^{**}).$

This provides a way to extend bounded holomorphic functions
$f\in\mathcal{H}^{\infty}(B_{X},Y)\leadsto\widetilde{f}\in\mathcal{H}^{\infty}(B_{X^{**}},Y^{**})$
and we know from [20] that this extension is an isometry:
$\|f\|=\|\widetilde{f}\|$.

Recall that $X$ is said to be regular if every continuous bilinear mapping
$A:X\times X\to\mathbb{C}$ is Arens regular. That is, the following two
extensions of $A$ to $X^{**}\times X^{**}\to\mathbb{C}$ coincide:

$\lim_{\alpha}\lim_{\beta}A(x_{\alpha},y_{\beta})\quad\textrm{ and
}\quad\lim_{\beta}\lim_{\alpha}A(x_{\alpha},y_{\beta}),$

where $(x_{\alpha})$ and $(y_{\beta})$ are nets in $X$ converging weak-star to
points $x_{0}^{**}$ and $y_{0}^{**}$ in $X^{**}$. The space $X$ is
symmetrically regular if the above holds for every continuous symmetric
bilinear form. Equivalently, $X$ is (symmetrically) regular if any continuous
(symmetric) linear mapping $T:X\to X^{*}$ is weakly compact. Several
equivalent characterizations of this notion can be seen in [8, Th. 8.3] and
some interesting properties appeared in [9, Section 1]. As examples of non
reflexive regular (and hence, symmetrically regular) Banach spaces we have,
for instance, those that satisfy property (V) of Pełczyński, like $c_{0}$,
$C(K)$ or $\mathcal{H}^{\infty}(\mathbb{D})$ while typical non symmetrically
regular spaces are $\ell_{1}$ and $X\oplus X^{*}$, for any non reflexive space
$X$. Also, Leung [35, Th. 12] provided an example of a symmetrically regular
space that is not regular and in [9] it is showed that $c_{0}(\ell_{1}^{n})$
is regular but its bidual $\ell_{\infty}(\ell_{1}^{n})$ is not symmetrically
regular.

We now want to work with the AB extension for elements in
$\mathcal{H}L_{0}(B_{X})$. For $f\in\mathcal{H}L_{0}(B_{X})$, in order to
compute the Lipschitz constant of $\tilde{f}$ we need to deal with the
differential of the AB extension, $d\widetilde{f}$ which belongs to
$\mathcal{H}(B_{X^{**}},X^{***})$. Instead, we do know the norm of the AB
extension of the differential $\ThisStyle{\stackengine{-.1\LMpt}{$\SavedStyle
df$}{\stretchto{\scaleto{\SavedStyle\mkern
0.2mu\AC}{.5150}}{.6}}{O}{c}{F}{T}{S}}\in\mathcal{H}^{\infty}(B_{X^{**}},X^{***})$.
Fortunately, on symmetrically regular spaces they coincide:

######  Proposition 4.3.

If $X$ is symmetrically regular and $f\in\mathcal{H}L_{0}(B_{X})$ then
$d\widetilde{f}=\ThisStyle{\stackengine{-.1\LMpt}{$\SavedStyle
df$}{\stretchto{\scaleto{\SavedStyle\mkern
0.2mu\AC}{.5150}}{.6}}{O}{c}{F}{T}{S}}$.

###### Proof.

If $f=\sum_{k=0}^{\infty}P^{k}f(0)$ then the series expansion of $df$ at $0$
is given by $df=\sum_{k=0}^{\infty}dP^{k}f(0)$. Thus,
$\ThisStyle{\stackengine{-.1\LMpt}{$\SavedStyle
df$}{\stretchto{\scaleto{\SavedStyle\mkern
0.2mu\AC}{.5150}}{.6}}{O}{c}{F}{T}{S}}=\sum_{k=0}^{\infty}\ThisStyle{\stackengine{-.1\LMpt}{$\SavedStyle(dP^{k}f(0))$}{\stretchto{\scaleto{\SavedStyle\mkern
0.2mu\AC}{.5150}}{.6}}{O}{c}{F}{T}{S}}$. On the other hand,
$\widetilde{f}=\sum_{k=0}^{\infty}\widetilde{P^{k}f(0)}$ and so
$d\widetilde{f}=\sum_{k=0}^{\infty}d(\widetilde{P^{k}f(0)})$.

Therefore, the result is proved once we show that for any given
$m\in\mathbb{N}$ and any $P\in\mathcal{P}(^{m}X)$,
$\widetilde{dP}=d\widetilde{P}$. Note that in this case
$\widetilde{P}\in\mathcal{P}(^{m}X^{**})$, $dP\in\mathcal{P}(^{m-1}X,X^{*})$
while both $\widetilde{dP}$ and $d\widetilde{P}$ belong to
$\mathcal{P}(^{m-1}X^{**},X^{***})$.

When $X$ is symmetrically regular, it follows from [8, Th. 8.3] that
$\widetilde{\widecheck{P}}=\widecheck{\widetilde{P}}$. The argument is now
complete because, for each $x^{**},y^{**}\in X^{**}$ we have
$\widetilde{dP}(x^{**})(y^{**})=m\widetilde{\widecheck{P}}(x^{**},\dots,x^{**},y^{**})$
and
$d\widetilde{P}(x^{**})(y^{**})=m\widecheck{\widetilde{P}}(x^{**},\dots,x^{**},y^{**})$.
∎

######  Proposition 4.4.

If $X$ is symmetrically regular then the AB extension mapping

$\displaystyle E:\mathcal{H}L_{0}(B_{X})$
$\displaystyle\to\mathcal{H}L_{0}(B_{X^{**}})$ $\displaystyle f$
$\displaystyle\mapsto\widetilde{f}$

is an isometry.

###### Proof.

If $f\in\mathcal{H}L_{0}(B_{X})$ then its norm is given by $\|df\|$. By [20],
$\|df\|=\|\ThisStyle{\stackengine{-.1\LMpt}{$\SavedStyle
df$}{\stretchto{\scaleto{\SavedStyle\mkern
0.2mu\AC}{.5150}}{.6}}{O}{c}{F}{T}{S}}\|$. Also, by the previous proposition
we know that $d\widetilde{f}=\ThisStyle{\stackengine{-.1\LMpt}{$\SavedStyle
df$}{\stretchto{\scaleto{\SavedStyle\mkern
0.2mu\AC}{.5150}}{.6}}{O}{c}{F}{T}{S}}$. So, we obtain that
$\|df\|=\|d\widetilde{f}\|$, meaning that $\widetilde{f}$ does indeed belong
to $\mathcal{H}L_{0}(B_{X^{**}})$ and that the mapping $f\mapsto\widetilde{f}$
is an isometry. ∎

In the previous result symmetric regularity is used to obtain that
$d\widetilde{f}=\ThisStyle{\stackengine{-.1\LMpt}{$\SavedStyle
df$}{\stretchto{\scaleto{\SavedStyle\mkern
0.2mu\AC}{.5150}}{.6}}{O}{c}{F}{T}{S}}$. Actually we only need the identity of
their norms:
$\|d\widetilde{f}\|=\|\ThisStyle{\stackengine{-.1\LMpt}{$\SavedStyle
df$}{\stretchto{\scaleto{\SavedStyle\mkern
0.2mu\AC}{.5150}}{.6}}{O}{c}{F}{T}{S}}\|$. We do not know if this equality
holds in general.

######  Corollary 4.5.

If $X$ is symmetrically regular then
$\mathcal{G}_{0}(B_{X})\subset\mathcal{G}_{0}(B_{X^{**}})$.

Note that in the above corollary the hypothesis of symmetric regularity is not
a necessary condition since, for example, for $X=\ell_{1}$ the result holds
due to Proposition 4.1.

A generalization of this procedure (which, however, uses the AB extension in
its definition) is when there exists an isometric extension morphism $s\colon
X^{*}\to Y^{*}$. This happens, for instance, when $X$ is an M-ideal in $Y$.
More generally, if $X\subset Y$ then the existence of an isometric extension
morphism $s\colon X^{*}\to Y^{*}$ is equivalent to $X^{**}$ being
1-complemented in $Y^{**}$. Actually, the existence of an isometric extension
morphism $s\colon X^{*}\to Y^{*}$ is equivalent to $X$ being 1-locally
complemented in $Y$ (see the definition in the next section and the comment
before Corollary 5.5).

Note that $s(x^{*})(x)=x^{*}(x)$ for all $x\in X$, $x^{*}\in X^{*}$ and that
$\|s(x^{*})\|=\|x^{*}\|$. This extension transfers to
$\mathcal{H}^{\infty}(B_{X})$ in the following way:

$\displaystyle\overline{s}:\mathcal{H}^{\infty}(B_{X})$
$\displaystyle\to\mathcal{H}^{\infty}(B_{Y})$ $\displaystyle f$
$\displaystyle\mapsto\widetilde{f}\circ s^{*}\circ i_{Y},$

where $i_{Y}:Y\to Y^{**}$ is the canonical inclusion.

The mapping $\overline{s}$ is an isometric extension from
$\mathcal{H}^{\infty}(B_{X})$ to $\mathcal{H}^{\infty}(B_{X^{**}})$. Again, to
work in $\mathcal{H}L_{0}(B_{X})$ we require a symmetrically regular
hypothesis.

######  Proposition 4.6.

If $X$ is symmetrically regular, $X\subset Y$ and there is an isometric
extension morphism $s:X^{*}\to Y^{*}$ then

$\displaystyle\overline{s}:\mathcal{H}L_{0}(B_{X})$
$\displaystyle\to\mathcal{H}L_{0}(B_{Y})$ $\displaystyle f$
$\displaystyle\mapsto\widetilde{f}\circ s^{*}\circ i_{Y}$

is an isometric extension.

###### Proof.

For any $P\in\mathcal{P}(^{m}X)$ we have that
$\overline{s}(P)\in\mathcal{P}(^{m}Y)$ and
$d(\overline{s}(P))\in\mathcal{P}(^{m-1}Y,Y^{*})$. Now, for $y,z\in B_{Y}$,

$\displaystyle d(\overline{s}(P))(y)(z)$
$\displaystyle=m\widecheck{(\overline{s}(P))}(y,\dots,y,z)=m\widecheck{\widetilde{P}}\left(s^{*}(i_{Y}(y)),\dots,s^{*}(i_{Y}(y)),s^{*}(i_{Y}(z))\right)$
$\displaystyle=d\widetilde{P}(s^{*}(i_{Y}(y)))(s^{*}(i_{Y}(z)))=(i_{Y}^{*}\circ
s^{**}\circ d\widetilde{P}\circ s^{*}\circ i_{Y})(y)(z).$

This says that $d(\overline{s}(P))=i_{Y}^{*}\circ s^{**}\circ
d\widetilde{P}\circ s^{*}\circ i_{Y}$ for every polynomial
$P\in\mathcal{P}(^{m}X)$. Then, the same equality holds for every
$f\in\mathcal{H}L_{0}(B_{X})$:

$d(\overline{s}(f))=i_{Y}^{*}\circ s^{**}\circ d\widetilde{f}\circ s^{*}\circ
i_{Y}.$

Since $X$ is symmetrically regular, by Proposition 4.4 we obtain that
$\|d(\overline{s}(f))\|\leq\|d\widetilde{f}\|=\|df\|$. Also, note that for
$x\in B_{X}$, we have $s^{*}\circ i_{Y}(x)=i_{X}(x)$. This implies that
$d\widetilde{f}(s^{*}(i_{Y}(x))=i_{X^{*}}(df(x))$. Therefore,

$d(\overline{s}(f))(x)=i_{Y}^{*}\circ
s^{**}\left(i_{X^{*}}(df(x))\right)=s(df(x)).$

This equality and the fact that $s$ is an isometry allow us to derive the
other inequality:

$\displaystyle\|d(\overline{s}(f))\|$ $\displaystyle\geq\sup_{x\in
B_{X}}\|d(\overline{s}(f))(x)\|=\sup_{x\in B_{X}}\|s(df(x))\|$
$\displaystyle=\sup_{x\in B_{X}}\|df(x)\|=\|df\|,$

which concludes the proof. ∎

######  Corollary 4.7.

If $X$ is symmetrically regular, $X\subset Y$ and there is an isometric
extension morphism $s\colon X^{*}\to Y^{*}$ then
$\mathcal{G}_{0}(B_{X})\subset\mathcal{G}_{0}(B_{Y})$.

###  4.1. Dual isometric spaces

It is known that there exist non isomorphic Banach spaces with isomorphic
duals. Attending to that, Díaz and Dineen [22] posed the following question:
if $X$ and $Y$ are Banach spaces such that $X^{*}$ and $Y^{*}$ are isomorphic,
under which conditions is it true that $\mathcal{P}(^{n}X)$ and
$\mathcal{P}(^{n}Y)$ are isomorphic for every $n\geq 1$? That is, if $X^{*}$
and $Y^{*}$ are isomorphic (i. e. the spaces of 1-homogeneous polynomials are
isomorphic) does it imply that the spaces of $n$-homogeneous polynomials are
isomorphic for every $n$? They also gave a partial answer to this question.
Later, a relaxation of the conditions was obtained by Cabello-Sánchez,
Castillo and García [15, Th. 1] and Lassalle and Zalduendo [34, Th. 4]
independently, proving that the answer is affirmative whenever $X$ and $Y$ are
symmetrically regular. We present here a version of this result for
holomorphic Lipschitz functions on the ball. Since we need to remain inside
the ball when changing the space we have to restrict ourselves to the case of
isometric isomorphisms.

######  Proposition 4.8.

If $X$ and $Y$ are symmetrically regular Banach spaces such that $X^{*}$ and
$Y^{*}$ are isometrically isomorphic then $\mathcal{H}L_{0}(B_{X})$ and
$\mathcal{H}L_{0}(B_{Y})$ are isometrically isomorphic as well.

###### Proof.

Let us denote by $s\colon X^{*}\to Y^{*}$ the isometric isomorphism and
consider the mapping
$\overline{s}:\mathcal{H}L_{0}(B_{X})\to\mathcal{H}L_{0}(B_{Y})$ as in
Proposition 4.6. By the proof of that proposition we derive that
$\overline{s}$ is continuous and $\|\overline{s}\|\leq 1$. Since $Y$ is
symmetrically regular, we can use the same procedure for the mapping
$\overline{s^{-1}}:\mathcal{H}L_{0}(B_{Y})\to\mathcal{H}L_{0}(B_{X})$ leading
to $\|\overline{s^{-1}}\|\leq 1$. Finally, appealing to [34, Cor. 3] we obtain
that $\overline{s^{-1}}\circ\overline{s}(P)=P$ for every homogeneous
polynomial $P$ on $X$ and, hence, $\overline{s^{-1}}\circ\overline{s}(f)=f$
for every $f\in\mathcal{H}L_{0}(B_{X})$. Indeed, if $\sum_{k=0}^{\infty}P^{k}$
is the Taylor series expansion of a given $f\in\mathcal{H}L_{0}(B_{X})$, then
$\tilde{f}(z)=\sum_{k=0}^{\infty}\tilde{P}^{k}(z)$ for every $z\in
B_{X^{**}}$. Thus

$\overline{s}(f)(y)=\tilde{f}(s^{*}(i_{Y}(y)))=\sum_{k=0}^{\infty}\tilde{P}^{k}(s^{*}(y))=\sum_{k=0}^{\infty}\overline{s}(P^{k})(y),$

for every $y\in Y$. From here

$\displaystyle\overline{s^{-1}}(\overline{s}(f))(x)$ $\displaystyle=$
$\displaystyle\widetilde{\overline{s}(f)}\big{(}(s^{-1})^{*}(i_{X}(x))\big{)}=\sum_{k=0}^{\infty}\widetilde{\overline{s}(P^{k})}\big{(}(s^{-1})^{*}(i_{X}(x)\big{)}$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{\infty}\overline{s^{-1}}(\overline{s}(P^{k}))(x)=\sum_{k=0}^{\infty}P^{k}(x)=f(x),$

for every $x\in X$. Analogously one can check that
$\overline{s}\circ\overline{s^{-1}}(f)=f$ for every $f\in\mathcal{H}L(B_{Y})$.
∎

In the previous proposition we can change the hypothesis of $X$ and $Y$ being
symmetrically regular by $X$ or $Y$ being regular. Indeed, it is proved in
[34, Rmk. 2] (see also [15, Prop. 1]) that if $X^{*}$ and $Y^{*}$ are
isomorphic and $X$ is regular then so is $Y$.

###  4.2. Mapping between $\mathcal{G}_{0}(B_{X})$ and
$\mathcal{G}_{0}(B_{Y})$

Any linear mapping between $X$ and $Y$ produces a mapping between
$\mathcal{G}_{0}(B_{X})$ and $\mathcal{G}_{0}(B_{Y})$ by a canonical procedure
(actually, two canonical procedures depending on the norm of the mapping).

(i) Let $\psi\colon X\to Y$ a linear mapping with $\|\psi\|\leq 1$. Note that
$L(\psi)=\|\psi\|$ in this case. Since $\psi(B_{X})\subset B_{Y}$ we can
define the canonical mapping with norm $\leq 1$:

$\displaystyle\mathcal{H}L_{0}(B_{Y})$
$\displaystyle\to\mathcal{H}L_{0}(B_{X})$ $\displaystyle f$
$\displaystyle\mapsto f\circ\psi.$

Thus, the following also has norm $\leq 1$:

$\displaystyle T_{\delta_{Y}\circ\psi}\colon\mathcal{G}_{0}(B_{X})$
$\displaystyle\to\mathcal{G}_{0}(B_{Y})$ $\displaystyle\varphi$
$\displaystyle\mapsto\widehat{\varphi},$

where $\widehat{\varphi}(f)=\varphi(f\circ\psi)$.

(ii) When $\|\psi\|>1$ the previous construction does not work but we can
appeal to a linearization plus differentiation process (as we used to show
that $X$ is a 1-complemented subspace of $\mathcal{G}_{0}(B_{X})$).

Let $\psi\in\mathcal{L}(X,Y)$ so that
$\psi|_{B_{X}}\in\mathcal{H}L_{0}(B_{X},Y)$. We have the usual commutative
diagram:

$\textstyle{B_{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi|_{B_{X}}}$$\scriptstyle{\delta_{X}}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d\delta_{Y}(0)}$$\textstyle{\mathcal{G}_{0}(B_{X})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{T_{\psi}}$$\textstyle{\mathcal{G}_{0}(B_{Y})}$

where $T_{\psi}\in\mathcal{L}(\mathcal{G}_{0}(B_{X}),Y)$.

Applying the differential at 0 to the equality
$\psi|_{B_{X}}=T_{\psi}\circ\delta_{X}$ we get the commutative diagram:

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi}$$\scriptstyle{d\delta_{X}(0)}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d\delta_{Y}(0)}$$\textstyle{\mathcal{G}_{0}(B_{X})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d\delta_{Y}(0)\circ
T_{\psi}}$$\scriptstyle{T_{\psi}}$$\textstyle{\mathcal{G}_{0}(B_{Y}).}$

Note that the linear mapping $d\delta_{Y}(0)\circ
T_{\psi}:\mathcal{G}_{0}(B_{X})\to\mathcal{G}_{0}(B_{Y})$ has norm less than
or equal to $\|\psi\|$.

##  5\. Local complementation in the bidual

In this section, we are interested in the relationship between
$\mathcal{G}_{0}(B_{X^{**}})$ and $\mathcal{G}_{0}(B_{X})^{**}$ under the
hypothesis of $X^{**}$ having the MAP, in the spirit of what is done in [16].

We begin with a result about a special approximation behavior in the case that
the bidual space has the MAP.

######  Proposition 5.1.

Let $X,Y$ be Banach spaces such that $X^{**}$ has the MAP. For each
$f\in\mathcal{H}L_{0}(B_{X^{**}},Y)$ with $L(f)=1$ there exists a net
$(Q_{\alpha})\subset\mathcal{P}_{f,0}(X,Y)$ with $L(Q_{\alpha}|_{B_{X}})\leq
1$ satisfying $\widetilde{Q}_{\alpha}(x^{**})\to f(x^{**})$ for all $x^{**}\in
B_{X^{**}}$.

###### Proof.

By Proposition 3.1 it is enough to consider $f=P\in\mathcal{P}_{0}(X^{**},Y)$
with $L(P|_{B_{X^{**}}})\leq 1$. If $X^{**}$ has the MAP we can appeal to [16,
Cor. 1] to obtain a net of finite rank mappings
$(t_{\alpha})\subset\mathcal{L}(X,X^{**})$ with $\|t_{\alpha}\|\leq 1$ and
$t_{\alpha}^{**}(x^{**})\to x^{**}$ for all $x^{**}\in X^{**}$. Now we define
$Q_{\alpha}=P\circ t_{\alpha}$, which clearly belongs to
$\mathcal{P}_{f,0}(X,Y)$. Note that, for any $x,y\in B_{X}$,

$\|Q_{\alpha}(x)-Q_{\alpha}(y)\|=\|P(t_{\alpha}(x))-P(t_{\alpha}(y))\|\leq
L(P|_{B_{X^{**}}})\|t_{\alpha}\|\|x-y\|\leq\|x-y\|.$

Then, $L(Q_{\alpha}|_{B_{X}})\leq 1$. Since $t_{\alpha}$ is a finite rank
mapping, we have that $t_{\alpha}^{**}\in\mathcal{L}(X^{**},X^{**})$. Hence,
$\widetilde{Q}_{\alpha}=\widetilde{P}\circ t_{\alpha}^{**}=P\circ
t_{\alpha}^{**}$. As a consequence,
$\widetilde{Q}_{\alpha}(x^{**})=P(t_{\alpha}^{**}(x^{**}))\to P(x^{**})$ for
all $x^{**}\in B_{X^{**}}$.

∎

For a symmetrically regular space $X$, we consider the following mapping

$\displaystyle\Theta:B_{X^{**}}$
$\displaystyle\to\mathcal{G}_{0}(B_{X})^{**}=\mathcal{H}L_{0}(B_{X})^{*}$
$\displaystyle x^{**}$
$\displaystyle\mapsto[f\in\mathcal{H}L_{0}(B_{X})\mapsto\widetilde{f}(x^{**})].$

######  Proposition 5.2.

If $X$ is symmetrically regular then $\Theta$ belongs to
$\mathcal{H}L_{0}(B_{X^{**}},\mathcal{G}_{0}(B_{X})^{**})$ with $L(\Theta)=1$.

###### Proof.

If $X$ is symmetrically regular, by Proposition 4.4, the AB extension is an
isometry from $\mathcal{H}L_{0}(B_{X})$ into $\mathcal{H}L_{0}(B_{X^{**}})$,
so $\Theta$ is well defined. For any $f\in\mathcal{H}L_{0}(B_{X})$, we have
$\Theta(\cdot)(f)=\widetilde{f}$, meaning that $\Theta$ is weak-star
holomorphic and thus, it is holomorphic. Also, $\Theta(0)=0$ and for any
$x^{**},y^{**}\in B_{X^{**}}$, once again by the symmetric regularity of $X$
we have

$\|\Theta(x^{**})-\Theta(y^{**})\|=\sup_{f\in
B_{\mathcal{H}L_{0}(B_{X})}}\|\widetilde{f}(x^{**})-\widetilde{f}(y^{**})\|\leq\|x^{**}-y^{**}\|.$

This means that
$\Theta\in\mathcal{H}L_{0}(B_{X^{**}},\mathcal{G}_{0}(B_{X})^{**})$ with
$L(\Theta)\leq 1$. On the other hand,

$\|\Theta(x^{**})-\Theta(y^{**})\|\geq\sup_{x^{*}\in
B_{X^{*}}}|x^{**}(x^{*})-y^{**}(x^{*})|=\|x^{**}-y^{**}\|.$

Therefore, $L(\Theta)=1$. ∎

As a consequence of the previous proposition, if $X$ is symmetrically regular
we can linearize the mapping $\Theta$:

$\textstyle{B_{X^{**}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Theta\
}$$\scriptstyle{\delta_{X^{**}}}$$\textstyle{\mathcal{G}_{0}(B_{X})^{**}}$$\textstyle{\mathcal{G}_{0}(B_{X^{**}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{T_{\Theta}}$

This produces a linear mapping
$T_{\Theta}\in\mathcal{L}(\mathcal{G}_{0}(B_{X^{**}}),\mathcal{G}_{0}(B_{X})^{**})$
with $\|T_{\Theta}\|=L(\Theta)=1$.

Motivated by the Principle of Local Reflexivity, Kalton [32] introduced the
following definition:

######  Definition 5.3.

Given Banach spaces $X\subset Y$ we say that $X$ is 1-locally complemented in
$Y$ if for every $\varepsilon>0$ and every finite dimensional subspace $F$ of
$Y$ there exist a linear mapping $T:F\to X$ such that $\|T\|\leq
1+\varepsilon$ and $T(x)=x$ for all $x\in F\cap X$.

Note that the Principle of Local Reflexivity says that $X$ is 1-locally
complemented in $X^{**}$, for any Banach space $X$.

######  Theorem 5.4.

If $X$ is symmetrically regular and $X^{**}$ has the MAP then $T_{\Theta}$
embeds $\mathcal{G}_{0}(B_{X^{**}})$ as a 1-locally complemented subspace of
$\mathcal{G}_{0}(B_{X})^{**}$. In particular, $T_{\Theta}$ is an isometry.

###### Proof.

We know that the mapping $\delta_{X^{**}}$ belongs to
$\mathcal{H}L_{0}(B_{X^{**}},\mathcal{G}_{0}(B_{X^{**}}))$ with
$L(\delta_{X^{**}})=1$. Thus, we can apply Proposition 5.1 to get a net
$(Q_{\alpha})\subset\mathcal{P}_{f,0}(X,\mathcal{G}_{0}(B_{X^{**}}))$ with
$L(Q_{\alpha}|_{B_{X}})\leq 1$ such that
$\widetilde{Q}_{\alpha}(x^{**})\to\delta_{X^{**}}(x^{**})$ for all $x^{**}\in
B_{X^{**}}$.

Consider the following two commutative diagrams:

$\textstyle{B_{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{Q_{\alpha}|_{B_{X}}}$$\scriptstyle{\delta_{X}}$$\textstyle{\mathcal{G}_{0}(B_{X^{**}})}$$\textstyle{\mathcal{G}_{0}(B_{X})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{T_{Q_{\alpha}}}$
$\textstyle{B_{X^{**}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\widetilde{Q}_{\alpha}|_{B_{X^{**}}}}$$\scriptstyle{\delta_{X^{**}}}$$\textstyle{\mathcal{G}_{0}(B_{X^{**}})}$$\textstyle{\mathcal{G}_{0}(B_{X^{**}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{T_{\widetilde{Q}_{\alpha}}}$

Note that, since $X$ is symmetrically regular we have

$\|T_{Q_{\alpha}}\|=L(Q_{\alpha}|_{B_{X}})=L(\widetilde{Q}_{\alpha}|_{B_{X}})=\|T_{\widetilde{Q}_{\alpha}}\|\leq
1.$

For each $\alpha$, since $T_{Q_{\alpha}}$ is a finite rank operator we have
that $T_{Q_{\alpha}}^{**}$ belongs to
$\mathcal{L}(\mathcal{G}_{0}(B_{X})^{**},\mathcal{G}_{0}(B_{X^{**}}))$. Thus,
we have the following diagram

The space $\mathcal{G}_{0}(B_{X^{**}})$ has the MAP witnessed by the net
$(T_{\widetilde{Q}_{\alpha}})$ thanks to (the proof of) Theorem 3.2. Appealing
to [16, Lem. 4], the proof will be completed once we check that the previous
diagram is commutative. For this, it is enough to prove that
$T_{\widetilde{Q}_{\alpha}}(\delta_{X^{**}}(x^{**}))=T_{Q_{\alpha}}^{**}\circ
T_{\Theta}(\delta_{X^{**}}(x^{**}))$ for every $x^{**}\in B_{X^{**}}$.

On the one hand we know that
$T_{\widetilde{Q}_{\alpha}}(\delta_{X^{**}}(x^{**}))=\widetilde{Q}_{\alpha}(x^{**})$.
On the other hand, $T_{Q_{\alpha}}^{**}\circ
T_{\Theta}(\delta_{X^{**}}(x^{**}))=T_{Q_{\alpha}}^{**}(\Theta(x^{**}))$. To
understand this element of $\mathcal{G}_{0}(B_{X^{**}})$ let us see how it
acts on any $f\in\mathcal{H}L_{0}(B_{X^{**}})$:

(5) $\langle
T_{Q_{\alpha}}^{**}(\Theta(x^{**})),f\rangle=\langle\Theta(x^{**}),T_{Q_{\alpha}}^{*}(f)\rangle.$

Now, $T_{Q_{\alpha}}^{*}(f)$ belongs to $\mathcal{H}L_{0}(B_{X})$ and for any
$x\in B_{X}$ satisfies

$T_{Q_{\alpha}}^{*}(f)(x)=\langle
T_{Q_{\alpha}}^{*}(f),\delta_{X}(x)\rangle=\langle
f,T_{Q_{\alpha}}(\delta_{X}(x))\rangle=\langle
f,Q_{\alpha}(x)\rangle=(T_{f}\circ Q_{\alpha})(x).$

Then, $T_{Q_{\alpha}}^{*}(f)=T_{f}\circ Q_{\alpha}$. Replacing this equality
in (5) and using the definition of $\Theta$ and the fact that the range of
$\widetilde{Q}_{\alpha}$ is contained in $\mathcal{G}_{0}(B_{X^{**}})$ we
derive

$\displaystyle\langle T_{Q_{\alpha}}^{**}(\Theta(x^{**})),f\rangle$
$\displaystyle=\langle\Theta(x^{**}),T_{f}\circ
Q_{\alpha}\rangle=\widetilde{T_{f}\circ
Q}_{\alpha}(x^{**})=T_{f}^{**}\circ\widetilde{Q}_{\alpha}(x^{**})$
$\displaystyle=T_{f}(\widetilde{Q}_{\alpha}(x^{**}))=\langle\widetilde{Q}_{\alpha}(x^{**}),f\rangle,\quad\textrm{for
all }f\in\mathcal{H}L_{0}(B_{X^{**}}).$

Therefore,
$T_{Q_{\alpha}}^{**}(\Theta(x^{**}))=\widetilde{Q}_{\alpha}(x^{**})$ and thus
$T_{Q_{\alpha}}^{**}\circ
T_{\Theta}(\delta_{X^{**}}(x^{**}))=T_{\widetilde{Q}_{\alpha}}(\delta_{X^{**}}(x^{**}))$
for every $x^{**}\in B_{X^{**}}$, which finishes the proof.

∎

It is known (see, for instance, [16, Lem. 3] or [32, Th. 3.5]) that $X$ is
1-locally complemented in $Y$ if and only if $X^{*}$ is 1-complemented in
$Y^{*}$ (with projection the restriction mapping). This is also equivalent to
$X^{**}$ being 1-complemented in $Y^{**}$ (under the natural embedding).

######  Corollary 5.5.

If $X$ is symmetrically regular and $X^{**}$ has the MAP then
$\mathcal{H}L_{0}(B_{X^{**}})$ is isometric to a 1-complemented subspace of
$\mathcal{H}L_{0}(B_{X})^{**}$.

Under the same conditions of the previous results we can also obtained a
version for holomorphic Lipschitz functions of the following characterization
of unique norm preserving extensions to the bidual proved by Godefroy in [29].

######  Lemma 5.6.

Let $X$ be a Banach space and $x^{*}\in X^{*}$ with $\|x^{*}\|=1$. The
following are equivalent:

  * $(i)$

$x^{\ast}$ has a unique norm preserving extension to a functional on
$X^{\ast\ast}$.

  * $(ii)$

The function
$Id_{\overline{B}_{X^{\ast}}}:(\overline{B}_{X^{\ast}},w^{\ast})\longrightarrow(\overline{B}_{X^{\ast}},w)$
is continuous at $x^{\ast}$.

Aron, Boyd and Choi [7] gave a version of this result for homogeneous
polynomials. Later, other extensions appeared (for instance, in [24] for
ideals of homogeneous polynomials and in [23] for bilinear mappings in
operator spaces).

Now, the statement of the theorem in our setting is the following:

######  Theorem 5.7.

Suppose $X$ is symmetrically regular and $X^{**}$ has the MAP. Consider a
function $f\in\mathcal{H}L_{0}(B_{X})$ with $L(f)=1$. Then, the following are
equivalent:

  1. $(i)$

$f$ has a unique norm preserving extension to $\mathcal{H}L_{0}(B_{X^{**}})$.

  2. $(ii)$

The AB extension from $(\overline{B}_{\mathcal{H}L_{0}(B_{X})},w^{*})$ to
$(\overline{B}_{\mathcal{H}L_{0}(B_{X^{**}})},w^{*})$ is continuous at $f$.

  3. $(iii)$

If the net $(f_{\alpha})\subset\overline{B}_{\mathcal{H}L_{0}(B_{X})}$
converges pointwise to $f$, then
$(\widetilde{f}_{\alpha})\subset\overline{B}_{\mathcal{H}L_{0}(B_{X^{**}})}$
converges pointwise to $\widetilde{f}$.

###### Proof.

$(i)\Rightarrow(ii)$ Let
$(f_{\alpha})\subset\overline{B}_{\mathcal{H}L_{0}(B_{X})}$ be a net weak-star
convergent to a function $f\in\overline{B}_{\mathcal{H}L_{0}(B_{X})}$. By the
weak-star compactness of the ball
$\overline{B}_{\mathcal{H}L_{0}(B_{X^{**}})}$ there is a subnet
$(\widetilde{f}_{\beta})$ weak-star convergent to a function
$g\in\overline{B}_{\mathcal{H}L_{0}(B_{X^{**}})}$. Since for each $x\in
B_{X}$, $\widetilde{f}_{\alpha}(x)=f_{\alpha}(x)\to f(x)$ we derive that
$g|_{B_{X}}=f$. Also, since $L(g)\leq 1=L(f)$, it follows that $L(g)=L(f)$,
which means that $g$ is a norm preserving extension of $f$. By $(i)$ and
Proposition 4.4 we obtain that $g=\widetilde{f}$. Now, a standard subnet
argument shows that the whole net $(\widetilde{f}_{\alpha})$ must converge
weak-star to $\widetilde{f}$.

$(ii)\Rightarrow(iii)$ It is clear due to Proposition 2.5 $(d)$.

$(iii)\Rightarrow(i)$ Let $g\in\overline{B}_{\mathcal{H}L_{0}(B_{X^{**}})}$ be
a norm preserving extension of $f$. By Proposition 5.1 there is a net
$(Q_{\alpha})\subset\mathcal{P}_{f,0}(X,Y)$ with $L(Q_{\alpha}|_{B_{X}})\leq
1$ satisfying $\widetilde{Q}_{\alpha}(x^{**})\to g(x^{**})$ for all $x^{**}\in
B_{X^{**}}$. But for any $x\in B_{X}$ we have
$\widetilde{Q}_{\alpha}(x)=Q_{\alpha}(x)\to g(x)=f(x)$. Now, $(iii)$ clearly
implies that $g=\widetilde{f}$. ∎

All the numbered results of Sections 4 and 5 have easily adapted analogous
versions for $\mathcal{G}$ and $\mathcal{H}L$ instead of $\mathcal{G}_{0}$ and
$\mathcal{H}L_{0}$.

###  5.1. The case of $\mathcal{H}^{\infty}(B_{X})$ and
$\mathcal{G}^{\infty}(B_{X})$

The arguments of this section can be canonically translated to prove analogous
results for the case of $\mathcal{G}^{\infty}$ instead of $\mathcal{G}_{0}$
(and $\mathcal{H}^{\infty}$ instead of $\mathcal{H}L_{0}$). Moreover, for this
case the hypothesis of symmetrical regularity is unnecessary. Let us state the
results without proofs, since they are similar to the previous arguments.

######  Theorem 5.8.

If $X^{**}$ has the MAP then $\mathcal{G}^{\infty}(B_{X^{**}})$ is isometric
to a 1-locally complemented subspace of $\mathcal{G}^{\infty}(B_{X})^{**}$ and
$\mathcal{H}^{\infty}(B_{X^{**}})$ is isometric to a 1-complemented subspace
of $\mathcal{H}^{\infty}(B_{X})^{**}$.

The following question is posed in [16]: when $X^{**}$ has the BAP, is it true
that $\mathcal{H}^{\infty}(B_{X^{**}})$ is isomorphic to a complemented
subspace of $\mathcal{H}^{\infty}(B_{X})^{**}$? Note that the previous theorem
answers affirmatively this open question for the case $X^{**}$ having MAP.

######  Theorem 5.9.

Suppose $X^{**}$ has the MAP. Consider a function
$f\in\mathcal{H}^{\infty}(B_{X})$ with $\|f\|=1$. Then, the following are
equivalent:

  1. (i)

$f$ has a unique norm preserving extension to
$\mathcal{H}^{\infty}(B_{X^{**}})$.

  2. (ii)

The AB extension from $(\overline{B}_{\mathcal{H}^{\infty}(B_{X})},w^{*})$ to
$(\overline{B}_{\mathcal{H}^{\infty}(B_{X^{**}})},w^{*})$ is continuous at
$f$.

  3. (iii)

If the net $(f_{\alpha})\subset\overline{B}_{\mathcal{H}^{\infty}(B_{X})}$
converges pointwise to $f$, then
$(\widetilde{f}_{\alpha})\subset\overline{B}_{\mathcal{H}^{\infty}(B_{X^{**}})}$
converges pointwise to $\widetilde{f}$.

##  6\. Appendix

Finally we will prove the following result as promised in Section 2.

######  Theorem 6.1.

There exists an isomorphism into
$F\colon\ell_{\infty}\to\mathcal{H}_{0}^{\infty}(\mathbb{D})$ such that
$F(\ell_{\infty}\setminus\\{0\\})\subset
H_{0}^{\infty}(\mathbb{D})\setminus\mathcal{H}L_{0}(\mathbb{D})$ and
$F(c_{0}\setminus\\{0\\})\subset\mathcal{A}(\mathbb{D})\setminus\mathcal{H}L_{0}(\mathbb{D})$.

Note that one can easily prove a version for holomorphic functions on $B_{X}$
for any $X$ using the same ideas as in the proof of Theorem 2.4.

In what follows, we will use the function
$\varphi_{\lambda}\colon\mathbb{C}\to\mathbb{C}$ given by

$\varphi_{\lambda}(z)=\frac{\overline{\lambda}z+1}{2}.$

It is a standard fact that

(6) $\varphi_{\lambda}(\lambda)=1,\quad|\varphi_{\lambda}(z)|<1\text{ for all
}z\in\overline{\mathbb{D}}\setminus\\{\lambda\\}.$

We also need the following technical lemma, which in particular provides
another example of a non-Lipschitz function in the disc algebra
$\mathcal{A}(\mathbb{D})$.

######  Lemma 6.2.

Fix $\lambda\in\mathbb{C}$ with $|\lambda|=1$ and define
$f_{\lambda}\colon\mathbb{C}\to\mathbb{C}$ by

$f_{\lambda}(z)=\begin{cases}1+(\overline{\lambda}z-1)e^{1/(\overline{\lambda}z-1)}&\text{if
$z\neq\lambda$}\\\ 1&\text{if $z=\lambda$}.\end{cases}$

Then

  1. (a)

$f_{\lambda}$ is holomorphic in $\mathbb{C}\setminus\\{\lambda\\}$.

  2. (b)

The restriction of $f_{\lambda}$ to $\overline{\mathbb{D}}$ belongs to
$\mathcal{A}(\mathbb{D})\setminus\mathcal{H}L(\mathbb{D})$.

  3. (c)

$|f_{\lambda}(z)|\leq 3$ for all $z\in\overline{\mathbb{D}}.$

  4. (d)

If $0<s<1$, then $|f^{\prime}_{\lambda}(z)|\leq\frac{s+1}{s}$ for all
$z\in\mathbb{D}$ such that $|z-\lambda|\geq s$.

  5. (e)

Given $k\in\mathbb{N}$ and $0<\delta<1$, we have that

$\sup_{z\in\mathbb{D}(\lambda,\delta)\cap\mathbb{D}}|\big{(}f_{\lambda}\cdot\varphi_{\lambda}^{k}\big{)}^{\prime}(z)|=+\infty.$

###### Proof.

A standard computation shows that a holds. Now, to prove the rest of the
claims it is enough to consider the case $\lambda=1$. Denote $f=f_{1}$ and
take $z=a+ib\in\overline{\mathbb{D}}\setminus\\{1\\}$, with
$a,b\in\mathbb{R}$. We have that

$\Big{|}e^{\frac{1}{z-1}}\Big{|}=e^{\operatorname{Re}\frac{1}{z-1}}=e^{\frac{a-1}{(a-1)^{2}+b^{2}}}\leq
e^{0}=1.$

Hence $f$, defined as $f(z)=1+(z-1)e^{\frac{1}{z-1}}$ is holomorphic on
$\mathbb{C}\setminus\\{1\\}$ and continuously extends to
$\overline{\mathbb{D}}$. Further $|f(z)|\leq 3$ for every
$z\in\overline{\mathbb{D}}$. Let us show that $f$ is not a Lipschitz function.
For that, it is enough to check that $f^{\prime}$ is not bounded on
$\mathbb{D}$. Taking a null sequence $0<\theta_{n}<1$ and setting
$z_{n}:=\cos\theta_{n}(\cos\theta_{n}+i\sin\theta_{n})$, we obtain that the
sequence $(z_{n})\subset\mathbb{D}$ converges to $1$ and

$|f^{\prime}(z_{n})|=\Big{|}\frac{z_{n}-2}{z_{n}-1}\Big{|}e^{\operatorname{Re}\Big{(}\frac{1}{z_{n}-1}\Big{)}}=\Big{|}\frac{z_{n}-2}{z_{n}-1}\Big{|}e^{-1}.$

Consequently, $\lim_{n\to+\infty}|f^{\prime}(z_{n})|=+\infty$. Thus far we
have proved a, b and c. Let’s check d. We have

$|f^{\prime}(z)|=\Big{|}\frac{z-2}{z-1}\Big{|}\cdot\Big{|}e^{\frac{1}{z-1}}\Big{|}\leq
1+\frac{1}{|z-1|},$

for all $z\in\mathbb{D}$. Hence, if $0<s<1$ and $z\in\mathbb{D}$ with
$|z-1|\geq s$ we have that $|f^{\prime}(z)|\leq\frac{s+1}{s}$.

Finally e is a consequence of
$(f\varphi^{k})^{\prime}(z)=f^{\prime}(z)\varphi^{k}(z)+f(z)(\varphi^{k})^{\prime}(z)$
for all $z\in\mathbb{C}\setminus\\{1\\}$. ∎

###### Proof of Theorem 6.1.

To begin with, we choose a sequence
$(\lambda_{n})\subset\mathbb{C}\setminus\\{1\\}$ convergent to $1$ with
$|\lambda_{n}|=1$ and $\lambda_{n}\neq\lambda_{m}$ for every $n\neq m$.
Consider the function $\Phi\colon\mathbb{C}^{2}\to\mathbb{C}$ and
$\varphi_{n}\colon\mathbb{C}\to\mathbb{C}$,
$\varphi_{n}(z):=\Phi(z,\lambda_{n})$ defined as

$\Phi(z,\lambda)=\frac{\overline{\lambda}z+1}{2}$

and, for each $p\in\mathbb{N}$, the compact subset of $\mathbb{C}^{2}$

$K_{p}=\\{(\lambda_{p},\lambda_{n}):n\in\mathbb{N},n\neq
p\\}\cup\\{(\lambda_{p},1)\\}.$

We have $\big{|}\Phi(z,\lambda)\big{|}<1$ for every $(z,\lambda)\in K_{p}$ by
(6), and $\Phi$ is continuous on $\mathbb{C}^{2}$. Hence, there exists
$0<s_{p}<1$ such that $\big{|}\Phi(z,\lambda)\big{|}<1$ for every
$(z,\lambda)\in K_{p}+\overline{\mathbb{D}}((0,0),s_{p})$. In particular,

(7) $|\varphi_{n}(z)|=\big{|}\Phi(z,\lambda_{n})\big{|}<1,$

for all $z\in\overline{\mathbb{D}}(\lambda_{p},s_{p})$ and all $n\neq p$.

Now, since the sequence $(\lambda_{n})$ is convergent to $1$ we can find a
sequence of positive numbers $(r_{n})$ that tends to $0$ such that
$0<2r_{n}<s_{n}$ for all $n\in\mathbb{N}$ and such that
$\overline{\mathbb{D}}(\lambda_{n},2r_{n})\cap\overline{\mathbb{D}}(\lambda_{p},2r_{p})=\emptyset,$
for all $n\neq p$. Moreover, as $(r_{n})$ converges to 0, for each
$n\in\mathbb{N}$ the set

$L_{n}:=\bigcup_{p\neq
n}\overline{\mathbb{D}}(\lambda_{p},2r_{p})\cup\\{1\\},$

is also a compact subset of $\mathbb{C}$, (although it is not a subset of
$\overline{\mathbb{D}}$) and $|\varphi_{n}(z)|<1$ for all $z\in L_{n}$. Since
$|\varphi_{n}|$ is continuous on $\mathbb{C}$ we obtain that

$\max\\{|\varphi_{n}(z)|\,:\,z\in C_{n}\cup L_{n}\\}<1,$

for all $n$, where
$C_{n}=\overline{\mathbb{D}}\setminus\mathbb{D}(\lambda_{n},r_{n})$. As a
consequence, for each $n$ the sequence
$\Big{(}\varphi_{n}^{k}\Big{)}_{k=1}^{\infty}$ converges uniformly to 0 on
$C_{n}\cup L_{n}$ and we can find a $k_{n}\in\mathbb{N}$ such that

(8) $|\varphi_{n}^{k_{n}}(z)|<\frac{r_{n}}{3^{n+1}},$

for every $z\in C_{n}\cup L_{n}$.

We denote $f_{n}:=f_{\lambda_{n}}$, for $n\in\mathbb{N}$ and we define
$F\colon\ell_{\infty}\longrightarrow\mathcal{H}^{\infty}(\mathbb{D})$ by

$F(a_{n}):=\sum_{n=1}^{\infty}a_{n}f_{n}\varphi_{n}^{k_{n}},$

For each $(a_{n})\in\ell_{\infty}$ the series $F(a_{n})(z)$ is convergent for
each $z\in\overline{\mathbb{D}}$. To see this, we first suppose that

a)
$z\in\overline{\mathbb{D}}\setminus\left(\bigcup_{n=1}^{\infty}\overline{\mathbb{D}}(\lambda_{n},r_{n})\right)$.
In that case, by (8) and Lemma 6.2. c,

(9)
$\sum_{n=1}^{\infty}|a_{n}f_{n}(z)\varphi_{n}^{k_{n}}(z)|\leq\sum_{n=1}^{\infty}3|a_{n}|\frac{r_{n}}{3^{n+1}}\leq\frac{1}{2}\|(a_{n})\|_{\infty}.$

Hence $F(a_{n})(z)$ converges. Moreover, the series $F(a_{n})$ converges
absolutely and uniformly on the open set
$\mathbb{D}\setminus\Big{(}\bigcup_{n=1}^{\infty}\overline{\mathbb{D}}(\lambda_{n},r_{n})\Big{)}$.
Thus $F(a_{n})$ is holomorphic in that open set.

If this does not occur, then it must be that we have:

b) There exists a unique $n_{0}\in\mathbb{N}$ such that
$z\in\mathbb{D}(\lambda_{n_{0}},2r_{n_{0}})$. By (8), for every
$u\in\mathbb{D}(\lambda_{n_{0}},2r_{n_{0}})$ we have that

$|a_{n}f_{n}(u)\varphi_{n}^{k_{n}}(u)|\leq
3|a_{n}|\frac{r_{n}}{3^{n+1}}<\frac{|a_{n}|}{3^{n}},$

for all $n\neq n_{0}$ and

$|a_{n_{0}}f_{n_{0}}(u)\varphi_{n_{0}}^{k_{n_{0}}}(u)|\leq 3|a_{n_{0}}|.$

Hence,

(10) $\sum_{n=1}^{\infty}|a_{n}f_{n}(z)\varphi_{n}^{k_{n}}(z)|\leq
4\|(a_{n})\|_{\infty},$

and we have obtained that for every
$z\in\mathbb{D}(\lambda_{n_{0}},2r_{n_{0}})$, $F(a_{n})(z)$ exists and in fact
$|F(a_{n})(z)|\leq 4\|(a_{n})\|_{\infty}$. But our argument shows that the
series $F(a_{n})$ is absolutely and uniformly convergent in the open disc
$\mathbb{D}(\lambda_{n_{0}},2r_{n_{0}})$. Hence, $F(a_{n})$ is holomorphic on
$\mathbb{D}\cup\bigcup_{n=1}^{\infty}\mathbb{D}(\lambda_{n},2r_{n})$ and
$F\colon\ell_{\infty}\to\mathcal{H}^{\infty}(\mathbb{D})$ is a continuous
linear mapping since $\|F(a_{n})\|\leq 4\|(a_{n})\|_{\infty}$ for all
$(a_{n})\in\ell_{\infty}$.

Now we check that $F$ is bounded below. We already know that for each
$(a_{n})\in\ell_{\infty}$, the function $F(a_{n})$ is holomorphic on
$\mathbb{D}\cup\bigcup_{n=1}^{\infty}\mathbb{D}(\lambda_{n},2r_{n})$ and
bounded on $\mathbb{D}$. Thus, using (8) and the fact that
$\lambda_{p}\in\overline{\mathbb{D}}$, we get

$\displaystyle\|F(a_{n})\|$
$\displaystyle=\sup_{z\in\mathbb{D}}|F(a_{n})(z)|\geq\sup_{p\in\mathbb{N}}|F(a_{n})(\lambda_{p})|\geq\sup_{p\in\mathbb{N}}\left\\{|a_{p}|-\sum_{n\neq
p}3|a_{n}|\frac{r_{n}}{3^{n+1}}\right\\}$
$\displaystyle\geq\sup_{p\in\mathbb{N}}\left\\{|a_{p}|-\frac{\|(a_{n})\|_{\infty}}{2}\right\\}=\frac{\|(a_{n})\|_{\infty}}{2}$

for every $(a_{n})\in\ell_{\infty}$.

Let’s check that if $(b_{n})\in c_{0}$, then $F(b_{n})$ belongs to
$\mathcal{A}(\mathbb{D})$. Given $\varepsilon>0$, there exists
$n_{1}\in\mathbb{N}$ such that $|b_{n}|<\frac{\varepsilon}{3}$, for every
$n\geq n_{1}$. Thus, if $z\in\overline{\mathbb{D}}$.

(11) $\sum_{n=n_{1}}^{\infty}|b_{n}f_{n}(z)\varphi_{n}^{k_{n}}(z)|\leq
3\varepsilon\sum_{n=n_{1}}^{\infty}|\varphi_{n}^{k_{n}}(z)|.$

Now if,
$z\in\overline{\mathbb{D}}\setminus\Big{(}\bigcup_{n=1}^{\infty}\overline{\mathbb{D}}(\lambda_{n},r_{n})\Big{)}$,
then by(8), $|\varphi_{n}^{k_{n}}(z)|\leq\frac{r_{n}}{3^{n+1}}$. Hence, by
(11),

$\sum_{n=n_{1}}^{\infty}|b_{n}f_{n}(u)\varphi_{n}^{k_{n}}(z)|<\varepsilon.$

Otherwise, if
$z\in\overline{\mathbb{D}}\cap\Big{(}\bigcup_{n=1}^{\infty}\overline{\mathbb{D}}(\lambda_{n},r_{n})\Big{)}$,
there is a unique $n_{0}\in\mathbb{N}$ such that
$z\in\overline{\mathbb{D}}(\lambda_{n_{0}},r_{n_{0}})$ and

$\sum_{n=n_{1}}^{\infty}|b_{n}f_{n}(z)\varphi_{n}^{k_{n}}(z)|\leq\varepsilon+\sum_{\stackrel{{\scriptstyle
n=n_{1}}}{{n\neq
n_{0}}}}^{\infty}\varepsilon\frac{r_{n}}{3^{n+1}}<2\varepsilon.$

Consequently the series
$\sum_{n=1}^{\infty}b_{n}f_{n}(z)\varphi_{n}^{k_{n}}(z)$ converges absolutely
and uniformly on $\overline{\mathbb{D}}$ and $F_{|c_{0}}\colon
c_{0}\to\mathcal{A}(\mathbb{D})$ is a well-defined continuous linear mapping.

Consider $(a_{n})\in\ell_{\infty}\setminus\\{0\\}$. There exists $n_{0}$ such
that $a_{n_{0}}\neq 0$. We are going to show that $F(a_{n})^{\prime}(z)$ is
not bounded on
$\mathbb{D}(\lambda_{n_{0}},\frac{r_{n_{0}}}{3})\cap\mathbb{D}$.

By the Weierstrass theorem,

$F(a_{n})^{\prime}(z)=\sum_{n=1}^{+\infty}a_{n}\big{(}f_{n}\varphi_{n}^{k_{n}}\big{)}^{\prime}(z),$

for every
$z\in\mathbb{D}\cup\bigcup_{n=1}^{\infty}\mathbb{D}(\lambda_{n},2r_{n})$. If
$n\neq n_{0}$, then by the Cauchy integral formula

$\big{(}\varphi_{n}^{k_{n}}\big{)}^{\prime}(z)=\frac{1}{2\pi
i}\int_{C(\lambda_{n_{0}},r_{n_{0}})}\frac{\varphi_{n}^{k_{n}}(u)}{(u-z)^{2}}du,$

for every $z\in\mathbb{D}(\lambda_{n_{0}},\frac{r_{n_{0}}}{3})$. Thus, by (7)
and (8), we obtain

$\begin{split}\sup_{z\in\mathbb{D}(\lambda_{n_{0}},\frac{r_{n_{0}}}{3})}|\big{(}\varphi_{n}^{k_{n}}\big{)}^{\prime}(z)|&\leq\frac{r_{n_{0}}}{(\frac{2}{3}r_{n_{0}})^{2}}\sup_{|u-\lambda_{n_{0}}|=r_{n_{0}}}|\varphi_{n}^{k_{n}}(u)|<\frac{9}{4r_{n_{0}}}\frac{r_{n}}{3^{n+1}}<\frac{1}{r_{n_{0}}}\frac{1}{3^{n}},\end{split}$

and we get

$|\big{(}f_{n}\varphi_{n}^{k_{n}}\big{)}^{\prime}(z)|\leq|f^{\prime}_{n}(z)||\varphi_{n}^{k_{n}}(z)|+|f_{n}(z)||(\varphi_{n}^{k_{n}})^{\prime}(z)<\frac{1}{3^{n}}+\frac{1}{r_{n_{0}}}\frac{1}{3^{n-1}},$

where in the second inequality we have applied, (7), (8) and the properties of
$f_{n}$ and $f^{\prime}_{n}$ given in Lemma 6.2. Hence,

$|F(a_{n})^{\prime}(z)|\
\geq|a_{n_{0}}||\big{(}f_{n_{0}}\varphi_{n_{0}}^{k_{n_{0}}}\big{)}^{\prime}(z)|-\|(a_{n})\|_{\infty}\big{(}\frac{1}{2}+\frac{3}{2r_{n_{0}}}\big{)},\\\
$

for every $z\in\mathbb{D}(\lambda_{n_{0}},\frac{r_{n_{0}}}{3})$. Finally, by
Lemma 6.2.e, we have that $F(a_{n})^{\prime}$ is unbounded on
$\mathbb{D}(\lambda_{n_{0}},\frac{r_{n_{0}}}{3})\cap\mathbb{D}$ and hence,
$F(a_{n})$ does not belong to $\mathcal{H}L(\mathbb{D})$.

Finally, if we define
$F_{1}\colon\ell_{\infty}\to\mathcal{H}^{\infty}_{0}(\mathbb{D})$ by
$F_{1}(a_{n})(z):=zF(a_{n})(z)$ for $(a_{n})\in\ell_{\infty}$ and
$z\in\mathbb{D}$, it is clear that $F_{1}$ is an isomorphism onto its image
and that
$F_{1}(\ell_{\infty}\setminus\\{0\\})\subset\mathcal{H}^{\infty}_{0}(\mathbb{D})\setminus\mathcal{H}L(\mathbb{D})$.

∎

Finally, we note that if we are only interested in $\mathcal{A}(\mathbb{D})$
there are known results related to Theorem 6.1. Indeed, in three relevant
papers [10, 11, 12], L. Bernal et al. have obtained many results on the
existence of large subspaces of functions that belong to
$\mathcal{A}(\mathbb{D})\setminus\mathcal{H}L(\mathbb{D})\cup\\{0\\}$. In
particular, in [10, Th. 4.1.c] the authors show that there exists an infinite
dimensional Banach space $X$ contained in $\mathcal{A}(\mathbb{D})$ such that
any non-null function in $X$ is not differentiable on any point of a fixed
dense subset of $\mathbb{T}$. Also, in [12, Th. 3.4], they prove that there
exists an infinite dimensional Banach space $X$, contained in
$\mathcal{A}(\mathbb{D})$, (which, however, is endowed with a stronger norm
than the one inherited from $\mathcal{A}(\mathbb{D})$) such that if $f\in X$,
then the restriction of $f$ to $\mathbb{T}$ is nowhere Hölder on $\mathbb{T}$.

### Acknowledgments

This paper had its start during the summer, 2022 in Banff, thanks to the BIRS
Research in Teams program. The research was also partially supported by
Agencia Estatal de Investigación and EDRF/FEDER “A way of making Europe”
(MCIN/AEI/10.13039/501100011033) through grants PID2021-122126NB-C32 (García-
Lirola) and PID2021-122126NB-C33 (Aron and Maestre). The research of García-
Lirola was also supported by DGA project E48-20R, Fundación Séneca - ACyT
Región de Murcia project 21955/PI/22 and Generalitat Valenciana project
CIGE/2022/97. The research of Dimant was partially supported by CONICET PIP
11220200101609CO and ANPCyT PICT 2018-04104. The research of Maestre also was
partially supported by PROMETEU/2021/070. We are very grateful to A. Procházka
for his comments that allowed us to shorten the proof of Theorem 3.10
significantly, and to A. Chávez-Domínguez for helpful conversations about
Section 3.

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