# Remark on the Daugavet property for complex Banach spaces

Han Ju Lee  Department of Mathematics Education, Dongguk University - Seoul,
04620 (Seoul), Republic of Korea<EMAIL_ADDRESS>and  Hyung-Joon Tag
Department of Mathematics Education, Dongguk University - Seoul, 04620
(Seoul), Republic of Korea<EMAIL_ADDRESS>
###### Abstract.

In this article, we study the Daugavet property and the diametral diameter two
properties in complex Banach spaces. The characterizations for both Daugavet
and $\Delta$-points are revisited in the context of complex Banach spaces. We
also provide relationships between some variants of alternative convexity and
smoothness, nonsquareness, and the Daugavet property. As a consequence, every
strongly locally uniformly alternatively convex or smooth (sluacs) Banach
space does not contain $\Delta$-points from the fact that such spaces are
locally uniformly nonsquare. We also study the convex diametral local diameter
two property (convex-DLD2P) and the polynomial Daugavet property in the
vector-valued function space $A(K,X)$. From an explicit computation of the
polynomial Daugavetian index of $A(K,X)$, we show that the space $A(K,X)$ has
the polynomial Daugavet property if and only if either the base algebra $A$ or
the range space $X$ has the polynomial Daugavet property. Consequently, we
obtain that the polynomial Daugavet property, Daugavet property, diameteral
diameter two properties, and property ($\mathcal{D}$) are equivalent for
infinite-dimensional uniform algebras.

###### Key words and phrases:

Daugavet points, $\Delta$-points, alternative convexity or smoothness,
nonsquareness, polynomial Daugavet property

###### 2010 Mathematics Subject Classification:

Primary 46B20; Secondary 46B04, 46E40, 46J10

The first author was supported by Basic Science Research Program through the
National Research Foundation of Korea(NRF) funded by the Ministry of
Education, Science and Technology [NRF-2020R1A2C1A01010377].

The second author was supported by Basic Science Research Program through the
National Research Foundation of Korea(NRF) funded by the Ministry of
Education, Science and Technology [NRF-2020R1A2C1A01010377].

##  1\. Introduction

In the theory of Banach spaces, various properties that are related to certain
behaviors of vector measures and bounded linear operators have been studied
from a geometrical point of view. We focus on the Daugavet property and
diametral diameter two properties in this article. Let $X$ be a Banach space
on $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$. For a given $\epsilon>0$, a slice
$S(x^{*},\epsilon)$ of the unit ball $B_{X}$ is defined by
$S(x^{*},\epsilon)=\\{x\in B_{X}:\text{Re}\,x^{*}x>1-\epsilon\\}$ and a
weak∗-slice $S(x,\epsilon)$ of $B_{X^{*}}$ by $S(x,\epsilon)=\\{x^{*}\in
B_{X^{*}}:\text{Re}\,x^{*}x>1-\epsilon\\}$.

Many results on Banach spaces with the Radon-Nikodým property have been
obtained from this perspective. Even though its definition was originally
given in terms of vector measures, it is now well-known that the Banach spaces
with the Radon-Nikodým property can be characterized by the existence of
slices with arbitrarily small diameter as well as the existence of denting
points. For more details on the geometrical aspect of the Radon-Nikodým
property and its application to other research topics in Banach spaces, we
refer to [11].

A Banach space $X$ is said to have the Daugavet property if every rank-one
operator $T:X\rightarrow X$ satisfies the following equation:

$\|I+T\|=1+\|T\|.$

We call this equation the Daugavet equation. The space $C(K)$, where $K$ does
not have isolated points, $L_{1}(\mu)$, and $L_{\infty}(\mu)$ with a nonatomic
measure $\mu$ are classical examples with the Daugavet property. The infinite-
dimensional uniform algebras also have the Daugavet property if and only if
their Shilov boundaries do not have isolated points [38, 27]. Moreover, the
Daugavet property in Musielak-Orlicz spaces [24], in Lipschitz-free spaces
[17], and in rearrangement-invariant Banach function lattices [3, 21] have
been examined. It is well-known that every slices of $B_{X}$ with diameter two
if $X$ has the Daugavet property, which tells us that Banach spaces with this
property are on the opposite spectrum to the Radon-Nikodým property.

The following characterization allows us to observe the Daugavet property with
slices.

######  Lemma 1.1.

[22, Lemma 2.2] The following are equivalent.

  1. (i)

A Banach space $(X,\|\cdot\|)$ has the Daugavet property,

  2. (ii)

For every slice $S=S(x^{*},\epsilon)$ where $x^{*}\in S_{X^{*}}$, every $x\in
S_{X}$ and every $\epsilon>0$, there exists $y\in S_{X}\cap S$ such that
$\|x+y\|>2-\epsilon$,

  3. (iii)

For every weak∗-slice $S^{*}=S(x,\epsilon)$ where $x\in S_{X}$, every
$x^{*}\in S_{X^{*}}$ and every $\epsilon>0$, there exists $y^{*}\in
S_{X^{*}}\cap S^{*}$ such that $\|x^{*}+y^{*}\|>2-\epsilon$,

Later, the diametral diameter two properties (diametral D2Ps), the property
($\mathcal{D}$), and the convex diameter local diameter two property have
gained attention [1, 37] by many researchers. They are known to be weaker than
the Daugavet property.

######  Definition 1.2.

  1. (i)

A Banach space $X$ has the property ($\mathcal{D}$) if for every rank-one,
norm-one projection $P:X\rightarrow X$ satisfies $\|I-P\|=2$.

  2. (ii)

A Banach space $X$ has the diametral local diameter two property (DLD2P) if
for every slice $S$ of the unit ball, every $x\in S\cap S_{X}$, and every
$\epsilon>0$ there exists $y\in B_{X}$ such that $\|x-y\|\geq 2-\epsilon$.

  3. (iii)

A Banach space $X$ has the diametral diameter two property (DD2P) if for every
nonempty weakly open subset $W$ of the unit ball, every $x\in W\cap S_{X}$,
and every $\epsilon>0$, there exists $y\in B_{X}$ such that $\|x-y\|\geq
2-\epsilon$.

  4. (iv)

A Banach space $X$ has the convex diametral local diameter two property
(convex-DLD2P) if $\overline{conv}\Delta_{X}=B_{X}$.

The first known example that possesses the property ($\mathcal{D}$) is a
certain subspace of $L_{1}$ constructed with martingales [4]. Later on, this
space is shown to have the Daugavet property [23]. In view of [15], every
rank-one projection $P$ on a Banach space $X$ with the DLD2P satisfies
$\|I-P\|\geq 2$, and so the DLD2P implies the property ($\mathcal{D}$). As a
matter of fact, the property ($\mathcal{D}$) was thought to be equivalent to
the DLD2P. However, since a scalar multiple of a projection is not a
projection [1], the validity of the equivalence is not clear up to now. The
implication (iii) $\implies$ (ii) holds because every slice is a weakly open
subset of the unit ball.

The DLD2P and the Daugavet property can be also considered from a local
perspective by using $\Delta$-points and Daugavet points. Let
$\Delta_{\epsilon}(x)=\\{y\in B_{X}:\|x-y\|\geq 2-\epsilon\\}$.

######  Definition 1.3.

  1. (i)

A point $x\in S_{X}$ is a $\Delta$-point if
$x\in\overline{conv}\Delta_{\epsilon}(x)$ for every $\epsilon>0$

  2. (ii)

A point $x\in S_{X}$ is a Daugavet point if
$B_{X}=\overline{conv}\Delta_{\epsilon}(x)$ for every $\epsilon>0$.

Notice that the set $\Delta_{\epsilon}(x)$ is defined independently of the
scalar fields $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$ on Banach spaces. Hence
we may use the same definitions for $\Delta$-points and Daugavets points for
complex Banach spaces. It is well-known that a real Banach space $X$ has the
Daugavet property (resp. the DLD2P) if and only if every point on the unit
sphere is a Daugavet point (resp. a $\Delta$-point).

We mention that many recent results on the Daugavet property, the diametral
D2Ps, Daugavet points, and $\Delta$-points have been mostly revolved around
real Banach spaces. But there are several results concerning these concepts in
complex Banach spaces; see [18, 32]. For a real Banach space $X$, it is well-
known that $\Delta$-points are connected to certain behaviors of slices of the
unit ball and of rank-one projections.

######  Theorem 1.4.

[1] Let $X$ be a real Banach space. Then the following statements are
equivalent.

  1. (i)

$x\in S_{X}$ is a $\Delta$-point.

  2. (ii)

For every slice $S$ of $B_{X}$ with $x\in S\cap S_{X}$ and $\epsilon>0$, there
exists $y\in S$ such that $\|x-y\|\geq 2-\epsilon$.

  3. (iii)

For every rank-1 projection $P=x^{*}\otimes x$ with $x^{*}x=1$, we have
$\|I-P\|\geq 2$.

Even though the complex analogue of this relationship may be well-known to the
specialists, we will state and prove them for completeness. In addition, while
the Daugavet property for complex Banach spaces can be examined through rank-
one real-linear operators [19], it has not been known whether we can examine
the DLD2P in a similar spirit with rank-one real-projections. We also study
this here.

Since a denting point is always contained in a slice of arbitrarily small
diameter, such a point cannot be either a $\Delta$-point or a Daugavet point.
This implies that any (locally) uniformly rotund real Banach spaces cannot
have $\Delta$-points. Recently, identifying the Banach spaces that do not
contain these points has been an active research topic. For example, it is
shown in [2] that every uniformly nonsquare real Banach space does not have
$\Delta$-points. Furthermore, a locally uniformly nonsquare real Banach space
does not have $\Delta$-points [25]. We will examine strongly locally uniformly
alternatively convex or smooth (sluacs) Banach spaces in this article. We
mention that alternative convexity or smoothness are related to the anti-
Daugavet property and the nonsquareness property [16, 36].

Banach spaces that satisfy the Daugavet equation for weakly compact
polynomials are also studied in [7, 8, 30]. For Banach spaces $X,Y$, let
$\mathcal{L}(^{k}X;Y)$ be the space of bounded $k$-linear mappings from $X$ to
$Y$ and let $\Delta_{k}:X\rightarrow X^{k}$ be a diagonal mapping defined by

$\Delta_{k}(x)=\underset{k\,\,\,\text{times}}{\underbrace{(x,x,\dots,x)}}.$

A mapping is called a bounded $k$-homogeneous polynomial from $X$ to $Y$ if it
is the composition of $\Delta_{k}$ with an element in $\mathcal{L}(^{k}X;Y)$.
We denote the set of all bounded $k$-homogenous polynomials from $X$ to $Y$ by
$\mathcal{P}(^{k}X;Y)$. A polynomial is a finite sum of bounded homogenous
polynomials from $X$ to $Y$. We also denote the set of all polynomials from
$X$ to $Y$ by $\mathcal{P}(X;Y)$ and the set of all scalar-valued continuous
polynomials by $\mathcal{P}(X)$. We endow the space $\mathcal{P}(X;X)$ (resp.
$\mathcal{P}(X)$) with the norm $\|P\|=\sup_{x\in B_{X}}\|Px\|_{X}$ (resp.
$\|P\|=\sup_{x\in B_{X}}|Px|$).

We say a polynomial $P\in\mathcal{P}(X;Y)$ is weakly compact if $P(B_{X})$ is
a relatively weakly compact subset of $Y$. A Banach space $X$ is said to have
the polynomial Daugavet property if for every weakly compact polynomial
$P\in\mathcal{P}(X,X)$ satisfies

$\|I+P\|=1+\|P\|.$

If $X$ has the polynomial Daugavet property, then the space also has the
Daugavet property. It is also well-known that the polynomial Daugavet property
can be described in terms of scalar-valued polynomials.

######  Theorem 1.5.

[7, Corollary 2.2] Let $X$ be a real or complex Banach space. Then the
following statements are equivalent:

  1. (1)

$X$ has the polynomial Daugavet property.

  2. (2)

For every $p\in\mathcal{P}(X)$ with $\|p\|=1$, every $x_{0}\in S_{X}$, and
every $\epsilon>0$, there exists $\omega\in S_{\mathbb{C}}$ and $y\in B_{X}$
such that $\text{Re}\,p(y)>1-\epsilon$ and $\|x_{0}+\omega y\|>2-\epsilon$.

  3. (3)

For every $p\in\mathcal{P}(X)$ and every $x_{0}\in X$, the polynomial
$p\otimes x_{0}$ satisfies the Daugavet equation.

In this article, we will look at the polynomial Daugavet property in a
function space $A(K,X)$ over the base algebra $A$, which will be defined
later. This class of function spaces includes uniform algebras and the space
of Banach space-valued continuous functions on a compact Hausdorff space. The
Daugavet property and the diametral D2Ps of the vector-valued function spaces
$A(K,X)$ are studied in [27]. From the same article, assuming the uniform
convexity of the range space $X$ and $A\otimes X\subset A(K,X)$, it is shown
that the space $A(K,X)$ has the Daugavet property if and only if its base
algebra also has the Daugavet property. It is also shown in [6] that if $X$
has the Daugavet property then $A(K,X)$ has the Daugavet property. Here we
attempt to find a necessary and sufficient condition for $A(K,X)$ with the
polynomial Daugavet property.

The article consists of three parts. In section 2, we revisit well-known facts
about $\Delta$-points and Daugavet points in the context of complex Banach
spaces. Like the Daugavet property, the DLD2P can be also analyzed by using
rank-one real-projections (Theorem 2.2). In Section 3, we examine the
relationship between alternative convexity or smoothness and nonsquareness.
From the fact that strongly locally uniformly alternatively convex or smooth
(sluacs) Banach spaces are locally uniformly nonsquare (Proposition 3.8.(i)),
every sluacs Banach space does not have a $\Delta$-point (Theorem 3.4). In
section 4, we study the polynomial Daugavet property of the space $A(K,X)$.
Here we explicitly compute the polynomial Daugavetian index of the space
$A(K,X)$ (Theorem 4.11). The space $A(K,X)$ has a bicontractive projection if
the Shilov boundary of the base algebra $A$ has isolated points (Proposition
4.9). As a consequence, we will show that $A(K,X)$ has the polynomial Daugavet
property if and only if either the base algebra $A$ or the range space $X$ has
the polynomial Daugavet property (Corollary 4.12).

##  2\. Delta-points and Daugavet points in complex Banach spaces

In this section, we study $\Delta$-points and Daugavet points for complex
Banach spaces. Although one may find that a certain portion of the proofs are
similar to the real case, we include them in this article for completeness.
However, we mention that the complex scalar field $\mathbb{C}$ provides
something more, namely, a tool to analyze the Daugavet property and the DLD2P
for complex Banach spaces through rank-one real-linear operators and rank-one
real-projections, respectively. We recall the following useful lemma:

######  Lemma 2.1.

[15, Lemma 1.4] Let $x^{*}\in S_{X^{*}}$, $\epsilon>0$. Then for every $x\in
S(x^{*},\epsilon)$ and every $\delta\in(0,\epsilon)$ there exists $y^{*}\in
S_{X^{*}}$ such that $x\in S(y^{*},\delta)$ and $S(y^{*},\delta)\subset
S(x^{*},\epsilon)$.

######  Theorem 2.2.

Let $X$ be a complex Banach space. Then the following statements are
equivalent.

  1. (i)

$x\in S_{X}$ is a $\Delta$-point.

  2. (ii)

For every slice $S$ of $B_{X}$ with $x\in S\cap S_{X}$ and $\epsilon>0$, there
exists $y\in S$ such that $\|x-y\|\geq 2-\epsilon$.

  3. (iii)

For every rank-1 projection $P=x^{*}\otimes x$ with $x^{*}x=1$, we have
$\|I-P\|\geq 2$.

  4. (iv)

For every rank-1 real-projection $P=\text{Re}\,x^{*}\otimes x$ with
$x^{*}x=1$, we have $\|I-P\|\geq 2$.

###### Proof.

The implications (i) $\iff$ (ii) and (ii) $\iff$ (iv) come from modifying the
proofs given in [33, Proposition 1.4.5] for Daugavet points and [15].

(i) $\implies$ (ii): Assume to the contrary that there exist a slice
$S=S(x^{*},\alpha)$ containing $x$ and $\alpha>0$ such that $\|x-y\|<2-\alpha$
for every $y\in S$. This implies that $S\cap\Delta_{\alpha}(x)=\emptyset$.
Since $x$ is a $\Delta$-point and $x\in S$, we see that
$S\cap\overline{conv}\Delta_{\alpha}(x)\neq\emptyset$. Choose $y\in S$ such
that $\text{Re}\,x^{*}y>1-\alpha+\delta$ for sufficiently small $\delta>0$.
Then there exist $y_{1},y_{2},\dots,y_{n}\in\Delta_{\alpha}(x)$ such that

$\text{Re}\,x^{*}y-\frac{1}{n}\sum_{i=1}^{n}\text{Re}\,x^{*}y_{i}\leq\left\|y-\frac{1}{n}\sum_{i=1}^{n}y_{i}\right\|<\delta.$

From the fact that $y_{i}$’s are not in the slice $S$, we have

$\displaystyle
1-\alpha<\text{Re}\,x^{*}y-\delta=\text{Re}\,x^{*}y-\frac{1}{n}\sum_{i=1}^{n}\text{Re}\,x^{*}y_{i}+\frac{1}{n}\sum_{i=1}^{n}\text{Re}\,x^{*}y_{i}-\delta$
$\displaystyle<$ $\displaystyle\frac{1}{n}\sum_{i=1}^{n}\text{Re}\,x^{*}y_{i}$
$\displaystyle<$ $\displaystyle 1-\alpha,$

which leads to contradiction.

(ii) $\implies$ (i): Suppose that $x\notin\overline{conv}\Delta_{\epsilon}(x)$
for some $\epsilon>0$. Notice that a singleton $\\{x\\}$ is convex as well as
$\overline{conv}\Delta_{\epsilon}(x)$. Moreover, the set $\\{x\\}$ is compact.
So by Hahn-Banach separation theorem, there exist $x^{*}\in S_{X^{*}}$ and
$\alpha>0$ such that for every $z\in\overline{conv}\Delta_{\epsilon}(x)$, we
have $\text{Re}\,x^{*}z<\alpha<\text{Re}x^{*}x\leq 1$. Hence, we see that
$z\notin S(x^{*},1-\alpha)$ for every
$z\in\overline{conv}\Delta_{\epsilon}(x)$, in particular, for every
$z\in\Delta_{\epsilon}(x)$. This leads to a contradiction from our assumption
(ii).

(iii) $\implies$ (ii): Consider a slice $S(x^{*},\delta)$, an element $x\in
S(x^{*},\delta)$ and $\epsilon>0$. Then there exist $\delta_{1}>0$ such that
$\frac{\sqrt{2\delta_{1}}}{1-\delta_{1}}<\frac{\epsilon}{4}$ and a bounded
linear functional $y^{*}\in S_{X^{*}}$ such that $x\in
S(y^{*},\delta_{1})\subset S(x^{*},\delta)$.

Consider a rank-one projection $P:y\mapsto y^{*}y\frac{x}{y^{*}x}$. Then by
the assumption (iii), for every $\beta<\frac{\epsilon}{2}$, there exists $y\in
S_{X}$ such that

(1) $\|y-Py\|=\left\|y-y^{*}y\frac{x}{y^{*}x}\right\|=\left\|\gamma
y-\text{Re}\,y^{*}(\gamma y)\frac{x}{y^{*}x}\right\|>2-\beta,$

where $\gamma\in\mathbb{T}$ such that $|y^{*}y|=\gamma y^{*}y=y^{*}(\gamma
y)=\text{Re}\,y^{*}(\gamma y)$.

Moreover, we also see that $\gamma y\in S(y^{*},\delta_{1})$. Let
$\tilde{y}=\gamma y$. Then by (1), we obtain

$\displaystyle\|\tilde{y}-x\|$ $\displaystyle=$ $\displaystyle\left\|\gamma
y-\text{Re}\,y^{*}(\gamma y)\frac{x}{y^{*}x}+\text{Re}\,y^{*}(\gamma
y)\frac{x}{y^{*}x}-x\right\|$ $\displaystyle\geq$ $\displaystyle\left\|\gamma
y-\text{Re}\,y^{*}(\gamma
y)\frac{x}{y^{*}x}\right\|-\left\|x-\text{Re}\,y^{*}(\gamma
y)\frac{x}{y^{*}x}\right\|$ $\displaystyle>$ $\displaystyle
2-\beta-\left|1-\frac{\text{Re}\,y^{*}(\gamma y)}{y^{*}x}\right|.$

Since $|y^{*}x|\geq\text{Re}\,y^{*}x>1-\delta_{1}$, we can see that

$\left|1-\frac{\text{Re}\,y^{*}(\gamma
y)}{y^{*}x}\right|=\frac{|y^{*}x-\text{Re}\,y^{*}(\gamma
y)|}{|y^{*}x|}\leq\frac{|1-y^{*}x|+(1-\text{Re}\,y^{*}(\gamma
y))}{1-\delta_{1}}$

and

$(\text{Im}\,y^{*}x)^{2}=|y^{*}x|^{2}-(\text{Re}\,y^{*}x)^{2}<1-(1-\delta_{1})^{2}<\delta_{1},$

Hence, we have

$\displaystyle|1-y^{*}x|=\sqrt{(1-\text{Re}\,y^{*}x)^{2}+(\text{Im}\,y^{*}x)^{2}}$
$\displaystyle<$ $\displaystyle\sqrt{(1-\text{Re}\,y^{*}x)^{2}+\delta_{1}}$
$\displaystyle<$
$\displaystyle\sqrt{\delta_{1}^{2}+\delta_{1}}<\sqrt{2\delta_{1}}.$

These consequently show that

$\|\tilde{y}-x\|>2-\beta-\frac{2\sqrt{2\delta_{1}}}{1-\delta_{1}}>2-\epsilon.$

(ii) $\implies$ (iii): Every rank-one projection is of the form
$P=x^{*}\otimes x$, where $\|x^{*}\|\geq 1,\|x\|=1$, and $x^{*}x=1$. Define an
operator $T\in L(X)$ by $T(y)=\frac{x^{*}y}{\|x^{*}\|}\cdot x$ and consider a
slice $S=\\{y\in B_{X}:\text{Re}\,\frac{x^{*}}{\|x^{*}\|}y\geq
1-\frac{\epsilon}{2}\\}$ containing $x$. Since
$\left|\frac{x^{*}}{\|x^{*}\|}y\right|\geq\text{Re}\,\frac{x^{*}}{\|x^{*}\|}y>1-\frac{\epsilon}{2}$,
we know that
$\left(\text{Im}\,\frac{x^{*}}{\|x^{*}\|}y\right)^{2}=\left|\frac{x^{*}}{\|x^{*}\|}y\right|^{2}-\left(\text{Re}\frac{x^{*}}{\|x^{*}\|}y\right)^{2}<1-(1-\frac{\epsilon}{2})^{2}$.
Then

$\left|1-\frac{x^{*}}{\|x^{*}\|}y\right|=\sqrt{\left(1-\text{Re}\,\frac{x^{*}}{\|x^{*}\|}y\right)^{2}+\left(\text{Im}\,\frac{x^{*}}{\|x^{*}\|}y\right)^{2}}<\sqrt{\frac{\epsilon^{2}}{4}+1-\left(1-\frac{\epsilon}{2}\right)^{2}}<\sqrt{\epsilon}.$

Moreover, we see that

$\displaystyle\|(I-T)y\|\geq\|y-x\|-\left\|x-\frac{x^{*}}{\|x^{*}\|}y\cdot
x\right\|$ $\displaystyle>$ $\displaystyle
2-\frac{\epsilon}{2}-\left|1-\frac{x^{*}}{\|x^{*}\|}y\right|$ $\displaystyle>$
$\displaystyle 2-\frac{\epsilon}{2}-\sqrt{\epsilon}.$

Hence, $\|I-T\|\geq 2$.

Now define a function $\varphi(\lambda)=\|I-\lambda T\|$ where
$\lambda\in[0,\infty)$. It is easy to show that the function $\varphi$ is a
convex function on $[0,\infty)$. Also, $\varphi(0)=1$ and
$\varphi(1)=\|I-T\|\geq 2$. Furthermore,

$0<\frac{\varphi(1)-\varphi(0)}{1-0}\leq\frac{\varphi(s)-\varphi(1)}{s-1}\,\,\,\text{for
all}\,\,\,s\geq 1.$

So, for every $s\geq 1$ we see that $\varphi(s)>\varphi(1)\geq 2$. In
particular, if $s=\|x^{*}\|\geq 1$ we obtain that
$\varphi(\|x^{*}\|)=\|I-\|x^{*}\|\cdot T\|=\|I-P\|\geq 2$, which proves (ii)
$\implies$ (iii). ∎

Hence we can verify the relationship between the $\Delta$-points and the DLD2P
as well as the space with bad projections [15] for complex Banach spaces.

######  Corollary 2.3.

Let $X$ be a complex Banach space. The following statements are equivalent:

  1. (i)

The space $X$ has the DLD2P.

  2. (ii)

Every point on the unit sphere $S_{X}$ is a $\Delta$-point.

  3. (iii)

Every rank-one projection $P$ has $\|I-P\|\geq 2$, i.e. $X$ is the space with
bad projections.

  4. (iv)

Every rank-one real-projection $P$ has $\|I-P\|\geq 2$.

The following statement about Daugavet points is a compliation of well-known
results, but we include them for completeness.

######  Theorem 2.4.

Let $X$ be a complex Banach spaces. Then the following statements are
equivalent.

  1. (i)

$x\in S_{X}$ is a Daugavet point.

  2. (ii)

For every slice $S$ of $B_{X}$ and $\epsilon>0$, there exists $y\in S$ such
that $\|x-y\|\geq 2-\epsilon$.

  3. (iii)

For every rank-one operator $T=x^{*}\otimes x$ satisfies $\|I-T\|=1+\|T\|$.

  4. (iv)

For every rank-one operator $T=x^{*}\otimes x$ with norm-one satisfies
$\|I-T\|=2$.

  5. (v)

For every rank-one real-linear operator $T=\text{Re}\,x^{*}\otimes x$
satisfies $\|I-T\|=1+\|T\|$.

  6. (vi)

For every rank-one real-linear operator $T=\text{Re}\,x^{*}\otimes x$ with
norm-one satisfies $\|I-T\|=2$.

###### Proof.

One may see the proof for the real case in [33, Proposition 1.4.5].

(i) $\implies$ (ii): Assume to the contrary that there exist a slice
$S=S(x^{*},\alpha)$ containing $x$ and $\alpha>0$ such that
$\|x-y\|<2-\epsilon$ for every $y\in S$. This implies that
$S\cap\Delta_{\epsilon}(x)=\emptyset$. Since $x$ is a Daugavet point and $x\in
S$, we see that $S\subset\overline{conv}\Delta_{\epsilon}(x)$. Choose $y\in S$
such that $\text{Re}\,x^{*}y>1-\alpha+\delta$ for sufficiently small
$\delta>0$. Then there exist $y_{1},y_{2},\dots,y_{n}\in\Delta_{\epsilon}(x)$
such that

$\text{Re}\,x^{*}y-\frac{1}{n}\sum_{i=1}^{n}\text{Re}\,x^{*}y_{i}\leq\left\|y-\frac{1}{n}\sum_{i=1}^{n}y_{i}\right\|<\delta.$

From the fact that $y_{i}$’s are not in the slice $S$, we have

$\displaystyle
1-\alpha<\text{Re}\,x^{*}y-\delta=\text{Re}\,x^{*}y-\frac{1}{n}\sum_{i=1}^{n}\text{Re}\,x^{*}y_{i}+\frac{1}{n}\sum_{i=1}^{n}\text{Re}\,x^{*}y_{i}-\delta$
$\displaystyle<$ $\displaystyle\frac{1}{n}\sum_{i=1}^{n}\text{Re}\,x^{*}y_{i}$
$\displaystyle<$ $\displaystyle 1-\alpha,$

which leads to contradiction.

(ii) $\implies$ (i): Suppose that $x\in
B_{X}\neq\overline{conv}\Delta_{\epsilon}(x)$ for some $\epsilon>0$. Notice
that a singleton $\\{x\\}$ is convex as well as
$\overline{conv}\Delta_{\epsilon}(x)$. Moreover, the set $\\{x\\}$ is compact.
So by Hahn-Banach separation theorem, there exist $x^{*}\in S_{X^{*}}$ and
$\alpha>0$ such that for every $z\in\overline{conv}\Delta_{\epsilon}(x)$, we
have $\text{Re}\,x^{*}(z)<\alpha<\text{Re}x^{*}y\leq 1$. Hence, we see that
$z\notin S(x^{*},1-\alpha)$ for every
$z\in\overline{conv}\Delta_{\epsilon}(x)$. This leads to a contradiction from
our assumption (ii).

(iii) $\implies$ (iv) is clear, and (iv) $\implies$ (iii) comes from the fact
that for Daugavet property considering rank-one operators with norm-one is
enough [37]. The equivalence (ii) $\iff$ (v) $\iff$ (vi) is a well-known
result in [1].

(iv) $\implies$ (ii): Let $x^{*}\in S_{X^{*}}$ and $S=\\{y\in
B_{X}:\text{Re}\,x^{*}y>1-\frac{\epsilon}{2}\\}$ containing $x\in S_{X}$. Let
$T\in L(X)$ be a rank-one operator with norm one of the form $T(y)=x^{*}y\cdot
x$ where $\|x^{*}\|=\|x\|=1$. By the assumption (iv), for every $\epsilon>0$,
there exists $y\in S_{X}$ such that $\|y-Ty\|=\|y-x^{*}y\cdot
x\|>\frac{\epsilon}{2}$. Notice that $|x^{*}y|=\text{Re}\,x^{*}(\gamma y)$ for
some $\gamma\in\mathbb{T}$, and so

$1+\text{Re}\,x^{*}(\gamma y)\geq\|y-x^{*}y\cdot x\|>2-\frac{\epsilon}{2}.$

Hence $\gamma y\in S$. Let $\tilde{y}=\gamma y$. Then we have

$\|\tilde{y}-x\|\geq\|y-\text{Re}\,x^{*}(\gamma y)\cdot
x\|-1+\text{Re}\,x^{*}(\gamma y)>2-\epsilon.$

Therefore, we see that (ii) holds.

(ii) $\implies$ (iv): Every rank-one operator with norm-one is of the form
$T=x^{*}\otimes x$ where $\|x^{*}\|=\|x\|=1$. Let $S=\\{y\in
B_{X}:\text{Re}\,x^{*}y>1-\frac{\epsilon}{2}\\}$ be a slice containing $x$. By
the assumption (ii), there exists $y\in B_{X}$ such that
$\|x-y\|>2-\frac{\epsilon}{2}$. Notice that
$1\geq|x^{*}y|^{2}=(\text{Re}\,x^{*}y)^{2}+(\text{Im}\,x^{*}y)^{2}$. Hence we
have

$|1-x^{*}y|=\sqrt{(1-\text{Re}\,x^{*}y)^{2}+(\text{Im}\,x^{*}y)^{2}}<\sqrt{\frac{\epsilon^{2}}{4}+1-\left(1-\frac{\epsilon}{2}\right)^{2}}<\sqrt{\epsilon}.$

Moreover,

$\displaystyle\|(I-T)y\|=\|y-x^{*}yx\|\geq\|y-x\|-\|x-x^{*}y\cdot x\|$
$\displaystyle>$ $\displaystyle 2-\frac{\epsilon}{2}-|1-x^{*}y|$
$\displaystyle>$ $\displaystyle
2-\frac{\epsilon}{2}-\sqrt{\epsilon}>2-\epsilon.$

Since $\epsilon>0$ is arbitrary, we obtain $\|I-T\|\geq 2$. Then (iv) holds
immediately from the fact that $\|I-T\|\leq 1+\|T\|=2$. ∎

######  Corollary 2.5.

Let $X$ be a complex Banach space. Then the following statements are
equivalent:

  1. (i)

The space $X$ has the Daugavet property.

  2. (ii)

Every point on the unit sphere $S_{X}$ is a Daugavet point.

Now, we make a similar observation on rank-one, norm-one projections.

######  Proposition 2.6.

Let $X$ be a complex Banach space. The following statements are equivalent:

  1. (i)

Every rank-one projection $P=x^{*}\otimes x$ of norm-one on $X$ satisfies
$\|I-P\|=2$.

  2. (ii)

Every rank-one real-projection $P=\text{Re}\,x^{*}\otimes x$ of norm-one on
$X$ satisfies $\|I-P\|=2$.

###### Proof.

(i) $\implies$ (ii): Let $P=\text{Re}\,x^{*}\otimes x$, where $x^{*}\in
S_{X^{*}}$ and $x\in S_{X}$ satisfying $\text{Re}\,x^{*}x=x^{*}x=1$. Then for
every $\epsilon>0$, there exists $y\in S_{X}$ such that $\|y-x^{*}y\cdot
x\|\geq 2-\epsilon$ in view of (i). Now take $\gamma\in\mathbb{T}$ such that
$|x^{*}y|=\gamma x^{*}y$. This implies that $x^{*}(\gamma
y)=\text{Re}\,x^{*}(\gamma y)$. Hence we see that

$\|I-P\|\geq\|\gamma y-\text{Re}\,x^{*}(\gamma y)\cdot
x\|=\|\gamma(y-x^{*}y\cdot x)\|=\|y-x^{*}y\cdot x\|\geq 2-\epsilon.$

Since $\epsilon>0$ is arbitrary, we obtain $\|I-P\|=2$.

(ii) $\implies$ (i): Let $P=x^{*}\otimes x$ , where $x^{*}\in S_{X^{*}}$ and
$x\in S_{X}$ satisfying $x^{*}x=1$. From (ii), we see that for every
$\epsilon>0$, there exists $y\in S_{X}$ such that $\|y-\text{Re}\,x^{*}y\cdot
x\|\geq 2-\frac{\epsilon}{2}$. Then we have $|\text{Re}\,x^{*}y|\geq
1-\frac{\epsilon}{2}$. Notice that

$(\text{Im}\,x^{*}y)^{2}=|x^{*}y|^{2}-(\text{Re}\,x^{*}y)^{2}<1-\left(1-\frac{\epsilon}{2}\right)^{2}<\epsilon.$

Hence, for $y\in S_{X}$, we obtain

$\|I-P\|\geq\|y-x^{*}y\cdot x\|\geq\|y-\text{Re}\,x^{*}y\cdot
x\|-|\text{Im}\,x^{*}y|\geq 2-2\epsilon.$

Therefore, since $\epsilon>0$ is arbitrary, we obtain $\|I-P\|=2$. ∎

##  3\. Alternative convexity or smoothness, nonsquareness, and the Daugavet
property

In this section, we study the relationship between alternative convexity or
smoothness, nonsquareness, and the Daugavet property. First, we recall various
nonsquareness properties, in the sense of James [16, 36]. The uniform
nonsquareness has been examined on both real and complex Banach spaces via
Jordan-von Neumann constants [26].

######  Definition 3.1.

  1. (i)

A Banach space $X$ is uniformly nonsquare (UNSQ) if there exists $\delta>0$
such that for every $x,y\in S_{X}$, $\min\\{\|x\pm y\|\\}\leq 2-\delta$.

  2. (ii)

A Banach space $X$ is locally uniformly nonsquare (LUNSQ) if for every $x\in
S_{X}$, there exists $\delta>0$ such that $\min\\{\|x\pm y\|\\}\leq 2-\delta$
for every $y\in S_{X}$.

  3. (iii)

A Banach space $X$ is nonsquare (NSQ) if for every $x,y\in S_{X}$,
$\min\\{\|x\pm y\|\\}<2$.

Here we call each point $x\in S_{X}$ in (ii) a locally uniformly nonsqure
point (or uniformly non-$\ell_{1}^{2}$ point). We have the following
implication for these classes:

$\text{UNSQ}\,\,\,\implies\,\,\,\text{LUNSQ}\,\,\,\implies\,\,\,\text{NSQ}.$

It has been recently shown that UNSQ real Banach spaces do not have
$\Delta$-points [2] at all. We extend this result for the class of LUNSQ
Banach spaces on the complex scalar fields. Let us start with an improvement
of Theorem 2.2. For the proof on the real case, we refer to [17].

######  Lemma 3.2.

Let $X$ be a complex Banach space and $x\in S_{X}$ be a $\Delta$-point. For
every $\epsilon>0$, $\frac{\alpha}{1-\alpha}<\epsilon$, and every slice
$S=S(x^{*},\alpha)$ containing $x$, there exists a slice $S(z^{*},\alpha_{1})$
of $B_{X}$ such that $S(z^{*},\alpha_{1})\subset S(x^{*},\alpha)$ and
$\|x-y\|>2-\epsilon$ for all $y\in S(z^{*},\alpha_{1})$.

###### Proof.

Let $x^{*}\in S_{X^{*}}$ and $S=S(x^{*},\alpha)$ be a slice containing $x$.
First choose $\eta>0$ such that
$\eta<\min\left\\{1-\frac{1-\alpha}{\text{Re}\,x^{*}(x)},\epsilon-\frac{\alpha}{1-\alpha}\right\\}$.
Since $x\in S_{X}$ is a $\Delta$-point, for a projection
$P(y)=\frac{\text{Re}\,x^{*}y}{\text{Re}\,x^{*}x}\cdot x$ we have $\|I-P\|\geq
2$ by Theorem 2.2. Then there exists $y^{*}\in S_{X^{*}}$ such that
$\|y^{*}-P^{*}y^{*}\|\geq 2-\eta$. Now define
$z^{*}=\frac{P^{*}y^{*}-y^{*}}{\|P^{*}y^{*}-y^{*}\|}\in S_{X^{*}}$ and
$\alpha_{1}=1-\frac{2-\eta}{\|P^{*}y^{*}-y^{*}\|}$ where
$P^{*}y^{*}=\frac{y^{*}x}{\text{Re}\,x^{*}x}\cdot\text{Re}\,x^{*}$. For every
$y\in S(z^{*},\alpha_{1})$ notice that

$\frac{\text{Re}\,x^{*}y}{\text{Re}\,x^{*}x}\cdot\text{Re}\,y^{*}x-\text{Re}\,y^{*}y=\text{Re}\,z^{*}y\cdot\|P^{*}y^{*}-y^{*}\|>2-\eta.$

Hence we see that
$\frac{\text{Re}\,x^{*}y}{\text{Re}\,x^{*}x}\cdot\text{Re}\,y^{*}x>1-\eta$.
Since $\text{Re}\,y^{*}x$ cannot be zero, without loss of generality, assume
that $\text{Re}\,y^{*}x>0$. Then we have
$\text{Re}\,x^{*}y>(1-\eta)\cdot\text{Re}\,x^{*}x>1-\alpha$, which shows that
$S(z^{*},\alpha_{1})\subset S(x^{*},\alpha)$. Furthermore, notice that
$\text{Re}\,x^{*}y\leq|x^{*}y|\leq 1$, and so

$\left\|\frac{x}{\text{Re}\,x^{*}x}-y\right\|\geq\frac{\text{Re}\,y^{*}x}{\text{Re}\,x^{*}x}-\text{Re}\,y^{*}y\geq\frac{\text{Re}\,x^{*}y}{\text{Re}\,x^{*}x}\cdot\text{Re}\,y^{*}x-\text{Re}\,y^{*}y>2-\eta.$

Therefore, we obtain

$\|x-y\|\geq\left\|\frac{x}{\text{Re}\,x^{*}x}-y\right\|-\left(\frac{1}{\text{Re}\,x^{*}x}-1\right)>(2-\eta)-\left(\frac{\alpha}{1-\alpha}\right)>2-\epsilon.$

∎

Applying Lemma 3.2 again, we can also show the converse, the same proof in
[17, Lemma 2.2] transfers to complex Banach spaces.

######  Corollary 3.3.

Let $X$ be a complex Banach space. Then $x\in S_{X}$ is a $\Delta$-point if
and only if for every $\epsilon>0$ and every slice $S=S(x^{*},\alpha)$
containing $x\in S$, there exists a slice $S(z^{*},\alpha_{1})$ such that
$S(z^{*},\alpha_{1})\subset S(x^{*},\alpha)$ and $\|x-y\|>2-\epsilon$ for all
$y\in S(z^{*},\alpha_{1})$.

As a consequence, we obtain the relationship between the locally uniformly
nonsquare points and the $\Delta$-points on both real and complex Banach
spaces.

######  Proposition 3.4.

Let $X$ be a complex Banach space. A locally uniformly nonsquare point $x\in
S_{X}$ is not a $\Delta$-point of $X$.

###### Proof.

We show that a $\Delta$-point $x\in S_{X}$ cannot be a locally uniformly
nonsquare point. Let $\epsilon>0$ and $\eta\in(0,\frac{\epsilon}{2})$. By
Lemma 3.2, for every $\alpha>0$ where $\frac{\alpha}{1-\alpha}<\eta$ and every
slice $S=S(x^{*},\alpha)$ containing $x$, there exists a slice
$S(z^{*},\alpha_{1})\subset S$ such that $\|x-y\|>2-\eta>2-\epsilon$ for all
$y\in S(z^{*},\alpha_{1})$. In particular, we have
$\frac{\text{Re}\,z^{*}y}{\|y\|}>\frac{1-\alpha_{1}}{\|y\|}\geq 1-\alpha_{1}$
for every $y\in S(z^{*},\alpha_{1})$. Hence $y^{\prime}=\frac{y}{\|y\|}\in
S(z^{*},\alpha_{1})$ and $\|x-y^{\prime}\|>2-\epsilon$ for $y\in
S(z^{*},\alpha_{1})$.

Moreover, by the fact that $\alpha<\frac{\alpha}{1-\alpha}<\eta$ and $x,y\in
S$, we have
$\|x+y^{\prime}\|\geq\text{Re}\,x^{*}x+\text{Re}\,x^{*}y^{\prime}>2-2\alpha>2-2\eta>2-\epsilon$.
Thus, for every $\epsilon>0$ there exists $y^{\prime}\in S_{X}$ such that
$\min\\{\|x+y^{\prime}\|,\|x-y^{\prime}\|\\}>2-\epsilon$. This shows that
$x\in S_{X}$ is not a locally uniformly nonsquare point. ∎

######  Corollary 3.5.

Let $X$ be a complex Banach space. If $X$ is LUNSQ, then the space does not
admit $\Delta$-points. As a consequence, every LUNSQ space does not satisfy
the Daugavet property, DD2P, and DLD2P.

A Banach space $X$ is said to have the anti-Daugavet property for a class of
operators $\mathcal{M}$ if the following equivalence holds:

$\|I+T\|=1+\|T\|\iff\|T\|\in\sigma(T),$

where $\sigma(T)$ is the spectrum of $T\in\mathcal{M}$. If $\mathcal{M}=L(X)$,
we simply say the space $X$ satisfies the anti-Daugavet property. We mention
that the only if part always hold for any bounded linear operators.

It is well-known that any uniformly rotund or uniformly smooth Banach spaces
have the anti-Daugavet property. Moreover, this property is connected to the
alternative convexity or smoothness properties that are introduced in [22]:

######  Definition 3.6.

  1. (i)

A Banach space $X$ is uniformly alternatively convex or smooth (uacs) if for
all sequences $(x_{n}),(y_{n})\subset S_{X}$ and $(x_{n}^{*})\subset
S_{X^{*}}$, $\|x_{n}+y_{n}\|\rightarrow 2$ and $x_{n}^{*}(x_{n})\rightarrow 1$
implies $x_{n}^{*}(y_{n})\rightarrow 1$.

  2. (ii)

A Banach space $X$ is strongly locally uniformly alternatively convex or
smooth (sluacs) if for every $x\in S_{X}$, $(x_{n})\subset S_{X}$ and
$(x^{*}_{n})\subset S_{X^{*}}$, $\|x_{n}+x\|\rightarrow 2$ and
$x_{n}^{*}(x_{n})\rightarrow 1$ implies $x_{n}^{*}(x)\rightarrow 1$.

  3. (iii)

A Banach space $X$ is alternatively convex or smooth (acs) if for all $x,y\in
S_{X}$ and $x^{*}\subset S_{X^{*}}$, $\|x+y\|=2$ and $x^{*}(x)=1$ implies
$x^{*}(y)=1$.

Any uniformly convex (resp. locally uniformly rotund, rotund) and uniformly
smooth (resp. uniformly Gateaux-smooth, smooth) Banach spaces are known to be
uacs (resp. sluacs, acs) [14].

Even though it is mentioned in [22] that alternative convexity or smoothness
for complex Banach spaces can be defined in a similar fashion, any recent
investigation on this property also assumes the scalar field to be
$\mathbb{R}$. Hence, we provide equivalent definitions that only involve with
the real part of bounded linear functionals which enable us to consider
complex Banach spaces.

######  Proposition 3.7.

  1. (i)

A Banach space $X$ is uacs if and only if for all sequence
$(x_{n}),(y_{n})\subset S_{X}$ and $(x_{n}^{*})\subset S_{X^{*}}$,
$\|x_{n}+y_{n}\|\rightarrow 2$ and $\text{Re}\,x_{n}^{*}x_{n}\rightarrow 1$
implies $\text{Re}\,x_{n}^{*}y_{n}\rightarrow 1$.

  2. (ii)

A Banach space $X$ is sluacs if and only if for every $x\in S_{X}$,
$(x_{n})\subset S_{X}$ and $(x_{n}^{*})\subset S_{X^{*}}$,
$\|x_{n}+y_{n}\|\rightarrow 2$ and $\text{Re}\,x_{n}^{*}x_{n}\rightarrow 1$
implies $\text{Re}\,x_{n}^{*}x\rightarrow 1$.

  3. (iii)

A Banach space $X$ is acs if and only if for all $x,y\in S_{X}$ and $x^{*}\in
S_{X^{*}}$, $\|x+y\|=2$ and $\text{Re}\,x^{*}(x)=1$ implies
$\text{Re}\,x^{*}(y)=1$.

###### Proof.

We assume that $X$ is a complex Banach space. Since the proofs for (i) and
(ii) are similar, we only prove (i).

Suppose that for every $(x_{n}),(y_{n})\subset S_{X}$ and $(x_{n}^{*})\subset
S_{X^{*}}$, $x_{n}^{*}x_{n}\rightarrow 1$ and $\|x_{n}+y_{n}\|\rightarrow 2$.
Then for every $\epsilon>0$, there exists $N_{1}\in\mathbb{N}$ such that for
every $n\geq N_{1}$, $|x_{n}^{*}x_{n}|>1-\frac{\epsilon}{2}$. We see that

$1\geq(\text{Re}\,x_{n}^{*}x_{n})^{2}+(\text{Im}\,x_{n}^{*}x_{n})^{2}=|x_{n}^{*}x_{n}|^{2}>\left(1-\frac{\epsilon}{2}\right)^{2}+(\text{Im}\,x_{n}^{*}x_{n})^{2}.$

Hence $\text{Im}\,x_{n}^{*}x_{n}<\sqrt{\epsilon}$, which in turn implies that
$\text{Re}\,x_{n}^{*}x_{n}\rightarrow 1$. Then by the assumption, we obtain
$\text{Re}\,x_{n}^{*}y_{n}\rightarrow 1$. Again, for every $\epsilon>0$, there
exists $N_{2}\in\mathbb{N}$ such that
$\text{Re}\,x_{n}^{*}y_{n}>1-\frac{\epsilon}{2}$. This implies that

$1\geq|x_{n}^{*}y_{n}|^{2}=(\text{Re}\,x_{n}^{*}y_{n})^{2}+(\text{Im}\,x_{n}^{*}y_{n})^{2}\geq(1-\epsilon)^{2}+(\text{Im}\,x_{n}^{*}y_{n})^{2},$

and so $\text{Im}\,x_{n}^{*}y_{n}\rightarrow 0$ as $n\rightarrow\infty$.
Therefore, we see that
$x_{n}^{*}y_{n}=\text{Re}\,x_{n}^{*}y_{n}+i\text{Im}\,x_{n}^{*}y_{n}\rightarrow
1$.

For (iii), if for all $x,y\in S_{X}$ and $x^{*}\in S_{X^{*}}$ we have
$\|x+y\|=2$ and $x^{*}x=\text{Re}\,x^{*}x=1$, then $\text{Re}\,x^{*}y=1$ by
the assumption. Hence, we see that

$1\geq|x^{*}y|=(\text{Re}\,x^{*}y)^{2}+(\text{Im}\,x^{*}y)^{2}=1+(\text{Im}\,x^{*}y)^{2}.$

Therefore, $x^{*}y=\text{Re}\,x^{*}y=1$. ∎

Even though every uacs Banach spaces are UNSQ [14, 22], there have been no
explicit description on the relationship between the sluacs and LUNSQ (resp.
acs and NSQ) Banach spaces. As a matter of fact, the similar statement also
holds for both sluacs and acs Banach spaces.

######  Proposition 3.8.

  1. (i)

Every sluacs space is LUNSQ.

  2. (ii)

Every acs space is NSQ.

###### Proof.

(i) Suppose that a sluacs space $X$ is not LUNSQ. Then there exists $x\in
S_{X}$ such that for every $\delta>0$, there exists $y\in S_{X}$ such that
$\|x+y\|>2-\delta$ and $\|x-y\|>2-\delta$. So choose a sequence
$(x_{n})_{n=1}^{\infty}$ such that $\|x+x_{n}\|>2-\frac{1}{2^{n}}$ and
$\|x-x_{n}\|>2-\frac{1}{2^{n}}$. In view of Hahn-Banach theorem, we can also
find a sequence $(x_{n}^{*})_{n=1}^{\infty}$ such that

$\text{Re}\,x_{n}^{*}x_{n}-\text{Re}\,x_{n}^{*}x=\|x-x_{n}\|\,\,\,\text{and}\,\,\,\text{Re}\,x_{n}^{*}x_{n}+\text{Re}\,x_{n}^{*}x=\|x+x_{n}\|.$

Then we see that

$2-\frac{1}{2^{n}}<\text{Re}\,x_{n}^{*}x_{n}+\text{Re}\,x_{n}^{*}x\leq\|x\|+\text{Re}\,x_{n}^{*}x_{n}=1+\text{Re}\,x_{n}^{*}x_{n},$

and so $\text{Re}\,x_{n}^{*}x_{n}\rightarrow 1$ as $n\rightarrow\infty$. Since
the space $X$ is assumed to be sluacs, $x_{n}^{*}x\rightarrow 1$ as
$n\rightarrow\infty$. However, by repeating the same argument to
$\|x-x_{n}\|$, we also obtain that $-\text{Re}\,x_{n}^{*}x\rightarrow 1$. This
leads to a contradiction.

(ii) Suppose that an acs space $X$ is not nonsquare. Then there exist $x,y\in
S_{X}$ such that $\|x+y\|=\|x-y\|=2$. Let $x^{*}\in S_{X^{*}}$ such that
$\text{Re}\,x^{*}(x)+\text{Re}\,x^{*}(y)=\|x+y\|$. From the fact that

$2=\text{Re}\,x^{*}(x)+\text{Re}\,x^{*}(y)\leq\text{Re}\,x^{*}(x)+\|y\|=\text{Re}\,x^{*}(x)+1,$

we have $\text{Re}\,x^{*}x=1$. This also shows that $\text{Re}\,x^{*}y=1$.
However, since $\|x-y\|=2$ and the space $X$ is acs, we have
$-\text{Re}\,x^{*}x=1$, which is a contradiction. Therefore, the space $X$
must be nonsquare. ∎

We mention that any locally uniformly rotund (LUR) Banach spaces does not have
$\Delta$-points. As a matter of fact, we can show further that every sluacs
Banach space does not have $\Delta$-points based on our observation.

######  Corollary 3.9.

Let $X$ be a Banach space. Every sluacs Banach space does not contain
$\Delta$-points.

###### Proof.

Every sluacs Banach space is LUNSQ by Proposition 3.8.(i). Then by Proposition
3.4, we see that the space does not contain $\Delta$-points. ∎

There has been a long standing open problem whether a Banach space with the
Daugavet property can be rotund. While there is a rotund normed space (not
complete) with the Daugavet property [20], the existence has not been verified
for Banach spaces. Since every rotund Banach space is nonsquare, it will be
interesting to know the answer to the following question that may help us to
prove or disprove the open problem.

######  Problem 3.10.

Does a NSQ Banach space $X$ contain $\Delta$-points?

##  4\. Remarks on the Daugavet property of $A(K,X)$

Let $K$ be a compact Hausdorff space. The space $C(K)$ is the set of all
complex-valued continuous functions over $K$ endowed with the supremum norm
$\|\cdot\|_{\infty}$. A uniform algebra $A$ is a closed subalgebra of $C(K)$
that separates points and contains constant functions. For a compact subset
$K\subset\mathbb{C}$, the space $P(K)$ (resp. $R(K)$) of continuous functions
that can be approximated uniformly on $K$ by polynomials in $z$ (resp. by
rational functions with poles off $K$) and the space $A(K)$ of continuous
functions that are analytic on the interior of $K$ are well-known examples of
uniform algebras. When $K=\overline{\mathbb{D}}$, the corresponding uniform
algebra $A(K)=A(\overline{\mathbb{D}})$ is the disk algebra. We refer to [9,
28] for more details on uniform algebras.

For a complex Banach space $X$, let $C(K,X)$ be the set of all vector-valued
continuous functions over $K$ equipped with the supremum norm. We recall the
definition of the vector-valued function space $A(K,X)$.

######  Definition 4.1.

Let $K$ be a compact Hausdorff space and $X$ be a Banach space. The space
$A(K,X)$ is called a function space over the base algebra $A$ if it is a
subspace of $C(K,X)$ that satisfies:

  1. (i)

The base algebra $A:=\\{x^{*}\circ f:x^{*}\in X^{*},f\in A(K,X)\\}$ is a
uniform algebra.

  2. (ii)

$A\otimes X\subset A(K,X)$.

  3. (iii)

For every $g\in A$ and every $f\in A(K,X)$, we have $g\cdot f\in A(K,X)$.

If $X=\mathbb{F}$, then the space $A(K,X)$ becomes the uniform algebra $A$ on
a compact Hausdorff space $K$. It is clear that $C(K,X)$ is a function space
over a base algebra $C(K)$. As a nontrivial example, for given Banach spaces
$X$ and $Y$, let $A_{w^{*}}(B_{X^{*}},Y)$ be the space of all weak∗-to-norm
continuous functions on the closed unit ball $B_{X^{*}}$ that are holomorphic
on the interior of $B_{X^{*}}$. It is a closed subspace of $C(B_{X^{*}};Y)$,
where $B_{X^{*}}$ is the weak∗-compact set. Then $A_{w^{*}u}(B_{X^{*}};Y)$ is
a function space over base algebra $A_{w^{*}u}(B_{X^{*}})$.

A subset $L\subset K$ is said to be a boundary for $A$ if for every $f\in A$
there exists $t\in L$ such that $f(t)=\|f\|_{\infty}$. The smallest closed
boundary for $A$ is called the Shilov boundary denoted by $\Gamma$. A point
$x\in K$ is a strong boundary point for a uniform algebra $A$ if for every
open subset $U\subset K$ containing $x$, there exists $f\in A$ such that
$\|f\|_{\infty}=|f(t_{0})|=1$ and $\sup_{t\in K\setminus U}|f(t)|<1$. For a
compact Hausdorff space $K$, the set of all strong boundary points on $A$ is
coincides with the Choquet boundary $\Gamma_{0}$, that is, the set of all
extreme points on the set $K_{A}=\\{\lambda\in
A^{*}:\|\lambda\|=\lambda(1_{A})=1\\}$ [9, Theorem 4.3.5]. Moreover, the
closure of $\Gamma_{0}$ is $\Gamma$ in this case [9, Corollary 4.3.7.a]. For
instance, the Shilov boundary of the disk algebra $A(\overline{\mathbb{D}})$
is the unit circle $\partial\overline{\mathbb{D}}$.

To study various geometrical properties of $A(K,X)$ and Bishop-Phelps-Bollobás
property for Asplund operators which range space is a uniform algebra, a
Urysohn-type lemma has played an important role. Here we use a stronger
version of the lemma provided in [5].

######  Lemma 4.2.

[6, Lemma 3.10] Let $K$ be a compact Hausdorff space. If $t_{0}$ is a strong
boundary point for a uniform algebra $A\subset C(K)$, then for every open
subset $U\subset K$ containing $t_{0}$ and $\epsilon>0$, there exists
$\phi=\phi_{U}\in A$ such that $\phi(t_{0})=\|\phi\|_{\infty}=1$,
$\sup_{K\setminus U}|\phi(t)|<\epsilon$ and

$|\phi(t)|+(1-\epsilon)|1-\phi(t)|\leq 1$

for every $t\in K$.

We can also construct a Urysohn-type function at an isolated point in the
Shilov boundary.

######  Lemma 4.3.

[27, Lemma 2.5] Let $A$ be a uniform algebra on a compact Hausdorff space $K$
and let $t_{0}$ be an isolated point of the Shilov boundary $\Gamma$ of $A$.
Then there exists a function $\phi\in A$ such that $\phi(t_{0})=\|\phi\|=1$
and $\phi(t)=0$ for $t\in\Gamma\setminus\\{t_{0}\\}$.

The next statement is in the proof of the case (iii) for [27, Theorem 4.2],
but we state it explicitly here.

######  Lemma 4.4.

Let $K$ be a compact Hausdorff space and $\Gamma$ be the Shilov boundary of
the base algebra for the space $A(K,X)$. Suppose that $\Gamma$ has an isolated
point. Then, $A(K,X)$ is isometrically isomorphic to $X\oplus_{\infty}Y$ where
$Y$ is $A(K,X)$ restricted to $K\setminus\\{t_{0}\\}$.

###### Proof.

Let $t_{0}\in\Gamma$ be an isolated point. By Lemma 4.3, there exists $\phi\in
A$ such that $\phi(t_{0})=\|\phi\|_{\infty}=1$ and $\phi(t)=0$ for
$t\in\Gamma\setminus\\{t_{0}\\}$. Let $\tilde{K}=K\setminus\\{t_{0}\\}$.
Define a norm-one projection $P:A(K,X)\rightarrow A(K,X)$ by $Pf=\phi\cdot f$
and denote $Y$ be the restriction of $A(K,X)$ to $\tilde{K}$. As a matter of
fact, the image $P(A(K,X))$ is isometrically isomorphic to $X$. Indeed, define
a linear operator $\Psi:P(A(K,X))\rightarrow X$ by $\Psi(Pf)=f(t_{0})$. Then
$\|\Psi(Pf)\|_{X}=\|f(t_{0})\|_{X}=\|Pf\|$. Moreover, we see that for every
$x\in X$ there exists $f\in A(K,X)$ such that $f(t_{0})=x$. Hence, $\Psi$ is
surjective, which in turn implies that the operator $\Psi$ is an isometric
isomorphism on $P(A(K,X))$.

Now, we claim that the space $A(K,X)$ is isometrically isomorphic to
$X\oplus_{\infty}Y$. For $f\in A(K,X)$, define a bounded linear operator
$\Phi:A(K,X)\rightarrow X\oplus_{\infty}Y$ by $\Phi f=(Pf,f_{|\tilde{K}})$.
Then we see that

$\|\Phi
f\|=\max\\{\|Pf\|,\|f_{|\tilde{K}}\|\\}=\max\left\\{\|f(t_{0})\|_{X},\sup_{t\in\tilde{K}}\|f(t)\|_{X}\right\\}=\|f\|.$

Notice that for a given $(f,g)\in X\oplus_{\infty}Y$, there exist
$f_{1},f_{2}\in A(K,X)$ such that $f=Pf_{1}$ and $g=f_{2|\tilde{K}}$. Let
$h=Pf_{1}+f_{2}-Pf_{2}\in A(K,X)$. Then we have
$\Phi(h)=(Pf_{1},f_{2|\tilde{K}})$. Hence, the operator $\Phi$ is also
surjective, and so it is an isometric isomorphism between $A(K,X)$ and
$X\oplus_{\infty}Y$. ∎

The following lemma will be useful for later.

######  Lemma 4.5.

Let $X$ be a Banach space. Suppose that $L$ is a closed boundary for $A$. The
space of restrictions of elements of $A(K,X)$ to $L$ is denoted by $A(L,X)$
and the restrictions of elements of $A$ to $L$ is denoted by $A(L)$. Then
$A(L,X)$ is isometrically isomorphic to $A(K,X)$.

###  4.1. The polynomial Daugavet property in $A(K,X)$

First we provide a sufficient condition for $A(K,X)$ with the polynomial
Daugavet property. We mention that the proof method is inspired by [8, Theorem
2.7].

######  Theorem 4.6.

Let $K$ be a compact Hausdorff space and let $\Gamma$ be the Shilov boundary
of the base algebra $A$ of $A(K,X)$. If $\Gamma$ does not have isolated
points, then $A(K,X)$ has the polynomial Daugavet property.

###### Proof.

In view of [7, Corollary 2.2], it suffices to show that for every
$p\in\mathcal{P}(X)$ with $\|p\|=1$, every $x_{0}\in S_{X}$, and every
$\epsilon>0$, there exist $\alpha\in S_{\mathbb{C}}$ and $y\in B_{X}$ such
that

$\text{Re}\,\alpha p(y)>1-\epsilon\,\,\,\text{and}\,\,\,\|x_{0}+\alpha
y\|>2-\epsilon.$

Let $0<\epsilon<1$, $P\in\mathcal{P}(X)$ with $\|P\|=1$, and let $f_{0}\in
S_{A(K,X)}$. Choose $h\in S_{A(K,X)}$ and $\alpha\in\mathbb{T}$ such that
$|P(h)|>1-\frac{\epsilon}{2}$ and $\text{Re}\,\alpha
P(h)>1-\frac{\epsilon}{2}$. Also choose $t_{0}\in\Gamma_{0}$ such that
$\|f_{0}(t_{0})\|_{X}>1-\frac{\epsilon}{8}$. Let $U=\\{t\in
K:\|f_{0}(t)-f_{0}(t_{0})\|_{X}<\frac{\epsilon}{8}\,\,\,\text{and}\,\,\,\|h(t)-h(t_{0})\|_{X}<\frac{\epsilon}{8}\\}$
be a nonempty open subset of $K$. We consider two cases.

Case 1: Suppose that there exists $(t_{i})_{i=1}^{\infty}\subset U$ such that
$\|\alpha^{-1}f_{0}(t_{i})-h(t_{i})\|\rightarrow 0$. Then we have

$\displaystyle\|f_{0}+\alpha h\|$ $\displaystyle\geq$
$\displaystyle\sup_{i}\|f_{0}(t_{i})+\alpha h(t_{i})\|$ $\displaystyle\geq$
$\displaystyle\sup_{i}(2\|f_{0}(t_{0})\|_{X}-2\|f_{0}(t_{0})-f_{0}(t_{i})\|_{X}-\|f_{0}(t_{i})-\alpha
h(t_{i})\|_{X}$ $\displaystyle\geq$ $\displaystyle
2-\frac{\epsilon}{4}-\frac{\epsilon}{4}-\frac{\epsilon}{4}>2-\epsilon.$

Case 2: Now suppose that there exists $\eta>0$ such that
$\|\alpha^{-1}f_{0}(t)-h(t)\|>\eta$ for every $t\in U$. Since $\Gamma$ is
perfect, we see that the strong boundary point $t_{0}\in U$ is not an isolated
point. Let $\\{U_{i}\\}_{i=1}^{\infty}$ be a collection of pairwise disjoint
open subsets of $U$ such that $\cup_{i=1}^{\infty}U_{i}\subset U$. From the
fact that the Choquet boundary $\Gamma_{0}$ is dense in $\Gamma$, there exist
strong boundary points $t_{i}\in U_{i}$ for each $i\in\mathbb{N}$. Then by
Lemma 4.2, there exists $\phi_{i}\in A$ such that

(2) $\phi_{i}(t_{i})=1,\,\,\,,\sup_{K\setminus
U_{i}}|\phi_{i}(t)|<\frac{\epsilon}{2^{i+3}},\,\,\,\text{and}\,\,\,|\phi_{i}(t)|+\left(1-\frac{\epsilon}{2^{i+3}}\right)|1-\phi_{i}(t)|\leq
1\,\,\,\text{for every}\,\,\,t\in K.$

Let $h_{i}=h+\phi_{i}(\alpha^{-1}f_{0}(t_{i})-h(t_{0}))\in A(K,X)$. Then for
every $t\in\cup_{i=1}^{\infty}U_{i}$, by (2), we have

$\displaystyle\|h_{i}(t)\|_{X}$ $\displaystyle=$
$\displaystyle\|h(t)+\phi_{i}(t)\alpha^{-1}f_{0}(t_{i})-\phi_{i}(t)h(t_{i})\|_{X}$
$\displaystyle\leq$
$\displaystyle\|h(t)-h(t_{i})\|_{X}+\|h(t_{i})-\phi_{i}(t)h(t_{i})\|_{X}+\|\phi_{i}(t)\alpha^{-1}f_{0}(t_{i})\|_{X}$
$\displaystyle\leq$
$\displaystyle\|h(t)-h(t_{0})\|_{X}+\|h(t_{0})-h(t_{i})\|_{X}+|1-\phi_{i}(t)|+|\phi_{i}(t)|$
$\displaystyle\leq$
$\displaystyle\frac{\epsilon}{4}+\left(1-\frac{\epsilon}{2^{i+3}}\right)|1-\phi(t)|+\frac{\epsilon}{2^{i+3}}|1-\phi_{i}(t)|+|\phi_{i}(t)|$
$\displaystyle\leq$
$\displaystyle\frac{\epsilon}{4}+1+\frac{\epsilon}{2^{i+2}}<1+\frac{\epsilon}{2}.$

On the other hand, for every $t\in K\setminus\cup_{i=1}^{\infty}U_{i}$,

(3)
$\|h_{i}(t)\|_{X}\leq\|h(t)\|_{X}+|\phi_{i}(t)|\|\alpha^{-1}f_{0}(t_{i})-h(t_{i})\|_{X}\leq
1+\frac{\epsilon}{2^{i+3}}\cdot 2<1+\frac{\epsilon}{2}.$

Moreover, we see that

(4)
$\|h_{i}\|\geq\|h_{i}(t_{i})\|_{X}=\|f_{0}(t_{i})\|_{X}\geq\|f_{0}(t_{0})\|-\|f_{0}(t_{0})-f_{0}(t_{i})\|\geq
1-\frac{\epsilon}{2}.$

Now, let $g_{i}=\frac{h_{i}}{\|h_{i}\|}$. By (4) we obtain

$\|h_{i}-g_{i}\|=|1-\|h_{i}\||<\frac{\epsilon}{2}.$

For every $(\beta_{i})\in\ell_{\infty}$, notice that

$\displaystyle\sup_{n}\left\|\sum_{i=1}^{n}\beta_{i}\phi_{i}(\alpha^{-1}f_{0}(t_{i})-h(t_{i}))\right\|$
$\displaystyle\leq$ $\displaystyle\sup_{n}\sup_{t\in
K}\sum_{i=1}^{n}|\beta_{i}||\phi_{i}(t)|\|\alpha^{-1}f_{0}(t_{i})-h(t_{i})\|_{X}$
$\displaystyle\leq$ $\displaystyle\sup_{n}\sup_{t\in
K}\sum_{i=1}^{n}2|\beta_{i}||\phi_{i}(t)|$ $\displaystyle\leq$ $\displaystyle
2\sup_{i}|\beta_{i}|\left(1+\frac{\epsilon}{2^{4}}+\frac{\epsilon}{2^{5}}\cdots\right)=2\left(1+\frac{\epsilon}{8}\right)\sup_{i}|\beta_{i}|$

Hence by [10, Theorem V.6], the series
$\sum_{i=1}^{n}\beta_{i}\phi_{i}(\alpha^{-1}f_{0}(t_{i})-h(t_{i}))$ is weakly
unconditionally Cauchy. Since we assumed that
$\|\alpha^{-1}f_{0}(t)-h(t)\|>\eta$, there exists a basic subsequence
$(\phi_{\sigma(i)}(\alpha^{-1}f_{0}(t_{\sigma(i)})-h(t_{\sigma(i)}))$ that is
equivalent to the basis $(e_{i})$ in $c_{0}$ by using Bessaga-Pełczynski
principle [10, pg. 45]. From the fact that a polynomial on a bounded subset of
$c_{0}$ is weakly continuous [12, Proposition 1.59] , we have
$\text{Re}\,\alpha P(h_{\sigma(i)})\rightarrow\text{Re}\,\alpha P(h)$ as
$i\rightarrow\infty$.

Choose $k\in\mathbb{N}$ such that $\text{Re}\,\alpha
P(h_{k})>1-\frac{\epsilon}{2}$. Then we have

$\text{Re}\,\alpha P(g_{k})=\frac{\text{Re}\,\alpha
P(h_{k})}{\|h_{k}\|}>\frac{1-\epsilon/2}{1+\epsilon/2}\geq 1-\epsilon.$

Therefore, we finally obtain

$\displaystyle\|f_{0}+\alpha g_{k}\|\geq\|f_{0}+\alpha
h_{k}\|-\|g_{k}-h_{k}\|$ $\displaystyle\geq$
$\displaystyle\|f_{0}(t_{i})+\alpha h_{k}(t_{i})\|-\frac{\epsilon}{2}$
$\displaystyle=$ $\displaystyle 2\|f_{0}(t_{i})\|-\frac{\epsilon}{2}$
$\displaystyle\geq$ $\displaystyle
2\|f_{0}(t_{0})\|-2\|f_{0}(t_{i})-f_{0}(t_{0})\|-\frac{\epsilon}{2}$
$\displaystyle\geq$ $\displaystyle 2\|f_{0}(t_{0})\|-\frac{3\epsilon}{4}$
$\displaystyle\geq$ $\displaystyle 2-\epsilon.$

∎

Let $\mathcal{P}_{K}(X,X)$ be the set of all compact polynomials from $X$ to
itself. For $P\in\mathcal{P}_{K}(X,X)$, the numerical range $V(P)$ is defined
by

$V(T)=\\{x^{*}(Tx):x^{*}\in S_{X^{*}}\,\,\,\text{and}\,\,\,x\in
S_{X}\,\,\,\text{where}\,\,\,x^{*}(x)=1\\}.$

Now, we recall the polynomial Daugavetian index.

######  Definition 4.7.

[35] For a Banach space $X$, the Daugavetian index $\text{Daug}\,(X)$ is
defined by

$\displaystyle\text{Daug}_{p}\,(X)$ $\displaystyle=$
$\displaystyle\max\\{m\geq 0:\|I+P\|\geq 1+m\|P\|,\,\,\,\text{for
every}\,\,\,P\in\mathcal{P}_{K}(X,X)\\}$ $\displaystyle=$
$\displaystyle\inf\\{\omega(P):P\in\mathcal{P}_{K}(X,X),\|P\|=1\\},$

where $\omega(P)=\sup\text{Re}\,V(T)$.

It is well-known that $\text{Daug}_{p}\,(X)\in[0,1]$ and
$\text{Daug}_{p}(X)\leq\text{Daug}(X)$, where $\text{Daug}(X)$ is the
Daugavetian index introduced in [29]. A Banach space $X$ has the polynomial
Daugavet property if and only if $\text{Daug}_{p}(X)=1$. This comes from the
fact that a Banach space $X$ satisfies the Daugavet equation for every rank-
one polynomials if and only if $X$ satisfies the same equation for every
weakly compact polynomials (see Theorem 1.5). We recall the following lemma
that will be useful for later.

######  Lemma 4.8.

[35, Proposition 2.2, 2.3] Let $\\{X_{\lambda}\\}_{\lambda\in\Lambda}$ be a
family of infinite-dimensional Banach spaces and let $Z$ be the $c_{0}$\- or
$\ell_{\infty}$-sum of the family. Then

$\text{Daug}_{p}\,(Z)=\inf\\{\text{Daug}_{p}\,(X_{\lambda}):\lambda\in\Lambda\\}.$

If there exists a finite-rank projection on $X$ such that $\|P\|=\|I-P\|=1$,
then $\text{Daug}\,(X)=0$ [29]. Hence $\text{Daug}_{p}\,(X)=0$ in this case.
Examples of such spaces are $C(K)$ where $K$ has isolated points and Banach
spaces $X$ with an 1-unconditional basis [29, pp. 635]. Similar to the space
$C(K)$, we can also construct such a projection for uniform algebras.

######  Proposition 4.9.

Let $A$ be a uniform algebra on a compact Hausdorff space $K$ and let $t_{0}$
be an isolated point of the Shilov boundary $\Gamma$ of $A$. Then, there
exists $P:A\rightarrow A$ defined by $P=\delta_{t_{0}}\otimes\phi$, where
$\delta_{t_{0}}\in S_{A^{*}}$ and $\phi\in S_{A}$, such that $P$ is a
projection and $\|P\|=\|I-P\|=1$.

###### Proof.

Let $t_{0}\in K$ be an isolated point of $\Gamma$. Then, by Lemma 4.3, there
exists a function $\phi\in A$ such that $\phi(t_{0})=\|\phi\|=1$ and
$\phi(t)=0$ for $t\in\Gamma\setminus\\{t_{0}\\}$. Now, define
$P=\delta_{t_{0}}\otimes\phi$ where $\delta_{t_{0}}$ is the pointwise
evaluation at $t_{0}$. Since
$P^{2}f=f(t_{0})\phi(t_{0})\cdot\phi=f(t_{0})\phi=Pf$, the rank-one operator
$P$ is a projection on $A$.

Let $f\in S_{A}$. Then we have $|Pf(t)|=|f(t_{0})\phi(t)|=0$ for $t\in
K\setminus\\{t_{0}\\}$ and $|Pf(t_{0})|=|f(t_{0})|$. Hence,
$\|Pf\|_{\infty}=|f(t_{0})|\leq 1$. However, we see that
$\|P\phi\|_{\infty}=1$. So $\|P\|=1$. Similarly,
$|[(I-P)f](t)|=|f(t)-f(t_{0})\phi(t)|=|f(t)|$ for $t\in K\setminus\\{t_{0}\\}$
and $|[(I-P)f](t_{0})|=0$. Let $U$ be an open set in $K$ that contains
$t_{1}\in\Gamma_{0}$. Then there exists $\tilde{\phi}\in A$ such that
$\tilde{\phi}(t_{1})=1$ and $\sup_{K\setminus U}|\tilde{\phi}|<1$. We see that
$\|(I-P)\tilde{\phi}\|_{\infty}=\|\tilde{\phi}\|_{\infty}=1$. Therefore, we
obtain $\|I-P\|=1$. ∎

######  Corollary 4.10.

Let $K$ be a compact Hausdorff space and let $A$ be a uniform algebra on $K$.
If the Shilov boundary of $A$ contains an isolated point, then
$\text{Daug}_{p}(A)=0$.

###### Proof.

This is an immediate consequence of Proposition 4.9. ∎

######  Theorem 4.11.

Let $X$ be a complex Banach space and let $K$ be a compact Hausdorff space.
Then

$\text{Daug}_{p}\,(A(K,X))=\max\\{\text{Daug}_{p}(A),\text{Daug}_{p}(X)\\}.$

###### Proof.

Let $P\in\mathcal{P}_{K}(A(K,X))$. We first show that

$\|I+P\|\geq 1+\text{Daug}_{p}(X)\|P\|.$

For a given $\epsilon>0$ there exists $f_{0}\in S_{A(K,X)}$ and
$t_{0}\in\Gamma_{0}$ such that
$\|P(f_{0})(t_{0})\|_{X}\geq\|P\|-\frac{\epsilon}{2}$. Since $P$ is continuous
at $f_{0}$, there exists $\delta>0$ such that

(5)
$\text{If}\,\,\,\|f_{0}-g\|<\delta,\,\,\,\text{then}\,\,\,\|P(f_{0})-P(g)\|\leq\frac{\epsilon}{2}.$

Now, consider $U=\\{t\in K:\|f_{0}(t)-f_{0}(t_{0})\|_{X}<\frac{\delta}{4}\\}$.
Since the set $U$ is a nonempty open subset of $K$ that contains the strong
boundary point $t_{0}$, by Lemma 4.2, there exists $\phi\in A$ such that

(6) $\phi(t_{0})=1,\,\,\,\sup_{K\setminus
U}|\phi(t)|<\frac{\delta}{8},\,\,\,\text{and}\,\,\,|\phi(t)|+\left(1-\frac{\delta}{8}\right)|1-\phi(t)|\leq
1\,\,\,\text{for every}\,\,\,t\in K.$

Fix $x_{0}\in S_{X}$ such that $f_{0}(t_{0})=\|f_{0}(t_{0})\|\cdot x_{0}$ and
define $\Psi:\mathbb{C}\rightarrow A(K,X)$ by

$\Psi(z)=\left(1-\frac{\delta}{8}\right)(1-\phi)f_{0}+\phi\cdot x_{0}\cdot z.$

Then we have

$\displaystyle\Psi(\|f_{0}(t_{0})\|_{X})(t)-f_{0}(t)$ $\displaystyle=$
$\displaystyle\left(1-\frac{\delta}{8}\right)(1-\phi(t))f_{0}(t)+\phi(t)f_{0}(t_{0})-f_{0}(t)$
$\displaystyle=$
$\displaystyle\phi(t)(f_{0}(t)-f_{0}(t_{0}))-\frac{\delta(1-\phi(t))f_{0}(t)}{8}.$

In view of (6), notice that

$\left\|\phi(t)(f_{0}(t)-f_{0}(t_{0}))-\frac{\delta(1-\phi(t))f_{0}(t)}{8}\right\|_{X}\leq\|\phi\|_{\infty}\cdot\|(f_{0}(t)-f_{0}(t_{0}))\|_{X}+\frac{\delta}{4}<\frac{\delta}{2}$

for every $t\in U$ and that

$\left\|\phi(t)(f_{0}(t)-f_{0}(t_{0}))-\frac{\delta(1-\phi(t))f_{0}(t)}{8}\right\|_{X}<2\cdot\frac{\delta}{4}=\frac{\delta}{2}$

for every $t\in K\setminus U$. Hence we can see that
$\|\Psi(\|f_{0}(t_{0})\|_{X})-f_{0}\|<\delta$, and so

$\|P(\Psi(\|f_{0}(t_{0})\|_{X}))(t_{0})-P(f_{0})(t_{0})\|_{X}\leq\|P(\Psi(\|f_{0}(t_{0})\|_{X}))-P(f_{0})\|<\frac{\epsilon}{2}$

by (5). This implies that

$\|P(\Psi(\|f_{0}(t_{0})\|_{X}))(t_{0})\|_{X}>\|P(f_{0})(t_{0})\|_{X}-\frac{\epsilon}{2}>\|P\|-2\cdot\frac{\epsilon}{2}=\|P\|-\epsilon.$

In view of Hahn-Banach theorem, there exists $x_{0}^{*}\in S_{X^{*}}$ such
that

$x_{0}^{*}\left(P(\Psi(\|f_{0}(t_{0})\|_{X}))(t_{0})\right)=\|P(\Psi(\|f_{0}(t_{0})\|_{X}))(t_{0})\|_{X}>\|P\|-\epsilon.$

Notice that the function $f(z)=x_{0}^{*}\left(P(\Psi(z))(t_{0})\right)$ is
holomorphic. Hence, by the maximum modulus theorem, there exists
$z_{0}\in\mathbb{T}$ such that

$\|P(\Psi(z_{0}))(t_{0})\|_{X}\geq
x_{0}^{*}\left(P(\Psi(\|f_{0}(t_{0})\|_{X}))(t_{0})\right)>\|P\|-\epsilon.$

Take $x_{1}=z_{0}x_{0}\in S_{X}$ and let $x_{1}^{*}\in S_{X^{*}}$ such that
$x_{1}^{*}x_{1}=1$. Define a function $\Phi:X\rightarrow A(K,X)$ by

$\Phi(x)=x_{1}^{*}x\left(1-\frac{\delta}{8}\right)(1-\phi)f_{0}+\phi\cdot x.$

We see that $\|\Phi(x)\|\leq 1$ for every $x\in B_{X}$ from (6). In
particular, $\Phi(x_{1})=\Psi(z_{0})$. Hence
$\|P(\Phi(x_{1}))(t_{0})\|>\|P\|-\epsilon$. Consider $Q\in\mathcal{P}_{K}(X)$
defined by $Q(x)=P(\Phi(x))(t_{0})$. Notice that

$\|Q\|\geq\|Qx_{1}\|_{X}=\|(P(\Phi(x_{1}))(t_{0})\|_{X}>\|P\|-\epsilon.$

This implies that $\|I+Q\|\geq
1+\text{Daug}_{p}(X)\|Q\|>1+\text{Daug}_{p}(X)(\|P\|-\epsilon)$. Now choose
$x_{2}\in B_{X}$ such that
$\|x_{2}+Qx_{2}\|>1+\text{Daug}_{p}(X)(\|P\|-\epsilon)$ and let
$g=\Phi(x_{2})$. Then we obtain

$\displaystyle\|I+P\|\geq\|g+Pg\|$ $\displaystyle\geq$
$\displaystyle\|g(t_{0})+P(g)(t_{0})\|_{X}$ $\displaystyle\geq$
$\displaystyle\left\|x_{1}^{*}x_{2}\left(1-\frac{\delta}{8}\right)(1-\phi(t_{0})f(t_{0})+\phi(t_{0})x_{2}-Q(x_{2})\right\|_{X}$
$\displaystyle=$
$\displaystyle\|x_{2}+Q(x_{2})\|>1+\text{Daug}_{p}(X)(\|P\|-\epsilon)$

As $\epsilon\rightarrow 0$, we have $\|I+P\|\geq 1+\text{Daug}_{p}(X)\|P\|$.
This consequently shows that $\text{Daug}_{p}(A(K,X))\geq\text{Daug}_{p}(X)$.

If $\Gamma$ does not have isolated points, then $A(K,X)$ has the polynomial
Daugavet property by Theorem 4.6. This implies that

$\text{Daug}_{p}(A(K,X))=\text{Daug}_{p}(A)=1,$

and so we have
$\text{Daug}_{p}(A(K,X))=\max\\{\text{Daug}_{p}(A),\text{Daug}_{p}(X)\\}$.

If $\Gamma$ has isolated points, then $A(K,X)=X\oplus_{\infty}Y$ by Lemma 4.4
and $\text{Daug}(A)=0$ by Corollary 4.10. From Lemma 4.8, we see that
$\text{Daug}_{p}(A(K,X))\leq\text{Daug}_{p}(X)$. Therefore, we also obtain
$\text{Daug}_{p}(A(K,X))=\max\\{\text{Daug}_{p}(A),\text{Daug}_{p}(X)\\}$. ∎

######  Corollary 4.12.

Let $K$ be a compact Hausdorff space. Then the space $A(K,X)$ has the
polynomial Daugavet property if and only if either the base algebra $A$ or $X$
has the polynomial Daugavet property.

###  4.2. Remarks on the property $(D)$ and the convex diametral local
diameter two property in $A(K,X)$

Since the equivalence between the property ($\mathcal{D}$) and the DLD2P is
not clear, it is natural to explore various Banach spaces that potentially
distinguish these properties. However, we show that this is not the case for
$A(K,X)$. Under the additional assumption that $X$ is uniformly convex, the
space $A(K,X)$ has the Daugavet property if and only if the Shilov boundary of
the base algebra does not have isolated points [27, Theorem 5.6]. Moreover,
the Daugavet property of $A(K,X)$ is equivalent to all diametral D2Ps under
the same assumption. In fact, carefully inspecting the proof of [27, Theorem
5.4], we see that the rank-one projection constructed in there has norm-one.
With the aid of our previous observations, we can see that the DLD2P is also
equivalent to the property ($\mathcal{D}$) for $A(K,X)$.

######  Proposition 4.13.

[27, Theorem 5.4] Let $X$ be a uniformly convex Banach space, $K$ be a compact
Hausdorff space, $\Gamma$ be the Shilov boundary of the base algebra $A$ of
$A(K,X)$, and $f\in S_{A(K,X)}$. Then the following statements are equivalent:

  1. (i)

$f$ is a Daugavet point.

  2. (ii)

$f$ is a $\Delta$-point.

  3. (iii)

Every rank-one, norm-one projection $P=\psi\otimes f$, where $\psi\in
A(K,X)^{*}$ with $\psi(f)=1$, satisfies $\|I-P\|=2$.

  4. (iv)

there is a limit point $t_{0}$ of $\Gamma$ such that $\|f\|=\|f(t_{0})\|_{X}$.

###### Proof.

(i) $\implies$ (ii) is clear. The implication (ii) $\implies$ (iii) comes from
Theorem 2.2. Indeed, for a Banach space $Y$, a point $f\in S_{Y}$ is a
$\Delta$-point if and only if every rank-one projection of the form
$P=\psi\otimes f$, where $\psi\in Y^{*}$ with $\psi(f)=1$, satisfies
$\|I-P\|\geq 2$. Hence, we immediately have (iii) if this projection $P$ has
the norm one. (iii) $\implies$ (iv) and (iv) $\implies$ (i) are identical to
the proof of (ii) $\implies$ (iii) and (iii) $\implies$ (i) in [27, Theorem
5.4], respectively. ∎

######  Corollary 4.14.

[27, Corollary 5.5] Let $K$ be a compact Hausdorff space, $\Gamma$ be the
Shilov boundary of $A(K)$, and $f\in S_{A(K)}$. Then the following statements
are equivalent:

  1. (i)

$f$ is a Daugavet point.

  2. (ii)

$f$ is a $\Delta$-point.

  3. (iii)

Every rank-one, norm-one projection $P=\psi\otimes f$, where $\psi\in
A(K)^{*}$ with $\psi(f)=1$, satisfies $\|I-P\|=2$.

  4. (iv)

there is a limit point $t_{0}$ of $\Gamma$ such that
$\|f\|_{\infty}=|f(t_{0})|$.

As a consequence, we obtain the following characterizations for the space
$A(K,X)$ and infinite-dimensional uniform algebras.

######  Proposition 4.15.

[27, Theorem 5.6] Let $X$ be a uniformly convex Banach space, let $K$ be a
compact Hausdorff space, and let $\Gamma$ be the Shilov boundary of the base
algebra $A$ of $A(K,X)$. Then the following statements are equivalent:

  1. (i)

$A(K,X)$ has the polynomial Daugavet property.

  2. (ii)

$A(K,X)$ has the Daugavet property.

  3. (iii)

$A(K,X)$ has the DD2P.

  4. (iv)

$A(K,X)$ has the DLD2P.

  5. (v)

$A(K,X)$ has the property ($\mathcal{D}$).

  6. (vi)

The Shilov boundary $\Gamma$ does not have isolated points.

###### Proof.

(i) $\implies$ (ii) $\implies$ (iii) $\implies$ (iv) $\implies$ (v) is clear
from their definitions. The implication (vi) $\implies$ (i) can be shown by
Theorem 4.6. Showing (v) $\implies$ (vi) is also identical to the proof of
[27, Theorem 5.6] with Corollary 4.13. ∎

######  Corollary 4.16.

[27, Corollary 5.7] Let $K$ be a compact Hausdorff space and let $\Gamma$ be
the Shilov boundary of a uniform algebra $A(K)$. Then the following are
equivalent:

  1. (i)

$A(K)$ has the polynomial Daugavet property.

  2. (ii)

$A(K)$ has the Daugavet property.

  3. (iii)

$A(K)$ has the DD2P.

  4. (iv)

$A(K)$ has the DLD2P.

  5. (v)

$A(K)$ has the property ($\mathcal{D}$).

  6. (vi)

The Shilov boundary $\Gamma$ does not have isolated points.

In view of Lemma 4.2, we can also show that the sufficient condition for the
convex-DLD2P in [27, Theorem 5.9] can be described with strong boundary
points.

######  Theorem 4.17.

Let $K$ be a compact Hausdorff space, $X$ be a uniformly convex Banach space,
and let $\Gamma$ be the Shilov boundary of the base algebra of $A(K,X)$.
Denote by $\Gamma^{\prime}$ the set of limit points of the Shilov boundary. If
$\Gamma^{\prime}\cap\Gamma_{0}\neq\emptyset$, then $A(K,X)$ has the convex-
DLD2P.

###### Proof.

In view of Lemma 4.5, we assume that $K=\Gamma$. Denote the set of all
$\Delta$-points of $A(K,X)$ by $\Delta$ and the base algebra of $A(K,X)$ by
$A$. We claim that $S_{A(K,X)}\subset\overline{\text{conv}}\Delta$.

Let $f\in S_{A(K,X)}$. Choose a point $t_{0}\in\Gamma^{\prime}\cap\Gamma_{0}$
and let $\lambda=\frac{1+\|f(t_{0})\|_{X}}{2}$. For $\epsilon>0$, let $U$ be
an open subset of $K$ such that $\|f(t)-f(t_{0})\|_{X}<\epsilon$. Then by
Lemma 4.2, there exists $\phi\in A$ such that
$\|\phi\|_{\infty}=\phi(t_{0})=1$, $\sup_{t\in K\setminus
U}|\phi(t)|<\epsilon$, and

$|\phi(t)|+(1-\epsilon)|1-\phi(t)|\leq 1$

for every $t\in K$.

Choose a norm-one vector $v_{0}\in X$ and let

$x_{0}=\begin{cases}\frac{f(t_{0})}{\|f(t_{0})\|_{X}}&\mbox{if }f(t_{0})\neq
0\\\ v_{0}&\mbox{if }f(t_{0})=0.\end{cases}$

Now, define

$\displaystyle f_{1}(t)$
$\displaystyle=(1-\epsilon)(1-\phi(t))f(t)+\phi(t)x_{0}$ $\displaystyle
f_{2}(t)$ $\displaystyle=(1-\epsilon)(1-\phi(t))f(t)-\phi(t)x_{0},\ \ \ t\in
K.$

Notice that $f_{1},f_{2}\in A(K,X)$ because $A\otimes X\subset A(K,X)$.
Moreover,

$\displaystyle\|f_{1}(t)\|_{X}$
$\displaystyle=\left\|(1-\epsilon)(1-\phi(t))f(t)+\phi(t)x_{0}\right\|_{X}$
$\displaystyle\leq(1-\epsilon)|1-\phi(t)|+|\phi(t)|\leq 1,$

for every $t\in K$. In particular, we have $\|f_{1}(t_{0})\|_{X}=1$, and so
$\|f_{1}(t_{0})\|_{X}=\|f_{1}\|=1$. By the same argument, we also have
$\|f_{2}(t_{0})\|_{X}=\|f_{2}\|=1$. Thus, $f_{1},f_{2}\in\Delta$ by
Proposition 4.13. Let $g(t)=\lambda f_{1}(t)+(1-\lambda)f_{2}(t)$. We need to
consider two cases.

Case 1: Suppose $f(t_{0})\neq 0$. Then
$g(t)=(1-\epsilon)(1-\phi(t))f(t)+\phi(t)f(t_{0})$. We see that

$\displaystyle\|g(t)-f(t)\|_{X}$ $\displaystyle=$
$\displaystyle\|(1-\epsilon)(1-\phi(t))f(t)+\phi(t)f(t_{0})-f(t)\|_{X}$
$\displaystyle=$
$\displaystyle\|(1-\epsilon)(1-\phi(t))f(t)+\phi(t)f(t_{0})-(1-\epsilon)f(t)-\epsilon
f(t)\|_{X}$ $\displaystyle=$
$\displaystyle\|(1-\epsilon)(-\phi(t))f(t)+(1-\epsilon)\phi(t)f(t_{0})+\epsilon\phi(t)f(t_{0})-\epsilon
f(t)\|_{X}$ $\displaystyle=$
$\displaystyle\|(1-\epsilon)\phi(t)(f(t_{0})-f(t))+\epsilon\phi(t)f(t_{0})-\epsilon
f(t)\|_{X}$ $\displaystyle\leq$
$\displaystyle(1-\epsilon)|\phi(t)|\cdot\|f(t)-f(t_{0})\|_{X}+\epsilon|\phi(t)|\cdot\|f(t_{0})\|_{X}+\epsilon\|f(t)\|_{X}$
$\displaystyle\leq$
$\displaystyle(1-\epsilon)|\phi(t)|\cdot\|f(t)-f(t_{0})\|_{X}+2\epsilon.$

For $t\in U$, we see that
$(1-\epsilon)|\phi(t)|\cdot\|f(t)-f(t_{0})\|_{X}\leq(1-\epsilon)\epsilon<\epsilon$.
On the other hand, for $t\in K\setminus U$, we have
$(1-\epsilon)|\phi(t)|\cdot\|f(t)-f(t_{0})\|_{X}\leq
2(1-\epsilon)\epsilon<2\epsilon$. Hence, $\|g-f\|<4\epsilon$, and so
$f\in\overline{conv}\Delta$.

Case 2: Now, suppose $f(t_{0})=0$. Then we have $\|f(t)\|_{X}<\epsilon$ for
every $t\in U$. Moreover, notice that $\lambda=\frac{1}{2}$ and
$g(t)=(1-\epsilon)(1-\phi(t))f(t)$. This implies that

$\displaystyle\|g(t)-f(t)\|_{X}$ $\displaystyle=$
$\displaystyle\|(1-\epsilon)(1-\phi(t))f(t)-(1-\epsilon)f(t)-\epsilon
f(t)\|_{X}$ $\displaystyle\leq$
$\displaystyle(1-\epsilon)|\phi(t)|\cdot\|f(t)\|_{X}+\epsilon\|f(t)\|_{X}\leq(1-\epsilon)|\phi(t)|\cdot\|f(t)\|_{X}+\epsilon.$

Notice that
$(1-\epsilon)|\phi(t)|\cdot\|f(t)\|_{X}\leq(1-\epsilon)\epsilon<\epsilon$ for
every $t\in U$. From the fact that $\sup_{t\in K\setminus
U}|\phi(t)|<\epsilon$ for $t\in K\setminus U$, we have
$(1-\epsilon)|\phi(t)|\cdot\|f(t)\|_{X}\leq(1-\epsilon)\epsilon<\epsilon$.
This shows that $\|g-f\|<2\epsilon$, and so
$f\in\overline{\text{conv}}\Delta$.

Since $f\in S_{A(K,X)}$ is arbitrary, we see that
$S_{X}\subset\overline{\text{conv}}\Delta$. Therefore, the space $A(K,X)$ has
the convex-DLD2P. ∎

######  Corollary 4.18.

Let $K$ be a compact Hausdorff space and $\Gamma^{\prime}$ be the set of limit
points in the Shilov boundary of a uniform algebra. If
$\Gamma^{\prime}\cap\Gamma_{0}\neq\emptyset$, Then the uniform algebra has the
convex-DLD2P.

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