2. Notation and preliminaries

In this short section, we recall a few classical notions from Banach space geometry that we will be using throughout the text. The notation of this paper is standard, following [Reference Fabian, Habala, Hájek, Montesinos and Zizler11].

Let X be a Banach space and let A be a non-empty closed convex bounded subset of X. We denote by $\operatorname{span}(A)$, $\operatorname{conv}(A)$, $\overline{\operatorname{span}}(A)$ and $\overline{\operatorname{conv}}(A)$ the linear span and convex hull of A, as well as their respective closures. The following notions are well-studied properties for points of A. We say that a point x in A is

1. (1) an extreme point of A (writing $x\in \operatorname{ext}(A)$) if x does not belong to the interior of any segment of A;

2. (2) a point of continuity of A (writing $x\in \operatorname{PC}(A)$) if the identity mapping $\operatorname{Id}\colon (A,w)\to(A,\left\lVert\cdot\right\rVert)$ is continuous at x (that is, if x is contained in a relatively weakly open subset of A of arbitrarily small diameter);

3. (3) a denting point of A (writing $x\in \operatorname{dent}(A)$) if x is contained in slices of A of arbitrarily small diameter;

4. (4) a strongly exposed point of A (writing $x\in \operatorname{str-exp}(A)$) if there exists $x^*\in X^*$ such that for all sequences $(x_n)\subset A$, $\operatorname{Re}\,x^\ast(x_n)\rightarrow \operatorname{Re}\,x^\ast(x)$ if and only if $x_n\rightarrow x$ in norm.

Clearly, strongly exposed points are denting, and denting points are both extreme points and points of continuity. In fact, denting points are precisely those points which are simultaneously extreme points and points of continuity (see [Reference Fabian, Habala, Hájek, Montesinos and Zizler11, Exercise 3.146], for instance). For the case of $A=B_X$, a stronger version of strongly exposed point is the one of LUR point. An element $x\in B_X$ is a LUR point if for every $\varepsilon$ > 0 there is δ > 0 such that the implication

\begin{equation*} \left\lVert\frac{x+y}{2}\right\rVert \gt 1-\delta \implies \left\lVert x-y\right\rVert\leqslant \varepsilon \end{equation*}

holds for every $y\in B_X$. If every element of S_X is LUR, we say that the space X is LUR.

Recall that a Banach space X has the RNP if every closed convex bounded subset of X has a denting point. Also, a Banach space X has the point of continuity property (PCP for short), respectively, the CPCP, if every closed bounded subset, respectively every closed convex bounded subset, of A has a point of continuity. It follows that in an RNP space X, B_X is the closed convex hull of the set of all denting points of B_X, and that in spaces X with the CPCP, the set of all points of continuity of B_X is weakly dense in B_X, see [Reference Ghoussoub, Godefroy, Maurey and Schachermayer12] for more information and background. A related isometric notion is the following: a Banach space has the Kadec property if the identity map $\operatorname{Id}\colon (B_X,w)\to (B_X,\left\lVert\cdot\right\rVert)$ is continuous on the whole S_X (in other words, if $S_X\subset \operatorname{PC}(B_X)$), see e.g. [Reference Deville, Godefroy and Zizler10, Section II.1]. Finite-dimensional Banach spaces trivially satisfy the Kadec property, and also uniformly convex spaces or, more generally, those Banach spaces for which $\operatorname{dent}(B_X)=S_X$, such as LUR spaces. Moreover, asymptotically uniformly convex spaces satisfy a uniform version of the Kadec property (see e.g. the remark following [Reference Johnson, Lindenstrauss, Preiss and Schechtman15, Proposition 2.6], or the discussion at the end of Section 2 in [Reference Abrahamsen, Lima, Martiny and Perreau4]). Recall that these latter spaces include in particular all $\ell_p$-sums of finite-dimensional spaces, $1\leqslant p \lt \infty$.

Related to the differentiability of norms are the following notions. We say that a point x in a Banach space X is

1. (1) a point of Gâteaux differentiability if there exists a unique functional $f\in S_{X^*}$ such that $f(x)=\left\lVert x\right\rVert$;

2. (2) a point of Fréchet differentiability if there exists a functional $f\in S_{X^*}$ such that for every sequence $(f_n)$ in $S_{X^*}$, $f_n(x)\to \left\lVert x\right\rVert$ if and only if $f_n\to f$ in norm.

Recall that a Banach space X is Asplund if every continuous and convex function f from a non-empty open subset U of X into $\mathbb{K}$ is Fréchet differentiable on a G_δ subset of U. Equivalently, X is Asplund if and only if every separable subspace of X has a separable dual if and only if $X^*$ has the RNP.

From its various geometric characterizations, the DP admits several natural localizations to points of the unit sphere of Banach spaces, as we already mentioned in the introduction. Let us present two more notions, and refer to [Reference Haller, Langemets, Perreau and Veeorg13] for more information and background. Let X be a Banach space and $x\in S_X$. We say that x is

1. (1) a $\nabla$-point if for every slice S of B_X not containing x, we have $\sup\nolimits_{y\in S} \left\lVert x-y\right\rVert=2$.

2. (2) a Δ-point if for every slice S of B_X containing x, we have $\sup\nolimits_{y\in S} \left\lVert x-y\right\rVert=2$.

Observe that a point is a Daugavet point if and only if it is $\nabla$ and Δ simultaneously. Also, recall that finite-dimensional spaces contain no Daugavet points (see [Reference Abrahamsen, Aliaga, Lima, Martiny, Perreau, Prochazka and Veeorg2]).

Finally, we end the section with a brief recap about spear vectors. Let X be a Banach space. An element $x\in S_X$ is a spear vector (or just spear) [Reference Ardalani7, Reference Kadets, Martín, Merí and Pérez19] if $\max\limits_{\theta\in\mathbb{T}}\|x+\theta y\|=2$ for every $y\in S_X$; equivalently, if for every $x^*\in\operatorname{ext}(B_{X^*})$, we have $\left\lvert x^*(x)\right\rvert=1$, see [Reference Kadets, Martín, Merí and Pérez19] for more information and background. It is immediate that every spear is an extreme point of B_X, while the opposite is not always true. Examples of spears include the elements of the unit vector basis of $\ell_1(\Gamma)$, the extreme points of the unit ball of $\ell_\infty^n$ and $\ell_\infty$, or the constant 1 function in $C[0,1]$. Observe that, in the real case, two distinct spears are necessarily at distance 2 from one another (because they have to assume a different value on some extreme point). In the complex case, we have that $\operatorname{Spear}(X)$ is nowhere dense in S_X unless X is one-dimensional [Reference Kadets, Martín, Merí and Pérez19, Proposition 2.11]. From this discussion, the following result follows.

An easy convexity argument gives the following remark.

If X and Y are Banach spaces over $\mathbb{K}$, we write $X\oplus_p Y$ to denote the $\ell_p$-sum for $1\leqslant p\leqslant \infty$. For a generalization of the above, given a family $\{E_\gamma\colon \gamma\in \Gamma\}$ of Banach spaces, $\left[\bigoplus_{\gamma\in\Gamma}E_\gamma\right]_{\ell_p}$ represent the $\ell_p$-sum of the family for $1\leqslant p\leqslant \infty$ and $\left[\bigoplus_{\gamma\in\Gamma}E_\gamma\right]_{c_0}$ is the c ₀-sum of the family.

3. Super ADP for Banach spaces

Our goal here is to discuss the first examples and several properties of the super ADP.

Clearly, the DP implies the super ADP, and the super ADP implies the ADP. Our first result shows that the super ADP lies strictly between the ADP and the DP.

Recall that $\ell_1(\Gamma)$ has the Kadec property, since it is asymptotically uniformly convex. It follows that this space fails the super ADP, if the set Γ has more than one element. Moreover, it is known that these spaces have the ADP, providing infinite-dimensional examples separating these two properties.

A stronger result than the previous one will be provided in Theorem 3.9.

We will now show that even in infinite-dimensional Banach spaces, the super ADP is strictly weaker than the DP.

We provide here the following immediate lemma, which also has a local version, see Lemma 4.2.

Proof of Theorem 3.3

Let $(X,\left\lVert\cdot\right\rVert)$ be a Banach space with the DP, and fix $x_0\in S_{(X,\left\lVert\cdot\right\rVert)}$ and $\varepsilon\in(0,1)$. We consider the equivalent norm $|||\cdot|||$ on X whose unit ball is defined as

(3.1)\begin{equation} B_{(X,|||\cdot|||)} := \operatorname{conv}\Bigl(\mathbb{T} x_0\cup (1-\varepsilon)B_{(X,\left\lVert\cdot\right\rVert)}\Bigr). \end{equation}

Observe that this set is closed, since $\mathbb{T} x_0$ is compact and $(1-\varepsilon)B_{(X,|||\cdot|||)}$ is closed. Let

\begin{equation*} A:=\mathbb{T} x_0\cup (1-\varepsilon)B_{(X,\left\lVert\cdot\right\rVert)}, \end{equation*}

and observe that, using (3.1), we can compute the norm of $f\in (X^*,|||\cdot|||)$ as follows:

(3.2)\begin{equation} |||f||| = \sup_{x\in A}\left\lvert f(x)\right\rvert = \max\left\{\sup_{x\in(1-\varepsilon)B_{(X,\left\lVert\cdot\right\rVert)}}\left\lvert f(x)\right\rvert,\, \left\lvert f(x_0)\right\rvert\right\} = \max\{(1-\varepsilon)\left\lVert f\right\rVert, \left\lvert f(x_0)\right\rvert\}. \end{equation}

We claim that the Banach space $(X,|||\cdot|||)$ has the super ADP, but not the DP.

Notice that x ₀ is clearly a denting point in the unit ball $B_{(X,|||\cdot|||)}$ (see e.g. [Reference Becerra, López-Pérez and Rueda Zoca9, Lemma 2.1]). In particular, $(X, |||\cdot|||)$ fails the DP. To show that it has the super ADP, it suffices to find, given $x,y\in S_{(X,|||\cdot|||)}$ and δ > 0, a net $(y_{\alpha})$ which weakly converges to y and satisfies $\limsup_\alpha\max_{\theta\in\mathbb{T}}|||x+\theta y_{\alpha}|||\geqslant 2-\delta$, see Lemma 3.4. Actually, by the moreover part of Lemma 3.4, we can assume that

\begin{align*} x &= \lambda x_0 +(1-\lambda)(1-\varepsilon) u, \;\; \lambda\in(0,1), \; u\in B_{(X,||\cdot||)}; \\ y &= \mu \omega x_0 + (1-\mu)(1-\varepsilon)v, \;\; \mu\in(0,1),\; v\in B_{(X,||\cdot||)},\; \omega\in\mathbb{T}. \end{align*}

Fix $\eta\in(0,\delta/2)$. Since the space $(X,\left\lVert\cdot\right\rVert)$ has the DP, using [Reference Kadets, Shvidkoy, Sirotkin and Werner18, Lemma 2.8] and [Reference Shvydkoy26, Lemma 3], we can find a net $(v_{\alpha})_{\alpha\in I}$ weakly converging to $\omega^{-1} v$ and satisfying

(3.3)\begin{equation} \left\lVert k_1x_0+k_2u+k_3v_{\alpha}\right\rVert\geqslant (1-\eta)\big(\left\lVert k_1x_0+k_2u\right\rVert + \left\lvert k_3\right\rvert\big),\;\;\;\;k_1,k_2,k_3\in\mathbb{K},\;\;\alpha\in I. \end{equation}

We now take $f\in (X^*,|||\cdot|||)$, $|||f|||=1$, such that $f(x)=1$. By the representation of x above, this implies that $f(x_0)=1$ and $f((1-\varepsilon)u)=1$. For $\alpha\in I$, consider the linear functional $h_{\alpha}\colon \operatorname{span}\{x_0,u,v_{\alpha}\} \to \mathbb{K}$ defined as

\begin{equation*} k_1x_0+k_2u + k_3 v_{\alpha} \longmapsto k_1f(x_0)+k_2f(u) + k_3f(u) \end{equation*}

for $k_1,k_2,k_3\in\mathbb{K}$. Using (3.3), we obtain a bound for $||h_{\alpha}||$: for any $k_1x_0+k_2u+k_3v_{\alpha}$, we have

\begin{align*} \left\lvert h_{\alpha}(k_1x_0+k_2u+k_3v_{\alpha})\right\rvert &= \left\lvert f(k_1x_0+k_2u) + k_3f(u)\right\rvert \leqslant \left\lvert f(k_1x_0+k_2u)\right\rvert+ \left\lvert k_3\right\rvert\left\lVert f\right\rVert\\ &\leqslant \left\lVert f\right\rVert\left\lVert k_1x_0+k_2u\right\rVert + |k_3|\left\lVert f\right\rVert\leqslant \left\lVert f\right\rVert\big(\left\lVert k_1x_0+k_2u\right\rVert+\left\lvert k_3\right\rvert\big)\\ &\leqslant\frac{\left\lVert f\right\rVert}{1-\eta}\left\lVert k_1x_0+k_2u+k_3v_{\alpha}\right\rVert. \end{align*}

Consequently, $\left\lVert h_{\alpha}\right\rVert\leqslant \left\lVert f\right\rVert/(1-\eta)$. Moreover, (3.2) yields

\begin{equation*} |||h_{\alpha}|||= \max\bigl\{(1-\varepsilon)\left\lVert h_{\alpha}\right\rVert, |h(x_0)|\bigr\}\leqslant \max\Big\{\tfrac{1-\varepsilon}{1-\eta}\left\lVert f\right\rVert, 1\Big\}. \end{equation*}

Since $|||f|||=1$ and $\left\lvert f(x_0)\right\rvert=1$, we have $\left\lVert f\right\rVert\leqslant 1/(1-\varepsilon)$, so

\begin{equation*} |||h_{\alpha}|||\leqslant \max\Big\{\tfrac{1}{1-\eta},1\Big\}=\frac{1}{1-\eta}. \end{equation*}

For every $\alpha\in I$, we now extend h_α, using the Hahn–Banach Theorem, to the space $(X,|||\cdot|||)$. Therefore, we have $h_{\alpha}\in(X^*,|||\cdot|||)$ with $|||h_{\alpha}||| \lt 1/(1-\eta)$. Denote $y_{\alpha} :=\mu\omega x_0 + (1-\mu)(1-\varepsilon)\omega v_{\alpha}$ for every $\alpha\in I$. Since $(v_{\alpha})_{\alpha\in I}$ converges weakly to $\omega^{-1} v$, the net $(y_{\alpha})_{\alpha\in I}$ converges weakly to y. We conclude the proof by showing that, for every $\alpha\in I$, we have

\begin{equation*} \max_{\theta\in\mathbb{T}}|||x+\theta y_{\alpha}|||\geqslant |||x+\omega^{-1} y_{\alpha}||| \gt 2-\delta. \end{equation*}

Indeed, for a fixed $\alpha\in I$, we have

\begin{align*} |||x+\omega^{-1}y_{\alpha}|||&\geqslant \left\lvert\frac{h_{\alpha}}{|||h_{\alpha}|||}\Big(x+\omega^{-1}y_{\alpha}\Big)\right\rvert\\ &\geqslant (1-\eta)\Big|h_{\alpha}\big(\lambda x_0+(1-\lambda)(1-\varepsilon)u+\omega^{-1}(\mu\omega x_0 + (1-\mu)(1-\varepsilon)\omega v_{\alpha})\big)\Big|\\ &=(1-\eta)\Big|\lambda f(x_0)+(1-\lambda)(1-\varepsilon)f(u)+\mu f(x_0)+(1-\mu)(1-\varepsilon)f(u)\Big|\\ &=(1-\eta)\Big|\lambda+(1-\lambda)+\mu +(1-\mu)\Big|=2(1-\eta)=2-2\eta \gt 2-\delta. \end{align*}

Having established that the super ADP lies strictly between the ADP and the DP, it is natural to ask what kind of constraints the super ADP imposes on the underlying Banach space. Our first result in this line is that separable Banach spaces with the CPCP cannot have the super ADP.

Proof. Let x be a point of Gâteaux differentiability of the norm. Then there exists a unique $f\in S_{X^*}$ such that $f(x)=1$. Consider the relatively weakly open subset $W:=\{y\in B_X\colon \left\lvert f(y)\right\rvert \lt 1/2\}$ of B_X. Since $\dim(X)\geqslant 2$, W must intersect S_X. By the hypothesis, W must contain a point of continuity y of B_X that belongs to S_X. Let us now prove that $\rho:=\max_{\theta\in\mathbb{T}}\left\lVert x+\theta y\right\rVert \lt 2$. Indeed, if $\theta\in\mathbb{T}$ satisfies that $\left\lVert x+\theta y\right\rVert=2$, we pick $g\in S_{X^*}$ such that $\operatorname{Re}\, g(x+\theta y)=\|x+\theta y\|=2$, so $\operatorname{Re}\,g(x)=1$ and g = f by uniqueness of the supporting functional. But then,

\begin{equation*}\left\lVert x+\theta y\right\rVert=|f(x+\theta y)|\leqslant 1 + |f(y)| \lt 3/2\end{equation*}

as $y\in W$. Now, we use that $y\in \operatorname{PC}(B_X)$ to find a weakly open subset W ₁ of B_X, such that $y\in W_1$ (so W ₁ intersects S_X) and with $\operatorname{diam}(W_1) \lt (2-\rho)/2$. If $z\in W_1$, we have that

\begin{align*} \max_{\theta\in\mathbb{T}}\left\lVert x+\theta z\right\rVert &\leqslant \max_{\theta\in\mathbb{T}}\left\lVert x+\theta y\right\rVert + \|z-y\| \leqslant \rho +\operatorname{diam}(W_1) \leqslant 1+ \frac{\rho}{2} \lt 2. \end{align*}

Hence, $\sup_{z\in W_1}\max_{\theta\in\mathbb{T}}\left\lVert x+\theta z\right\rVert \lt 2$ and so X fails the super ADP.

We will now prove that the super ADP is separably determined by a.i. ideals. Let Y be a Banach space and X be a subspace of Y. Recall that X is said to be an almost isometric ideal (a.i. ideal for short) of Y if for every $\varepsilon$ > 0, and for every finite-dimensional subspace E of Y, there exists a bounded linear operator $T\colon E\to X$ satisfying:

1. (1) for every $e\in E\cap X$, $T(e)=e$;

2. (2) for every $e\in E$, $(1-\varepsilon)\left\lVert e\right\rVert\leqslant \left\lVert T(e)\right\rVert\leqslant (1+\varepsilon)\left\lVert e\right\rVert$.

Also recall that a bounded linear operator $\varphi\colon X^*\to Y^*$ is called a Hahn–Banach extension operator if $\left\lVert\varphi x^*\right\rVert=\left\lVert x^*\right\rVert$ for every $x^*\in X^*$ and $\varphi x^*(x)=x^*(x)$ for every $(x,x^*)\in X\times X^*$. The following result was proved in [Reference Abrahamsen, Lima and Nygaard6].

We now get the stability of the super ADP by a.i. ideals.

Proof. Take $x\in S_X$ and $\varepsilon$ > 0. Then let $x_0\in S_X$ and let W be a neighborhood of x ₀ in the relative weak topology of B_X. Without loss of generality, there exist $n\in\mathbb{N}$, $x_1^*,\dots, x_n^*$ in $X^*\setminus\{0\}$ and δ > 0 such that

\begin{equation*} W:=\bigcap_{i=1}^n\{x\in B_X\colon \left\lvert x^*_i(x-x_0)\right\rvert \lt \delta\}. \end{equation*}

Let $\varphi\colon X^*\to Y^*$ be the Hahn–Banach extension operator given by the previous lemma, and consider the set

\begin{equation*} \widetilde{W}:=\bigcap_{i=1}^n\{y\in B_Y\colon \left\lvert\varphi x^*_i(y-x_0)\right\rvert \lt \delta/2\}. \end{equation*}

Since x is a super AD point in Y, there exist $y\in \widetilde{W}$ and $\theta\in\mathbb{T}$ such that $\left\lVert x+\theta y\right\rVert \gt 2-\varepsilon$. Let η > 0 to be chosen later. Applying the above with $E:=\operatorname{span}\{x_0,x,y\}$ and $F:=\operatorname{span}\{x_1^*,\ldots, x_n^*\}$, we get a bounded linear operator $T\colon E\to X$ satisfying:

1. (1) $Tx_0=x_0$ and Tx = x;

2. (2) for every $e\in E$ and for every $i\in\{1,\dots,n\}$, $\varphi \bigl(x_i^*(e)\bigr)=x_i^*(Te)$;

3. (3) for every $e\in E$, $(1-\eta)\left\lVert e\right\rVert\leqslant \left\lVert Te\right\rVert\leqslant (1+\eta)\left\lVert e\right\rVert$.

For simplicity, let us now distinguish two cases. First, assume that $\left\lVert Ty\right\rVert \gt 1$ and let $z:=\frac{Ty}{\left\lVert Ty\right\rVert}$. By assumption, we have that $\left\lVert z-Ty\right\rVert=\left\lVert Ty\right\rVert-1\leqslant \eta$. Thus, for every $i\in\{1,\dots,n\}$, we have

\begin{align*} \left\lvert x_i^*(z-x_0)\right\rvert\leqslant \left\lVert z-Ty\right\rVert\cdot\left\lVert x_i^*\right\rVert+\left\lvert x_i^*(Ty-x_0)\right\rvert =& \left\lVert z-Ty\right\rVert\cdot\left\lVert x_i^*\right\rVert+\left\lvert\varphi x_i^*(y-x_0)\right\rvert \\ & \lt \eta\cdot\left\lVert x_i^*\right\rVert+\delta/2. \end{align*}

Furthermore,

\begin{equation*} \left\lVert x+\theta z\right\rVert\geqslant \left\lVert x+\theta Ty\right\rVert-\left\lVert z-Ty\right\rVert=\left\lVert Tx+\theta Ty\right\rVert-\left\lVert z-Ty\right\rVert\geqslant (1-\eta)(2-\varepsilon)-\eta. \end{equation*}

So if η was initially chosen so that $\eta\cdot\left\lVert x_i^*\right\rVert \lt \delta/2$ for every i and so that $(1-\eta)(2-\varepsilon)-\eta \gt 2-2\varepsilon$, then we would get that $z\in W$ and $\left\lVert x+\theta z\right\rVert \gt 2-2\varepsilon$. The conclusion follows. The case $\left\lVert Ty\right\rVert\leqslant 1$ can be dealt with analogously, and with simpler computations, because we can then simply take z = Ty.

The separable determination of the super ADP now immediately follows from [Reference Abrahamsen1, Theorem 1.5], which ensures that given a Banach space and a separable subspace, there is a separable a.i. of the space containing the subspace.

Since the CPCP clearly passes to subspaces, we immediately get the following.

Finally, we will show that spaces with the super ADP fail to be Asplund in a rather strong way. Recall that the norm of a Banach space X is said to be ρ-rough for $0 \lt \rho\leqslant 2$ if

\begin{equation*} \limsup_{\|h\|\to 0} \frac{\|x+h\|+\|x-h\|-2\|x\|}{\|h\|}\geqslant \rho \end{equation*}

for all $x\in X$, see [Reference Deville, Godefroy and Zizler10, Section III.1] for background. Observe that the roughness of the norm is the extreme opposite of the Fréchet differentiability. It is known that the norm of X is ρ-rough if and only if every weak $^*$ slice of $B_{X^*}$ has diameter greater than or equal to ρ [Reference Deville, Godefroy and Zizler10, Proposition I.1.11]. Observe that spaces admitting a ρ-rough norm cannot be Asplund, as this norm is not Fréchet differentiable at any point. In particular, the following shows that super ADP spaces fail to be Asplund.

Proof. Fix $x\in S_X$ and δ > 0 and consider the weak $^*$ slice $S(B_{X^*},x,\delta)$. Let $x^*\in S(B_{X^*},x,\delta)$, pick $\varepsilon\in(0,\delta)$, and let $W:=\{y\in B_X\colon \left\lvert x^*(y)\right\rvert \lt \varepsilon\}$ (observe that W intersects S_X since $\dim(X) \gt 1$). As X has the super ADP and W is balanced (i.e. is such that $\mathbb{T} W=W$), there exists $y\in W$ such that $\left\lVert x+y\right\rVert \gt 2-\varepsilon$. Let $y^*\in S_{X^*}$ be such that $\operatorname{Re} y^*(x+y) \gt 2-\varepsilon$. On the one hand, we have $\operatorname{Re} y^*(x) \gt 1-\varepsilon \gt 1-\delta$, so $y^*\in S(B_{X^*},x,\delta)$. On the other hand,

\begin{equation*} \left\lVert x^*-y^*\right\rVert\geqslant \operatorname{Re} \left\langle x^*-y^*,-y\right\rangle\geqslant \operatorname{Re} y^*(y)- |x^*(y)| \gt 1-2\varepsilon. \end{equation*}

4. Super alternative Daugavet points and related notions

We devote this section to give the main examples and properties of super AD points, obtaining also results for AD points. This section is divided into two subsections: the first one is devoted to providing examples and comparisons with other localization notions; the second subsection will deal with the implications of super AD points on the geometry of the underlying Banach space.

Let us first provide a geometric characterization of the AD points and super AD points, which we will use throughout the section. The following characterizations of AD points follow directly from the proof of the global characterization of the ADP given in [Reference Martín and Oikhberg23, Proposition 2.1]. For every $x\in S_X$ and $\varepsilon$ > 0, let

\begin{equation*} \Delta_\varepsilon(x):=\{y\in B_X\colon \left\lVert x-y\right\rVert \gt 2-\varepsilon\}. \end{equation*}

We do not know if there is an operator characterization for super AD points (which would give a super AD points version of (ii) and (iii) above). However, let us point out that the natural analogue to condition (iv) above for super AD points (replacing slices with weak open sets) is immediate, because the sets of the form $\{y\in B_X\colon \left\lVert x+\theta y\right\rVert \gt 2-\varepsilon\}$ are relatively weakly open. On the other hand, condition (v) above should be replaced with a super AD point x with the weak denseness of the set $\mathbb{T}\Delta_{\varepsilon}(x)$. Equivalently, we can provide a net characterization of super AD points, which will be useful.

4.1. Examples and comparison with other localization notions

Our first goal is to discuss the relation between spears and super AD points. It is immediate that spear vectors are super AD points. The converse holds for spaces with the Kadec property (as it follows from the proof of Proposition 3.1).

Proposition 4.3. Let $X$ be a Banach space and let $x\in S_X$. If $x\in \operatorname{Spear}(X)$, then $x$ is a super AD point. Moreover, if $X$ has the Kadec property (in particular, if $X$ is finite-dimensional or $X$ is asymptotically uniformly convex), then the set of super AD points coincides with $\operatorname{Spear}(X)$.

This result easily allows us to determine the super AD points in some concrete spaces, by using the description of the set of spear vectors given in [Reference Kadets, Martín, Merí and Pérez19, Example 2.12]. Recall that the space $\ell_1(\Gamma)$ is asymptotically uniformly convex, and hence has the Kadec property for every non-empty set Γ.

Example 4.4. We give the description of the super AD points in some spaces:

1. (1) The super AD points of $\ell_\infty^m$ ( $m\in \mathbb{N}$) are the elements whose coordinates have all modulus one.

2. (2) The super AD points of $\ell_1(\Gamma)$ (in particular, of $\ell_1^m$ ( $m\in \mathbb{N}$) and of $\ell_1$) are the elements whose coordinates are zero except for one, which has modulus one (that is, all rotations of the elements of the canonical basis).

Recall that, since $\ell_\infty^m$ and $\ell_1(\Gamma)$ have the ADP, all norm one elements of these spaces are AD points.

The following example shows that the ADP does not necessarily imply the existence of any super AD point.

Example 4.5. The space c ₀ has the ADP [Reference Martín and Oikhberg23], and hence all points on its unit sphere are AD points. However, there are no super AD points in c ₀.

Proof. Indeed, fix $x_0=(x_n)_{n\in\mathbb{N}} \in S_{c_0}$ and find an index $N\in\mathbb{N}$ such that $\left\lvert x_n\right\rvert \lt 1/4$ whenever $n\geqslant N$. Consider the relatively weakly open set

\begin{equation*} W:= \{(y_n)\in B_{c_0}\colon \left\lvert y_n\right\rvert \lt 1/4, \; n=1,2,\dots,N\} \end{equation*}

and notice that for every $\theta \in\mathbb{T}$ and every element $y\in W$, we have

\begin{align*} \left\lVert x_0+\theta y\right\rVert = \sup_n \left\lvert x_n+\theta y_n\right\rvert \leqslant \sup_n \left\lvert x_n\right\rvert+\left\lvert y_n\right\rvert \lt 1+ \frac{1}{4} \lt 2. \end{align*}

(This is true, since if n < N, we have $\left\lvert y_n\right\rvert \lt 1/4$, and if $n\geqslant N$, then $\left\lvert x_n\right\rvert \lt 1/4$.) Hence x cannot be super AD.

We next would like to clarify the relations between AD and super AD points and the various localizations of the DP defined previously in the literature.

Clearly, every super Daugavet point is a super AD point, and every Daugavet point is an AD point. Let us note that the latter implication actually holds for $\nabla$-points.

Proposition 4.6. Let $X$ be a Banach space. If $x\in S_X$ is a $\nabla$-point, then $x$ is an AD point.

Proof. First, assume $X$ is a real Banach space. Consider a slice $S$ of $B_X$ and $\varepsilon$ > 0. We want to find $y\in S$ such that $\max\{\left\lVert x-y\right\rVert, \left\lVert x+y\right\rVert\} \gt 2-\varepsilon$. There are two cases: if $x$ belongs to $S$, we can take $y:=x$. In the other case, we can find the desired $y\in S$ simply by the fact that $x$ is a $\nabla$-point. If $X$ is a complex Banach space, then every $\nabla$-point is already a Daugavet point [Reference Lee, Roldán and Tag22, Proposition 2.2], hence an AD point.

An overview of the relations of both new and already known notions is given in Figure 1. Examples in [Reference Haller, Langemets, Perreau and Veeorg13, Reference Martín, Perreau and Rueda Zoca24] as well as the fact that the super ADP lies strictly between the ADP and the DP (Example 3.2 and Theorem 3.3), show that none of the above implications reverse and that super AD points are not necessarily Daugavet points and vice versa. Besides, we will provide an example showing that super AD points are not necessarily $\nabla$-points (Example 5.11).

Figure 1.

Relations between the notions.

Our next aim is to illustrate the difference between super AD and super Daugavet points, showing that the latter are super AD for all directions at the same time.

Proposition 4.7. Let $x\in S_X$ be a super Daugavet point. Then, for every $\varepsilon$ > 0 and every non-empty relatively weakly open subset $W$ of $B_X$, there exists $y\in W$ such that $\left\lVert x+\theta y\right\rVert \gt 2-\varepsilon$ for every $\theta\in\mathbb{T}$.

Proof. From the definition of super Daugavet points, it is immediate to show that for every $\theta\in\mathbb{T}$, there exists $y_\theta\in W$ such that $\left\lVert x+\theta y_\theta\right\rVert \gt 2-\varepsilon$ (it is enough to use that $\overline{\theta} x$ is super Daugavet). Let $\{\theta_i\colon i=1,\ldots,n\}$ be an $\varepsilon$-net in $\mathbb{T}$. Since the sets of the form

\begin{equation*} \Delta_{\varepsilon,\theta}(x):=\{y\in B_X\colon \left\lVert x+\theta y\right\rVert \gt 2-\varepsilon\} \end{equation*}

are relatively weakly open, by the argument above, we get that the set $W_1:=\Delta_{\varepsilon,\theta_1}(x)\cap W$ is a non-empty relatively weakly open subset of $B_X$. Hence the set $W_2:=\Delta_{\varepsilon,\theta_2}(x)\cap W_1$ is also a non-empty relatively weakly open subset, and iterating in this way, we get that the set $W\cap\bigcap_{i=1}^n\Delta_{\varepsilon,\theta_i}(x)$ is non-empty. Now, every y in this set satisfies $\left\lVert x+\theta_i y\right\rVert \gt 2-\varepsilon$ for every $i\in\{1,\dots,n\}$, hence $\left\lVert x+\theta y\right\rVert \gt 2-2\varepsilon$ for every $\theta\in \mathbb{T}$. The conclusion follows.

We can establish a similar statement for Daugavet points, which shows the difference with AD points.

Proposition 4.8. Let $X$ be a Banach space and let $x\in S_X$ be a Daugavet point. Then, for every slice $S\subset B_X$ and every $\varepsilon$ > 0, there exists a slice $T\subset S$ such that $\left\lVert x+\theta y\right\rVert \gt 2-\varepsilon$ for every $y\in T$ and every $\theta\in\mathbb{T}$.

Proof. Take a slice $S\subset B_X$, $\varepsilon$ > 0 and an $(\varepsilon/2)$-net $\{\theta_i\colon i=1,\ldots,n\}$ for $\mathbb{T}$. By [ Reference Jung and Zoca16, Remark 2.3] applied to $\overline{\theta_1} x$ which is a Daugavet point, we can find a slice $S_1\subset S$ satisfying $\left\lVert x+\theta_1y\right\rVert \gt 2-\varepsilon/2$ for any $y\in S_1$. Making use of [ Reference Jung and Zoca16, Remark 2.3] again, we can find for θ₂ a slice $S_2\subset S_1\subset S$ so that $\left\lVert x+\theta_2y\right\rVert \gt 2-\varepsilon/2$ for any $y\in S_2$. Using this method n times yields us a slice S_n such that

\begin{equation*} S_n\subset S_{n-1}\subset \dots \subset S_1\subset S \end{equation*}

and for any $y\in S_n$ we have $\left\lVert x+\theta_iy\right\rVert \gt 2-\varepsilon/2$. In conclusion, take $y\in T:=S_n$, $\theta\in\mathbb{T}$ and find $\theta_i\in\{1,\dots,n\}$ such that $\left\lvert\theta-\theta_i\right\rvert \lt \varepsilon/2$. Then,

\begin{equation*} \left\lVert x+\theta y\right\rVert\geqslant \left\lVert x+\theta_i y\right\rVert - \left\lVert\theta y-\theta_i y\right\rVert \gt 2-\frac{\varepsilon}{2}-\left\lvert\theta-\theta_i\right\rvert \gt 2-\varepsilon. \end{equation*}

According to [Reference Jung and Zoca16, Proposition 3.1] (respectively, [Reference Martín, Perreau and Rueda Zoca24, Lemma 3.7]), Daugavet points (respectively, super Daugavet points) are at distance 2 from every denting point (respectively, point of continuity) of the unit ball. Actually, the same is true for $\nabla$-points [Reference Haller, Langemets, Perreau and Veeorg13, Proposition 2.6]. With AD points, we have a similar (weaker) result whose proof is straightforward.

Lemma 4.9. Let $X$ be a Banach space and let $x\in S_X$ be an AD point. Then, for each $y\in \operatorname{dent}(B_X)$ there exists $\theta\in\mathbb{T}$ so that $\left\lVert x+\theta y\right\rVert=2$.

There is also a natural super AD counterpart of the previous lemma, also with straightforward proof.

Lemma 4.10. Let $X$ be a Banach space and let $x\in S_X$ be a super AD point. Then, for every $z\in \operatorname{PC}(B_X)\cap S_X$, there exists $\theta\in\mathbb{T}$ such that $\left\lVert x+\theta z\right\rVert=2$.

Note that the two previous results are actually characterizations of AD points and super AD points in spaces with the RNP and the CPCP, respectively. In the first case, this is because the unit ball is the closed convex hull of its denting points; in the latter case, because the points of continuity are weakly dense in the unit ball. We state these results here for future reference.

Corollary 4.11. Let $X$ be a Banach space.

1. (1) If $X$ has the RNP, then $x\in S_X$ is an AD point if (and only if) for every $y\in \operatorname{dent}(B_X)$ there exists $\theta\in\mathbb{T}$ so that $\left\lVert x+\theta y\right\rVert=2$.

2. (2) If $X$ has the CPCP, then $x\in S_X$ is a super AD point if (and only if) for every $z\in \operatorname{PC}(B_X)\cap S_X$, there exists $\theta\in\mathbb{T}$ such that $\left\lVert x+\theta z\right\rVert=2$.

We finish this subsection by proving that ‘ccs AD points’ are actually ccs Daugavet points. Therefore, the only interesting stronger version of AD point is the one of super AD point.

Proposition 4.12. Let $X$ be a Banach space and $x\in S_X$. Assume that for every convex combination of slices $C$ of $B_X$, we have that $\sup_{y\in C}\max_{\theta\in\mathbb{T}} \left\lVert x+\theta y\right\rVert=2$. Then, $x$ is a ccs Daugavet point.

Proof. Fix $\varepsilon$ > 0 and a convex combination of slices $C:=\sum_{i=1}^n\lambda_i S_i$, where $S_i=S(B_X,f_i,\alpha_i)$, $\lambda_i \gt 0$ for all $i=1,\ldots,n$, and $\sum_{i=1}^{n}\lambda_i=1$. It suffices to find $y\in C$ satisfying $\left\lVert x+y\right\rVert \gt 2-\varepsilon$. First, let us choose an $\varepsilon/2$-net $K:=\{\theta_1,\dots,\theta_m\}$ for $\mathbb{T}$. Consider the following convex combination of slices:

\begin{equation*} D:= \frac{1}{m}\sum_{j=1}^m\sum_{i=1}^n\lambda_i S(B_X,\theta_jf_i,\alpha_i). \end{equation*}

By hypothesis, we can find an element $d\in D$ and $\theta_0\in\mathbb{T}$, for which $\left\lVert x+\theta_0d\right\rVert \gt 2-\varepsilon/4m$. Write

\begin{equation*} d=\frac{1}{m}\sum_{j=1}^m\sum_{i=1}^n\lambda_i s^j_i, \end{equation*}

where $s^j_i\in S(B_X,\theta_jf_i,\alpha_i)$. Evidently, in this case, we have the inclusion $\overline{\theta_j}s^j_i\in S(B_X,f_i,\alpha_i)$. This implies that $\sum_i\lambda_i\overline{\theta_j}s^j_i\in C$ for each $j\in\{1,\dots,m\}$. Find $\theta_{N}\in K$ such that $\bigl|\theta_0-\overline{\theta_{N}}\bigr| \lt \varepsilon/2$. We now show that we can take $y:=\sum_i\lambda_i\overline{\theta_{N}}s^{N}_i\in C$, for which $\left\lVert x+y\right\rVert \gt 2-\varepsilon$. Assume, on the contrary, that $\left\lVert x+y\right\rVert\leqslant 2-\varepsilon$. Then

\begin{align*} 2-\frac{\varepsilon}{4m}& \lt \left\lVert x+\theta_0 d\right\rVert= \left\lVert x+\frac{\theta_0}{m}\sum_{j=1}^m\sum_{i=1}^n\lambda_i s^j_i\right\rVert\\ &\leqslant \frac{\left\lVert x+\theta_0\sum_{i=1}^n\lambda_is^{N}_i\right\rVert}{m}+\sum_{j\neq N}\frac{\left\lVert x+\theta_0\sum_{i=1}^n\lambda_is^j_i\right\rVert}{m}\\ &\leqslant \frac{\left\lVert x+\theta_0\theta_N\sum_{i=1}^n\lambda_i\overline{\theta_N}s^{N}_i\right\rVert}{m} +\frac{2(m-1)}{m}=\frac{\left\lVert x+\theta_0\theta_Ny\right\rVert}{m}+\frac{2(m-1)}{m}. \end{align*}

We can estimate the first term:

\begin{align*} \left\lVert x+\theta_0\theta_Ny\right\rVert&\leqslant \left\lVert x+y\right\rVert+\left\lVert\theta_0\theta_Ny-y\right\rVert\\ &\leqslant2-\varepsilon + \left\lvert\theta_0\theta_N-1\right\rvert=2-\varepsilon+\bigl|\theta_0-\overline{\theta_{N}}\bigr| \lt 2-\frac{\varepsilon}{2}. \end{align*}

Therefore, we conclude that

\begin{align*} 2-\frac{\varepsilon}{4m}& \lt \frac{\left\lVert x+\theta_0\theta_Ny\right\rVert}{m}+\frac{2(m-1)}{m}\\ & \lt \frac{2-\varepsilon/2}{m}+\frac{2(m-1)}{m}=\frac{2m}{m}-\frac{\varepsilon}{2m}= 2-\frac{\varepsilon}{2m}, \end{align*}

which is a contradiction.

We get the following characterization of the DP (compare it with the one in [Reference Kadets17]).

Corollary 4.13. Let $X$ be a Banach space. Suppose that for every $x\in S_X$ and every convex combination of slices $C$ of $B_X$ we have that $\sup_{y\in C}\max_{\theta\in\mathbb{T}} \left\lVert x+\theta y\right\rVert=2$. Then, $X$ has the DP.

Remark 4.14. If X has the strong diameter two property (that is, every convex combination of slices has diameter two), then every convex combination of slices almost reaches the unit sphere [Reference López, Martín and Rueda Zoca21, Theorem 3.1]. Therefore, every spear in X is a ‘ccs AD point’, hence a ccs Daugavet point by Proposition 4.12.

4.2. Implications of AD points and super AD points on the geometry of the underlying space

We first investigate the relations between rotundity properties of norms and AD points. To begin with, let us show that LUR points are never AD points.

Proposition 4.15. Let $X$ be a Banach space with $\dim(X) \gt 1$ and let $x\in S_X$ be a LUR point. Then, $x$ is not an AD point.

We need the following preliminary result.

Lemma 4.16. Let $X$ be a Banach space with $\dim(X) \gt 1$ and let $x\in S_X$. If there exist $\varepsilon,\delta \gt 0$ such that $\Delta_{\varepsilon}(x)\subset B(-x,1-\delta)$, then $x$ is not an AD point.

Proof. To show that x is not an AD point, it suffices to find $f\in S_{X^*}$ and $\eta,\xi \gt 0$ such that for all $y\in S_X$, if $\left\lvert f(y)\right\rvert \gt 1-\eta$, then $y\notin\Delta_\xi(x)$. Assume that we have $\varepsilon,\delta \gt 0$ such that $\Delta_{\varepsilon}(x)\subset B(-x,1-\delta)$. Since $\dim(X) \gt 1$, we can find $f\in S_{X^*}$ such that $f(x)=0$. Take $y\in S_X$ such that $\left\lvert f(y)\right\rvert \gt 1-\delta$. Then

\begin{equation*} \left\lVert-x-y\right\rVert\geqslant \left\lvert f(x+y)\right\rvert = \left\lvert f(y)\right\rvert \gt 1-\delta. \end{equation*}

This means that $y\not\in B(-x,1-\delta)$, hence $y\not\in \Delta_{\varepsilon}(x)$, which concludes the proof.

Proof of Proposition 4.15.

Fix $\varepsilon\in(0,1)$. By definition of LUR, there exists δ > 0 such that every $y\in B_X$ satisfies the implication

\begin{equation*} \left\lVert\frac{x+y}{2}\right\rVert \gt 1-\delta \implies \left\lVert x-y\right\rVert\leqslant \varepsilon. \end{equation*}

Let us show that Lemma 4.16 holds for x. Take any $\varepsilon_1\in(0,2\delta)$ and $\delta_1\in(0,1-\varepsilon)$. Then we claim that $\Delta_{\varepsilon_1}(x)\subset B(-x,1-\delta_1)$. Indeed, if $y\in B_X$ is such that $\left\lVert x-y\right\rVert \gt 2-\varepsilon_1$, then

\begin{equation*} \left\lVert\frac{x-y}{2}\right\rVert \gt 1-\frac{\varepsilon_1}{2} \gt 1-\delta, \end{equation*}

and therefore

\begin{equation*} \left\lVert-x-y\right\rVert= \left\lVert x+y\right\rVert\leqslant \varepsilon \leqslant 1-\delta_1. \end{equation*}

As a consequence, we get the following result, related to [Reference Kadets, Martín, Merí and Payá20, Proposition 2.4], where it is shown that spaces with the ADP do not contain LUR points.

Corollary 4.17. Let $X$ be a Banach space with $\dim(X) \gt 1$ which is LUR (in particular, if it is uniformly convex). Then, $X$ does not have any AD point.

We now deal with the relation between super AD points and differentiability of the norm. The following result can be immediately extracted from the proof of Proposition 3.5.

Proposition 4.18. Let $X$ be an infinite-dimensional Banach space such that $\operatorname{PC}(B_X)$ is weakly dense in $B_X$ (for instance, if $X$ has the CPCP). If $x\in S_X$ is a point of Gâteaux differentiability of the norm of $X$, then $x$ is not a super AD point.

Observe that the analogous result for AD points is not true: $\ell_1$ has the RNP (hence, the CPCP), and has a dense subset of points of Gâteaux differentiability of the norm (as being separable), and has the ADP, hence all elements in $S_{\ell_1}$ are AD points.

We will finally show that Example 4.5 can be generalized to any asymptotically smooth space, for which we will introduce a bit of notation. Let X be a Banach space. We denote by $\operatorname{cof}(X)$ the set of all subspaces of X of finite codimension. Recall that the modulus of asymptotic smoothness of X at a point $x\in S_X$ is the function $\overline{\rho}_{X}\left( x,\cdot \right)\colon \mathbb{R}^+\to \mathbb{R}^+$ given by

\begin{equation*} \overline{\rho}_{X}\left( x,t \right):=\inf_{Y\in\operatorname{cof}(X)}\sup_{y\in S_Y}\{\left\lVert x+ty\right\rVert-1\}. \end{equation*}

It is easy to check that for a given $t\in\mathbb{R}^+$, $\overline{\rho}_{X}\left( x,t \right)$ is the best constant ρ for which we have the following: for every weakly null net $(x_\alpha)$ in S_X, $\limsup\left\lVert x+t x_\alpha\right\rVert\leqslant 1+\rho$ (see e.g. [Reference Perreau25, Corollary 1.1.6]). The modulus of asymptotic smoothness of the space X is the function ${\overline\rho}_X\left(\cdot\right):\mathbb{R}^+\rightarrow{\mathbb{R}}^+$ given by

\begin{equation*} \overline{\rho}_{X}\left( t \right):=\sup_{x\in S_X}\overline{\rho}_{X}\left( x,t \right). \end{equation*}

A point $x\in S_X$ is said to be asymptotically smooth if $\lim_{t\to 0}\frac{\overline{\rho}_{X}\left( x,t \right)}{t}=0$. The space X is said to be asymptotically uniformly smooth (AUS, for short) if $\lim_{t\to 0}\frac{\overline{\rho}_{X}\left( t \right)}{t}=0$. Prototypical examples of AUS spaces are $\ell_p$-sums (p > 1) and c ₀-sums of finite-dimensional spaces. We get that super AD point fails to be asymptotically smooth in the strongest possible way.

Proposition 4.19. Let $X$ be an infinite-dimensional Banach space. If $x\in S_X$ is a super AD point, then $\overline{\rho}_{X}\left( x,t \right)=t$ for every $t\in (0,1)$, hence $x$ is not asymptotically smooth (in particular, $x$ is not a point of Fréchet differentiability of the norm).

Proof. Assume that x is a super AD point. Then, by Lemma 4.2 there exists a net $(x_\alpha)$ in S_X which converges weakly to 0 and such that $\lim_\alpha\left\lVert x+x_\alpha\right\rVert=2$. Then, by convexity, $\lim_\alpha\left\lVert x+tx_\alpha\right\rVert=1+t$ for every $t\in (0,1)$, and this implies that $\overline{\rho}_{X}\left( x,t \right)=t$. Hence, x is not an asymptotically smooth point.

As a consequence, we extend the class of Banach spaces that fail to have super AD points.

Corollary 4.20. Let $(E_n)$ be a sequence of finite-dimensional spaces and let $p\in(1,\infty)$. Then, the spaces $\left[\bigoplus_{n\in\mathbb{N}}E_n\right]_{\ell_p}$ and $\left[\bigoplus_{n\in\mathbb{N}}E_n\right]_{c_0}$ have no super AD points.

5. Super AD points and the super ADP in some classical Banach spaces

In this section we aim at characterizing super AD points and the super ADP in some Banach spaces, namely spaces $L_1(\mu)$ and C(K), as well as their vector-valued versions $L_1(\mu,X)$ and $C(K,X)$. We will first deal with super AD points in $\S$ 5.1 (for $\ell_1$- and $\ell_\infty$-sums) and 5.2 (for $L_1(\mu,X)$ and $C(K,X)$), and then apply our results to the global property in § 5.3.

5.1. Super AD points in $\ell_1$- and $\ell_\infty$-sums of Banach spaces

We start by gathering some results concerning super AD points in $\ell_1$- and $\ell_\infty$-sums of Banach spaces.

Proposition 5.1. Let $X$ and $Y$ be Banach spaces, let $x\in S_X$ and $y\in S_Y$ and let $a,b \gt 0$ with $a+b=1$. Then, we have the following implications:

1. (1) If $x$ is a super AD point in $X$, then $(x,0)$ is a super AD point in $X\oplus_1 Y$.

2. (2) If $x$ is a super AD point in $X$ and $y$ is a super Daugavet point in $Y$, then $(ax,by)$ is a super AD point in $X\oplus_1 Y$.

Proof. The first implication is straightforward, so we will only prove the second one. Assume that x is super AD and that y is super Daugavet, and pick $(u,v)\in B_{X\oplus_1 Y}$. We first suppose that $u,v\neq 0$. (If either u = 0 or v = 0, the proof is straightforward). Since x is super AD, there exist $\theta\in \mathbb{T}$ and a net $(u_\alpha)$ in B_X such that $(u_\alpha)$ converges weakly to $\frac{u}{\left\lVert u\right\rVert}$ and $\left\lVert x+\theta u_\alpha\right\rVert\to 2$. Since y is super Daugavet, we can find a net $(v_\alpha)$ in B_Y which converges weakly to $\frac{v}{\left\lVert v\right\rVert}$ and such that $\left\lVert y+\theta v_\alpha\right\rVert\to 2$. Then the net $(\left\lVert u\right\rVert u_\alpha,\left\lVert v\right\rVert v_\alpha)$ lives in $B_{X\oplus_1 Y}$, converges weakly to (u, v), and is such that

\begin{equation*} \bigl\|(ax,by)+\theta(\left\lVert u\right\rVert u_\alpha,\left\lVert v\right\rVert v_\alpha)\bigr\| =\left\lVert ax+\theta\left\lVert u\right\rVert u_\alpha\right\rVert+\left\lVert by+\theta\left\lVert v\right\rVert v_\alpha\right\rVert\to a+\left\lVert u\right\rVert+b+\left\lVert v\right\rVert=2 \end{equation*}

(where we have used Remark 2.2). Hence (ax, by) is a super AD point in $X\oplus_1 Y$ by Lemma 4.2.

Let us give a comment on the limitations of the above proof.

Remark 5.2. Since $\ell_1\equiv \ell_1 \oplus_1 \ell_1$, Example 4.4 shows that, in general, it is not enough to assume that x and y are super AD points in order to get that (ax, by) is a super AD point in $X\oplus_1 Y$. A simple example is given by $\ell_1^2\equiv \mathbb{K}\oplus_1 \mathbb{K}$.

Note that the reason why the above argument does not work in this context is that applying the super AD net characterization to $\{x,u\}$ and $\{y,v\}$ in the above proof can provide two different modulus one scalars $\theta,\omega\in\mathbb{T}$ for which the weakly convergent nets are far from the super AD points in the respective unit balls.

In the other direction, things behave as we expect them to.

Proposition 5.3. Let $X$ and $Y$ be Banach spaces, let $x\in S_X$ and $y\in S_Y$, and $a,b \gt 0$ with $a+b=1$. Then, we have the following implications:

1. (1) If $(x,0)$ is a super AD point, then $x$ is a super AD point.

2. (2) If $(ax; by)$ is a super AD point, then $x$ and $y$ are super AD points.

Proof. Assume that x is not super AD. Then there exists $u\in S_X$ such that for every net $(u_\alpha)$ in B_X which converges weakly to u and for every $\theta\in\mathbb{T}$, we have $\limsup\left\lVert x+\theta u_\alpha\right\rVert \lt 2$. Let $(u_\alpha,v_\alpha)$ be a net in $B_{X\oplus_1 Y}$ which converges weakly to $(u,0)$. Then, $(u_\alpha)$ converges weakly to u in B_X. Therefore, by the weak lower semi-continuity of the norm, we have that $\liminf\left\lVert u_\alpha\right\rVert\geqslant \left\lVert u\right\rVert=1$. Hence $\left\lVert v_\alpha\right\rVert\to 0$. Furthermore, by our assumption, we have that for every $\theta\in\mathbb{T}$, $\limsup\left\lVert x+\theta u_\alpha\right\rVert \lt 2$. So fix $\theta\in\mathbb{T}$. On one hand, we have

\begin{align*} \limsup \left\lVert(x,0)+\theta(u_\alpha,v_\alpha)\right\rVert & =\limsup (\left\lVert x+\theta u_\alpha\right\rVert+\left\lVert v_\alpha\right\rVert)\leqslant \limsup\left\lVert x+\theta u_\alpha\right\rVert\\ &\quad + \limsup\left\lVert v_\alpha\right\rVert \lt 2, \end{align*}

so $(x,0)$ is not a super AD point. On the other hand, we have

\begin{align*} \limsup \left\lVert(ax,by)+\theta(u_\alpha,v_\alpha)\right\rVert &=\limsup (\left\lVert ax+\theta u_\alpha\right\rVert+\left\lVert by+\theta v_\alpha\right\rVert) \\ &\leqslant (1-a)\limsup\left\lVert u_\alpha\right\rVert+a\limsup\left\lVert x+\theta u_\alpha\right\rVert+b+\limsup \\ & \left\lVert v_\alpha\right\rVert \lt (1-a)+2a+b=2(a+b)=2, \end{align*}

so (ax, by) is not a super AD point.

We now deal with super AD points in $\ell_\infty$-sums.

Proposition 5.4. Let $X$ and $Y$ be Banach spaces, and let $x\in S_X$ and $y\in S_Y$. If $x$ and $y$ are super AD points, then ( $x$, $y$) is a super AD point in $X\oplus_{\infty}Y$.

Proof. Let $(u,v)\in S_{X\oplus_\infty Y}$. Then $\left\lVert u\right\rVert=1$ or $\left\lVert v\right\rVert=1$. In the first case, since x is super AD, there exista net $(u_\alpha)$ in B_X and $\theta\in\mathbb{T}$ such that $(u_\alpha)$ converges weakly to u and $\left\lVert u+\theta u_\alpha\right\rVert\to 2$ by Lemma 4.2. Then $(u_\alpha,v)$ converges weakly to (u, v) in $B_{X\oplus_\infty Y}$ and $\left\lVert(u,v)+\theta(u_\alpha,v)\right\rVert\geqslant \left\lVert u+\theta u_\alpha\right\rVert\to 2$. The other case is analogous, using that y is super AD.

When one of the spaces is infinite-dimensional, being super AD point in the $\ell_\infty$-sum becomes easier.

Proposition 5.5. Let $X$ be an infinite-dimensional Banach space and let $Y$ be an arbitrary Banach space. If $x$ is a super AD point in $X$, then ( $x$, $y$) is a super AD point in $X\oplus_{\infty} Y$ for every $y\in B_Y$.

Proof. The proof is analogous to the above one, but it is not required here to work by cases, because the assumption $\dim(X)=\infty$ allows to produce the desired net in B_X for every $u\in B_X$ by the moreover part of Lemma 4.2, and not merely for unit sphere elements.

For the converse result, we have the following two versions. The first one deals with the case when one of the factors is finite-dimensional.

Proposition 5.6. Let $X$ and $Y$ be Banach spaces. If $(x,y)\in S_{X\oplus_\infty Y}$ is a super AD point in $X\oplus_{\infty}Y$ and $Y$ is finite-dimensional, then $x$ is a super AD point (and, in particular, $\|x\|=1$).

Proof. Assume that x is not a super AD point. Then there exists $u\in S_X$ such that for every net $(u_\alpha)$ in B_X which converges weakly to u and for every $\theta\in\mathbb{T}$, we have $\limsup_\alpha\left\lVert x+\theta u_\alpha\right\rVert \lt 2$. Pick a net $(u_\alpha,v_\alpha)$ in $B_{X\oplus_\infty Y}$ which converges weakly to $(u,0)$. Then $(u_\alpha)$ converges weakly to u, and $(v_\alpha)$ converges weakly to 0. Since $\dim(Y) \lt \infty$, it follows that $\left\lVert v_\alpha\right\rVert\to 0$. Furthermore, from the above, we have $\limsup_\alpha\left\lVert x+\theta u_\alpha\right\rVert \lt 2$ for every $\theta\in \mathbb{T}$. Hence

\begin{align*} \limsup_\alpha\left\lVert(x,y)+\theta(u_\alpha,v_\alpha)\right\rVert & =\limsup_\alpha\max\{\left\lVert x+\theta u_\alpha\right\rVert,\left\lVert y+\theta v_\alpha\right\rVert\} \\ &\leqslant \max\left\{\limsup_\alpha\left\lVert x+\theta u_\alpha\right\rVert,\left\lVert y\right\rVert\right\} \lt 2. \end{align*}

Therefore, (x, y) is not a super AD point in $X\oplus_\infty Y$.

Note that as a consequence, we can obtain the description of the super AD points in $\ell_\infty^m$ which we already obtained in Example 4.4 using spear vectors: the set of super AD points in $\ell_\infty^m$ is equal to the set $\mathbb{T}^m$ for every $m\in \mathbb{N}$.

In the general case for the converse result, we obtain that if a pair is a super AD point of an $\ell_\infty$-sum, then at least one of the coordinates has to be a super AD point. The proof is analogous to the one given for Proposition 5.6.

Proposition 5.7. Let $X$ and $Y$ be Banach spaces. If $(x,y)\in S_{X\oplus_\infty Y}$ is a super AD point in $X\oplus_{\infty}Y$, then $x$ is a super AD point in $X$ or $y$ is a super AD point in $Y$.

5.2. Super AD points in spaces of integrable and of continuous functions

As previously mentioned, in the classical function spaces $L_1(\mu)$ and C(K), every point in the unit sphere is an AD point (since these spaces have the ADP), but these points do not always satisfy any stronger notion. In this subsection, we study in detail super AD points in these spaces and in their vector-valued counterparts.

We start with the case of spaces of (vector-valued) integrable functions. Let $(S,\Sigma,\mu)$ be a measure space. Recall that a measurable set $A\subset S$ is called an atom for µ if $\mu(A) \gt 0$ and if $\mu(B)=0$ for every measurable subset $B\subset A$ such that $\mu(B) \lt \mu(A)$. It was proved in [Reference Martín, Perreau and Rueda Zoca24, Theorem 4.8] that Daugavet and super Daugavet points coincide in the space $L_1(\mu)$, and that these points are precisely the norm one functions in $L_1(\mu)$ whose support contains no atom. As we have already seen, the situation is different for super AD points, as $\ell_1$ contains points which are super AD but does not contain any Daugavet point. Let X be a Banach space. Recall that a measurable function f from S to X is almost everywhere constant on every atom. We denote by $L_1(\mu,X)$ the Banach space of all X-valued µ-integrable functions. We can decompose $L_1(\mu,X)$ as

(5.1)\begin{equation} L_1(\mu,X)=L_1(\nu,X)\oplus_1 \ell_1(\Gamma,X), \end{equation}

where ν is the continuous part of µ, Γ is the set of all atoms for µ (up to a measure 0 set), and $\ell_1(\Gamma,X)=\left[\bigoplus_{\gamma\in\Gamma}X\right]_{\ell_1}$ stands for the $\ell_1$-sum of Γ copies of the space X, see [Reference Johnson14, Theorem 2.1] for the scalar-valued case, the vector-valued case is a consequence of the latter result. The following result is an immediate corollary of [Reference Martín, Perreau and Rueda Zoca24, Theorem 4.11] together with the stability results under $\ell_1$-sums for Daugavet and super Daugavet points (see [Reference Abrahamsen, Haller, Lima and Pirk3, Section 4] and [Reference Martín, Perreau and Rueda Zoca24, Section 3.2]).

Proposition 5.8. Let $(S,\Sigma,\mu)$ be a measure space, let $X$ be a Banach space and let $f\in S_{L_1(\mu,X)}$. Then, $f$ is a Daugavet (respectively, super Daugavet) point if and only if $f(s)=0$ or $\frac{f(s)}{\left\lVert f(s)\right\rVert}$ is a Daugavet (respectively, super Daugavet) point almost everywhere on every atom.

From the results from the previous subsection, we deduce in a similar way the following characterization for super AD points.

Proposition 5.9. Let $(S,\Sigma,\mu)$ be a measure space, let $X$ be a Banach space and let $f\in S_{L_1(\mu,X)}$. Then, $f$ is a super AD point if and only if $f(s)=0$ or $\frac{f(s)}{\left\lVert f(s)\right\rVert}$ is a super Daugavet point almost everywhere on all but one atom, where $\frac{f(s)}{\left\lVert f(s)\right\rVert}$ is a super AD point.

Proof. Observe that the function f can be seen as the element $(f_c,(f_\gamma)_{\gamma\in \Gamma})$ in the decomposition (5.1), where f_c corresponds to the restriction of the function f to the continuous part of S and f_γ corresponds to the a.e. value of f on the atom γ for every $\gamma\in\Gamma$. Note that since ν is atomless, we have that either $f_c=0$ or $\frac{f_c}{\left\lVert f_c\right\rVert}$ is a super Daugavet point in $L_1(\nu,X)$. Therefore, using Propositions 5.1 and 5.3, we immediately infer that f is a super AD point if and only $f_\gamma=0$ or $\frac{f_\gamma}{\left\lVert f_\gamma\right\rVert}$ is a super Daugavet point for all $\gamma\in\Gamma$ except one $\gamma_0\in \Gamma$ for which $\frac{f_{\gamma_0}}{\left\lVert f_{\gamma_0}\right\rVert}$ is a super AD point.

In the scalar-valued case, as $X=\mathbb{K}$ contains no (super) Daugavet points, we get the following result.

Corollary 5.10. Let $(S,\Sigma,\mu)$ be a measure space and let $f\in S_{L_1(\mu)}$. Then, $f$ is a super AD point if and only if $f$ vanishes almost everywhere on all but one atom.

Note that this later result also allows us to give a concrete example of a super AD point which is not a $\nabla$-point.

Example 5.11. Let $(S,\Sigma,\mu)$ be a measure space which has non-trivial continuous and atomic parts, then the space $L_1(\mu)$ contains super AD points which are not $\nabla$-points. Indeed, take any norm one function f whose support contains a non-trivial continuous part and a single atom. On the one hand, f is a super AD point by the above corollary. On the other hand, in the real case, by [Reference Haller, Langemets, Perreau and Veeorg13, Proposition 3.7] $\nabla$-points in $L_1(\mu)$ are either Daugavet points (i.e. elements whose support contains no atoms [Reference Martín, Perreau and Rueda Zoca24, Theorem 4.8]), or elements of the form $f =\theta\mu(A)^{-1}\chi_A$ with $\theta\in\{-1,1\}$ and A is an atom for µ. In the complex case, $\nabla$-points are Daugavet points [Reference Lee, Roldán and Tag22, Proposition 2.2].

Now we move to the context of C(K)-spaces. Let K be an infinite compact Hausdorff space. From [Reference Martín, Perreau and Rueda Zoca24, Corollary 4.3], [Reference Martín, Perreau and Rueda Zoca24, Theorem 4.2] and [Reference Abrahamsen, Haller, Lima and Pirk3, Theorem 3.4] we get that the notions of Daugavet, super Daugavet and ccs Daugavet points are equivalent in C(K) and that a function f belong to this class if and only if it attains its norm at a cluster point of K. We first prove that, in this setting, super AD points also fall in the same category.

Proposition 5.12. Let $K$ be an infinite compact Hausdorff space. If $f\in S_{C(K)}$ does not attain its norm at a cluster point of $K$, then $f$ is not a super AD point.

Proof. Let $H:= \{t \in K\colon \left\lvert f(t)\right\rvert=1\}$. By assumption, every $t\in H$ is an isolated point of K, so we can find $\delta\in(0,1)$ such that $\left\lvert f(t)\right\rvert\leqslant 1-\delta$ whenever $t\in K\setminus H$ (otherwise we could construct some $t\in H$ as a cluster point of elements from $K\setminus H$). Furthermore, the singleton $\{t\}$ is an open subset of K for every $t\in H$ so, by compactness, it follows that the set H is finite. Hence, the set

\begin{equation*} W:= \{g\in B_{C(K)} \colon \left\lvert g(t)\right\rvert \lt \delta, \; t\in H\} \end{equation*}

is a relatively weakly open subset of $B_{C(K)}$. Take any $g\in W$ and $\theta \in \mathbb{T}$. Notice that, if $t\in H$, we have

\begin{equation*} \left\lvert f(t)+\theta g(t)\right\rvert\leqslant \left\lvert f(t)\right\rvert+\left\lvert g(t)\right\rvert \lt 1 +\delta \lt 2. \end{equation*}

On the other hand, if $t\in K\setminus{H}$:

\begin{equation*} \left\lvert f(t)+\theta g(t)\right\rvert\leqslant \left\lvert f(t)\right\rvert+\left\lvert g(t)\right\rvert \lt 1-\delta +1 = 2-\delta \lt 2. \end{equation*}

Therefore $\left\lVert f+\theta g\right\rVert \lt 2$, and f is not a super AD point.

Let K be an infinite compact Hausdorff space and let X be an infinite-dimensional Banach space, and let $C(K,X)$ be the space of all continuous X valued functions on K endowed with the supremum norm. Let $f\in S_{C(K,X)}$. By [Reference Martín, Perreau and Rueda Zoca24, Theorem 4.2], if the norm of f is attained at a cluster point of K, then f is a ccs Daugavet point so, in particular, a super AD point. Recall that if $t_0\in K$ is an isolated point of K, we obtain the decomposition $C(K,X) = C(K\setminus \{t_0\}, X)\oplus_{\infty} X$, where the isometry is given by the mapping $f \mapsto (f|_{K\setminus\{t_0\}},f(t_0))$. For functions attaining their norm only at isolated points of K, we have the following characterization of super AD points.

Proposition 5.13. Let $K$ be an infinite compact Hausdorff space and let $X$ be an infinite-dimensional Banach space. Assume that $f\in S_{C(K,X)}$ does not attain its norm at a cluster point of $K$. Then, $f$ is a super AD point if and only if $f(t_0)$ is a super AD point for some isolated point $t_0\in K$.

Proof. Observe that for every isolated point t ₀ in K, we can identify the function f with the element $(f|_{K\setminus\{t_0\}}, f(t_0))$ in the above decomposition. So from Proposition 5.5, we immediately get that if $f(t_0)$ is a super AD point, then f is also a super AD point. For the other direction note that, as in the previous proof, we can find $\delta\in(0,1)$ such that $\left\lVert f(t)\right\rVert\leqslant 1-\delta$ whenever $t\in K\setminus H$, and we have that the set $H:=\{t\in K\colon \left\lVert f(t)\right\rVert=1\}$ is finite. Iterating the above decomposition, we can write

\begin{equation*} C(K,X) = C(K\setminus H, X)\oplus_{\infty} Y , \end{equation*}

where $Y = \left[\bigoplus_{t\in H} X\right]_{\ell_{\infty}}$. In particular, f can be identified with the element $(f|_{K\setminus H},(f(t))_{t\in H})$ in this space. Since, $\left\lVert f|_{K\setminus H}\right\rVert \lt 1$, it follows from Proposition 5.7 that $(f(t))_{t\in H}$ is a super AD point in Y. Since H is finite, we get from the same proposition that $f(t_0)$ must be a super AD point for some $t_0\in H$.

6. Open questions

We end the paper with some related questions.

In Theorem 3.10, we proved that if an infinite-dimensional Banach space has the super ADP, then the norm is 1-rough. All the examples of super ADP spaces we know are actually 2-rough, hence we wonder if this is so in general.

By the geometric characterization of the DP in terms of convex combinations of slices, it follows that there is no Daugavet space which is strongly regular. In Theorem 3.9, we showed that there is no infinite-dimensional CPCP space with the super ADP property. This leaves open whether there could be a super ADP space, which is strongly regular.

Slicely countably determined Banach spaces (SCD for short) were introduced in [Reference Avilés, Kadets, Martín, Merí and Shepelska8] as a natural joint generalization of separable strongly regular (and in particular RNP) Banach spaces and separable Banach spaces not containing $\ell_1$ (and in particular separable Asplund spaces). It was proved there that separable Banach spaces with the DP fail to be SCD. So it is natural to ask the following.

As mentioned in the introduction, spaces with the DP do not embed into spaces with an unconditional basis. We do not know if this is the case for spaces with the super ADP.

Two more questions which remain open to the best of our knowledge are the following.

Observe that the first half of the question above is much weaker than Question 6.1, as octahedral spaces always contain copies of $\ell_1$, see e.g. [Reference Deville, Godefroy and Zizler10, Theorem III.2.5].