Summands in locally almost square and locally octahedral spaces

TL;DR

This paper investigates whether certain geometric properties of Banach spaces, such as local almost squareness and octahedrality, are inherited by component spaces from their absolute sum constructions.

Contribution

It provides new insights into the inheritance of geometric properties in Banach space sums, specifically analyzing conditions under which properties pass from sum spaces to summands.

Findings

Conditions identified for properties to pass from sum to summands

Counterexamples showing limitations of inheritance

Characterizations of spaces with these properties

Abstract

We study the question whether properties like local/weak almost squareness and local octahedrality pass down from an absolute sum $X\oplus_F Y$ to the summands $X$ and $Y$.

Equations

$$\Vert x+y\Vert \ge (1-\epsilon )(\Vert x\Vert +1)\forall x\in F.$$

$$\Vert sx+y\Vert \ge (1-\epsilon )(|s|\Vert x\Vert +1)\forall s\in \mathbb{R}.$$

$$\Vert x+y\Vert \ge (1-\epsilon )(|x^{\ast }(x)|+1)\forall x\in F.$$

$$1=F(|a|/(|a|+|b|),1-|a|/(|a|+|b|))=F(|a|/(|a|+|b|),|b|/(|a|+|b|)),$$

$$a,b\ge 0,F(a,b)=1\text{and}F(a,1)\le 1+\delta \Rightarrow b\ge 1-\epsilon .$$

$$a,b\ge 0,F(a,b)=1\text{and}F(a,1)\le 1+{\delta }_n\Rightarrow b\ge 1-2^{-n}.$$

$$F(\Vert u_n\Vert ,1)\le \frac{1}{2}(F(\Vert u_n\Vert ,\Vert y+v_n\Vert )+F(\Vert u_n\Vert ,\Vert y-v_n\Vert )\le 1+{\delta }_n$$

$$1+{\delta }_n\ge \Vert y+v_n\Vert \ge 2-\Vert y-v_n\Vert \ge 1-{\delta }_n\forall n\in \mathbb{N},$$

$$F(a,1)\le 1+\delta \Rightarrow b\ge 1-\epsilon ,$$

$$F(1,b)\le 1+\delta \Rightarrow a\ge 1-\epsilon .$$

$$r_F:=\inf \left\{a\in [0,1]:\exists b\ge 0F(a,b)=1\text{and}F(a+1,b)=2\right\}.$$

$$f(s)=h(s-1)\forall s\in [r_F,1],\text{where}h:=\frac{w}{r_F-1}.$$

$$\Vert x\pm z\Vert \ge \Vert x\pm u\Vert -\Vert u-u/\Vert u\Vert \Vert \ge 1+r_F-\frac{\epsilon }{2}-(1-\Vert u\Vert )\ge 2r_F-\epsilon .$$

$$s(X):=\sup \left\{s\in [0,2]:\forall x\in S_X\forall \epsilon >0\exists y\in S_X\Vert x\pm y\Vert \ge s-\epsilon \right\}.$$

$$F^{\ast }(c,d):=\sup \left\{|ac|+|bd|:(a,b)\in B_{({\mathbb{R}}^2,F)}\right\},$$

$$\Vert \frac{x_{i,k}}{\Vert x_{i,k}\Vert }+x_k\Vert \ge 2-2^{-k}\forall i=1,\dots ,n.$$

$$\Vert x_{i,k}+x_k\Vert \ge \Vert \frac{x_{i,k}}{\Vert x_{i,k}\Vert }+x_k\Vert -\Vert \frac{x_{i,k}}{\Vert x_{i,k}\Vert }-x_{i,k}\Vert \ge 2-2^{-k}-|1-\Vert x_{i,k}\Vert |.$$

$$\underset{k,\mathcal{U}}{\lim }\Vert x_i+y_k\Vert =\Vert x_i+y{\Vert }_{\mathcal{U}}\ge 2-\epsilon \forall i=1,\dots ,n.$$

$$\Vert x_i+\frac{y_{k_0}}{\Vert y_{k_0}\Vert }\Vert \ge \Vert x_i+y_{k_0}\Vert -\Vert y_{k_0}-\frac{y_{k_0}}{\Vert y_{k_0}\Vert }\Vert \ge 2-2\epsilon -|\Vert y_{k_0}\Vert -1|\ge 2-3\epsilon ,$$

$$\Vert z_i+tz{\Vert }_{\mathcal{U}}\ge |{\phi }_s(z_i)|+t\forall i\in \left\{1,\dots ,n\right\},\forall t>0.$$

$$d(X,Y):=\inf \left\{\Vert T\Vert \Vert T^{-1}\Vert :T:X\to Y\text{is an isomorphism}\right\}.$$

$$\Vert y_i+ty\Vert \ge (1-\delta )(|y^{\ast }(y_i)|/(1+\delta )+t)\forall i\in \left\{1,\dots ,n\right\},\forall t>0.$$

$$\Vert x_i+tz\Vert \ge \Vert x_i+t{T}^{-1}y\Vert -t|\Vert T^{-1}y\Vert -1|\ge \Vert y_i+ty\Vert -t\delta .$$

$$\Vert x_i+tz\Vert \ge \Vert (1+\delta )y_i+ty\Vert -\delta -t\delta \ge (1-\delta )(|x^{\ast }(x_i)|+t)-(1+t)\delta$$

$$\Vert x_i+tz\Vert \ge (1-\delta )|x^{\ast }(x_i)|+(1-\sqrt{\delta })t\ge (1-\epsilon )(|x^{\ast }(x_i)|+t).$$

Full text

Summands in locally almost square and locally octahedral spaces

Jan-David Hardtke

Abstract. We study the question whether properties like local/weak almost squareness and local octahedrality pass down from an absolute sum $X\oplus_{F}Y$ to the summands $X$ and $Y$.

††Keywords: absolute sums; almost square spaces; locally almost square spaces; octahedrality; local octahedrality; ultraproducts; Banach-Mazur distance††AMS Subject Classification (2010): 46B20

1 Introduction

First we fix some notation. Throughout this paper we denote by $X$, $Y$, etc. real Banach spaces. $X^{*}$ denotes the dual of $X$, $B_{X}$ its closed unit ball and $S_{X}$ its unit sphere.

Let us now begin by recalling the following definition (see [godefroy]): $X$ is called octahedral (OH) if the following holds: for every finite-dimensional subspace $F$ of $X$ and every $\varepsilon>0$ there is some $y\in S_{X}$ such that

[TABLE]

$\ell^{1}$ is the standard example of an octahedral space. In fact, a Banach space possesses an equivalent octahedral norm if and only if it contains an isomorphic copy of $\ell^{1}$ (see [deville]*Theorem 2.5, p. 106).

In the paper [haller3], the following weaker forms of octahedrality were introduced: $X$ is called locally octahedral (LOH) if for every $x\in X$ and every $\varepsilon>0$ there exists $y\in S_{X}$ such that

[TABLE]

$X$ is called weakly octahedral (WOH) if for every finite-dimensional subspace $F$ of $X$, every $x^{*}\in B_{X^{*}}$ and each $\varepsilon>0$ there is some $y\in S_{X}$ such that

[TABLE]

The motivation for these definitions was to give dual characterisations of the so called diameter-two-properties. For $x^{*}\in S_{X^{*}}$ and $\varepsilon>0$, the slice of $B_{X}$ induced by $x^{*}$ and $\varepsilon$ is the set $S(x^{*},\varepsilon):=\left\{z\in B_{X},x^{*}(z)>1-\varepsilon\right\}$. Following the terminology of [abrahamsen], a Banach space $X$ is said to have the local diameter-two-property (LD2P) if every slice of $B_{X}$ has diamter 2 and it is said to have the diameter-two-property (D2P) if every nonempty, relatively weakly open subset of $B_{X}$ has diameter 2. Finally, $X$ is said to have the strong diameter-two-property (SD2P) if every convex combination of slices of $B_{X}$ has diameter 2.

The following results were proved in [haller3]:

1. (a)

$X$ has the LD2P $\iff$ $X^{*}$ is LOH. 2. (b)

$X$ has the D2P $\iff$ $X^{*}$ is WOH. 3. (c)

$X$ has the SD2P $\iff$ $X^{*}$ is OH.

The result (c) was also proved independently in [becerra-guerrero2].

It is known that the three diameter-two-properties are indeed different. For example, it follows from the results on direct sums in [haller3] that $c_{0}\oplus_{2}c_{0}$ has the D2P but not the SD2P.

Concerning the nonequivalence of the LD2P and the D2P, it was proved in [becerra-guerrero4] that every Banach space containing an isomorphic copy of $c_{0}$ can be renormed such that the new space has the LD2P but its unit ball contains relatively weakly open subsets of arbitrarily small diameter.111Note that the abbreviation SD2P in [becerra-guerrero4] does not stand for “strong diameter-two-property” but for “slice diameter-two-property”, which coincides with the LD2P of [abrahamsen].

In [kubiak] it was proved that Cesàro function spaces have the D2P.

There are many equivalent formulations of the three octahedrality properties (see for instance [haller2, haller3]). We will recall only those which we need here (they can be found in [haller3]): a Banach space $X$ is octahedral if and only if for every $n\in\mathbb{N}$, all $x_{1},\dots,x_{n}\in S_{X}$ and every $\varepsilon>0$ there exists an element $y\in S_{X}$ such that $\lVert x_{i}+y\rVert\geq 2-\varepsilon$ for all $i=1,\dots,n$. $X$ is locally octahedral if and only if for every $x\in S_{X}$ and all $\varepsilon>0$ there exists $y\in S_{X}$ such $\lVert x\pm y\rVert\geq 2-\varepsilon$. We will use these characterisations later without further mention.

Now we come to the classes of almost square spaces and their relatives. In the paper [abrahamsen2], the following definitions were introduced. A real Banach space $X$ is said to be almost square (ASQ) if the following holds: for all $n\in\mathbb{N}$ and all $x_{1},\dots,x_{n}\in S_{X}$ there exists a sequence $(y_{k})_{k\in\mathbb{N}}$ in $B_{X}$ such that $\lVert y_{k}\rVert\to 1$ and $\lVert x_{i}+y_{k}\rVert\to 1$ for all $i=1,\dots,n$.

$X$ is called locally almost square (LASQ) if for every $x\in S_{X}$ there is a sequence $(y_{k})_{k\in\mathbb{N}}$ in $B_{X}$ such that $\lVert y_{k}\rVert\to 1$ and $\lVert x\pm y_{k}\rVert\to 1$.

$X$ is called weakly almost square (WASQ) if it fulfils the definition of an LASQ space with the additional condition that the sequence $(y_{k})_{k\in\mathbb{N}}$ converges weakly to zero.

Obviously, WASQ implies LASQ. It was shown in [abrahamsen2] that ASQ implies WASQ and that the converse of this statement does not hold, while it is not known whether LASQ is strictly weaker than WASQ.

The model example of an ASQ space is $c_{0}$ and it was further proved in [abrahamsen2] that every ASQ space contains an isomorphic copy of $c_{0}$ and, conversely, every separable Banach space containing an isomorphic copy of $c_{0}$ can be equivalently renormed to become ASQ. In [becerra-guerrero3] it was shown that the same holds true also for nonseparable spaces. Also, it was proved in [abrahamsen2] that if $X$ is ASQ, than $X^{*}$ is OH (i. e. $X$ has the SD2P). By [kubiak]*Propositions 2.5 and 2.6, LASQ spaces have the LD2P and WASQ spaces have the D2P.

Next we will recall the necessary basics on absolute sums. A norm $F$ on $\mathbb{R}^{2}$ is called absolute if $F(a,b)=F(\lvert a\rvert,\lvert b\rvert)$ for all $(a,b)\in\mathbb{R}^{2}$ and it is called normalised if $F(1,0)=1=F(0,1)$. If $F$ is an absolute, normalised norm on $\mathbb{R}^{2}$, and $X$ and $Y$ are two Banach spaces, then the absolute sum of $X$ and $Y$ with respect to $F$, denoted by $X\oplus_{F}Y$, is defined as the direct product $X\times Y$ equipped with the norm $\lVert(x,y)\rVert_{F}=F(\lVert x\rVert,\lVert y\rVert)$. $X\oplus_{F}Y$ is again a Banach space.

For every $1\leq p\leq\infty$, the $p$-norm $\lVert\cdot\rVert_{p}$ on $\mathbb{R}^{2}$ is an absolute, normalised norm and the corresponding sum is just the usual $p$-direct sum of two Banach spaces. We also note the following important facts (see for instance [bonsall]*p. 36, Lemmas 1 and 2): if $F$ is an absolute, normalised norm on $\mathbb{R}^{2}$, then we have for all $a,b,c,d\in\mathbb{R}$

1. 1.)

$\lvert a\rvert\leq\lvert c\rvert\ \mathrm{and}\ \lvert b\rvert\leq\lvert d\rvert\ \Rightarrow\ F(a,b)\leq F(c,d)$, 2. 2.)

$\lvert a\rvert<\lvert c\rvert\ \mathrm{and}\ \lvert b\rvert<\lvert d\rvert\ \Rightarrow\ F(a,b)<F(c,d)$, 3. 3.)

$\lVert(a,b)\rVert_{\infty}\leq F(a,b)\leq\lVert(a,b)\rVert_{1}$.

It follows in particular that $\lvert a\rvert,\lvert b\rvert\leq F(a,b)$ holds for all $a,b\in\mathbb{R}$.

We will also need the following (see [hardtke]): for every $t\in(-1,1)$ there exists a unique $f(t)\in(0,1]$ such that $F(t,f(t))=1$. We will call the function $f$ the upper boundary curve of $B_{(\mathbb{R}^{2},F)}$. It is even, concave (hence continuous), decreasing on $[0,1)$ and increasing $(-1,0]$. Thus it can be extended to a concave, continuous, even function on $[-1,1]$, which will also be denoted by $f$.

Octahedrality properties in $p$-direct sums were already studied in [haller3]. The following results were proved:

1. (i)

If $X$ or $Y$ is LOH/WOH/OH, then $X\oplus_{1}Y$ is LOH/WOH/OH. 2. (ii)

If $X$ and $Y$ are LOH/WOH, then $X\oplus_{p}Y$ is LOH/WOH for every $1<p\leq\infty$. 3. (iii)

If $X$ and $Y$ are OH, then $X\oplus_{\infty}Y$ is OH. 4. (iv)

$X\oplus_{p}Y$ is never OH for $1<p<\infty$ (provided that $X$ and $Y$ are nontrivial). 5. (v)

If $X\oplus_{p}Y$ is LOH/WOH for $1<p\leq\infty$, then $X$ and $Y$ are LOH/WOH. 6. (vi)

If $X\oplus_{\infty}Y$ is OH, then $X$ and $Y$ are OH.

Using their duality results, the authors of [haller3] also obtained corresponding results for diameter-two-properties in $p$-direct sums (see also [abrahamsen, acosta, becerra-guerrero, haller, lopez-perez] for previous results on diameter-two-properties in $p$-direct sums based on different methods). In [acosta] it was also proved that the LD2P and the D2P are stable under all absolute sums, and that $X$ has the LD2P/D2P if $\text{dim}(X)=\infty$, $X\oplus_{F}Y$ has the LD2P/D2P for some Banach space $Y$ and $(1,0)$ is an extreme point of $B_{(\mathbb{R}^{2},F)}$. Moreover, it was proved in [acosta] that $X\oplus_{F}Y$ does not have the SD2P if $F(1,1)<2$ and $(1,0)$, $(0,1)$ are extreme points of $B_{(\mathbb{R}^{2},F)}$ (see also [oja]).

In the recent paper [haller4] the stability of average roughness (which is a generalisation of octahedrality) with respect to absolute sums is investigated. The authors of [haller4] also introduce the notion of positive octahedrality for an absolute, normalised norm $F$ on $\mathbb{R}^{2}$, meaning that there exist $c,d\geq 0$ with $F(c,d)=1$ and $F(c+1,d)=F(c,d+1)=2$. They prove that $X\oplus_{F}Y$ is octahedral whenever $X$ and $Y$ are octahedral and $F$ is positively octahedral, and, conversely, if $X\oplus_{F}Y$ is octahedral for some nontrivial Banach spaces $X$ and $Y$, then $F$ has to be positively octahedral. Analogous results for the SD2P in absolute sums are also obtained in [haller4].

In [abrahamsen2] it is proved that the properties LOH and LASQ are stable under arbitrary (even infinite) absolute sums, and that WOH and WASQ are stable under all absolute sums which fulfil a simple density assumption, including in particular all finite absolute sums.

Also, the following results were obtained in [abrahamsen2] for any two nontrivial Banach spaces $X$ and $Y$.

1. (i)

For $1\leq p<\infty$, $X\oplus_{p}Y$ is LASQ/WASQ if and only if $X$ and $Y$ are LASQ/WASQ. 2. (ii)

$X\oplus_{\infty}Y$ is LASQ/WASQ/ASQ if and only if $X$ or $Y$ is LASQ/WASQ/

ASQ. 3. (iii)

For $1\leq p<\infty$, $X\oplus_{p}Y$ is never ASQ.

The purpose of this note is to extend these results by showing that (i) and (iii) also hold if we replace $\lVert\cdot\rVert_{p}$ by any absolute, normalised norm $F\neq\lVert\cdot\rVert_{\infty}$. We will also prove some results on summands in LOH spaces, which imply in particular that $X$ and $Y$ are LOH whenever $X\oplus_{F}Y$ is LOH and $F$ is strictly convex.

Finally, we will also discuss some results on ultrapowers of LOH, LASQ, etc. spaces and the closedness of these classes with respect to the Banach-Mazur distance.

2 Results and proofs

We start with the following lemma, which is surely well-known, but since the author was not able to find it explicitly in the literature, a proof is included here for the reader’s convenience.

Lemma 2.1**.**

Let $F$ be an absolute, normalised norm on $\mathbb{R}^{2}$.

1. (a)

$F(1,1)=1$* $\Leftrightarrow$ $F=\lVert\cdot\rVert_{\infty}$.* 2. (b)

$F(1,1)=2$* $\Leftrightarrow$ $F=\lVert\cdot\rVert_{1}$.*

Proof.

(a) Assume that $F(1,1)=1$. Let $(a,b)\in\mathbb{R}^{2}$ such that $F(a,b)=1$. Then $\lvert a\rvert,\lvert b\rvert\leq 1$. If both $\lvert a\rvert<1$ and $\lvert b\rvert<1$, then we would have $F(a,b)<F(1,1)=1$ (by the general monotonicity properties of absolute norms listed in Section 1). It follows that $\lvert a\rvert=1$ or $\lvert b\rvert=1$, hence $\lVert(a,b)\rVert_{\infty}=1$.

Thus we have $S_{(\mathbb{R}^{2},F)}\subseteq S_{(\mathbb{R}^{2},\lVert\cdot\rVert_{\infty})}$, which implies $F=\lVert\cdot\rVert_{\infty}$.

(b) Suppose that $F(1,1)=2$, i. e. the midpoint of $(0,1)$ and $(1,0)$ lies on the unit sphere of $(\mathbb{R}^{2},F)$. It follows that the whole line segment from $(0,1)$ to $(1,0)$ lies on $S_{(\mathbb{R}^{2},F)}$, thus $F(t,1-t)=1$ for every $t\in[0,1]$.

Hence we have for every $(a,b)\neq(0,0)$

[TABLE]

i. e. $F(a,b)=\lVert(a,b)\rVert_{1}$. ∎

Before we can come to the first main result on sums of LASQ (etc.) spaces, we have to prove another auxiliary lemma.

Lemma 2.2**.**

Let $F$ be an absolute, normalised norm on $\mathbb{R}^{2}$ with $F\neq\lVert\cdot\rVert_{\infty}$ and let $\varepsilon>0$. Then there is a $\delta>0$ such that the following holds:

[TABLE]

Proof.

Denote by $f$ the upper boundary curve of $B_{(\mathbb{R}^{2},F)}$. If the claim was false, then we could find sequences $(a_{n})_{n\in\mathbb{N}}$, $(b_{n})_{n\in\mathbb{N}}$ in $[0,\infty)$ such that $F(a_{n},b_{n})=1$, $F(a_{n},1)\leq 1+1/n$ and $b_{n}<1-\varepsilon$ for each $n\in\mathbb{N}$.

Since $a_{n},b_{n}\leq 1$ for every $n\in\mathbb{N}$, we can find subsequences $(a_{n_{k}})$, $(b_{n_{k}})$ such that $a_{n_{k}}\to a$ and $b_{n_{k}}\to b$ for some $a,b\in[0,1]$. It follows that $F(a,b)=1=F(a,1)$ and $b\leq 1-\varepsilon$.

Since $F\neq\lVert\cdot\rVert_{\infty}$ it follows from Lemma 2.1 that $F(1,1)>1$ and hence $a<1$. But then $b=f(a)=1$, be definition of $f$. This is a contradiction since $b<1$. ∎

Now we can prove the first main result of this paper.

Proposition 2.3**.**

If $F$ is any absolute, normalised norm on $\mathbb{R}^{2}$ with $F\neq\lVert\cdot\rVert_{\infty}$ and $X$ and $Y$ are nontrivial Banach spaces, then the following holds:

1. (i)

If $X\oplus_{F}Y$ is LASQ, then $X$ and $Y$ are LASQ. 2. (ii)

If $X\oplus_{F}Y$ is WASQ, then $X$ and $Y$ are WASQ. 3. (iii)

$X\oplus_{F}Y$* is not ASQ.*

Note that the converses of (i) and (ii) also hold by the general results in [abrahamsen2].

Proof.

First we will prove statement (ii). So let $Z:=X\oplus_{F}Y$ be WASQ and let $y\in S_{Y}$. Then there is a weakly null sequence $(z_{n}=(u_{n},v_{n}))_{n\in\mathbb{N}}$ in $B_{Z}$ such that $\lVert z_{n}\pm(0,y)\rVert_{F}\to 1$ and $\lVert z_{n}\rVert_{F}\to 1$. Actually, we may assume that $\lVert z_{n}\rVert_{F}=F(\lVert u_{n}\rVert,\lVert v_{n}\rVert)=1$ for every $n\in\mathbb{N}$.

By Lemma 2.2 there exists a sequence $(\delta_{n})_{n\in\mathbb{N}}$ in $(0,\infty)$ such that $\delta_{n}\to 0$ and for every $n\in\mathbb{N}$ the following holds:

[TABLE]

By passing to a subsequence if necessary, we may assume that $F(\lVert u_{n}\rVert,\lVert y\pm v_{n}\rVert)\leq 1+\delta_{n}$ for every $n\in\mathbb{N}$. It follows that

[TABLE]

and hence $\lVert v_{n}\rVert\geq 1-2^{-n}$ for every $n$.

Since we also have $\lVert v_{n}\rVert\leq F(\lVert u_{n}\rVert,\lVert v_{n}\rVert)=1$ for each $n$, we obtain $\lVert v_{n}\rVert\to 1$. Also, $(v_{n})_{n\in\mathbb{N}}$ is a weakly null sequence in $Y$, since $(z_{n})_{n\in\mathbb{N}}$ is weakly null in $Z$.

We further have $\lVert y\pm v_{n}\rVert\leq F(\lVert u_{n}\rVert,\lVert y\pm v_{n}\rVert)\leq 1+\delta_{n}$ and thus

[TABLE]

which implies $\lVert y\pm v_{n}\rVert\to 1$. Thus $Y$ is WASQ.

Since $X\oplus_{F}Y\cong Y\oplus_{\tilde{F}}X$, where $\tilde{F}(a,b):=F(b,a)$, the same argument also shows that $X$ is LASQ. This completes the proof of (ii) and statement (i) is proved analogously.

Now we will prove (iii). Assume to the contrary that $X\oplus_{F}Y$ is ASQ. Since $F\neq\lVert\cdot\rVert_{\infty}$, we have $F(1,1)>1$ (Lemma 2.1). Choose $\varepsilon>0$ such that $(1-\varepsilon)F(1,1)>1$.

By Lemma 2.2 (applied to $F$ and $\tilde{F}$) there exists a $\delta>0$ such that for all $a,b\geq 0$ with $F(a,b)=1$ the following holds:

[TABLE]

Now let $x\in S_{X}$ and $y\in S_{Y}$. Since $X\oplus_{F}Y$ is ASQ, there exist $u\in X$, $v\in Y$ such that $F(\lVert u\rVert,\lVert v\rVert)=1$ and $F(\lVert x\pm u\rVert,\lVert v\rVert)\leq 1+\delta$, $F(\lVert u\rVert,\lVert y\pm v\rVert)\leq 1+\delta$.

A similar calculation as in the proof of (ii) shows that $F(\lVert u\rVert,1)\leq 1+\delta$ and $F(1,\lVert v\rVert)\leq 1+\delta$. It follows that $\lVert u\rVert,\lVert v\rVert\geq 1-\varepsilon$.

But then $1=F(\lVert u\rVert,\lVert v\rVert)\geq(1-\varepsilon)F(1,1)>1$ and with this contradiction the proof is finished. ∎

Next we turn our attention to LOH sums. First recall that a Banach space $X$ is strictly convex (SC) if $x,y\in S_{X}$ and $\lVert x+y\rVert=2$ imply $x=y$. The $p$-norms are strictly convex for $1<p<\infty$. We will call a point $x\in S_{X}$ an SC-point of $X$ if $\lVert x+y\rVert<2$ for every $y\in S_{X}$ with $y\neq x$. Thus $X$ is strictly convex if and only if every point of $S_{X}$ is an SC-point.

Given an absolute, normalised norm $F$ on $\mathbb{R}^{2}$, set

[TABLE]

The following lemma is intuitively clear, but we include a proof for the sake of completeness.

Lemma 2.4**.**

Let $F$ be an absolute, normalised norm on $\mathbb{R}^{2}$ with upper boundary curve $f$. Then

1. (i)

$r_{F}=1$*  $\Leftrightarrow$  $(1,0)$ is an SC-point of $(\mathbb{R}^{2},F)$ or $f(1)>0$;* 2. (ii)

$r_{F}=0$*  $\Leftrightarrow$  $F=\lVert\cdot\rVert_{1}$.*

Proof.

(i) If $r_{F}<1$, then there must be $a\in[0,1)$ and $b>0$ such that $F(a,b)=1$ and $F(a+1,b)=2$. Hence $(1,0)$ is not an SC-point of $(\mathbb{R}^{2},F)$.

Moreover, the whole line segment from $(1,0)$ to $(a,b)$ belongs to the unit sphere of $(\mathbb{R}^{2},F)$, which implies that $f(a+t(1-a))=(1-t)b$ for $t\in[0,1)$. Hence $f(s)=\frac{1-s}{1-a}b$ for $s\in[a,1)$. Thus $f(1)=0$. This shows “$\Leftarrow$”.

Now assume that $r_{F}=1$ and $(1,0)$ is not an SC-point of $(\mathbb{R}^{2},F)$. Then we can find $(a,b)\in\mathbb{R}^{2}$ such that $F(a,b)=1$ and $F(1+a,b)=2$ but $(a,b)\neq(1,0)$. Without loss of generality we may assume that $a,b>0$. Since $r_{F}=1$ it follows that $a=1$, hence $F(1,b)=1$.

If $f(1)=0$ there would be $s\in[0,1)$ such that $f(s)<b$. But then we obtain a contradiction since $1=F(s,f(s))<F(1,b)=1$. Thus we must have $f(1)>0$.

(ii) Suppose that $r_{F}=0$. This easily implies $F(1,1)=2$ and thus we have $F=\lVert\cdot\rVert_{1}$ by Lemma 2.1. The converse is clear. ∎

We need two more auxiliary lemmas.

Lemma 2.5**.**

Let $F$ be an absolute, normalised norm on $\mathbb{R}^{2}$. If $a,b\geq 0$ such that $F(a,b)=1$ and $0\leq c\leq 1+a$ such that $F(c,b)=2$, then $c=1+a$ and $a\geq r_{F}$.

Proof.

First note that under the above assumptions we have $2=F(c,b)\leq F(1+a,b)\leq F(1,0)+F(a,b)=2$. Hence, by definition of $r_{F}$, we must have $a\geq r_{F}$.

Now we denote again by $f$ the upper boundary curve of $F$ and distinguish two cases.

Case 1: $r_{F}=1$. Then $a=1$ and $1<c\leq 2$ (if $c\leq 1$ we would obtain $2=F(c,b)\leq F(1,b)=1$). Thus $a^{\prime}:=c-1\in(0,1]$ and $F(1+a^{\prime},b)=2$ as well as $1=F(1,b)\geq F(a^{\prime},b)\geq F(c,b)-F(1,0)=1$.

Since $r_{F}=1$ it follows that $a^{\prime}=1$, i. e. $c=2$.

Case 2: $r_{F}<1$. Then we have $r_{F}\leq a<c\leq 1+a$. Put $d:=c-a\in(0,1]$ and let $w>0$ such that $F(r_{F},w)=1$ and $F(1+r_{F},w)=2$.

Then the line segment from $(1,0)$ to $(r_{F},w)$ lies completely in $S_{(\mathbb{R}^{2},F)}$ and we obtain

[TABLE]

We also put $g(s):=(d+1)f(s/(d+1))$ for $s\in[0,d+1]$. Then $F(s,g(s))=d+1$ for every such $s$.

It is easy to see that $c/(d+1)\in[a,1]\subseteq[r_{F},1]$ and thus we have $g(c)=(d+1)f(c/(d+1))=h(c-(d+1))=h(a-1)$.

If $a<1$ we have $f(a)=b$ and since $a\geq r_{F}$ it follows that $f(a)=h(a-1)$, thus $g(c)=b$.

For $a=1$ we must have $b=0=g(c)$ (otherwise there is $s\in[0,1)$ such that $f(s)<b$ and hence $1=F(s,f(s))<F(1,b)=F(a,b)=1$).

Thus we always have $g(c)=b$, which imples that $2=F(c,b)=F(c,g(c))=d+1$. Hence $d=1$ and $c=1+a$. ∎

Lemma 2.6**.**

Let $F$ be an absolute, normalised norm on $\mathbb{R}^{2}$ and let $\varepsilon>0$. Then there exists $\delta>0$ such that the following holds: whenever $a,b\geq 0$ with $F(a,b)=1$ and $0\leq c\leq 1+a$ with $F(c,b)\geq 2-\delta$, then $c\geq 1+r_{F}-\varepsilon$.

Proof.

This follows from Lemma 2.5 and a standard compactness argument. ∎

Now we can prove the main result on local octahedrality in absolute sums.

Proposition 2.7**.**

Let $F$ be an absolute, normalised norm on $\mathbb{R}^{2}$ and $X$ and $Y$ nontrivial Banach spaces such that $X\oplus_{F}Y$ is LOH. Then for every $x\in S_{X}$ and every $\varepsilon>0$ there is a $z\in S_{X}$ such that $\lVert x\pm z\rVert\geq 2r_{F}-\varepsilon$.

Proof.

Let $x\in S_{X}$ and $\varepsilon>0$. Choose $\delta>0$ according to Lemma 2.6 for the parameter $\varepsilon/2$. Since $X\oplus_{F}Y$ is LOH, we can find $(u,v)\in S_{X\oplus_{F}Y}$ such that $F(\lVert x\pm u\rVert,\lVert v\rVert)\geq 2-\delta$.

Because of $\lVert x\pm u\rVert\leq 1+\lVert u\rVert$ and $F(\lVert u\rVert,\lVert v\rVert)=1$ this implies $\lVert x\pm u\rVert\geq 1+r_{F}-\varepsilon/2$.

It follows that $\lVert u\rVert\geq\lVert x+u\rVert-1\geq r_{F}-\varepsilon/2$.

Now put $z:=u/\lVert u\rVert$. Then

[TABLE]

∎

It follows in particular from Proposition 2.7 that $X$ is LOH if $X\oplus_{F}Y$ is LOH and $r_{F}=1$ (which, by Lemma 2.4, is equivalent to the fact that $f(1)>0$ or $(1,0)$ is an SC-point of $(\mathbb{R}^{2},F)$).

More generally, for any Banach space $X$ we may define

[TABLE]

Then $X$ is LOH if and only if $s(X)=2$ and Proposition 2.7 reads: if $X\oplus_{F}Y$ is LOH, then $s(X)\geq 2r_{F}$.

Note that $s(\mathbb{R})=0$, while it easily follows from Riesz’s Lemma that $s(X)\geq 1$ whenever $\text{dim}(X)\geq 2$. The following statements are also easy to verify: $s(\ell^{\infty})=s(c_{0})=1=s(\ell^{\infty}(n))$ for $n\geq 2$, where $\ell^{\infty}$ is the space of bounded sequences, $c_{0}$ is the space of null sequences (both equipped with the supremum norm), and $\ell^{\infty}(n)$ is the space $\mathbb{R}^{n}$ equipped with the maximum norm. It is also easy to prove that $s(H)=\sqrt{2}$ for any Hilbert space $H$ with $\text{dim}(H)\geq 2$.

Putting everything together, we obtain the following corollary (for (b) note that $r_{F}>0$ if $F\neq\lVert\cdot\rVert_{1}$ (Lemma 2.4)).

Corollary 2.8**.**

Let $F$ be an absolute, normalised norm on $\mathbb{R}^{2}$ and $X$ and $Y$ nontrivial Banach spaces. Then the following holds:

1. (a)

If $r_{F}=1$ and $X\oplus_{F}Y$ is LOH, then so is $X$. In particular, this holds if $F$ is strictly convex or $F=\lVert\cdot\rVert_{\infty}$. 2. (b)

If $F\neq\lVert\cdot\rVert_{1}$, then $\mathbb{R}\oplus_{F}Y$ is not LOH. 3. (c)

If $r_{F}>1/2$, then $\ell^{\infty}\oplus_{F}Y$, $c_{0}\oplus_{F}Y$ and $\ell^{\infty}(n)\oplus_{F}Y$ for $n\in\mathbb{N}$ are not LOH. 4. (d)

If $r_{F}>1/\sqrt{2}$ and $H$ is a Hilbert space, then $H\oplus_{F}Y$ is not LOH.

Of course, it is also possible to prove results analogous to Proposition 2.7 and Corollary 2.8 for the second summand by modifying the definition of $r_{F}$ accordingly (i. e. using instead $r_{\tilde{F}}$, where $\tilde{F}(a,b):=F(b,a)$).

The author does not know whether there are any analogous results for WOH spaces, but let us remark that the above proof-techniques could also be used to show that $X\oplus_{F}Y$ is not octahedral if $X$ and $Y$ are nontrivial Banach spaces, $F\neq\lVert\cdot\rVert_{\infty}$ and $r_{F}=1=r_{\tilde{F}}$. However, this result already follows from the more general results on octahedrality in absolute sums that were proved in [haller4] and that we have already mentioned in the introduction.

Let us note one more corollary concerning the LD2P (recall that a norm is smooth if it is Gâteaux-differentiable at each nonzero point).

Corollary 2.9**.**

If $F$ is a smooth, absolute, normalised norm on $\mathbb{R}^{2}$ and $X$ and $Y$ are nontrivial Banach spaces such that $X\oplus_{F}Y$ has the LD2P, then $X$ has the LD2P.

Proof.

It is well-known that a finite-dimensional Banach space is smooth if and only if its dual is strictly convex. If we put

[TABLE]

then $F^{*}$ is an absolute, normalised norm on $\mathbb{R}^{2}$ and $(X\oplus_{F}Y)^{*}\cong X^{*}\oplus_{F^{*}}Y^{*}$ (this is a standard fact from the theory of absolute sums, which is easy to prove). The claim now follows from Corollary 2.8 and the duality between LOH and LD2P ([haller3]). ∎

Next we will consider ultrapowers of OH/LOH and ASQ/LASQ spaces. First we recall the necessary definitions. Given a free ultrafilter $\mathcal{U}$ on $\mathbb{N}$ and a bounded sequence $(a_{n})_{n\in\mathbb{N}}$ of real numbers, there exists (by a compactness argument) a unique number $a\in\mathbb{R}$ such that for every $\varepsilon>0$ we have $\left\{n\in\mathbb{N}:\lvert a_{n}-a\rvert<\varepsilon\right\}\in\mathcal{U}$. It is called the limit of $(a_{n})_{n\in\mathbb{N}}$ along $\mathcal{U}$ and will be denoted by $\lim_{n,\mathcal{U}}a_{n}$.

For a Banach space $X$, denote by $\ell^{\infty}(X)$ the space of all bounded sequences in $X$ and set $\mathcal{N}_{\mathcal{U}}:=\left\{(x_{n})\in\ell^{\infty}(X):\lim_{n,\mathcal{U}}\lVert x_{n}\rVert=0\right\}$. The ultarpower $X^{\mathcal{U}}$ of $X$ with respect to $\mathcal{U}$ is the quotient space $\ell^{\infty}(X)/\mathcal{N}_{\mathcal{U}}$ equipped with the (well-defined) norm $\lVert[(x_{n})]\rVert_{\mathcal{U}}:=\lim_{n,\mathcal{U}}\lVert x_{n}\rVert$. $X^{\mathcal{U}}$ is again a Banach space (for more information on ultraproducts see for example [heinrich]).

We have the following observations concerning octahedrality in ultrapowers.

Proposition 2.10**.**

Let $X$ be a Banach space and $\mathcal{U}$ a free ultrafilter on $\mathbb{N}$. Then the following assertions are equivalent:

1. (i)

$X$* is octahedral.* 2. (ii)

For all $z_{1},\dots,z_{n}\in S_{X^{\mathcal{U}}}$ there exists an element $z\in S_{X^{\mathcal{U}}}$ such that $\lVert z_{i}+z\rVert_{\mathcal{U}}=2$ for every $i\in\left\{1,\dots,n\right\}$. 3. (iii)

$X^{\mathcal{U}}$* is octahedral.*

Likewise, the following statements are equivalent:

1. (i)

$X$* is locally octahedral.* 2. (ii)

For every $z\in S_{X^{\mathcal{U}}}$ there is some $\tilde{z}\in S_{X^{\mathcal{U}}}$ such that $\lVert z\pm\tilde{z}\rVert_{\mathcal{U}}=2$. 3. (iii)

$X^{\mathcal{U}}$* is locally octahedral.*

Proof.

We will only prove the statement for octahedral spaces. The proof for local octahedrality is completely analogous.

So let us first assume that $X$ is OH and let $z_{1},\dots,z_{n}\in S_{X^{\mathcal{U}}}$. Let $(x_{i,k})_{k\in\mathbb{N}}$ be a representative of $z_{i}$. We may assume that $x_{i,k}\neq 0$ for all $k\in\mathbb{N}$ and all $i=1,\dots,n$.

Since $X$ is octahedral we can find, for each $k\in\mathbb{N}$, an element $x_{k}\in S_{X}$ such that

[TABLE]

Let $z:=[(x_{k})_{k\in\mathbb{N}}]\in S_{X^{\mathcal{U}}}$. For each $k\in\mathbb{N}$ and every $i\in\{1,\dots,n\}$ we have

[TABLE]

Since $\lim_{k,\mathcal{U}}\lVert x_{i,k}\rVert=\lVert z_{i}\rVert_{\mathcal{U}}=1$ it follows that $\lVert z_{i}+z\rVert_{\mathcal{U}}=\lim_{k,\mathcal{U}}\lVert x_{i,k}+x_{k}\rVert=2$ for all $i\in\{1,\dots,n\}$. This proves (i) $\Rightarrow$ (ii).

(ii) $\Rightarrow$ (iii) is clear.

(iii) $\Rightarrow$ (i): Let $x_{1},\dots,x_{n}\in S_{X}$ and $\varepsilon>0$. We consider $X$ as a subspace of $X^{\mathcal{U}}$ (via the canonical embedding). Since $X^{\mathcal{U}}$ is octahedral, there exists $y=[(y_{k})_{k\in\mathbb{N}}]\in S_{X^{\mathcal{U}}}$ such that

[TABLE]

It follows that $B_{i}:=\{k\in\mathbb{N}:\lVert x_{i}+y_{k}\rVert\geq 2-2\varepsilon\}\in\mathcal{U}$ for all $i=1,\dots,n$. Since $\lim_{k,\mathcal{U}}\lVert y_{k}\rVert=\lVert y\rVert_{\mathcal{U}}=1$ we also have $A:=\{k\in\mathbb{N}:|\lVert y_{k}\rVert-1|\leq\varepsilon\}\in\mathcal{U}$. Hence $M:=A\cap B_{1}\cap\dots\cap B_{n}\in\mathcal{U}$ and in particular, $M\neq\emptyset$.

Now let $k_{0}\in M$. Then we have, for each $i\in\{1,\dots,n\}$,

[TABLE]

thus $X$ is octahedral. ∎

For weakly octahedral spaces, the situation seems to be more complicated. Let us first introduce one more notation: if $s=(x_{n}^{*})_{n\in\mathbb{N}}$ is a sequence in $S_{X^{*}}$, then we may define a norm-one functional $\varphi_{s}$ on $X^{\mathcal{U}}$ by $\varphi_{s}([(x_{n})]):=\lim_{n,\mathcal{U}}x_{n}^{*}(x_{n})$.

Using the characterisation for WOH spaces from [haller3]*Proposition 2.2, one can easily prove the following: if $X$ is WOH, then for all $z_{1},\dots,z_{n}\in S_{X^{\mathcal{U}}}$ and every sequence $s=(x_{n}^{*})_{n\in\mathbb{N}}$ in $S_{X^{*}}$ there exists $z\in S_{X^{\mathcal{U}}}$ such that

[TABLE]

However, it is not clear whether the converse of this statement also holds, nor whether it is equivalent to the weak octahedrality of $X^{\mathcal{U}}$.

Similarly to Proposition 2.10 one can also prove the following result for ASQ/LASQ spaces (we skip the details).

Proposition 2.11**.**

Let $X$ be a Banach space and $\mathcal{U}$ a free ultrafilter on $\mathbb{N}$. Then the following assertions are equivalent:

1. (i)

$X$* is ASQ.* 2. (ii)

For all $z_{1},\dots,z_{n}\in S_{X^{\mathcal{U}}}$ there exists an element $z\in S_{X^{\mathcal{U}}}$ such that $\lVert z_{i}+z\rVert_{\mathcal{U}}=1$ for every $i\in\left\{1,\dots,n\right\}$. 3. (iii)

$X^{\mathcal{U}}$* is ASQ.*

Likewise, the following statements are equivalent:

1. (i)

$X$* is LASQ.* 2. (ii)

For every $z\in S_{X^{\mathcal{U}}}$ there is some $\tilde{z}\in S_{X^{\mathcal{U}}}$ such that $\lVert z\pm\tilde{z}\rVert_{\mathcal{U}}=1$. 3. (iii)

$X^{\mathcal{U}}$* is LASQ.*

For WASQ spaces, the situation is again a bit more involved. First we note the following equivalent characterisation of WASQ spaces with separable dual (the proof is easy and will therefore be omitted).

Lemma 2.12**.**

Let $X$ be a Banach space. If $X$ is WASQ, then the following holds: for every $x\in S_{X}$, every $\varepsilon>0$ and all $x_{1}^{*},\dots,x_{n}^{*}\in S_{X^{*}}$ there exists a $y\in S_{X}$ such that $\lVert x\pm y\rVert\leq 1+\varepsilon$ and $x_{i}^{*}(y)\leq\varepsilon$ for every $i=1,\dots,n$.

If $X^{*}$ is separable, then the converse of this statement also holds.

Using this lemma, it is easy to show the next result (again the details are omitted).

Proposition 2.13**.**

Let $X$ be a Banach space and $\mathcal{U}$ a free ultrafilter on $\mathbb{N}$. If $X$ is WASQ, then for every $z\in S_{X^{\mathcal{U}}}$ and all double-sequences $(x_{ik}^{*})_{i,k\in\mathbb{N}}$ in $S_{X^{*}}$ there is some $\tilde{z}\in S_{X^{\mathcal{U}}}$ satisfying $\lVert z\pm\tilde{z}\rVert_{\mathcal{U}}=1$ and $\varphi_{s_{i}}(\tilde{z})=0$ for every $i\in\mathbb{N}$, where $s_{i}:=(x_{ik}^{*})_{k\in\mathbb{N}}$.

If $X^{*}$ is separable, then the converse also holds.

Again, it is not clear whether the property in Proposition 2.13 is equivalent to the weak almost squareness of $X^{\mathcal{U}}$.

Finally, we would like to show that the classes of OH/WOH/LOH/ASQ/

LASQ spaces and the classes of spaces with the SD2P/D2P/LD2P are closed with respect to the Banach-Mazur distance. Recall that this distance between two isomorphic Banach spaces $X$ and $Y$ is defined by

[TABLE]

Proposition 2.14**.**

Let $X$ be a Banach space such that for every $\delta>0$ there is some OH/WOH/LOH/ASQ/LASQ space $Y$ isomorphic to $X$ with $\operatorname{d}(X,Y)<1+\delta$. Then $X$ is also OH/WOH/LOH/ASQ/LASQ.

Proof.

The proofs are all similar, so we will only show the most complicated case of WOH spaces. Let $x_{1},\dots,x_{n}\in S_{X}$, $x^{*}\in B_{X^{*}}$ and $0<\varepsilon<1/2$. Choose $0<\delta<\varepsilon^{2}$ such that $\sqrt{\delta}/(1-2\sqrt{\delta})\leq\varepsilon$.

By assumption there is a WOH space $Y$ isomorphic to $X$ such that $\operatorname{d}(X,Y)<1+\delta$. Hence we can find an isomorphism $T:X\rightarrow Y$ such that $\lVert T\rVert=1$ and $\lVert T^{-1}\rVert<1+\delta$.

Put $y_{i}:=Tx_{i}\in B_{Y}\setminus\left\{0\right\}$ for $i=1,\dots,n$ and $y^{*}:=(T^{*})^{-1}x^{*}=x^{*}\circ T^{-1}\in(1+\delta)B_{Y^{*}}$.

Since $Y$ is WOH, there exists, by [haller3]*Proposition 2.2, an element $y\in S_{Y}$ such that

[TABLE]

Let $z:=T^{-1}y/\lVert T^{-1}y\rVert\in S_{X}$. Then we have for each $i\in\left\{1,\dots,n\right\}$ and every $t>0$

[TABLE]

Combining this with (2.1) and observing that $y^{*}(y_{i})=x^{*}(x_{i})$ we obtain

[TABLE]

for every $i\in\left\{1,\dots,n\right\}$ and every $t>0$.

Now if $t\geq\varepsilon$, then by the choice of $\delta$ we obtain $(\sqrt{\delta}-2\delta)t\geq\delta$, which implies $(1-2\delta)t-\delta\geq(1-\sqrt{\delta})t$.

Thus for every $t\geq\varepsilon$ and every $i\in\left\{1,\dots,n\right\}$ we have

[TABLE]

By [haller3]*Proposition 2.2 this implies that $X$ is WOH. ∎

Proposition 2.14 and the duality between diameter-two- and octahedrality properties, together with the fact that $\operatorname{d}(X^{*},Y^{*})\leq\operatorname{d}(X,Y)$ holds for all Banach spaces $X$ and $Y$, immediately yield the following corollary.

Corollary 2.15**.**

Let $X$ be a Banach space such that for every $\delta>0$ there is a Banach space $Y$ isomorphic to $X$ which has the SD2P/D2P/LD2P and satisfies $\operatorname{d}(X,Y)<1+\delta$. Then $X$ also has the SD2P/D2P/LD2P.

Concerning WASQ spaces, using Lemma 2.12 one can show the following result (again an easy proof is omitted).

Proposition 2.16**.**

Let $X$ be a Banach space such that for every $\delta>0$ there is some WASQ space $Y$ isomorphic to $X$ with $\operatorname{d}(X,Y)<1+\delta$. If $X^{*}$ is separable, then $X$ is also WASQ.