1 Introduction

Our main goal in this short paper is to show that the set of continuous quasinorms defined on \(\mathbb R^n\) for some \(n\ge 2\) has a, somehow, canonical structure of Banach space after quotienting by the proportional quasinorms.

For this to make sense, we first need to endow this set with a vector space structure, this will be done by means of something that everyone can expect to represent the mean of two quasinorms: \(\sqrt{\Vert \cdot \Vert _X\Vert \cdot \Vert _Y}\) for each pair of quasinorms \(\Vert \cdot \Vert _X, \Vert \cdot \Vert _Y\). Once the mean is given, we just need to choose the element of the space which will play the rôle of the origin in order to determine a vector space structure, in the present paper we have chosen \((\mathbb R^n,\Vert \cdot \Vert _2)\). Of course, this may seem anything but canonical. On the bright side, the choice of an origin will not affect any property of the newly defined vector (or Banach) space. For example, we may consider C[0, 1] endowed with the scalar multiplication \((\uplambda \star f)(x)=\uplambda (f(x)-1)\) and the addition \((f\oplus g)(x)=f(x)+g(x)-1\), of course, the same can be done with any other function in C[0, 1] instead of 1. Now, if we define a norm in \((C[0,1],\oplus ,\star )\) as \(\Vert f\Vert =\max \{|f(x)-1|\}\) then we have a Banach space structure \((C[0,1],\oplus ,\star ,\Vert \cdot \Vert )\) that is indistinguishable from the usual \((C[0,1],+,\cdot ,\Vert \cdot \Vert _{\infty })\), in the sense that the map

$$\begin{aligned} (C[0,1],+,\cdot ,\Vert \cdot \Vert _{\infty })\rightarrow (C[0,1],\oplus ,\star ,\Vert \cdot \Vert ),\quad f\mapsto f+1 \end{aligned}$$

is a linear isometry. What we have done is equivalent to considering the affine structure of C[0, 1] and taking two different choices for the origin. This is doable because every norm gives a translation invariant metric.

Once the operations are given, we have to define the norm. This idea is not ours, but taken from A. Khare’s preprint [9]. Given two continuous quasinorms \(\Vert \cdot \Vert _X, \Vert \cdot \Vert _Y\), the distance between them is defined as

$$\begin{aligned} \mathrm{{d}}(\Vert \cdot \Vert _X,\Vert \cdot \Vert _Y)=\min \{\mu :\Vert \cdot \Vert _X\le \uplambda \Vert \cdot \Vert _Y\le \mu \Vert \cdot \Vert _X\text {\ for\ some\ }\uplambda >0\}, \end{aligned}$$

where the order relation is the pointwise order: \(\Vert \cdot \Vert _X\le \uplambda \Vert \cdot \Vert _Y\) means \(\Vert x\Vert _X\le \uplambda \Vert x\Vert _Y\) for every \(x\in \mathbb R^n\). Of course, two quasinorms are proportional if and only if the distance between them is 1, so we must take the reasonable quotient

$$\begin{aligned} \spadesuit \quad \Vert \cdot \Vert _X\sim \Vert \cdot \Vert _Y\text { if and only if } \Vert \cdot \Vert _X=\uplambda \Vert \cdot \Vert _Y\text { for some }\uplambda \in (0,\infty ) \end{aligned}$$

to make d an actual (multiplicative) metric. So, defining

$$\begin{aligned} \mathrm{{d}}([\Vert \cdot \Vert _X],[\Vert \cdot \Vert _Y])=\min \{\mu :\Vert \cdot \Vert _X\le \uplambda \Vert \cdot \Vert _Y\le \mu \Vert \cdot \Vert _X\text {\ for\ some\ }\uplambda >0\} \end{aligned}$$
(1)

we have a distance between the equivalence classes of quasinorms that turns out to induce a norm when we endow \(\{\text {Continuous quasinorms on }\mathbb R^n\}/\negthinspace \sim \) with the above-explained operations.

This paper is far from being the first one in which the sets of (quasi) norms are endowed with some structure. The best known structure given to the set of norms on a finite-dimensional space is the Banach–Mazur pseudometric defined as

$$\begin{aligned} \mathrm{{d}}(\Vert \cdot \Vert _X,\Vert \cdot \Vert _Y)=\inf \{\Vert T\Vert \Vert T^{-1}\Vert \}, \end{aligned}$$
(2)

where T runs over the set of linear isomorphisms \(T:(\mathbb R^n,\Vert \cdot \Vert _X)\rightarrow (\mathbb R^n,\Vert \cdot \Vert _Y).\) It is well known that, after taking the appropriate quotient, this pseudometric turns out to be a metric that makes the space to be compact, there is still significant interest on this topic, see, e.g. [1, 18, 19].

The present paper is neither the first one about, say, mixing pairs of norms to obtain something new. In this setting, interpolation of (quasi) normed spaces—or even more general spaces—has been the main topic for at least half a century, see [8, 12, 17]. For the reader interested in interpolation, we suggest [3] and the very interesting [15]. A nice paper on interpolation in quasinormed spaces is [16]. To the best of our knowledge, this paper is the first where someone considers the kind of interpolation that we have in Definition 3.3, that is

$$\begin{aligned} \Vert \cdot \Vert _{(X,Y)_\theta }=\Vert \cdot \Vert _X^\theta \Vert \cdot \Vert _Y^{1-\theta }. \end{aligned}$$

There is a very good reason to avoid this kind of interpolation in the normed space literature. Namely, in Remark 3.8 we provide an example to show that the mean of a pair of norms on \(\mathbb R^2\) does not need to be a norm but a quasinorm.

2 Notations and Preliminary Results

We will consider some positive integer n fixed throughout the paper. Every vector space will be over \(\mathbb R\); observe that any \(\mathbb C^n\) can be seen as \(\mathbb R^{2n}\). Moreover, we will consider from now on the vector space \(\mathbb R^n\) endowed with its only topological vector space structure, i.e. the one given by \(\Vert \cdot \Vert _2\).

Definition 2.1

A map \(\Vert \cdot \Vert :\mathbb R^n\rightarrow [0,\infty )\) is a quasinorm if the following conditions hold:

  1. (1)

    \(\Vert x\Vert =0\) if and only if \(x=0.\)

  2. (2)

    \(\Vert \uplambda x\Vert =|\uplambda |\Vert x\Vert \) for every \(\uplambda \in \mathbb R, x\in \mathbb R^n.\)

  3. (3)

    There exists \(k>0\) such that \(\Vert x+y\Vert \le k(\Vert x\Vert +\Vert y\Vert )\) for every \(x,y\in \mathbb R^n\).

If the map \(\Vert \cdot \Vert \) is continuous then we say that it is a continuous quasinorm. If k can be chosen to be 1, then \(\Vert \cdot \Vert \) is a norm and it is continuous.

2.2

As is customary, given a quasinormed space \((\mathbb R^n,\Vert \cdot \Vert _X)\), we will denote its unit (closed) ball as \(B_X\), its unit sphere as \(S_X\).

Definition 2.3

Some subset \(B\subset \mathbb R^n\) is bounded if, for every neighbourhood U of 0 there is \(M\in (0,\infty )\) such that \(B\subset MU\). \(B\subset \mathbb R^n\) is balanced when \(\uplambda B\subset B\) for every \(\uplambda \in [-1,1]\).

Definition 2.4

For \(B\subset \mathbb R^n\), the Minkowski functional of B is \(\rho _B(x)=\inf \{\uplambda \in (0,\infty ):x\in \uplambda B\}\).

It is quite well known that the quasinorms on a topological vector space are in correspondence with the bounded, balanced neighbourhoods of the origin, see the beginning of Section 2 in [7], and for a proof of such a key result the reader may check [6, Theorem 4]. The version that we will use is the following, where we use that \(\Vert \cdot \Vert _2\) gives the only topological vector space structure to \(\mathbb R^n\) and \(B_2\) denotes the Euclidean unit ball of \(\mathbb R^n\):

Theorem 2.5

The Minkowski functional \(\rho _B\) of a given subset \(B\subset \mathbb R^n\) is a quasinorm if and only if B fulfils the following:

  • B contains \(\varepsilon B_2\) for some \(\varepsilon >0\).

  • For every \(\uplambda \in [-1,1]\) one has \(\uplambda B\subset B\), i.e. B is balanced.

  • B is contained in \(MB_2\) for some \(M>0\).

In this case, \(\rho _B\) is a continuous quasinorm if and only if B is closed. Moreover, \(\rho _B\) is a norm if and only if the above hold and \(\frac{x+y}{2}\in B\) for any pair \(x,y\in B.\)

We could even replace the first and third items in 2.5 by

“If \(B'\) is the unit ball of some quasinorm on \(\mathbb R^n\) then there are \(\varepsilon , M>0\) such that \(\varepsilon B\subset B' \subset MB\)”. Observe that this implies that the constants \(\uplambda , \mu \) in (1) actually exist.

We will deal in this note with \(\mathcal Q_0=\{\text {Continuous quasinorms\ defined\ on\ }\mathbb R^n\}\) and \(\mathcal Q=\mathcal Q_0/\negthinspace \sim \), where two quasinorms are equivalent if and only if they are proportional, endowed with the multiplicative distance on \(\mathcal Q\) defined in [9] by A. Khare and given by

$$\begin{aligned} \mathrm{{d}}([\Vert \cdot \Vert _X],[\Vert \cdot \Vert _Y])=\min \{\mu :\Vert \cdot \Vert _X\le \uplambda \Vert \cdot \Vert _Y\le \mu \Vert \cdot \Vert _X\text {\ for\ some\ }\uplambda >0\}. \end{aligned}$$
(3)

In the same paper, it is shown that d endows \(\mathcal N=\{\text {Norms\ defined\ on\ }\mathbb R^n\}/\negthinspace \sim \) with a complete metric space structure. To keep the notations consistent, we will write \(\mathcal N_0\) for \(\{\text {Norms\ defined\ on\ }\mathbb R^n\}\). The infimum in (3) exists because in \(\mathbb R^n\) every pair of continuous quasinorms are equivalent and, moreover, by the continuity of the quasinorms, it is pretty clear that the minimum is attained. A nice feature of Khare’s distance is that, in \(\mathbb R^2\), it distinguishes the \(\max \)-norm from \(\Vert \cdot \Vert _1\). In some sense, these norms are as different as two norms can be, but the usual distances between norms, such as the Banach–Mazur or the Gromov–Hausdorff, make them indistinguishable.

3 The Main Result

Throughout this section, we will only consider continuous quasinorms.

Our first goal is to show that d is actually a multiplicative distance on \(\mathcal Q\). For this, the following lemma will be useful.

Lemma 3.1

Take any pair of quasinorms \(\Vert \cdot \Vert _X,\Vert \cdot \Vert _Y\), \(\uplambda >0\) and \(\mu \ge 1\) such that \(\Vert \cdot \Vert _X\le \uplambda \Vert \cdot \Vert _Y\le \mu \Vert \cdot \Vert _X\). Then, \(\mu \) is minimal if and only if both \(S_X\cap \uplambda ^{-1} S_Y\) and \(\uplambda ^{-1} S_Y\cap \mu ^{-1} S_X\) are non-empty.

Moreover, the distance between \([\Vert \cdot \Vert _X]\) and \([\Vert \cdot \Vert _Y]\) is \(\mu \) if and only if there are representatives \(\Vert \cdot \Vert _X\) and \(\Vert \cdot \Vert _Y\) such that

  1. (1)

    One has \(\Vert \cdot \Vert _X\le \Vert \cdot \Vert _Y\le \mu \Vert \cdot \Vert _X\).

  2. (2)

    There are \(x\in S_X\cap S_Y\) and \(y\in S_Y\cap \mu ^{-1}S_X\).

In particular, the distance \(\mu \) is always attained.

Proof

The chain of inequalities in the statement is equivalent to the chain of inclusions \(\mu ^{-1}B_X\subset \uplambda ^{-1}B_Y\subset B_X\), so suppose that \(S_X\cap \uplambda ^{-1} S_Y=\emptyset \). The distance between the compact sets \(\uplambda ^{-1}B_Y\) and \(S_X\) is attained, so if they do not meet, then the distance between them is strictly positive and we can multiply the sets \(\uplambda ^{-1}B_Y\) and \(\mu ^{-1}B_X\) by \(1+\varepsilon \) for some \(\varepsilon >0\) and the contentions are still fulfilled. So, if we define \(\mu '=\dfrac{\mu }{1+\varepsilon }\) we obtain \(\mu '^{-1}B_X\subset (1+\varepsilon )\uplambda ^{-1}B_Y\subset B_X\). So, \(\mu \) would not be minimal because \(\mu '<\mu \). The case \(\uplambda ^{-1} S_Y\cap \mu ^{-1} S_X=\emptyset \) is analogous.

The other implication is clear. \(\square \)

Proposition 3.2

The function d defined in  (3) is a multiplicative distance.

Proof

We need to show that d fulfils the following:

  1. (1)

    \(\mathrm{{d}}([\Vert \cdot \Vert _X],[\Vert \cdot \Vert _Y])=1\) if and only if \(\Vert \cdot \Vert _X\) and \(\Vert \cdot \Vert _Y\) are proportional.

  2. (2)

    \(\mathrm{{d}}([\Vert \cdot \Vert _X],[\Vert \cdot \Vert _Y])=\mathrm{{d}}([\Vert \cdot \Vert _Y],[\Vert \cdot \Vert _X])\).

  3. (3)

    \(\mathrm{{d}}([\Vert \cdot \Vert _X],[\Vert \cdot \Vert _Y])\le \mathrm{{d}}([\Vert \cdot \Vert _X],[\Vert \cdot \Vert _Z])\mathrm{{d}}([\Vert \cdot \Vert _Z],[\Vert \cdot \Vert _Y])\).

The first item is obvious since we have taken the quotient exactly for this.

It is clear that

$$\begin{aligned} \begin{aligned} \mathrm{{d}}([\Vert \cdot \Vert _X],[\Vert \cdot \Vert _Y])=&\min \{\mu :\Vert \cdot \Vert _X\le \Vert \cdot \Vert _Y\le \mu \Vert \cdot \Vert _X\} \\ =&\min \{\mu :\Vert \cdot \Vert _X\le \Vert \cdot \Vert _Y\le \mu \Vert \cdot \Vert _X\le \mu \Vert \cdot \Vert _Y\} \\ =&\min \{\mu :\Vert \cdot \Vert _Y\le \mu \Vert \cdot \Vert _X\le \mu \Vert \cdot \Vert _Y\} \\ =&\mathrm{{d}}([\Vert \cdot \Vert _Y],[\Vert \cdot \Vert _X]), \end{aligned} \end{aligned}$$
(4)

so the second item also holds.

For the third item, let \(\mu =\mathrm{{d}}([\Vert \cdot \Vert _X],[\Vert \cdot \Vert _Z]), \mu '=\mathrm{{d}}([\Vert \cdot \Vert _Z],[\Vert \cdot \Vert _Y])\). There exist \(\uplambda , \uplambda '\) such that

$$\begin{aligned} \Vert \cdot \Vert _X\le \uplambda \Vert \cdot \Vert _Z\le \mu \Vert \cdot \Vert _X\text {\quad and\quad }\Vert \cdot \Vert _Z\le \uplambda '\Vert \cdot \Vert _Y\le \mu '\Vert \cdot \Vert _Z. \end{aligned}$$

Joining these inequalities, we obtain \(\Vert \cdot \Vert _X\le \uplambda \Vert \cdot \Vert _Z\le \uplambda \uplambda '\Vert \cdot \Vert _Y\le \uplambda \mu '\Vert \cdot \Vert _Z\le \mu \mu '\Vert \cdot \Vert _X.\) This readily implies that \(\mathrm{{d}}([\Vert \cdot \Vert _X],[\Vert \cdot \Vert _Y])\le \mu \mu '=\mathrm{{d}}([\Vert \cdot \Vert _X],[\Vert \cdot \Vert _Z])\mathrm{{d}}([\Vert \cdot \Vert _Z],[\Vert \cdot \Vert _Y])\). \(\square \)

In order to define the operations in \(\mathcal Q\), we need the following:

Definition 3.3

Let us denote \(X=(\mathbb R^n,\Vert \cdot \Vert _X)\) and \(Y=(\mathbb R^n,\Vert \cdot \Vert _Y)\) and let \(\theta \in [0,1]\). We will call the space \(\mathbb R^n\) endowed with the quasinorm

$$\begin{aligned} \Vert \cdot \Vert _{(X,Y)_\theta }=\Vert \cdot \Vert _X^\theta \Vert \cdot \Vert _Y^{1-\theta } \end{aligned}$$

the interpolated space between X and Y at \(\theta \) and will denote it as \((X,Y)_\theta \).

3.4

Observe that this kind of interpolation cannot be applied directly to infinite-dimensional spaces unless we consider only equivalent quasinorms on a given space.

3.5

When dealing with vector spaces, it is customary to have clear which vector is the origin of the space, in function spaces it is the 0 function, in spaces of sequences it is the sequence \((0,0,\ldots )\). But we are giving a vector space structure to a set without a clear 0, so we need to choose it. The idea behind this work is that we have been given a kind of mean of two norms in a quite intuitive way, for our purposes, the most suitable candidate to be the mean of \(\Vert \cdot \Vert _X\) and \(\Vert \cdot \Vert _Y\) is

$$\begin{aligned} \Vert \cdot \Vert _{(X,Y)_{1/2}}=\Vert \cdot \Vert _X^{1/2}\Vert \cdot \Vert _Y^{1/2}. \end{aligned}$$

Of course, this means that when we choose the origin of our space, we are given the opposite \(\Vert \cdot \Vert _{\tilde{X}}\) for each \(\Vert \cdot \Vert _X\). The central rúle that the Euclidean norm plays in the classical analysis could be enough for it to be our origin, but there is another reason for choosing it. When we think of a non-strictly convex norm, it seems that it is, in some faint sense, an extreme point of a segment. A visual way to explain this is the curve \(\{[\Vert \cdot \Vert _p]:p\in [1,\infty ]\}\). If you reach a non-strictly convex norm like \([\Vert \cdot \Vert _1]\) or \([\Vert \cdot \Vert _\infty ]\) and you keep going in the same direction you will find that what you are dealing with is not convex any more. In this sense, the Euclidean norm is the most convex norm and it deserves to be the centre of our vector space. The space \((\mathbb R^n,\Vert \cdot \Vert _2)\) is, up to isometric isomorphism, the only homogeneous n-dimensional space and so, the one with the greatest group of isometries. So, we have defined our vector space as follows:

Definition 3.6

Let \(n\in \mathbb N\) and consider \(\mathcal Q\) as the quotient of the set of quasinorms on \(\mathbb R^n\) by the equivalence relation of dilating quasinorms. We consider \(\left[ \Vert \cdot \Vert _2\right] \) as the origin of our space and the mean of two classes of quasinorms as

$$\begin{aligned} ([\Vert \cdot \Vert _X],[\Vert \cdot \Vert _Y])_{1/2}=\left[ \Vert \cdot \Vert _X^{1/2}\Vert \cdot \Vert _Y^{1/2}\right] , \end{aligned}$$

so the opposite of some \([\Vert \cdot \Vert _X]\) is \([\Vert \cdot \Vert _{\tilde{X}}]\), where

$$\begin{aligned} \Vert \cdot \Vert _{\tilde{X}}=\frac{\Vert \cdot \Vert ^2_2}{\Vert \cdot \Vert _X} \end{aligned}$$

on \(\mathbb R^n\setminus \{0\}\) and \(\Vert 0\Vert _{\tilde{X}}=0\); the scalar multiplication is given by

$$\begin{aligned} \theta \star \left[ \Vert \cdot \Vert _X\right] =\left[ \Vert \cdot \Vert _X^\theta \Vert \cdot \Vert _2^{1-\theta }\right] ,\quad -\theta \star \left[ \Vert \cdot \Vert _X\right] =\left[ \Vert \cdot \Vert _{\widetilde{X}}^\theta \Vert \cdot \Vert _{2}^{1-\theta }\right] \end{aligned}$$

for \(\theta \in [0,\infty )\); and the addition of two classes of quasinorms by

$$\begin{aligned} \left[ \Vert \cdot \Vert _X\right] \oplus \left[ \Vert \cdot \Vert _Y\right] =2\star \left[ \Vert \cdot \Vert _{(X,Y)_{1/2}}\right] . \end{aligned}$$

Theorem 3.7

With the above operations, \(\mathcal Q\) is a linear space. If we, moreover, define

$$\begin{aligned} \big |\!\big |\!\big | \Vert \cdot \Vert _X \big |\!\big |\!\big |=\log _2(\mathrm{{d}}(\Vert \cdot \Vert _X,\Vert \cdot \Vert _2)), \end{aligned}$$

then \(\left( \mathcal Q,\big |\!\big |\!\big |\cdot \big |\!\big |\!\big |\right) \) is a Banach space where the set of equivalence classes of norms in \(\mathbb R^n\) is closed.

Proof

It is easy to see that, whenever \(\Vert \cdot \Vert _X\) and \(\Vert \cdot \Vert _Y\) are quasinorms over a finite-dimensional space \(\mathbb R^n\) and \(\theta >0\), the subset

$$\begin{aligned} B_\theta =\{x\in \mathbb R^n:\Vert x\Vert _X^\theta \Vert x\Vert _Y^{1-\theta }\le 1\} \end{aligned}$$

is bounded, absorbing, and balanced, and its boundary is bounded away from 0, so Theorem 2.5 implies that \(\Vert \cdot \Vert _X^\theta \Vert \cdot \Vert _Y^{1-\theta }\) is a quasinorm, and it is clear that it is continuous. So, this kind of extrapolation of quasinorms is well defined. In order to show that \(\mathcal Q\) is a linear space we need to show that the scalar multiplication and the addition are well defined. On the one hand, it is clear that the operations do not depend on the representative of any class of quasinorms. On the other hand, all the expressions in Definition 3.6 give rise to a continuous quasinorm.

In [9, Theorem 1.18] it is seen that the distance we are dealing with is complete on \(\mathcal N\), and this implies that \(\mathcal N\) is closed in any metric space where it is isometrically embedded, in particular in \(\mathcal Q\). Anyway, it is not hard to see that its complement \(\mathcal Q\setminus \mathcal N\) is open.

Now, we need to show that d is absolutely homogeneous and additively invariant.

For the homogeneity, let \(\theta \in (0,\infty )\) and take any \(\Vert \cdot \Vert _X\) such that \(\Vert \cdot \Vert _X\ge \Vert \cdot \Vert _2\) and \(S_X\cap S_2\ne \emptyset \). Then, \((\Vert \cdot \Vert _X,\Vert \cdot \Vert _2)_\theta \) fulfils the same, i.e. \((\Vert \cdot \Vert _X,\Vert \cdot \Vert _2)_\theta \ge \Vert \cdot \Vert _2\) and \(S_{(\Vert \cdot \Vert _X,\Vert \cdot \Vert _2)_\theta }\cap S_2\ne \emptyset \). Moreover, if we take \(y\in S_2\) such that

$$\begin{aligned} \mathrm{{d}}([\Vert \cdot \Vert _X],[\Vert \cdot \Vert _2])=\Vert y\Vert _X, \end{aligned}$$

then it is quite clear that

$$\begin{aligned} \mathrm{{d}}([(\Vert \cdot \Vert _X,\Vert \cdot \Vert _2)_\theta ],[\Vert \cdot \Vert _2])=\Vert y\Vert _X^\theta . \end{aligned}$$

For negative values of \(\theta \) we only need to see what happens when \(\theta =-1\), but it is easily seen that \(\mathrm{{d}}([\Vert \cdot \Vert _X],[\Vert \cdot \Vert _2])=\mathrm{{d}}([\Vert \cdot \Vert _{\tilde{X}}],[\Vert \cdot \Vert _2])\).

To see that d is additively invariant, take \(\Vert \cdot \Vert _X,\Vert \cdot \Vert _Y,\Vert \cdot \Vert _Z\). We just need to show that

$$\begin{aligned} \mathrm{{d}}([(\Vert \cdot \Vert _X,\Vert \cdot \Vert _Z)_{1/2}],[(\Vert \cdot \Vert _Y,\Vert \cdot \Vert _Z)_{1/2}])=\mathrm{{d}}([\Vert \cdot \Vert _X],[\Vert \cdot \Vert _Y])^{1/2}. \end{aligned}$$

For any \(z\in \mathbb R^n, z\ne 0\) one has

$$\begin{aligned} \frac{\Vert z\Vert _{(X,Z)_{1/2}}}{\Vert z\Vert _{(Y,Z)_{1/2}}}= \frac{\Vert z\Vert _X^{1/2}\Vert z\Vert _Z^{1/2}}{\Vert z\Vert _Y^{1/2}\Vert z\Vert _Z^{1/2}}= \left( \frac{\Vert z\Vert _X}{\Vert z\Vert _Y}\right) ^{1/2}. \end{aligned}$$
(5)

Let \(\mu =\mathrm{{d}}([\Vert \cdot \Vert _X],[\Vert \cdot \Vert _Y]).\) Applying Lemma 3.1 we may suppose that \(\Vert \cdot \Vert _X\le \Vert \cdot \Vert _Y\le \mu \Vert \cdot \Vert _X\) and choose x, y such that \(\Vert x\Vert _X=\Vert x\Vert _Y=1,\) \(\Vert y\Vert _X=1/\mu \) and \(\Vert y\Vert _Y=1.\) Now (5) implies that

$$\begin{aligned} \begin{aligned} \mathrm{{d}}([\Vert \cdot \Vert _X],[\Vert \cdot \Vert _Y])^{1/2}&= \mu ^{1/2}= \left( \frac{\Vert x\Vert _X}{\Vert x\Vert _Y}\frac{\Vert y\Vert _Y}{\Vert y\Vert _X}\right) ^{1/2}= \frac{\Vert x\Vert _{(X,Z)_{1/2}}}{\Vert x\Vert _{(Y,Z)_{1/2}}} \frac{\Vert y\Vert _{(Y,Z)_{1/2}}}{\Vert y\Vert _{(X,Z)_{1/2}}} \\&\le \mathrm{{d}}([(\Vert \cdot \Vert _X,\Vert \cdot \Vert _Z)_{1/2}],[(\Vert \cdot \Vert _Y,\Vert \cdot \Vert _Z)_{1/2}]) \end{aligned} \end{aligned}$$

Applying Lemma 3.1 to \((\Vert \cdot \Vert _X,\Vert \cdot \Vert _Z)_{1/2}\) and \((\Vert \cdot \Vert _Y,\Vert \cdot \Vert _Z)_{1/2}\) we see that the symmetric inequality also holds.

It remains to show the completeness of our norm. Take a Cauchy sequence

$$\begin{aligned} \big (\big [\Vert \cdot \Vert ^1\big ],\big [\Vert \cdot \Vert ^2\big ],\ldots ,\big [\Vert \cdot \Vert ^k\big ],\ldots \big )\subset \mathcal Q. \end{aligned}$$

We may choose a representative of each class, so we may suppose that \(\Vert \cdot \Vert ^k(e_1)=1\) for every \(k\in \mathbb N.\) Every Cauchy sequence is bounded, so we may take \(\varepsilon , M>0\) such that

$$\begin{aligned} \varepsilon \Vert \cdot \Vert _2\le \Vert \cdot \Vert ^k\le M\Vert \cdot \Vert _2 \text { for every }k. \end{aligned}$$
(6)

With this in mind, the very definition of \(\big |\!\big |\!\big |\cdot \big |\!\big |\!\big |\) implies that for every \(x\in \mathbb R^n\) the sequence \(\Vert x\Vert ^k\) is also Cauchy, so we may define \(\Vert x\Vert _X\) as the limit of \(\Vert x\Vert ^k\) as \(k\rightarrow \infty \). By (6) we have that \(B_X\) is a bounded, balanced, neighbourhood of 0, so \(\Vert \cdot \Vert _X\) is a quasinorm and it is continuous because, locally, it is the uniform limit of continuous quasinorms. It is easy see that it is the limit of the sequence, and this implies that \(\big |\!\big |\!\big |\cdot \big |\!\big |\!\big |\) is complete on \(\mathcal Q\). \(\square \)

Remark 3.8

The set of norms is not convex in \(\mathcal Q\). In fact, if we define \(\Vert (a,b)\Vert _X=2|a|+|b|/2,\) \(\Vert (a,b)\Vert _Y=2|b|+|a|/2\) then we have

$$\begin{aligned} \Vert (1,0)\Vert _X=2=\Vert (0,1)\Vert _Y,\quad \Vert (0,1)\Vert _X=1/2=\Vert (1,0)\Vert _Y, \end{aligned}$$

but \(\Vert (1,1)\Vert _X=5/2=\Vert (1,1)\Vert _Y\), which implies that

$$\begin{aligned} \Vert (1,1)\Vert _{(X,Y)_{1/2}}>\Vert (1,0)\Vert _{(X,Y)_{1/2}}+\Vert (0,1)\Vert _{(X,Y)_{1/2}}. \end{aligned}$$
Fig. 1
figure 1

The spheres of the three quasinorms of Remark 3.8

Full size image

3.9

We can describe the space \(\mathcal Q\) as some C(K). Namely, let \(\mathbb P_{n-1}\) be the projective space of dimension \(n-1\), i.e.

$$\begin{aligned} \mathbb P_{n-1}=\big (\mathbb R^n\setminus \{0\}\big )/\negthinspace \sim , \text { with } x\sim y \text { if and only if } x=\uplambda y \text { for some }\uplambda \in \mathbb R\setminus \{0\}, \end{aligned}$$

endowed with the quotient topology relative to the projection \(\mathbb R^n\setminus \{0\}\rightarrow \big (\mathbb R^n\setminus \{0\}\big )/\negthinspace \sim ,\ x\mapsto [x]\). In the sequel, we will think the projective space of dimension \(n-1\) as the quotient \(S^{n-1}/\negthinspace \sim \), where \(x\sim y\) if and only if \(x=\pm y\) and \(S^{n-1}\) denotes the sphere of \((\mathbb R^n,\Vert \cdot \Vert _2)\).

This is essentially the same idea as that reflected in Subsection 3.2 (Proof of the main result) in [9].

As every quasinorm is absolutely homogeneous, i.e. \(\Vert \uplambda x\Vert =|\uplambda |\Vert x\Vert \) for any \(\uplambda \in \mathbb R, x\in X\), \(\Vert \cdot \Vert \) is always determined by its value at every point of, say, the Euclidean sphere \(S^{n-1}\). Take into account now the universal property of the quotient, that assures that any continuous function \(f:S_X\rightarrow \mathbb R\) such that \(f(x)=f(-x)\) gives rise to a well-defined and continuous \(\tilde{f}:\mathbb P_{n-1}\rightarrow \mathbb R,\ \tilde{f}([x])=f(x)\). With this in mind, it is clear that each continuous quasinorm \(\Vert \cdot \Vert _X\) defines a continuous function \(f_X:\mathbb P_{n-1}\rightarrow (0,\infty )\).

Recall that \(\mathbb P_{n-1}\) is compact—it is the continuous image of a compact space—so every continuous \(f:\mathbb P_{n-1}\rightarrow (0,\infty )\) is bounded from above and bounded away from 0, and we can define a quasinorm on \(\mathbb R^n\) as \(\Vert \uplambda x\Vert _f=|\uplambda |f([x])\) for every \(\uplambda \in \mathbb R,\ x\in S_2\).

It is clear that this is a one-to-one correspondence between the space of continuous quasinorms \(\Vert \cdot \Vert :\mathbb R^n\rightarrow \mathbb R\) and the space of positive continuous functions \(\mathbb P_{n-1}\rightarrow (0,\infty )\). If we consider again the equivalence relation \(\Vert \cdot \Vert _X\sim \Vert \cdot \Vert _Y\iff \Vert \cdot \Vert _X=\uplambda \Vert \cdot \Vert _Y\), \(\uplambda \in \mathbb R\setminus \{0\}\), then the correspondence still holds if we consider \(C(\mathbb P_{n-1})\) endowed with the equivalence relation \(f\equiv g\iff f=\uplambda g\), \(\uplambda \in \mathbb R\setminus \{0\}\). So, we have a bijection \(\mathcal Q\longleftrightarrow C(\mathbb P_{n-1},(0,\infty ))/\!\equiv \). To end the description of \(\mathcal Q\) we just need to consider \(\log :C(\mathbb P_{n-1},(0,\infty ))\rightarrow C(\mathbb P_{n-1})\), endow this space with the equivalence relation \(f\sim g\iff f=\uplambda +g\) for some \(\uplambda \in \mathbb R\)—to preserve the bijection with the former space—and, for any \([f],[g]\in C(\mathbb P_{n-1})/\negthinspace \sim \) define the metric

$$\begin{aligned}&\mathrm{{d}}([f],[g])=\max _{x\in \mathbb P_{n-1}}\{f(x)-g(x)\}-\min _{x\in \mathbb P_{n-1}}\{f(x)-g(x)\}, \text { where } f,g\\&\quad \text { are any representatives of }[f], [g]. \end{aligned}$$

Observe that this value is the range of \(f-g\) and that we can rewrite this as

$$\begin{aligned}&\mathrm{{d}}([f],[g])=\max _{x\in \mathbb P_{n-1}}\{f(x)-g(x)\}+\max _{x\in \mathbb P_{n-1}}\{g(x)-f(x)\}, \text { where } f,g \\&\quad \text { are any representatives of }[f], [g]. \end{aligned}$$

With this, we have \([\Vert \cdot \Vert _2]\in \mathcal Q\mapsto [0]\in C(\mathbb P_{n-1})/\negthinspace \sim \) (see, again, [9], Subsection 3.2). We also have that the map between \((\mathcal Q,\big |\!\big |\!\big |\cdot \big |\!\big |\!\big |)\) and \(C(\mathbb P_{n-1})/\negthinspace \sim \) is an onto isometry. Moreover, if \(e_1\) is the first vector of the usual basis of \(\mathbb R^n\), then we can see the latter space as

$$\begin{aligned} C_0(\mathbb P_{n-1})=\{f\in C(\mathbb P_{n-1}):f([e_1])=0\}, \end{aligned}$$

whose bijection with some space of quasinorms arises from considering only the quasinorms in \(\mathcal Q_0\) that take value 1 at \(e_1\).

The reader interested in Projective Geometry can check out [4, 13].

4 The Banach–Mazur Compactum

4.1

The space of \(n\times n\) real matrices will be denoted as \(\mathcal {M}_n\).

Every time we write isometry we will mean linear isometry. This, by the Mazur–Ulam Theorem, means just that we will consider only isometries sending 0 to 0.

As we will deal just with finite-dimensional spaces, we can fix the standard basis of \(\mathbb R^n\), so that each operator \(T:(\mathbb R^n,\Vert \cdot \Vert _X)\rightarrow (\mathbb R^n,\Vert \cdot \Vert _Y)\) can be seen as a matrix \(A\in \mathcal {M}_n\). We will use \(AB_X=\{Ax:x\in B_X\}\) and \(TB_X\) indistinctly.

We will intertwine operators and norms and will need some notation for the norm whose value at each x is \(\Vert Ax\Vert _X\) (respectively, \(\Vert Tx\Vert _X\)), where \(A\in {\text {GL}}(n)\) (respectively, T is a linear isomorphism). This will be written as \(A^*\Vert \cdot \Vert _X\) (respectively, \(T^*\Vert \cdot \Vert _X\)).

Now that we have determined the structure of \(\mathcal Q\), we may relate it to the well-known Banach–Mazur compactum. This compactum is obtained by endowing the set \(\mathcal {N}_0\) of norms defined on \(\mathbb R^n\) with the pseudometric

$$\begin{aligned} \mathrm{{d}}_{BM}(\Vert \cdot \Vert _X,\Vert \cdot \Vert _Y)=\min \left\{ \big \Vert T\big \Vert \big \Vert T^{-1}\big \Vert \right\} , \end{aligned}$$

where the minimum is taken in

$$\begin{aligned} \{T:(\mathbb R^n,\Vert \cdot \Vert _X)\rightarrow (\mathbb R^n,\Vert \cdot \Vert _Y)\text { is a linear isomorphism}\}. \end{aligned}$$

This pseudometric does not distinguish between isometric norms, so the quotient needed to turn it into a metric is by the equivalence relation

$$\begin{aligned} \Vert \cdot \Vert _X\equiv \Vert \cdot \Vert _Y\text { when there is a linear isometry }T:(\mathbb R^n,\Vert \cdot \Vert _X)\rightarrow (\mathbb R^n,\Vert \cdot \Vert _Y). \end{aligned}$$

As we are dealing with finite-dimensional spaces, the isomorphism T can be seen as an invertible matrix of order n, i.e. T is associated to some \(A\in {\text {GL}}(n)\). Conversely, every invertible matrix gives an isomorphism, so the Banach–Mazur distance can be seen as

$$\begin{aligned} \mathrm{{d}}_{BM}(\Vert \cdot \Vert _X,\Vert \cdot \Vert _Y)=\min \{\mu :B_X\subset AB_Y\subset \mu B_X\text { for some }A\in {\text {GL}}(n)\} \end{aligned}$$

and the quotient as

$$\begin{aligned} \Vert \cdot \Vert _X\equiv \Vert \cdot \Vert _Y\text { if and only if there is } A\in {\text {GL}}(n) \text { such that } AB_X=B_Y. \end{aligned}$$

As the equivalence relation \(\sim \) that defines \(\mathcal Q\) can obviously be seen as

$$\begin{aligned} \Vert \cdot \Vert _X\sim \Vert \cdot \Vert _Y\text { if and only if there is } \uplambda \ne 0 \text { such that } \uplambda B_X=B_Y, \end{aligned}$$

if we denote \(\mathbb R^*=\mathbb R\setminus \{0\}\), then the relation between both spaces seems to be given by \({\text {PGL}}(n,\mathbb R)={\text {GL}}(n)/\mathbb R^*\). It is, however, a little more complex.

Let us study the fibres in \(\mathcal N=\{\text {Norms\ defined\ on\ }\mathbb R^n\}/\negthinspace \sim \) of each element of \(BM=(\mathcal N_0/\equiv )=(\mathcal N/\equiv )\). Suppose we are given a norm \(\Vert \cdot \Vert _X\) whose group of autoisometries is trivial, i.e. the only (linear) isometries \((\mathbb R^n,\Vert \cdot \Vert _X)\rightarrow (\mathbb R^n,\Vert \cdot \Vert _X)\) are the identity and its opposite. Then, \(AB_X= CB_X\) implies \(A=\pm C\) and this means that the fibre of \([\Vert \cdot \Vert _X]\in BM\) in \(\mathcal N\) is indeed \(\{A^*\Vert \cdot \Vert _X:A\in {\text {GL}}(n)\}/\mathbb R^*\). However, if \(\Vert \cdot \Vert _X\) has non-trivial group of autoisometries then \(AB_X=AGB_X\) whenever \(G:(\mathbb R^n,\Vert \cdot \Vert _X)\rightarrow (\mathbb R^n,\Vert \cdot \Vert _X)\) is an isometry. Denoting as \({\text {Iso}}_X\) this group of autoisometries for each \(\Vert \cdot \Vert _X\) we obtain a one-to-one relation

$$\begin{aligned} \mathcal N\longleftrightarrow \{(\{[\Vert \cdot \Vert _X]\}\times {\text {PGL}}(n,\mathbb R))/{\text {Iso}}_X:[\Vert \cdot \Vert _X]\in BM\}. \end{aligned}$$

Before we proceed with the main result in this section we need a couple of results about the group \({\text {Iso}}_X\). As is customary, the distance between two linear operators \(F,G:(\mathbb R^n,\Vert \cdot \Vert _X)\rightarrow (\mathbb R^n,\Vert \cdot \Vert _Y)\) is defined as the operator norm of its difference:

$$\begin{aligned} \mathrm{{d}}(F,G)=\Vert F-G\Vert _Y=\max \{\Vert F(x)-G(x)\Vert _Y:x\in B_X\}. \end{aligned}$$

The first result we need is as follows:

Lemma 4.2

Let \(F:(\mathbb R^n,\Vert \cdot \Vert _X)\rightarrow (\mathbb R^n,\Vert \cdot \Vert _X)\) be an isometry. Then there are linearly independent \(u,v\in \mathbb R^n\) such that the plane \(\langle u,v\rangle \) is invariant for F and such that the matrix of the restriction of F to \(\langle u,v\rangle \) with respect to the basis \(\{u,v\}\) is one of the following:

$$\begin{aligned} \left( \begin{array}{c c} \cos (\alpha )&{}-\sin (\alpha )\\ \sin (\alpha )&{}\cos (\alpha ) \end{array} \right) , \qquad \left( \begin{array}{c c} 1 &{} 0\\ 0 &{} -1 \end{array} \right) , \end{aligned}$$
(7)

where \(\alpha \in (-\pi ,\pi ]\).

Proof

If \(F=\pm {\text {Id}}\) then the results holds with \(\alpha =0,\pi \), so we assume henceforth that this is not the case. It is well known that every linear endomorphism \(F:(\mathbb R^n,\Vert \cdot \Vert _X)\rightarrow (\mathbb R^n,\Vert \cdot \Vert _X)\) has at least one complex eigenvalue \(\uplambda \) (see, e.g. [2, 9.8]) and non-real eigenvalues occur in conjugate pairs. If \(\uplambda =a+bi\not \in \mathbb R\), then there are \(u,v\in X\setminus \{0\}\) such that \(F(u)=au-bv\) and \(F(v)=av+bu\). Let \(|\uplambda |=\sqrt{a^2+b^2}\) and H be the plane generated by u and v endowed by the basis \(\{u,v\}\), observe that \(F(H)=H\). Then, there is \(\alpha \in (-\pi ,\pi )\) such that the matrix of the restriction of F to H is

$$\begin{aligned} |\uplambda | \left( \begin{array}{c c} \cos (\alpha )&{}-\sin (\alpha )\\ \sin (\alpha )&{}\cos (\alpha ) \end{array} \right) . \end{aligned}$$

As F is an isometry, one has \(\Vert F^k(u)\Vert _X=1\) for every \(k\in \mathbb N\), so the sequence \((F^k(u))_k\) is bounded (with respect to every norm) and it is clear that this implies that \(|\uplambda |=1\).

Suppose, now, that every eigenvalue is real, and let \(\uplambda \in \mathbb R\) and \(u\in S_X\) be such that \(F(u)=\uplambda u\). It is obvious that, again, \(|\uplambda |=1\). If F has at least two different eigenvalues then we may suppose \(\uplambda =1\) and the other eigenvalue must be \(-1\), so let v be such that \(F(v)=-v\). With respect to the basis \(\{u,v\}\) the matrix of the restriction of F to \(\langle u,v\rangle \) is

$$\begin{aligned} \left( \begin{array}{c c} 1 &{} 0\\ 0 &{} -1 \end{array} \right) . \end{aligned}$$

This leaves the case where all the eigenvalues of F are real and are the same. We may suppose that all of them equal 1. As we are assuming that \(F\ne {\text {Id}}\), it is clear that the dimension of \(\ker (F-{\text {Id}})\) is at most \(n-1\) and the Cayley–Hamilton Theorem implies that \(\dim (\ker (F-{\text {Id}})^2)>\dim (\ker (F-{\text {Id}}))\). This means that we may find \(v\in S_X,\) \(u\in X\setminus \{0\},\) such that \(u=F(v)-v\ne 0,\) and \( F(u)-u=(F-{\text {Id}})(u)=(F-{\text {Id}})^2(v)=0\). Thus, we have \(F(v)=u+v\) and \(F(u)=u\) and this implies that in the plane \(\langle u,v\rangle \) endowed with the basis \(\{u,v\}\), the matrix of the restriction of F is

$$\begin{aligned} \left( \begin{array}{c c} 1 &{} 1\\ 0 &{} 1 \end{array} \right) , \end{aligned}$$
(8)

which leads to the matrix of \(F^k\):

$$\begin{aligned} \left( \begin{array}{c c} 1 &{} k\\ 0 &{} 1 \end{array} \right) . \end{aligned}$$
(9)

So, the sequence \((F^k(v))_k=(ku+v)_k\) is unbounded and we are done. \(\square \)

Remark 4.3

It is easy to see that the same computation as the one at the end of the proof of Lemma 4.2 rules out the option that F has some Jordan block of the form:

$$\begin{aligned} \left( \begin{array}{c c c c} \cos (\alpha ) &{} -\sin (\alpha ) &{} 1 &{} 0 \\ \sin (\alpha ) &{} \cos (\alpha ) &{} 0 &{} 1 \\ 0 &{} 0 &{} \cos (\alpha ) &{} -\sin (\alpha ) \\ 0 &{} 0 &{} \sin (\alpha ) &{} \cos (\alpha ) \end{array} \right) , \end{aligned}$$
(10)

so every isometry is diagonalizable over \(\mathbb C\).

Remark 4.4

We have not used the fact that F is an isometry, we merely needed that the sequence \((\Vert F^k\Vert _X)_k\) is bounded and bounded away from 0.

Proposition 4.5

Whenever \(\Vert \cdot \Vert _X\) has non-trivial group of isometries, there is some autoisometry \(F:(\mathbb R^n,\Vert \cdot \Vert _X)\rightarrow (\mathbb R^n,\Vert \cdot \Vert _X)\) such that \(\min \{\Vert F+{\text {Id}}\Vert _X,\Vert F-{\text {Id}}\Vert _X\}\ge 1\).

Proof

Let \(F\in {\text {Iso}}_X, F\ne \pm {\text {Id}}\). Then \(\max \{\Vert F(x)+x\Vert _X,\Vert F(x)-x\Vert _X\}\le 2\) for every \(x\in S_X\), so \(\max \{\Vert F+{\text {Id}}\Vert _X,\Vert F-{\text {Id}}\Vert _X\}\le 2\). If all the eigenvalues of F are real, then the proof of Lemma 4.2 implies that there are \(u,v\in S_X\) such that \(\Vert F(u)+u\Vert _X=2, \Vert F(v)-v\Vert _X=2\), so we actually have \(\Vert F+{\text {Id}}\Vert _X=\Vert F-{\text {Id}}\Vert _X=2\).

If some eigenvalue is not real, say \(\uplambda =a+bi, b\ne 0\), then we know by Lemma 4.2 that \(|\uplambda |=1\). Let \(u\in S_X,v\in X\setminus \{0\}\) be such that the matrix of the restriction of F to \(H=\langle u,v\rangle \) is, with respect to the basis \(\{u,v\}\), the rotation of angle \(\alpha \)

$$\begin{aligned} \left( \begin{array}{c c} \cos (\alpha )&{}-\sin (\alpha )\\ \sin (\alpha )&{}\cos (\alpha ) \end{array} \right) \end{aligned}$$

for some \(\alpha \in (-\pi ,\pi ]\), the existence of such a basis is outlined in the proof of Lemma 4.2. If \(\alpha >\pi /2\) (respectively, \(\alpha \le -\pi /2\)) then we may compose with \(-{\text {Id}}\) and get the rotation of angle \(-\pi +\alpha \) (respectively, \(\pi +\alpha \)), so we may suppose \(\alpha \in (-\pi /2,\pi /2]\). If \(\alpha <0\) then the inverse of \(F_{|H}\) is the rotation of angle \(-\alpha \), so we only need to deal with \(\alpha \in [0,\pi /2]\). As \(\alpha =0\) gives the identity, what we have is \(\alpha \in (0,\pi /2]\). Now we have to break down the different options.

If \(\alpha =\pi /m\) for some \(m\in \mathbb N\), then \(F_{|H}^m=-{\text {Id}}_{|H}\). Consider the half-orbit of u, \(\{x_0=u,x_1=F(u),\ldots ,x_m=F^m(u)=-u\}\).

If \(m\in 2\mathbb {Z}\), then \(x_{m/2}=v\) is at the same distance from u and \(-u\) because \(F^{m/2}(u)=v\) and \(F^{m/2}(v)=-u\). Indeed,

$$\begin{aligned} \Vert v-u\Vert _X=\Vert F^{m/2}(u)-u\Vert _X=\Vert F^{m/2}(F^{m/2}(u)-u)\Vert _X=\Vert -u-v\Vert _X. \end{aligned}$$

This readily implies that

$$\begin{aligned} \min \{\Vert F^{m/2}-{\text {Id}}\Vert _X, \Vert F^{m/2}+{\text {Id}}\Vert _X\}\ge \Vert v-u\Vert _X=\Vert v+u\Vert _X\ge 1, \end{aligned}$$
(11)

where the last inequality holds because of the triangular inequality:

$$\begin{aligned} 2=\Vert v+v\Vert _X\le \Vert v-u\Vert _X+\Vert v+u\Vert _X=2\Vert v-u\Vert _X. \end{aligned}$$

For \(m\not \in 2\mathbb N\) we are going to restrict every coordinate-wise computation to the plane H, the difference would be a certain amount of zeroes after the two first coordinates. Taking coordinates with respect to \(\{u,v\}\), we have

$$\begin{aligned} x_{(m-1)/2}= & {} \left( \cos \left( \frac{(m-1)\pi }{2m}\right) \!\!, \sin \left( \frac{(m-1)\pi }{2m}\right) \right) ,\\ x_{(m+1)/2}= & {} \left( \cos \left( \frac{(m+1)\pi }{2m}\right) \!\!, \sin \left( \frac{(m+1)\pi }{2m}\right) \right) . \end{aligned}$$

We are going to show that \(\Vert x_{(m-1)/2}-u\Vert _X\ge 1\). For this, we first need to check that the segment whose endpoints are \(x_{(m\pm 1)/2}\) equals the intersection of the line that contains both of them with the unit ball \(B_X\). Observe that the first coordinate of \(x_{(m+1)/2}\) is the opposite of the first coordinate of \(x_{(m-1)/2}\) and that the second coordinates of both points agree. So, if we denote by r the horizontal line whose height is \(\sin ((m+1)\pi /2m)\), we have \(x_{(m\pm 1)/2}\in r\cap S_X\). The convexity of \(B_X\) implies that if there are three collinear points in \(S_X\), then the segment determined by them is included in \(S_X\), too. In particular, if there is some \(y\in (S_X\cap r)\setminus \{x_{(m\pm 1)/2}\}\), then the segment whose endpoints are \(x_{(m\pm 1)/2}\) is included in \(S_X\). On the one hand, this means that the Euclidean regular 2m-gon with vertices in every \(x_k\) is included in \(S_X\) because each segment of the 2m-gon is the image of this segment by some \(F^k\). On the other hand, under these circumstances it is clear that this 2m-gon is \(S_X\cap H\), so in any case, \((t,\sin ((m-1)\pi /2m))\in B_X\) if and only if

$$\begin{aligned} t\in [-\cos ((m-1)\pi /2m)),\cos ((m-1)\pi /2m))]. \end{aligned}$$
(12)

If \(m=3\), then \(x_1=(1/2,\sqrt{3}/2)\) and \(x_2=(-1/2,\sqrt{3}/2)\), so \(x_1-u=x_2\). This implies that \(\Vert x_1-u\Vert _X=\Vert x_2+u\Vert _X=1\), and also \(\Vert x_1+u\Vert _X=\Vert x_2-u\Vert _X>1\). So, \(\min \{\Vert F+{\text {Id}}\Vert _X,\Vert F-{\text {Id}}\Vert _X\}\ge 1\).

If \(m\ge 5\), then \(0<\cos ((m-1)\pi /2m))<\cos (\pi /3)=1/2\) and this, along with (12), implies that

$$\begin{aligned} x_{(m-1)/2}-u=(\cos ((m-1)\pi /2m))-1,\sin ((m-1)\pi /2m))) \end{aligned}$$

lies outside the unit ball, so \(\Vert F^{(m-1)/2}\pm {\text {Id}}\Vert _X>1\).

If \(\alpha =\frac{p}{q}\pi \) for some coprime \(p,q\in \mathbb N\), then the Chinese Remainder Theorem implies that the rotation of angle \(\pi /q\) is also an isometry and we are in the previous case.

If \(\alpha \not =\frac{p}{q}\pi \) for any \(p,q\in \mathbb N\), then the orbit of u is dense in \(S_H\) and, actually, in the sphere of \(\Vert \cdot \Vert _2\), i.e. in \(\{\uplambda u+\mu v\in H:\uplambda ^2+\mu ^2=1\}\). The continuity of \(\Vert \cdot \Vert _X\) with respect to any norm defined over H implies that \(\Vert \cdot \Vert _X\) restricted to H is \(\Vert \cdot \Vert _2\) and so, any map \(F^k\) that sends u close enough to v has distance to \(\pm {\text {Id}}\) close to \(\sqrt{2}>1\). \(\square \)

Definition 4.6

Let \(\Vert \cdot \Vert _X\) be a norm defined over \(\mathbb R^n\). We say that \(\Vert \cdot \Vert _X\) is a polyhedral norm or, equivalently, that \((\mathbb R^n,\Vert \cdot \Vert _X)\) is a polyhedral space, if its closed unit ball is a polytope.

Definition 4.7

Given a normed space \((X,\Vert \cdot \Vert _X)\), we say that \(x\in S_X\) is an exposed point if there is \(f\in X^*\) such that \(f(x)=1\) and \(f(y)<1\) for every \(y\in S_X, y\ne x\). We say that \(x\in S_X\) is an extreme point if it does not lie in the interior of a segment included in \(S_X\).

4.8

It is clear that if \(B_X\) is a polytope, then \(x\in B_X\) is extreme if and only if it is exposed.

We will need the following weak version of the Krein–Milman Theorem, see [10]:

Theorem 4.9

(Krein–Milman) The unit ball of every finite-dimensional normed space is the convex hull of its subset of extreme points.

In the proof of Theorem 4.11 we will also use the Brouwer fixed-point Theorem, see [11, Theorem 6] or directly, [14]:

Theorem 4.10

If C is a closed convex subset of a Banach space, then every compact continuous map \(f : C\rightarrow C\) has a fixed point. In particular, if C is convex and compact, then every continuous map \(f : C\rightarrow C\) has a fixed point.

Now we can proceed with the main result in this section.

Theorem 4.11

Let \(U=\{[\Vert \cdot \Vert _X]\in \mathcal N:{\text {Iso}}_X=\{{\text {Id}},-{\text {Id}}\}\}.\) Then, U is a dense open subset of \(\mathcal N\).

Proof

To see that U is dense we need the following fact:

The subset of equivalence classes of polyhedral norms is dense in \(\mathcal N\). This is clear from [5, Theorem 1.1].

With this fact in mind, and given some polyhedral norm \(\Vert \cdot \Vert _X\), we are going to sketch how to construct a norm with trivial group of isometries and whose distance to \(\Vert \cdot \Vert _X\) is as small as we want. The Krein–Milman Theorem implies that there is a basis \(\mathcal B=\{x_1,\ldots ,x_n\}\) such that every \(x_i\) is an exposed point of \(B_X\). Given \(\delta >0\) we may consider

$$\begin{aligned} x_{n+1}=\frac{1+\delta }{\Vert (1,\ldots ,1)\Vert _X}(1,\ldots ,1) \end{aligned}$$

and the norm \(\Vert \cdot \Vert _{X'}\) whose unit ball is the convex hull of \(B_X\cup \{\pm x_{n+1}\}\). On the one hand, this norm is as close as we want to \(\Vert \cdot \Vert _X\), so we just need to approximate \(\Vert \cdot \Vert _{X'}\). On the other hand, every \(x_i\) with \(i=1,\ldots ,n+1\) is exposed in \(B_{X'}\). Indeed, for each \(i\in \{1,\ldots ,n\}\), consider some linear \(f_i:\mathbb R^n\rightarrow \mathbb R\) such that \(f_i(x_i)=1\) and \(f_i(y)<1\) for every \(y\in S_X, y\ne x_i\). It is clear that there exist \(\alpha _1,\ldots ,\alpha _n\in (0,\infty )\) such that \(f_i(x_{n+1})<(1-\alpha _i)\). As there are finitely many \(\alpha _i\) we can choose \(\delta >0\) so that \(f_i(x_{n+1})<(1-\alpha _i)(1+\delta )<1\) for every \(i=1,\ldots ,n\). This means that, when \(\delta >0\) is small enough, \(f_i(x_{n+1})<1\). The only points in \(B_{X'}\) that do not belong to \(B_X\) are \(x_{n+1}\) and convex combinations \(\uplambda x_{n+1}+(1-\uplambda )z\) with \(z\in B_X\) and \(\uplambda \in (0,1)\). This clearly implies that \(f_i(y)<1\) for every \(y\in B_{X'}\), \(y\ne x_i\), with \(i=1,\ldots ,n.\)

Now, consider some linear \(f_{n+1}:\mathbb R^n\rightarrow \mathbb R\) such that \(f_{n+1}(x_{n+1})=1\) and \(f_{n+1}(y)<1\) for every \(y\in S_X, y\ne x_{n+1}\). Choose a basis \(\mathcal B^i=\{u^i_1,\ldots ,u^i_n\}\subset S_X\) with \(u^i_1=x_i\) and \(u^i_j\in \ker (f_i)\) when \(j\ne i\). Given some \(M_i>0\) and \(1>\varepsilon _i>0\) we may define \(\Vert \cdot \Vert _i\) as

$$\begin{aligned}&\Vert \uplambda _1u^i_1+\cdots +\uplambda _{n}u^i_{n}\Vert _i=\big (|(1+\varepsilon _i)\uplambda _1|^{2i+2}\\&\quad +(|\uplambda _2|/M_i)^{2i+2}+\cdots +(|\uplambda _{n}|/M_i)^{2i+2}\big )^{1/{(2i+2)}}. \end{aligned}$$

If we take \(\varepsilon _i\) small enough and \(M_i\) big enough, then the norm \(\Vert \cdot \Vert _Y=\max \{\Vert \cdot \Vert _{X'},\Vert \cdot \Vert _1,\ldots ,\Vert \cdot \Vert _n\}\) equals \(\Vert \cdot \Vert _{X'}\) in every point of \(S_{X'}\) except for small neighbourhoods of \((1-\varepsilon )x_1,\ldots ,(1-\varepsilon )x_{n+1}\), say \(V_1,\ldots ,V_{n+1}\), where the sphere takes the form of a variant of the p-norm with \(p={2i+2}\). Observe that we may take each \(e_{i+1}\) small enough and \(M_{i+1}\) big enough to make the diameter of \(V_{i+1}\) strictly smaller than that of \(V_i\) and, moreover, we may suppose that the diameter of every \(V_i\) is strictly smaller than the distance between any pair of \(V_j, V_k\), with \(i,j,k\in 1,\ldots ,{n+1}\).

Claim

Reducing \(\varepsilon \) and increasing M if necessary, we may suppose that every collection \(y_1\in V_1,\ldots , y_{n+1}\in V_{n+1}\) are in general position, i.e. no hyperplane contains n of them.

Proof

This is clear from the following facts:

  1. (1)

    \(x_1,\ldots ,x_{n+1}\) are in general position.

  2. (2)

    An n-tuple \(\{u_1,\ldots ,u_n\}\) lies in the same hyperplane if and only if every skew-symmetric linear n-form vanishes when applied to it, i.e. \(\omega (u_1,\ldots ,u_n)=0\) for every (some) \(\omega :(\mathbb R^n)^n:\rightarrow \mathbb R, \omega \ne 0.\)

  3. (3)

    Any skew-symmetric linear n-form is continuous.

\(\square \)

This new norm \(\Vert \cdot \Vert _Y\) has trivial autoisometry group. Indeed, the points in \(V_1,\ldots ,V_{n+1}\) are the only exposed points where \(S_{Y}\) is smooth, besides \(-V_1,\ldots ,-V_{n+1}\). So, as being exposed and being smooth are properties preserved by linear isometries, \((\bigcup V_i)\bigcup (\bigcup -V_i)\) is invariant for any autoisometry \(F:(\mathbb R^n,\Vert \cdot \Vert _Y)\rightarrow (\mathbb R^n,\Vert \cdot \Vert _Y)\). It is clear that every \(V_i\) is connected, so its image by F (or any other continuous function) is also connected. This implies that for each i there is some j such that \(F(V_i)\subset V_j\). This is also true for \(F^{-1}\), so \(F^{-1}(V_j)\subset V_i\) and we have the equality \(F(V_i)=V_j\). There is no way that \((V_i,\Vert \cdot \Vert _{Y})\) is isometric to \((V_j,\Vert \cdot \Vert _{Y}), j\ne i,\) because their diameters are different, so every \(V_i\bigcup (-V_i)\) is invariant for F.

Let us denote by \({\text {ch}}(V_i)\) the convex hull of \(V_i\), analogously \({\text {ch}}(-V_i)\). As F is linear, \({\text {ch}}(V_i)\bigcup {\text {ch}}(-V_i)\) is invariant for F, too. Now, either F or \(-F\) sends \(V_i\) onto itself. Thus, the Brouwer fixed-point Theorem implies that either F or \(-F\) has some fixed point \(y_i\in {\text {ch}}(V_i)\). In any case, \(\{F(y_i),F(-y_i)\}=\{y_i,-y_i\}\) for every \(i=1,\ldots ,n+1\).

So, in the basis \(\{y_1,\ldots ,y_n\}\), the matrix of F is diagonal, and all the diagonal entries are \(\{\pm 1\}\), say the k-th is \(\delta _k\). In this basis, we have \(y_{n+1}=(\uplambda _1,\ldots ,\uplambda _n)\), with \(\uplambda _1\cdots \uplambda _n\ne 0\)—recall that \(\{y_1,\ldots ,y_{n+1}\}\) are in general position—and \(\{\pm y_{n+1}\}\) is also invariant, say \(F(y_{n+1})=\delta _{n+1}y_{n+1}\). As F is linear we have

$$\begin{aligned} \delta _{n+1}(\uplambda _1,\ldots ,\uplambda _n)=\delta _{n+1}y_{n+1}= F(\uplambda _1,\ldots ,\uplambda _n)=(\delta _1\uplambda _1,\ldots ,\delta _n\uplambda _n). \end{aligned}$$

So, \(\delta _1=\cdots =\delta _n=\delta _{n+1}\) and this implies that F is either the identity or \(-{\text {Id}}\).

With \(\varepsilon \) close enough to 0 and M great enough, \(\Vert \cdot \Vert _X'\) is as close to \(\Vert \cdot \Vert _X\) as we want, so U is dense.

To show that U is open, let \(([\Vert \cdot \Vert ^k])_k\subset U^c\) be a convergent sequence. We need to show that \([\Vert \cdot \Vert ]=\lim ([\Vert \cdot \Vert ^k])\) has non-trivial group of isometries, i.e. that \(U^c\) is closed. As the sequence of norms converges, in particular it is bounded, so there exists \(R\in (1,\infty )\) such that \(\mathrm{{d}}([\Vert \cdot \Vert ]^k,[\Vert \cdot \Vert ]_2)\le R\) for every \(k\in \mathbb N\). So, for each k, we may take representatives \(\Vert \cdot \Vert ^k\) such that

$$\begin{aligned} \Vert \cdot \Vert _2\le \Vert \cdot \Vert ^k\le R\Vert \cdot \Vert _2, \end{aligned}$$
(13)

and also \(\Vert \cdot \Vert _2\le \Vert \cdot \Vert \le R\Vert \cdot \Vert _2\). Suppose that for every \(\Vert \cdot \Vert ^k\) there exists \(T_k\in {\text {Iso}}_{X_k} \setminus \{{\text {Id}},-{\text {Id}}\}\). By (13), \(1\le \Vert T_kx\Vert _2\le R\) for every \(k\in \mathbb N\) and \(x\in S_{X_k}\), so \((T_k)\) is uniformly bounded in \(\mathcal {M}_n\) endowed with the Euclidean matrix norm. This implies that \((T_k)_k\) must have some accumulation point T; we will suppose that T is the limit of the sequence. We need to see that T is an autoisometry for \(\Vert \cdot \Vert \) and that it can be chosen to be neither \({\text {Id}}\) nor \(-{\text {Id}}\).

For the first part, applying the triangle inequality to \(\big |\!\big |\!\big |\cdot \big |\!\big |\!\big |\) gives

$$\begin{aligned} \begin{aligned} \big |\!\big |\!\big |T^*\Vert \cdot \Vert -\Vert \cdot \Vert \big |\!\big |\!\big |&\le \big |\!\big |\!\big |T^*\Vert \cdot \Vert -T^*_k\Vert \cdot \Vert \big |\!\big |\!\big |+\big |\!\big |\!\big |T^*_k\Vert \cdot \Vert -T^*_k\Vert \cdot \Vert ^k\big |\!\big |\!\big |\\&\quad +\,\big |\!\big |\!\big |T^*_k\Vert \cdot \Vert ^k-\Vert \cdot \Vert ^k\big |\!\big |\!\big |+\big |\!\big |\!\big |\Vert \cdot \Vert ^k-\Vert \cdot \Vert \big |\!\big |\!\big |. \end{aligned} \end{aligned}$$
(14)

The third term in the sum is 0 for every k and the fourth term tends to 0 when \(k\rightarrow \infty \), so we need to show that it is also the case for the first two terms, or, equivalently, that the map

$$\begin{aligned} (T,\Vert \cdot \Vert ) \mapsto T^*\Vert \cdot \Vert \end{aligned}$$

—that assigns to each linear operator \(T:\mathbb R^n\rightarrow \mathbb R^n\) and each norm \(\Vert \cdot \Vert :\mathbb R^n\rightarrow \mathbb R\) the norm \(T^*\Vert \cdot \Vert \) defined as \(T^*\Vert x\Vert =\Vert Tx\Vert \)—is continuous. For the sake of clarity, we will denote \(S=S_{X_k}\) for the remainder of the proof. We need to show that

$$\begin{aligned} \lim _k\max _{y\in S}\left\{ \frac{T^*_k\Vert y\Vert }{T^*\Vert y\Vert }\right\} \max _{y\in S}\left\{ \frac{T^*\Vert y\Vert }{T^*_k\Vert y\Vert }\right\} =1. \end{aligned}$$

Given \(k\in \mathbb N\), Lemma 3.1, implies that we may take \(y_k, z_k\) such that

$$\begin{aligned} \max _{y\in S}\left\{ \frac{T^*_k\Vert y\Vert }{T^*\Vert y\Vert }\right\} \max _{y\in S}\left\{ \frac{T^*\Vert y\Vert }{T^*_k\Vert y\Vert }\right\} = \frac{T^*_k\Vert y_k\Vert }{T^*\Vert y_k\Vert }\frac{T^*\Vert z_k\Vert }{T^*_k\Vert z_k\Vert } \end{aligned}$$

and one has

$$\begin{aligned} \lim _k\max _{y\in S}\left\{ \frac{T^*_k\Vert y\Vert }{T^*\Vert y\Vert }\right\} \max _{y\in S}\left\{ \frac{T^*\Vert y\Vert }{T^*_k\Vert y\Vert }\right\} = \lim _k\frac{T^*_k\Vert y_k\Vert }{T^*\Vert y_k\Vert } \frac{T^*\Vert z_k\Vert }{T^*_k\Vert z_k\Vert } =\lim _k\frac{\Vert T_ky_k\Vert }{\Vert Ty_k\Vert } \frac{\Vert Tz_k\Vert }{\Vert T_kz_k\Vert }=1 \end{aligned}$$

since \(\Vert \cdot \Vert \) is continuous. Analogously we see that

$$\begin{aligned} \lim _k\max _{y\in S}\left\{ \frac{T^*_k\Vert y\Vert ^k}{T_k^*\Vert y\Vert }\right\} \max _{y\in S}\left\{ \frac{T^*_k\Vert y\Vert }{T_k^*\Vert y\Vert ^k}\right\} =1. \end{aligned}$$

So, taking logarithms, the right-hand side of the inequality (14) converges to 0 and this implies that \(\big |\!\big |\!\big |T^*\Vert \cdot \Vert -\Vert \cdot \Vert \big |\!\big |\!\big |=0\), so \(T^*\Vert \cdot \Vert =\Vert \cdot \Vert \) and T is an isometry.

Proposition 4.5 implies that we can choose every \(T_k\) at distance at least 1 from \(\pm {\text {Id}}\), so \(T\ne \pm {\text {Id}}\) and we are done. \(\square \)

Remark 4.12

In the previous proof we have seen that \((T,\Vert \cdot \Vert )\mapsto T^*\Vert \cdot \Vert \) is continuous when \(\Vert \cdot \Vert \) is a norm. This is not always true when \(\Vert \cdot \Vert \) is a quasinorm. Indeed, we just need to consider \(\mathbb R^2\) endowed with the quasinorm

$$\begin{aligned} \Vert (x,y)\Vert =\left\{ \begin{array}{r l} \Vert (x,y)\Vert _2 &{} \text { if } (x,y)\not \in \{(\uplambda ,0):\uplambda \in \mathbb R^*\} \\ \frac{1}{2}\Vert (x,y)\Vert _2 &{} \text { if } (x,y)\in \{(\uplambda ,0):\uplambda \in \mathbb R\} \end{array} \right. , \end{aligned}$$

define the operators

$$\begin{aligned} T_k(x,y)=\left( \!\! \begin{array}{c r} \cos (\pi /k) &{} -\sin (\pi /k) \\ \sin (\pi /k) &{} \cos (\pi /k) \end{array} \!\!\right) \left( \!\! \begin{array}{c } x \\ y \end{array} \!\!\right) \end{aligned}$$

and observe that \(\big |\!\big |\!\big |T^*_k\Vert \cdot \Vert -T^*_l\Vert \cdot \Vert \big |\!\big |\!\big |\) does not depend on \(k, l\in \mathbb N\) as long as they are different. Indeed, the operator \(T_k\) is the rotation of angle \(\pi /k\) and the only points in the sphere of \(\Vert \cdot \Vert \) outside the Euclidean sphere are \(\pm (2,0)\). So, \(T^*_k\Vert x\Vert =\Vert x\Vert _2\) for every \(k\in \mathbb N\) unless \(T_k(x)=(\uplambda ,0)\), in which case \(T^*_k\Vert x\Vert =\Vert x\Vert _2/2.\) So, if k and l are different then one has

$$\begin{aligned}&\big |\!\big |\!\big |T^*_k\Vert \cdot \Vert -T^*_l\Vert \cdot \Vert \big |\!\big |\!\big |\\&\qquad =\,\log _2(\mathrm{{d}}(T^*_k\Vert \cdot \Vert ,T^*_l\Vert \cdot \Vert ))= \log _2\left( \max _{x\in S}\frac{\Vert T^k(x)\Vert }{\Vert T^l(x)\Vert } \max _{x\in S}\frac{\Vert T^l(x)\Vert }{\Vert T^k(x)\Vert }\right) \\&\qquad =\,\log _2(4)=2, \end{aligned}$$

where S denotes the unit sphere of the norm \(\Vert \cdot \Vert .\)

In spite of this, it is quite clear that the proof of the continuity of \((T,\Vert \cdot \Vert )\mapsto T^*\Vert \cdot \Vert \) still works when we deal with continuous quasinorms.