Dynamics of the semigroup of contractive automorphisms of Banach spaces

This paper studies the “dynamics” of the semigroup of contractive automorphisms of finite dimensional normed spaces. Our interest in this subject, and even the topic itself, stems from Mazur rotations problem: Is every separable Banach space whose group of linear isometries acts transitively on the unit sphere isometric (or isomorphic) to a Hilbert space? (Cf. [1, remarque à la section 5 du chapitre XI].)

We are not going to go into this issue, firstly because there is a very recent survey paper on the subject and secondly because we believe that our study is (moderately) interesting in its own right. In any case, we recommend the reader to have a look at the papers [3, 8, 4, 7].

We consider only real spaces; most of the time we work in finite dimensions, often in the plane. In particular $\ell_{p}^{n}$ denotes $\mathbb{R}^{n}$ with the norm $\|x\|_{p}=\big{(}\sum_{i=1}^{n}|x_{i}|^{p}\big{)}^{1/p}$ for $1\leqslant p<\infty$, while $\|x\|_{\infty}=\max_{i\leqslant n}|x_{i}|$, and $e_{i}$ denotes the corresponding unit vector of $\mathbb{R}^{n}$.

Given a (real) normed space $X$, its (closed unit) ball is the set $B=\{x\in X:\|x\|\leqslant 1\}$ and $S=\{x\in X:\|x\|=1\}$ is the unit sphere. We write $L(X)$ for the space of all (bounded, linear) endomorphisms of $X$ with the operator norm $\|T\|=\sup_{\|x\|\leqslant 1}\|T(x)\|$. An operator $T$ on $X$ is an automorphism if there is $L\in L(X)$ such that $TL=LT={\bf I}_{X}$. An operator $T$ is contractive, or a contraction, if $\|T\|\leqslant 1$. This means that $TB\subset B$.

Clearly, the contractive automorphisms of $X$ form a semigroup of $L(X)$ that we will denote by $\operatorname{Aut}_{1}(X)$. We are interested in the “action” of $\operatorname{Aut}_{1}(X)$ on the unit sphere of $X$, especially in the size of the “orbits” $O(x)=\{y\in S:y=Tx\text{ for some }T\in\operatorname{Aut}_{1}(X)\}$ for $x\in S$. Note that if $\|y\|<\|x\|$ then there is always $T\in\operatorname{Aut}_{1}(X)$ such that $y=Tx$, so considering only points of $S$ is fine. Life without local convexity can be much harder, see [12] for examples of rigid quasi Banach spaces.

The space $\ell_{1}^{2}$ already provides a quite interesting example (see Figure 1 and the comments following Theorem 2.3):

• •

  $O(x)=S\iff x=\pm e_{i}$ for some $i=1,2$.

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  $O(x)=S\backslash\{\pm e_{1},\pm e_{2}\}$ for any other $x\in S$.

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  For each $x\in S$ the set $O(x)$ is a neighbourhood of $x$ relative to $S$.

A Banach space $X$ is said to be semitransitive (ST) if for every $x,y\in S$ there is $T\in\operatorname{Aut}_{1}(X)$ such that $y=Tx$. Hilbert spaces are ST: actually they are even transitive, that is, $T$ can be taken to be isometric, that is, $\|T\|=\|T^{-1}\|=1$. This quickly follows from the 2D case and the existence of orthogonal complements. Any other example in sight? Not yet. To understand what the issue is really about we need to introduce a couple of definitions.

Let $X$ be a finite dimensional (FD) space (or even a Banach space isomorphic to a Hilbert space) with unit ball $B$ and unit sphere $S$. An ellipsoid in $X$ is the closed unit ball (centered at the origin) of an Euclidean norm equivalent to the original norm of $X$. Ellipsoids are invariably assumed to be centered at the origin. If $X=\mathbb{R}^{n}$ with some norm, then a disc (centered at $x$) is a set of the form $D=\{y\in\mathbb{R}^{n}:\|y-x\|_{2}\leqslant r\}$, where $r>0$ and $\|\cdot\|_{2}$ is the usual Euclidean norm.

We say that an ellipsoid $E$ is inner at $x\in S$ if $E\subset B$ and $x\in E$. We say that $E$ is outer at $x$ if $B\subset E$ and $x\in\partial E$. The following duality argument will be used over and over without further mention: if $x\in X$ and $x^{*}\in X^{*}$ are such that $\|x\|=\|x^{*}\|=\langle x^{*},x\rangle=1$, then $E$ is inner (respectively, outer) at $x$ if and only if the dual ellipsoid $E^{*}=\{y^{*}\in X^{*}:|\langle y^{*},y\rangle|\leqslant 1\;\forall\,y\in E\}$ is outer (respectively, inner) at $x^{*}$.

Those (nonzero) points where the norm is (Gâteaux) differentiable are called smooth. This happens to $x$ if and only if there is exactly one functional $x^{*}$ such that $\|x\|=\|x^{*}\|$ and $\langle x^{*},x\rangle=\|x\|\|x^{*}\|$. Such an $x^{*}$ is called a support functional, necessarily agrees with $\tfrac{1}{2}d_{x}\|\cdot\|^{2}$, and is often denoted by $J(x)$. If the norm is differentiable off the origin the space $X$ itself is called smooth and the mapping $J:X\longrightarrow X^{*}$ sending each point into its unique support functional (and $0$ to $0$) is called the duality map.

It is clear that points that admit inner ellipsoids are smooth, while those that have outer ellipsoids are extreme. The following remark is obvious, yet very useful:

The following result is almost obvious, just using compactness:

See [6, Lemma 3.4] for an extension to infinite dimensional spaces. The following characterization is in turn a particularization of [6, Proposition 3.5].

The above criterion shows that for $n\geqslant 2$ and $p\neq 2$ the spaces $\ell_{p}^{n}$ are not ST (if $p<2$ the unit vectors do not have inner ellipsoids; if $p>2$ they do not have outer ellipsoids). The paper [6] contains examples of ST norms on the plane which were constructed with slightly different purposes, using sledgehammers to crack nuts. Example 2.7 below is much simpler; Example 2.8 exhibits a rather unexpected behaviour.

Let us call $y\in S$ flat if there exists some homogeneous hyperplane $H\subset X$ such that $(y+H)\cap S$ is a neighbourhood of $y$ relative to $S$ —note that $H=\ker J(y)$. If $y$ is flat, then $y\in O(x)$ for all $x\in S$. Indeed, let $x^{*}$ be a norm-one functional such that $\langle x^{*},x\rangle=1$, put $H_{x}=\ker x^{*}$, and let $L:H_{x}\longrightarrow H$ be any linear isomorphism. If $\varepsilon>0$ is sufficiently small, the automorphism $T$ given by $T(cx+u)=cy+\varepsilon L(u)$, where $c\in\mathbb{R},u\in H_{x}$, is contractive —and sends $x$ to $y$, of course.

Call $x\in S$ pilgrim if $O(x)$ is dense in $S$. It is clear from Lemma 2.2 that every FD normed space has pilgrim points. As for the “size” of the set of pilgrim points it is nearly obvious that $X$ is strictly convex and smooth, then the set of pilgrim points is dense in the unit sphere of $X$. See why? On the other hand, we have just seen that the orbit of any flat point consists precisely of the set of flat points of $S$ and this means that in any polyhedral space the set of pilgrim points is dense in the sphere. The same is true for spaces whose unit ball is the the intersection of two ellipsoids.

It is also clear that a FD space whose set of pilgrim points is dense in the unit sphere must must have some kind of regularity: if we mix a polyhedral norm with some strictly convex norm, we loose the density of the set of pilgrim points —think of the norm in $\mathbb{R}^{2}$ that agrees with $\|\cdot\|_{2}$ in the first quadrant and with $\|\cdot\|_{1}$ in the second one.

The following example shows that some natural conjectures that one might come up with are false:

The notion of curvature (of a plane curve) is very useful both for understanding the geometry of the norm and for checking the ST character of norms in the plane. As far as this paper is concerned, everything we need to know about curves can be found in the lovely book by the Bucks [5, §8.4]. Let $I$ be an interval and $\Phi:I\subset\mathbb{R}\longrightarrow\mathbb{R}^{n}$ be a rectifiable curve parameterized by arc length. If $x=\Phi(t)$, and $\Phi$ is twice differentiable at $t$, then the curvature of $\Phi$ at $x$ is defined as $\varkappa(\Phi,x)=\|\Phi^{\prime\prime}(t)\|_{2}.$ We are aware that there is a small problem here if $x$ is a double (or multiple) point of $\Phi$ since $\varkappa(\Phi,x)$ depends not only on $x$, but also on $t$. We will not insist on this point. Note also that this notion of curvature refers to the usual Euclidean norm, although other approaches are possible (see [9, §4.3]).

If $\Phi$ is the graph of a function $f:I\longrightarrow\mathbb{R}$, in the sense that $\Phi(t)=(t,f(t))$ for $t\in I$, then

where $x=(t,f(t))$, and we no longer assume $\Phi$ parameterized by its arc length (see Bucks’ [5, Exercise 5 on p. 416]). As a consequence:

Since a circle of radius $r$ has curvature $r^{-1}$ at all points the Lemma implies that if $S$ has finite curvature at $x$, then $x$ belongs to a disc $D$ which is contained in $B$. What is more interesting than it may seem:

Here is a simple example of ST norm on $\mathbb{R}^{2}$. Fix any $R>1$ and let $C$ be the circle of center $a=(0,R-1)$ and radius $R$. Let $b=(b_{1},b_{2})$ be any point of $C$ such that $b_{1}>0$ and $b_{2}<0$. Let $c$ be the intersection of the horizontal axis and the line joining $a$ and $b$. Let $C^{\prime}$ be the circle of center $c$ and radius $R^{\prime}=\|c-b\|_{2}$. Note that $C^{\prime}$ passes through $b$ and that $R^{\prime}<R$. Let $d$ be the rightmost point in the insersection of $C^{\prime}$ and the horizontal axis. Let $\Gamma$ be the curve (in the fourth quadrant) that agrees with $C$ from $(0,-1)$ to $b$ and with $C^{\prime}$ from $b$ to $d$.

It is clear that there is exactly one norm $\|\cdot\|$ on the plane such that $\|(x,y)\|=1\iff(|x|,-|y|)\in\Gamma$ and also that $\|\cdot\|$ is ST since every point in the unit sphere satisfies the criteria provided by Lemma 2.6.

Let $X$ be a FD space and assume that the orbit of $x$ is $S$. Put

Then $S=\bigcup_{n=1}^{\infty}O(x,n)$. An obvious “category” argument and the fact that the sets $\{T\in\operatorname{Aut}_{1}(X):\|T^{-1}\|\leqslant K\}$ are compact in $L(X)$ show that for every open set $U\subset S$ there is $K$ such that $U\cap O(x,K)$ has nonempty interior in $S$.

This fosters the idea that ST spaces could be boundedly semitransitive (BST) in the following sense: there is a constant $K$ such that, for every $x,y\in S$ there is $T\in\operatorname{Aut}_{1}(X)$ such that $y=Tx$ and $\|T^{-1}\|\leqslant K$.

Unfortunately this is not the case in general:

Recall that given a locally integrable function $k:I\longrightarrow[0,\infty)$, where $I$ is an interval containing the origin, the formulæ

define a $C^{1}$ plane curve $\Gamma(s)=(x(s),y(s))$ such that

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  $\|\Gamma^{\prime}(s)\|_{2}=1$ for all $s\in I$.

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  If $s$ is a Lebesgue point of $k$, then the curvature of $\Gamma$ at $s$ is $k(s)$. See [10, Definition 7.1.19]

Let us first define $k:[-\frac{\pi}{2},1]\longrightarrow(0,1]$ as follows (see Figure 5):

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  For each $n\in\mathbb{N}$ the value of $k$ on $[2^{-n},2^{-n}+2^{-n-2}]$ is $2^{-n}$.

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  Otherwise $k(s)=1$.

Then we consider the curve defined as above, with $x_{0}=-1,y_{0}=0$, and call it $\Gamma$. Now look at the endpoint $p=(x(1),y(1))$ of $\Gamma$ corresponding to $s=1$: a moment’s reflection suffices to realize that:

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  The initial point of $\Gamma$ is $(-1,0)$.

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  $0<x(1)<1,-1<y(1)<0$ since $(x(0),y(0))=(0,-1)$, both $x^{\prime}(s)$ and $y^{\prime}(s)$ are strictly positive for $0<s<1$ and $s$ parametrizes $\Gamma$ naturally.

• •

  The (semi) tangent line to $\Gamma$ at $p$ has strictly positive slope and, moreover, if $(a,0)$ is the only point of that line that lies on the “horizontal” axis, then $a>1$.

It follows (see Figure 4) that there exist a pair of circles $C,C^{\prime}$ such that $C$ is tangent to $L$ at $p$, $C^{\prime}$ is tangent to the line $x=1$ at $(1,0)$ and $C$ and $C^{\prime}$ are mutually tangent at some other intermediate point, say $q$.

Let $S^{-}$ be the curve that “agrees” with $\Gamma$ between $(-1,0)$ and $p$, with $C$ between $p$ and $q$ and with $C^{\prime}$ between $q$ and $(1,0)$. It is clear that there is exactly a norm, say $\|\cdot\|$, on $\mathbb{R}^{2}$ whose unit sphere $S$ contains $S^{-}$.

We want to see that the resulting space is ST. Observe that $S^{-}$ is a (“smooth”) countable union of arcs of circles: going from right to left we find first the arcs $C^{\prime}$ and $C$, then an arc of circle of radius 1 and lenght $\frac{1}{2}-\frac{1}{2^{3}}=\frac{3}{2^{3}}$, then an arc of circle of radius $2$ and lenght $\frac{1}{8}$, then another arc of circle of radius 1 and lenght $\frac{3}{2^{4}}$, then an arc of circle of radius $4$ and lenght $\frac{1}{16}$ and so on. These arcs accumulate at the point $(0,-1)$ in the obvious sense. Finally we have an arc of radius 1, centered at the origin, between $(-1,0)$ and $(0,-1)$.

Now it should be clear that all points of $S^{-}$ admit inner discs and also that all of them, except $(0,-1)$, have outer discs. But actually $(0,-1)$ admits an outer disc whose radius is not very large. Indeed for every $n$ the (Lebesgue) measure of the set $\{t\in[0,2^{-n}]:k(t)=1\}$ is at least $\frac{3}{4}2^{-n}$. Therefore, if we put $K(s)=\int_{0}^{s}k(t)dt$, then $\frac{3}{5}s\leqslant K(s)\leqslant s$ for $0\leqslant s\leqslant 1$. Since

it is clear that $\Gamma(s)$ cannot leave the disc of radius $\frac{5}{3}$ centered at $(0,-1)$ for any $0\leqslant s\leqslant 1$, which is enough.

The resulting normed plane, whose sphere is $S$, cannot be BST because the inverse of a contractive automorphism sending a point with small curvature to a point of large curvature must have large norm, see why? If you still do not see it clearly, take a look at the upcoming Proposition 3.1.

We now study BST spaces. It turns out that these admit a quite elegant geometric characterization. Before going any further, recall that the modulus of uniform convexity of a Banach space $X$ is the function $\delta_{X}:(0,2]\longrightarrow[0,1]$ defined by

Note that $1-\frac{1}{2}\left\|x+y\right\|$ represents the distance between the midpoint of $x$ and $y$ and its closest multiple in the unit sphere (the larger the MUC the more convex the ball is) and that one can replace all inequalities by equalities in (2) without altering the value of $\delta_{X}(\varepsilon)$. We say that $\delta_{X}$ is of power type 2 if $\delta_{X}(\varepsilon)\geqslant c\varepsilon^{2}$ for some $c>0$ and all $0<\varepsilon\leqslant 2$.

Hilbert spaces have “optimal” MUC given by

Condition (d) below is stated also in terms of the modulus of uniform smoothness (MUS, a measure of the “flatness” of the unit ball which we do not even define here). It will suffice to recall here that $X$ has MUS of power type 2 if and only if $X^{*}$ has MUC of power type 2, so that (d) is equivalent to: Both $X$ and $X^{*}$ have MUC of power type $2$. We refer the reader to [11, § 1.e] for the basics on uniform convexity and smoothness.

The remainder of the section is devoted to proving this result. Let’s get the boring parts out of the way now. The equivalence (a)$\iff$(b) can be proved as in Theorem 4.2 since taking Banach space adjoints preserves automorphisms, inverses, and the operator norm.

(a)$\implies$(c). Assume $X$ is BST with constant $\beta$. Fix $y\in S$ and let $E$ and $F$ be inner and outer ellipsoids at $y$, respectively, which exists by ST and Theorem 4.2. Clearly $F\subset\mu E$ for sufficiently large $\mu$. Pick $x\in S$ and then $T,L\in\operatorname{Aut}_{1}(X)$ such that $x=Ty,y=Lx$, with $\|T^{-1}\|,\|L^{-1}\|\leqslant\beta$. Then $E^{\prime}=TE$ is inner at $x$, $F^{\prime}=L^{-1}F$ is outer at $x$, and $F^{\prime}\subset\beta^{2}\mu E^{\prime}$.

The implication (c)$\implies$(a) can be proved as the implication (b)$\implies$(a) of Theorem 4.2, just following the track of the norms.

We now adress (c)$\implies$(d). The proof uses the so-called modulus of strong extremality, defined for $x\in S$ and $\varepsilon\in(0,1]$ as:

It is not hard to see that $\delta_{X}(2\varepsilon)=\inf_{\|x\|=1}\Delta_{X}(x,\varepsilon)$ —only the easiest part “$\geqslant$” will be used here.

Assume $x\in S$ admits an outer ellipsoid $F$ with $F\subset\alpha B$. This implies that if $|\cdot|$ denotes the Euclidean norm corresponding to $F$, then $|\cdot|\leqslant\|\cdot\|\leqslant\alpha|\cdot|$. An easy computation based on the fact that $S$ and $\partial F$ are tangent at $x$ shows that

where, as the reader may guess, $\Delta_{2}$ stands for the modulus of strong extremality of the Euclidean norms. Thus, (c) implies that $\delta_{X}(\varepsilon)\geqslant c\varepsilon^{2}$ where $c$ depends only on $\lambda$. The fact that $X$ has MUS of power type 2 follows from the fact that a finite dimensional space satisfies (c) if and only if its dual does since a Banach space has MUS of power type 2 if and only if its dual has MUC of power type 2.

It only remains to check that (d)$\implies$(c), which amounts to see that if $X$ has MUC of power type 2, then there is a constant $\alpha>0$ such that every $x\in S$ admits an outer ellipsoid $F$ such that $F\subset\alpha B$.

The proof is based on the following, surely well-known, observation:

To complete the proof, let $|\cdot|$ be an Euclidean norm on $X$ such that $|\cdot|\leqslant\|\cdot\|\leqslant C|\cdot|$ for some constant $C$, for instance one could take the norm associated to the ellipsoid of minimal volume containing $B$. We define a two-parameter family of Euclidean norms as follows: Given $x\in S$, let $x^{*}$ be the (or a fixed) support functional of $B$ at $x$ and set $H_{x}=\ker x^{*}$, so that each $z\in X$ can be written as $z=tx+u$, with $t\in\mathbb{R}$ and $u\in H_{x}$ — of course $t=\langle x^{*},z\rangle$ and $u=z-tx$. Then, for each $b>0$ we set

Note that $a<b\implies E_{b}^{x}\subset E_{a}^{x}$ for fixed $x\in S$ and that for each fixed $b>0$, the norms $|\cdot|_{b}^{x}$ are all uniformly equivalent and uniformly equivalent to $\|\cdot\|$. Hence it suffices to see that there exists $b>0$ such that $B\subset E_{b}^{x}$ for all $x\in S$. But if we assume $\delta_{X}(\varepsilon)\geqslant(c\varepsilon)^{2}$ for some $c>0$ it is clear from the Lemma that if $x\in S$ and $u\in H_{x}$ are such that $z=tx+u$ belongs to $B$, then

It follows that $|\cdot|^{x}_{c}\leqslant\|\cdot\|$ for all $x$ and since $|x|^{x}_{c}=1$ we have that the ellipsoid $E^{x}_{c}$ is outer at $x$, which is enough.

It is clear that UMST $\implies$ BST $\implies$ ST, see [7, Remark 2.4] or [4, Lemma 2.3].

The interest in UMST stems from the facts that all 1-complemented FD subspaces of a separable and transitive Banach space are UMST ([6, Theorem 3.2]) and that the only previously known UMST norms were the Euclidean norms, which made it conceivable that UMST would be a characteristic property of Hilbert spaces.

Unfortunately this is not the case: in this Section we prove that every $C^{2}$ norm on the plane whose dual is also $C^{2}$ turns out to be UMST.

The remainder of this section is devoted to proving the following.

The following “uniform” version of Lemma 2.5 is the key ingredient of the proof of Theorem 4.2. It will be applied in due course to compare the unit sphere of the given norm with its image under a slightly-changing automorphism.

Let $\|\cdot\|$ be a (smooth) norm on the plane and let $S$ be its unit sphere with the standard (counterclockwise) orientation. If $\Sigma:\mathbb{R}\longrightarrow\mathbb{R}^{2}$ is a regular parametrization of $S$ preserving the orientation and $a=\Sigma(t)$, put $a^{\perp}={\Sigma^{\prime}(t)}\big{/}{\|\Sigma^{\prime}(t)\|},$ that is, $a^{\perp}$ the only point in $S\cap H_{a}$ such that $a,a^{\perp},-a$ is positively oriented.

It is almost obvious that if $a,b$ are close in $S$ and $T\in\operatorname{Aut}_{1}(X)$ is close to the identity and maps $a$ to $b$, then $Ta^{\perp}=\lambda b^{\perp}$, with $\lambda\in(0,1]$. From now on we denote by $L^{ab}_{\varepsilon}$ the only linear endomorphism of $X$ that maps $a$ to $b$ and $a^{\perp}$ to $(1-\varepsilon)b^{\perp}$, with the subscript omitted if $\varepsilon=0$. This is actually an automorphism unless $\varepsilon=1$ and, quite clearly,

Note that if $X$ is assumed to be merely smooth, the proof yields $\|L^{ab}_{\varepsilon}-{\bf I}_{X}\|\longrightarrow 0$ as $\max\big{(}\|a-b\|,\varepsilon\big{)}\longrightarrow 0$.