2 Representation of countable groups on separable real Banach spaces

Countable groups of isometries on Banach spaces

Valentin Ferenczi and El´oi Medina Galego ferenczi@ime.usp.br, eloi@ime.usp.br

Abstract

A group G is representable in a Banach space X if G is isomorphic to the group of isometries on X in some equivalent norm. We prove that a countable group G is representable in a separable real Banach spaceX in several general cases, including whenG ' {−1,1} ×H,H finite and dimX ≥ |H|, or when Gcontains a normal subgroup with two elements andXis of the formc0(Y)or`p(Y),1≤p <+∞. This is a consequence of a result inspired by methods of S. Bellenot [2] and stating that under rather general conditions on a separable real Banach spaceX and a count- able bounded groupGof isomorphisms onX containing−Id, there exists an equivalent norm onXfor whichGis equal to the group of isometries on X.

We also extend methods of K. Jarosz [11] to prove that any complex Banach space of dimension at least2may be renormed with an equivalent complex norm to admit only trivial real isometries, and that any complexifi- cation of a Banach space may be renormed with an equivalent complex norm to admit only trivial and conjugation real isometries. It follows that every real Banach space of dimension at least4and with a complex structure may be renormed to admit exactly two complex structures up to isometry, and that every real cartesian square may be renormed to admit a unique complex structure up to isometry.¹²

1MSC numbers: 46B03, 46B04.

2Keywords: group of isometries on Banach spaces, group representable in a Banach space, complex structures up to isometry.

1 Introduction

What groups G may be seen as the group of isometries on a Banach space X?

This general question may be formulated by the following definition given by K.

Jarosz in 1988.

Definition 1 (Jarosz [11]) A group G is representable in a Banach space X if there exists an equivalent norm on X for which the group of isometries onX is isomorphic toG.

In [11], Jarosz stated as an open question which groups were representable in a given Banach space. The difference with the classical theory of representa- tion of groups on linear spaces is that here we require an isomorphism with the group of isometries on a Banach space, and not just some group of isometries or isomorphisms. Since {−Id, Id}is always a normal subgroup of the group of isometries on a real Banach space, it follows that a group which is representable in a real Banach space must always contain a normal subgroup with two elements.

Conversely, J. Stern [19] proved that for any group G which contains a normal subgroup with two elements, there exists a real Hilbert space H such that G is representable in H. Furthermore ifGis countable then H may be chosen to be separable.

For an arbitrary Banach space X it remains open which groups are repre- sentable in X. Jarosz proved that {−1,1} is representable in any real Banach space, and that the unit circleC1 is representable in any complex space (the sepa- rable real case had been solved previously by S. Bellenot [2]). He also proved that for any countable groupG,{−1,1} ×Gis representable inC([0,1]), and that for any groupGthere exists a complex spaceX such thatC1×Gis representable in X.

In a first section of this paper, we give a much more general answer to the question of representability by proving that:

• The group{−1,1} ×Gis representable inX wheneverGis a finite group andXa separable real spaceX such thatdimX ≥ |G|, Theorem 17,

• The group G is representable in X whenever G is a countable group ad- mitting a normal subgroup with two elements and X is a separable real Banach space with a symmetric decomposition either isomorphic to c0(Y) or to `_p(Y) for someY and1 ≤ p < +∞, or with the Radon-Nikodym Property, Theorem 19,

• The group {−1,1} ×Gis representable in X whenever G is a countable group andX an infinite-dimensional separable real Banach space contain- ing a complemented subspace with a symmetric basis, Theorem 21.

These results are partial answers to a conjecture of Jarosz who asked whether {−1,1} ×Gis representable inXfor any groupGand any real spaceXsuch that dimX≥ |G|.

As an application of our results we obtain that a countable groupGis repre- sentable inc₀ (resp. C([0,1]), `_p for1 ≤ p < +∞, L_p for1 ≤ p < +∞) if and only if it contains a normal subgroup with two elements, Corollary 20.

Our method is to ask, given a groupGof linear isomorphisms on a real Banach spaceX, whether there exists an equivalent norm onXfor whichGis the group of isometries onX. Once the problem of representability is reduced to representing a given group as some group of isomorphisms on a given Banach space, it is much simpler to address, and this leads to Theorem 17, Theorem 19, and Theorem 21. In other words, we explore in which respect the question of representability of groups in Banach spaces belongs to the renorming theory or rather may be reduced to the purely isomorphic theory.

If a group of isomorphisms is the group of isometries on a real (resp. com- plex) Banach space in some equivalent norm, then it must be bounded, contain

−Id (resp. λId for all λ ∈ C1), and be closed for the topology of the strong convergence ofT andT⁻¹. Therefore the question is:

Question 2 LetX be a real (resp. complex) Banach space and letGbe a group of isomorphisms onXwhich is bounded, contains−Id(resp.λIdfor allλ ∈C1), and is closed for the topology of the strong convergence ofT andT⁻¹. Does there exist an equivalent norm onXfor whichGis the group of isometries onX?

A positive answer was obtained by Y. Gordon and R. Loewy [9] whenX =Rⁿ andGis finite (in which case of course the boundedness and closedness hypothe- ses are automatically satisfied). This answered a question by J. Lindenstrauss. In this paper, we extend the methods of Bellenot and use a result of V. Kadec, see [4] p. 48, or a renorming method of G. Lancien [14] to considerably improve this result:

• Let X be a separable real Banach space. Then for any finite group G of isomorphisms on X which contains −Id, there exists an equivalent norm onXfor whichGis equal to the group of isometries onX, Theorem 11.

• LetX be a separable real Banach space with the Radon-Nikodym Property.

Then for any countable bounded group G of isomorphisms on X which contains−Idand is separated by some point with discrete orbit, there exists an equivalent norm onX for whichGis equal to the group of isometries on X, Theorem 12.

Therefore for separable real spaces and finite groups, the question of repre- sentability really does not belong to renorming theory. Also, note that a countable group of isomorphisms on X which is equal to the group of isometries in some equivalent norm must always be discrete for the topology of the strong conver- gence ofT andT⁻¹and admit a separating point, Lemma 13. It remains unknown however whether this implies the existence of a separating point with discrete orbit, that is, if the implication in Theorem 12 is an equivalence for countable groups.

To conclude that section we deduce Theorem 17, Theorem 19 and Theorem 21 essentially from Theorem 11 and Theorem 12. We also prove that Theorem 17 and Theorem 21 are optimal in the sense that there exists a real space in which representable finite groups are exactly those of the form{−1,1} ×G, Proposition 22, and a real space containing a complemented subspace with a symmetric basis in which representable countable groups are exactly those of the form{−1,1}×G, Proposition 23. On the other hand we have the classical examples ofc₀,C([0,1]),

`_p,1 ≤ p < +∞ and L_p,1 ≤ p < +∞ for which Corollary 20 states that representable countable groups are exactly those which admit a normal subgroup with two elements, and we also provide an intermediary example of a space in which the class of representable finite groups is strictly contained in between the above two classes, Proposition 24.

In a second section of this paper, we use the renorming methods of Jarosz in [11] to study complex structures on real Banach spaces up to isometry. Our results are actually related to the representability of the circle groupC1 and of the group of isometries onCas the group ofR-linear isometries on a complex Banach space.

We recall a few facts about complex structures. An introduction to this subject may also be found in [5]. Any complex Banach space is also a real Banach space, and conversely, the linear structure on a real Banach space X may be induced by a C-linear structure; the corresponding complex Banach space is said to be a complex structure on X in the isometric sense. It is clear that any complex structure onX is canonically associated to some R-linear map I onX such that I² = −Id and cosθId+ sinθI is an isometry for all θ, and which defines the

multiplication by the imaginary number i. Conversely for any such mapI, there exists an associated complex structure denotedX^I, defined fora, b ∈ Randx ∈ X by

(a+ib).x=ax+bIx.

The existing theory of complex structure, however, concerns existence and uniqueness of complex structure up to isomorphism. In this case, complex struc- tures correspond to real isomorphismsI of square−Id, and conversely such op- erators define a complex structure by the same procedure as above and under the equivalent norm k|.k| defined by k|xk| = max_θkcosθx+ sinθIxk. It is well- known that complex structures do not always exist up to isomorphism on a Banach space. By [3], [12] there exists real spaces with at least two complex structures up to isomorphism, and the examples of [3] and [1] (which are separable) actually admit a continuum of complex structures. By [6] for eachn ∈ N^∗ there exists a space with exactlyncomplex structures up to isomorphism. In [6] and [7] various examples of spaces different from the classical example of `₂ are also shown to have a unique complex structure up to isomorphism, including a HI example, a space with an unconditional basis, and aC(K)space defined by Plebanek.

It turns out actually that the classical spaces c₀, C([0,1]), `_p,1 ≤ p ≤ +∞

andL_p,1 ≤p < +∞also admit a unique complex structure up to isomorphism.

A nice and simple proof of this fact was given to us by N.J. Kalton after a first version of our paper was written and is included here, Theorem 28.

The isometric theory of complex structures turns out to be totally different from the isomorphic theory. For a very general class of Banach spacesX, we show that quite various situations may be obtained concerning existence and uniqueness of complex structures up to isometry on X by choosing different renormings on X. This may justify why the isometric theory of complex structures has not been investigated before, as it is unclear what other results one may want to obtain in that area.

We first prove that`₂ has a unique complex structure up to isometry, Propo- sition 26. On the other hand, since Jarosz [11] showed that every real Banach space may be renormed to admit only trivial isometries (i.e. the only isometries areIdand−Id), every real Banach space may be renormed not to admit complex structures in the isometric sense.

Extending the methods of Jarosz [11] we prove that:

• Any complex Banach space of dimension at least 2 may be renormed to admit only trivial real isometries, Corollary 45

• Any complex Banach space which is real isomorphic to a cartesian square, and whose complex law is the canonical one associated to the decompo- sition as a square, may be renormed to admit only trivial and conjugation isometries, Corollary 46.

It follows, Theorem 30:

• Every real Banach space of dimension at least4and with a complex struc- ture may be renormed to admit exactly two complex structures up to isom- etry,

• Every real cartesian square may be renormed to admit as unique complex structure up to isometry the canonical complex structure associated to its decomposition as a square.

In a last section we extend results of F. Rabiger and W.J. Ricker, [17], about isometries on complex spaces of the type of Gowers and Maurey, by proving that any isometry on a real Banach space on which any operator is a strictly singular perturbation of a multiple of the identity must be of the form±Id+K,Kcompact, Proposition 49. This applies to Gowers and Maurey’s hereditarily indecomposable spaceGM, [10].

For classical results in Banach space theory, such as, for example, results con- cerning the Radon-Nikodym Property or symmetric bases, we shall refer to [15];

for renorming questions in Banach spaces, we shall refer to [4].

2 Representation of countable groups on separable real Banach spaces

2.1 G-pimple norms on separable Banach spaces

In [2] Bellenot showed how to renorm a given separable real Banach space so that the group of isometries in the new norm is equal to {−Id, Id} . In this subsection we extend the construction of Bellenot to an arbitrary countable group of isometries containing −Id instead of {−Id, Id}. So in which follows, X is real separable, Gis a countable group of isometries on X containing−Id, and under certain conditions on G, we construct an equivalent norm on X for which Gis the group of isometries onX.

In [2], Bellenot renormsXwith an LUR norm and then defines, for somex₀in Xof norm1, a new unit ball (which he calls the ”pimple” ball) obtained by adding two small cones in x0 and −x0. Any isometry in the new norm must preserve the cones and therefore sendx₀ to±x₀. Repeating this for a sequence(x_n)_nwith dense linear span, chosen carefully so that one can add the cones ”independently”, and so that the sizes of the cones are ”sufficiently” different, any isometry sends x_n to±x_n. Finally, if eachx_n was chosen ”much closer” to x₀ than to−x₀, any isometry fixingx₀ must fix eachx_n and therefore any isometry is equal to Idor

−Id.

In our case, fixing somex₀ inXof norm1, one should obviously put cones of same size in eachgx₀, g ∈G, defining a ”G-pimple ball”, so that any isometry in the new norm preserves the orbitGx0. This first step could be realized even when Gis not countable. Then one repeats a similar procedure as in [2], adding other cones ingx_n, g ∈Gfor a sufficiently dense sequence(x_n)_n, so that any isometry preserves Gxn for alln. Thesexn’s forn ≥1are called of type1. Finally, a last step is added to only allow as isometries isomorphisms whose restriction toGx₀ is a permutation which corresponds to the action of someg ∈ GonGx₀. This is technically more complicated and is obtained by adding cones at some points of span Gx₀ which code the structure ofG and are called of type 2. The fact that Gis countable andXseparable guarantees that there are at most countably many cones, which will allow us to provide lower estimates for the distances between two such cones.

The reader may get a geometric feeling of this proof by looking at the group G = {±Id,±R} of R-linear isometries onC where R is the rotation of angle π/2. By adding cones “of type 1”on the unit ball at ±1 and±i, one allows the isometries inGbut also symmetries with respect to the axes. A way of correcting this is to add one well-placed smaller cone “of type 2” next to each element of {±1,±i}, for example in±e^iπ/6 and±e^i2π/3, so that the only isometries in the new norm are those ofG.

Definition 3 Let X be a real Banach space with normk.k, let G be a group of isometries onXsuch that−Id ∈G, and let(x_k)k∈Kbe a possibly finite sequence of normalized vectors ofX. Let Λ = (λ_k)k∈K be such that 1/2 < λ_k < 1for all k ∈K. TheΛ, G-pimple at(x_k)_kfork.kis the equivalent norm onX defined by

kyk_Λ,G = inf{X

[[y_i]]_Λ,G :y=X y_i},

where[[y]]_Λ,G =λ_kkyk,whenever y ∈ span(g.x_k)for somek ∈ K andg ∈ G, and[[y]]_Λ,G =kykotherwise.

In other words, the closed unit ball for k.k_Λ,G is the convexification of the union of the closed unit ball for k.k with line segments between gx_k/λ_k and

−gxk/λkfor eachk ∈Kandg ∈G.

In [2] Bellenot had defined the notion ofλ-pimple at x₀ ∈ X, which corre- sponds to(λ),{−Id, Id}-pimple in our terminology: that is, the closed unit ball of theλ-pimple atx₀ is the convexification of the closed unit ball fork.kand of a single line segment betweenx₀/λand−x₀/λ.

It may be observed that ( inf

k∈Kλ_k)k.k ≤ k.k_Λ,G ≤ k.k,

so that k.k_Λ,G is an equivalent norm on X, and that by definition, any g ∈ G remains an isometry in the normk.k_Λ,G.

Recall that a normk.kislocally uniformly rotund(or LUR) at some point x, kxk = 1, if for all ∈]0,2], there existsλ(x, ) <1such that wheneverkyk = 1 and kx−yk ≥ , it follows that

^x+y₂

≤ λ(x, ). Equivalently lim_nx_n = x whenever limnkxnk = kxk and limnkx+xnk = 2kxk. It is LUR when it is LUR at all points of the unit sphere. As in [2] we shall say that a normalized vectoryisextremalfor a normk.kif it is an extremal point of the closed unit ball fork.k, that is, if whenever kyk = kzk = 1and xbelongs to the segment[y, z], theny = x = z. A norm isstrictly convexat a point xof the unit sphere ifxis extremal. A norm is strictly convex when it is strictly convex at all points of the unit sphere. Note that if a norm is LUR atxthen it must be strictly convex at x.

More details about these notions may be found in [4].

We recall a crucial result from [2] (Proposition p.90)

Proposition 4 (Bellenot [2]) Let(X,k.k)be a real Banach space and letx₀ ∈X be normalized such that

(1) k.kis LUR atx₀, and

(2) there exists > 0 such that if kyk = 1 and kx₀−yk < , then y is an extremal point fork.k.

Then givenδ >0,b >0and0< m <1, there exists a real number1/2< λ₀ <1 such that wheneverλ₀ ≤λ <1andk.k_λ is theλ-pimple atx₀, then

(3) mk.k ≤ k.k_λ ≤ k.k,

(4) if1 =kyk>kyk_λ thenkx₀−yk< δ orkx₀+yk< δ,

(5) x_λ = λ⁻¹x₀ is the only isolated extremal point for k.k_λ which satisfies kx/kxk −x₀k< ,

(6) if w is a vector so that x_λ and x_λ +w are endpoints of a maximal line segment in the unit sphere ofk.k_λ, thenb ≥ kwk ≥λ⁻¹−1.

Our objective is to generalize this result to(Λ, G)-pimples in a natural manner which for theΛpart is directly inspired from[2].

In the following, k.k_Λ,G is as before the Λ, G-pimple associated to a given sequence (xk)k. We let B denote the closed unit ball for k.k, B_Λ^G denote the closed unit ball for the Λ, G-pimple at (x_k)_k, B_k^g denote the closed unit ball for theλ_k-pimplesk.k_λ

k,g atgx_k, andB₀ denote the union overk ∈ Kandg ∈ Gof the closed unit ballsB_k^g. It is clear thatB ⊂B0 ⊂B_Λ^G.

If Λ = (λ_k)k∈K, Λ⁰ = (λ⁰_k)k∈K, and c, d ∈ R, we write Λ < Λ⁰ to mean λ_k < λ⁰_kfor allk ∈K,c <Λto meanc < λ_k for allk ∈ K, andΛ < dto mean λk< dfor allk ∈K.

We shall first prove a lemma stating that forΛclose enough to1,B_Λ^Gis actually equal toB₀, Lemma 5. This will allow us to prove Proposition 6, our generaliza- tion of Bellenot’s results, by reducing essential parts of the argument to the simple application of Proposition 4 in eachgx_k, that is, to the case of a singleλ_k-pimple associated to the single pointgx_k.

Lemma 5 Let(X,k.k)be a real Banach space, letGbe a group of isometries on X containing−Idand let(x_k)k∈K be a finite or infinite sequence of normalized vectors ofX. Assume

(1) k.kis strictly convex onX, and LUR atx_kfor eachk∈K, and

(2) for allk ∈K,ck := inf{kxj −gxkk:j ∈K, g ∈G,(j, g)6= (k, Id)}>0.

Then there exists Λ₀ = (λ_0k)_k, with 1/2 < Λ₀ such that whenever Λ = (λ_k)_k satisfiesΛ₀ <Λ <1, then

(a) wheneverx∈B^g_k\Bandy∈B_l^h\Bwith(k, g)6= (l,±h), thenkx−yk ≥ c_min(k,l)/3,

(b) The closed unit ballB_Λ^Gof theΛ, G-pimple at(x_k)_kis equal to the unionB₀ overk ∈K andg ∈Gof the closed unit ballsB_k^gof theλ_k-pimplesk.k_λ

k,g

atgxk;

(c) The following equality holdsk.k_Λ,G = inf_k∈K,g∈Gk.k_λ

k,g.

(d) whenever kxk_Λ,G < kxk, there exists a unique pair {−g, g} ⊂ G and a uniquek ∈K such thatkxk_λ

k,g <kxk, and in this casekxk_Λ,G =kxk_λ

k,g. Furthermore for eachk,λ_0kdepends only onx_i, c_i,1≤i≤k.

Proof : Note that by (1), k.k is LUR at x_k for each k, so Proposition 4 (1) is satisfied for x0 = xk. By (1) again, k.k is strictly convex, and therefore any normalized vector y is extremal fork.k, so Proposition 4 (2) applies in (x_k)for any choice of > 0. We let_k =c_k/2, and fix a decreasing sequence(δ_k)_ksuch that for all k ≥ 1, δk ≤ck/4and ^4δ₃^k ≤ 1−λ(xk, ck), where λ(xk, .)is the LUR function associated to the normk.kinx_k.

For eachk, we letλ_0k be the realλ₀ associated by Proposition 4 tox₀ = x_k, = k, δ = δk, b = 1 and m = 1/2. Up to replacing each λ0k by a larger number in ]1/2,1[, we may assume that λ⁻¹_0k −1 ≤ δ_k/3for all k ∈ K and that limk→+∞λ_0k = 1ifK is infinite.

We letΛ = (λk)k be such thatΛ0 <Λ<1.

We first prove (a). Wheneverx∈ B_k^g \B we have, by Proposition 4 (4), that kz−gx_kk< δ_korkz+gx_kk< δ_k, wherez =x/kxk. Up to redefiningg as−g if necessary we may assume that the first holds. Then

kx−gx_kk ≤ kz−xk+kz−gx_kk<kxk −1 +δ_k ≤λ⁻¹_k −1 +δ_k ≤ 4δ_k 3 . Likewise ify ∈B_l^h\Bthen up to redefininghas−h,

ky−hxlk< 4δ_l 3 .

If nowx∈B_k^g\Bandy∈B_l^h\Bwith(k, g)6= (l,±h)andk ≤l, we have, that kx−yk ≥ kgx_k−hx_lk − kx−gx_kk − ky−hx_lk ≥c_k− 4

3δk

− 4 3δl

, so sinceδ_l≤δ_k,

kx−yk ≥ck− 8δ_k

3 ≥ck/3.

Therefore (a) is proved.

We now intend to prove b). First we observe that B0 is closed. Let indeed x be the limit of a convergent sequence (x_n) in B₀, and let k_n, g_n be such that x_n ∈ B_k^gn

n: we claim that x ∈ B₀. Indeed if x_n ∈ B for infinitely many n then x ∈ B and we are done, so we may assume that xn ∈ B_k^gn_n \B for eachn. If k_nis bounded then we may assume thatk_nis constantly equal to somek and that kx_m−x_nk < c_k/3 for alln, m. But then by a), B^g_kⁿ = B_k^gm for alln, m, sox also belongs to B_k^gn for any choice of n and therefore to B0, and we are done.

So we may assume thatk_nconverges to+∞; then sinceλ_0kn converges to1,λ_kn also converges to 1, and since kx_nk ≤ 1/λ_kn for each n, kxk ≤ 1. Therefore x∈B ⊂B0, which proves the claim. FinallyB0is closed.

Next we observe thatB₀ is convex. Assuming towards a contradiction that x, y ∈ B₀ and ^x+y₂ ∈/ B₀. Let(k, g)and(l, h)be such thatx ∈ B_k^g andy ∈ B_l^h, and without loss of generality assume that k ≤ l. By convexity of B_k^g andB_l^h, these two balls are different, otherwise ^x+y₂ would belong to either of them and therefore to B₀. Furthermore x ∈ B_k^g \B, otherwise x ∈ B ⊂ B_l^h and ^x+y₂ ∈ B_l^h ⊂ B0. In other wordskxk_λ

k,g <kxk. Likewisekyk_λ

l,h <kyk. Therefore by Proposition 4 (4) applied tox/kxkfor theλ_k-pimple atgx_k, and up to replacing g by−g if necessary,kgx_k−x/kxkk< δ_k. Then

kgx_k−xk ≤

gx_k− x kxk

+

x− x kxk

≤δ_k+λ⁻¹_k −1≤ 4δ_k 3 . Likewisekhxl−yk< ^4δ₃^l. Then

x+y

2 −gx_k+hx_l 2

≤ 2(δ_k+δ_l)

3 ≤ 4δ_k 3 . Sincekgxk−hxlk ≥ck by (2), it follows by LUR ofk.kingxkthat

gxk+hxl

2

≤λ(gxk, ck) = λ(xk, ck),

and

x+y 2

≤ 4δ_k

3 +λ(x_k, c_k)≤1,

a contradiction, since ^x+y₂ does not belong toB₀ and therefore neither toB. This contradiction proves thatB₀is convex.

Finally we have proved that B₀ is closed convex. Since it contains B and each segment[−gx_k/λ_k, gx_k/λ_k], it therefore contains B_Λ^G, and since alsoB₀ is included inB_Λ^G, (b) is proved.

The equality in (c),

k.k_Λ,G = inf

k∈K,g∈Gk.k_λ

k,g, follows immediately from (b).

To prove (d), letxbe such thatkxk_Λ,G <kxk. Then by (c) there existsλ_k, g such that kxk_λ

k,g < kxk. Thereforez = x/kxk_λ

k,g ∈ B_k^g \B. Since in (a), the realc_min(k,l)is positive, there exists no other B_l^h such thatz ∈ B_l^h \B. In other wordsz /∈B_l^hfor(k, g)6= (l,±h).

If we had thatkxk_λ

l,h <kxkfor some(l, h)6= (k,±g), thenz⁰ =x/kxk_λ

l,h

would belong toB_l^h \B. Ifkxk_λ

l,h ≤ kxk_λ

k,g then by convexity ofB_λ^h

l,z ∈B^h_λ

l

and soz ∈B_λ^h

l\B, a contradiction. Ifkxk_λ

l,h≥ kxk_λ

k,g, then we obtain a similar contradiction usingz⁰. Thereforekxk_λ

l,h ≥ kxk. Finally we have proved thatk and{−g, g}are unique so thatkxk_λ

k,g <kxk. From (c) we therefore deduce that kxk_Λ,G =kxk_λ

k,g. This concludes the proof of (d).

Summing up, we see that ifxis a point such thatkxk_Λ,G <kxk, thenkxk_Λ,G = kxk_λ

k,gfor some(g, λ_k)such thatkxk_λ

k,g <kxk,(λ_k, g)and(λ_k,−g)are the only to share this property, and for all other(λ_l, h),kxk_λ

l,h=kxk.

Proposition 6 Let(X,k.k)be a real Banach space, letGbe a group of isometries onXcontaining−Idand let(x_k)k∈Kbe a finite or infinite sequence of normalized vectors ofX. Assume

(1) k.kis strictly convex onXand LUR atxk for eachk ∈K, and

(2) for allk ∈K,c_k := inf{kx_j −gx_kk:j ∈K, g ∈G,(j, g)6= (k, Id)}>0.

Then given(δ_k)_k >0, (b_k)_k > 0and0 < m < 1, there existsΛ₀ = (λ_0k)_k, with 1/2<Λ₀ such that wheneverΛ = (λ_k)_ksatisfiesΛ₀ <Λ<1, then

(3’) mk.k ≤ k.k_Λ,G ≤ k.k,

(4’) if1 =kyk>kyk_Λ,Gthen∃g ∈G, k∈K :kgx_k−yk< δ_k

(5’) x_k,λ = λ⁻¹_k x_k is the only isolated extremal point ofk.k_Λ,G which satisfies kx/kxk −x_kk< c_k/2,

(6’) ifwis a vector so that x_k,λ andx_k,λ +ware endpoints of a maximal line segment in the unit sphere ofk.k_Λ,G, thenbk≥ kwk ≥λ⁻¹_k −1.

Furthermore for eachk,λ0kdepends only onmandxi,ci,δi,bi,1≤i≤k.

Proof : Fix G, (x_k)k∈K and (δ_k)_k > 0, (b_k)_k > 0 and 0 < m < 1 as in the hypotheses. We may again assume that(δk)k is decreasing and that for allk ≥1, δ_k ≤c_k/4and ^3δ₂^k ≤1−λ(x_k, c_k), and we let_k =c_k/2.

Note that as in Lemma 5, by (1), Proposition 4 (1) is satisfied forx₀ =x_kand Proposition 4 (2) applies in(xk)for any choice of >0.

Let therefore, for eachk, λ_0k be the λ₀ given by Proposition 4 forx₀ = x_k, = _k, δ = δ_k, b = b_k and m. Up to replacing each λ_0k by a larger number in ]1/2,1[, we may also assume that (a) to (d) of Lemma 5 are satisfied whenever Λ = (λ_k)_kis such thatΛ₀ <Λ<1.

We now fix someΛsuch thatΛ₀ <Λ<1and verify (3’) to (6’).

Affirmation (3’) is obvious from Proposition 4 (3) for each(λk, g), that is mk.k ≤ k.k_λ

k,g ≤ k.k, and from Lemma 5 (c), that is

k.k_Λ,G = inf

k∈K,g∈Gk.k_λ

k,g.

For (4’) assume1 = kyk > kyk_Λ,G. Then by Lemma 5 (d), there existg, k such that 1 = kyk > kyk_λ

k,g, so from Proposition 4 (4) applied for k.k_λ

k,g, kgxk−yk< δk ork−gxk−yk< δk. This proves (4’).

To prove (5’) we note that ifkx/kxk −x_kk < c_k/2then wheneverg 6= ±Id ork6=l,

kx/kxk −gx_lk>kgx_l−x_kk −c_k/2≥c_k/2≥δ_k, and likewise

kx/kxk+gxlk ≥δk. Applying Proposition 4 (4) toy=x/kxkandk.k_λ

l,hwe deduce that kxk=kxk_λ

l,h

whenever(l, h)6= (k,±g). From Lemma 5 (c) it follows that kxk_Λ,G =kxk_λ

k,g.

Therefore for all thosexsuch thatkx/kxk −x_kk < _k,xis an isolated extremal point fork.k_Λ,Gif and only if it is an isolated extremal point fork.k_λ

k,g, which, by an application of Proposition 4 (5) forx_k and _k = c_k/2, is equivalent to saying thatx=x_k,Λ. Therefore (5’) is proved.

The proof of (6’) is a little bit longer. We denote byS_k^g the unit sphere for k.k_λ

k,g, byS_Λ^Gthe unit sphere fork.k_Λ,G, bySthe unit sphere fork.k,S⁰the set of points ofSon whichk.k_Λ,G =k.k. By Lemma 5 (c)(d),S_Λ^G =S⁰∪(∪_k,g(S_k^g\S)).

We prove the following result:

Claim: A line segment inS_Λ^Gcannot contain points both inS_k^g\SandS_l^h\S withS_k^g 6=S_l^h.

Proof : First note that by Lemma 5 (a) , wheneverx ∈S_k^g \S, y ∈S_l^h \S, with S_k^g 6= S_l^h, it follows that d(x, y) ≥ c_min(k,l)/3. Therefore whenevery ∈ S_l^h \S, we have that

d(y, S_k^g\S)≥d_k/3,

whered_k:= minl≤kc_l >0. Assume a line segmentLinS_Λ,Gcontained a pointx inS_k^g\Sand a pointyinS_l^h\SwithS_l^h 6=S_k^g. By the aboved(y, S_k^g\S)≥d_k/3.

Since also d(x, S_k^g \S) = 0, we could define x⁰ 6= y⁰ in the line segment [x, y]

such thatd(z, S_k^g \S) ∈ [d_k/9,2d_k/9]wheneverz ∈ [x⁰, y⁰]. For any suchz we would have that

z 6∈S_k^g0⁰\S,

wheneverS_k^g0⁰ 6=S_k^g, becaused(z, S_k^g\S)≤2d_k/9<inf_w∈Sg0

k0\Sd(w, S_k^g\S), and also that

z 6∈S_k^g\S,

becaused(z, S_k^g\S)≥dk/9>0. Sincezbelongs toS_Λ^GandS_Λ^G =S⁰∪(∪k,g(S_k^g\ S)), this would mean thatzbelongs toS⁰and therefore toS. Then the non trivial line segment [x⁰, y⁰] would be included inS, but this would contradict the strict convexity ofk.k. This concludes the proof of the claim.

Going back tox_k,Λ, since this vector belongs toS_k^Id\S, we deduce from the claim and from the formulaS_Λ^G = S⁰ ∪(∪_k,g(S_k^g\S))that if[x_k,Λ, x_k,Λ +w]is a maximal line segment inS_Λ^G, it is a line segment in(S_k^Id\S)∪S⁰ ⊂ S_k^Id. We shall now prove that this segment cannot be extended inS_k^Id.

To show this note that for any strict extension[x_k,λ, y] of [x_k,Λ, x_k,Λ+w]in S_k^Id, either [x_k,λ, y] ⊂ S_k^Id \S ⊂ S_Λ^G by Lemma 5 (c)(d), and in this case the maximality in S_Λ^G is contradicted; or there exists a sequence (yn)n of distinct points converging tox_k,Λ+win the segment[x_k,λ, y]withy_n ∈S for alln, and thereforeS contains three different points of a same line segment, but this again contradicts the strict convexity ofk.k. Therefore[xk,Λ, xk,Λ+w]is a maximal line segment inS_k^Id.

By Proposition 4 (6) applied tok.k_λ

k,Id, we therefore deduce thatb_k≥ kwk ≥ λ⁻¹_k −1, which proves (6’) and concludes the proof.

We shall say that a vectorx ∈ X separates a bounded group G of isomor- phisms on X (or sometimes that X is separating for G) if for anyg 6= h in G, gx 6= hx, or equivalently, gx 6= xwhenever g 6= Id. Saying thatxseparatesG and that the orbitGxis discrete is easily equivalent toinfg6=Idkx−gxk>0.

We shall now use Proposition 6 to renorm a separable real space(X,k.k)with an equivalent norm k|.k| for which the group of isometries on X is some given groupGofk.k-isometries. This will of course depend on certain conditions onG andX. We begin with a lemma.

Lemma 7 Let X be a separable real Banach space with a norm k.k and let G be a countable group of isometries onX such that−Id ∈ G. Assume that there exists a normalized vectorx₀inXwhich separatesGand such that the orbitGx₀ is discrete. Letα ∈]0,infg6=Idkx₀−gx₀k)[. Then there exists a countable setK containing0, a partition({0}, T1, T2)ofK, and points(xk)k∈K in the unit sphere ofXsuch that

(a) Ifk∈K thenx_k =a_kx₀ +z_k witha_k >0andkz_kk ∈[α/10, α/5],

(b) The spaceXis the closed linear span of the set{gx_k :g ∈G, k∈ {0}∪T₁}, (c) If k ∈ T₂, then x_k = a_kx₀ +α_kg_kx₀, with a_k > 0andα_k ∈ [α/10, α/5],

g_k∈G\ {−Id, Id},

(d) for allk ∈K,inf{kx_j −gx_kk:j ∈K, g∈G,(j, g)6= (k, Id)}>0.

Pointsxkfork ∈T1 will be said to be oftype1, and pointsxkfork∈T2will be said to be oftype2. The pointx₀ will be the unique point oftype0.

Proof : We first define a finite or infinite sequence of vectors(y_n)n∈Nby induction.

LetV₀ = span{gx₀, g ∈G}and lety₀ =x₀. IfV₀ =Xthen we letN ={0}and

we are done. Otherwise we pick somey₁ ∈/ V₀ and considerV₁ = span{gy_n, n= 0,1, g ∈ G}. If V₁ = X then we let N = {0,1} and we are done. Otherwise repeating the procedure it is possible to pick a finite or infinite sequence(yn)n∈N

such that, if V_n := span {gy_k, k ≤ n, g ∈ G}, we have that y_n ∈/ Vn−1 for all n ≥1and∪n∈NV_nis dense inX.

We associate to(yn)n∈N an enumeration(un)n∈Kof{gx0, g ∈G\ {±Id}} ∪ {y_n, n ∈ N}. Note that K may be infinite (when G is infinite or X is infinite dimensional), in which case we shall assume K = N, or finite (when Gis finite andXis finite dimensional), in which case we shall assumeK ={0,1, . . . ,|K|−

1}. We shall also assume that the enumeration starts withy₀, i.e. u₀ =y₀ =x₀. We let

T1 ={n ∈K :un=ykfor some k ≥1}

and

T₂ ={n∈K;u_n =gx₀ for some g∈G\ {±Id}}

Note that ifG={−Id, Id}then(u_n)is simply an enumeration of(y_n), in which case T₂ = ∅and there will only be points of type 0 or1. Likewise, whenX = span {gx0, g ∈ G}, thenT1 = ∅and there will only be points of type 0or2. If T₁ =T₂ =∅, then we are in the extreme (and simplest) case whenG={−Id, Id}

andXis one dimensional, and the result is trivial - but the following proof is still valid.

Associated to (u_n)n∈K we define a sequence(x_n)n∈K of normalized vectors ofXas follows. Forn= 0,x₀is already defined, and, sincex₀ = (1−α/10)x₀+ (α/10)x0, satisfies (a).

The definition of x_n’s of type 1, that is for n ∈ T₁, is as follows. For any such n, let k ≥ 1 be such that u_n = y_k and let E_k = span(Vk−1, y_k). Pick some zn ∈ Ek such that kznk ∈ [α/10, α/5] and d(zn, Vk−1) = α/10, and let x_n =a_nx₀+z_nwherea_n >0is such thatkx_nk= 1. Obviously (a) is satisfied for suchx_n’s. To prove property (b), observe that V₀ ⊂span{gx₀, g ∈G}, and that ifVk−1 ⊂span{gxn, g ∈G, n∈ {0} ∪T1 :un=yj for somej ≤k−1}then

V_k ⊂Vk−1⊕span{gy_k, g ∈G} ⊂Vk−1⊕span{gx_n, g ∈G},

wheren ∈T₁is such thatu_n =y_k, sinceVk−1⊕spany_k =E_k =Vk−1⊕spanx_n and sinceVk−1⊕spangy_k =gE_k=Vk−1⊕spangx_n. Therefore by induction,

V_k ⊂span{gx_n, g ∈G, n∈ {0} ∪T₁ :u_n=y_j for somej ≤k}, and since∪_kV_kis dense inX, it follows that

X = span{gx_n, g ∈G, n∈ {0} ∪T₁}.

The definition ofx_n’s of type 2, that is for n ∈ T₂, is more refined and will require an induction. For any suchn, that is ifu_nis of the formg_nx₀ forg_n ∈G, we shall define xn as follows: we shall pick someαn ∈ [α/10, α/5]and define z_n =α_ng_nx₀,x_n =a_nx₀+z_nwitha_n >0andkx_nk= 1. That is (c) simply states the definition of such points, and assertion (a) follows immediately for points of type 2.

To conclude the proof the lemma, it therefore only remains to check that the points(x_n)may be chosen in such a way as to ensure that condition (d) is satisfied.

The difficult part is to verify condition (d) for points of type 2, and for this the choice of α_n in their definition will need to follow a precise procedure. Before writing this procedure, which is based on an induction onn ∈ T₂, we shall note a few estimates which only require knowing that αn ∈ [α/10, α/5]for suchn’s.

First of all it will be useful to observe that for all n, kz_nk ≤ α/5 and therefore a_n ∈[1−α/5,1 +α/5]. This holds forx₀ as well as for points of type1or2.

Our aim is therefore to obtain lower estimates for expressions of the form kx_n−gx_mk, when(n, Id)6= (m, g). We distinguish four cases.

First case:g 6=Id.

Ifg 6=Idthenkx_n−gx_mk=ka_nx₀+z_n−ga_mx₀−gz_mktherefore kx_n−gx_mk ≥ kx₀−gx₀k − |1−a_n| − |1−a_m| − kz_nk − kgz_mk ≥α/5.

Second case:g =Idandx_max(m,n)is of type1.

Without loss of generality assumen > m. Sincex_nis of type 1, ifk is such thatx_nis associated toy_k, the vectorx_m is inV_k−1and so

kxn−xmk ≥d(xn, Vk−1) = α/10.

Third case:g =Id,x_max(m,n)is of type2,x_min(m,n)is of type1.

Assumingn > m, sincex_nis of type2andx_m of type1then kx_n−x_mk ≥d(x_m, V₀)≥α/10.

Fourth case:g =Id,x_max(m,n)is of type2,x_min(m,n) is of type0or2.

We describe the induction we need to choose the xn’s of type 2, i.e. we de- scribe how to choose each corresponding α_n ∈ [α/10, α/5] in the definition of x_n, n∈T₂ to obtain good estimates forkx_n−x_mkin this fourth case.

LetT₂ ={m₁, . . . , m|T₂|}, whenGis finite, orT₂ ={m₁, m₂, . . .}, whenGis infinite, be the increasing enumeration of T₂, and letm₀ = 0. Form ∈ T₂, recall thatgmdenotes the element ofGsuch thatum =gmx0, and that the corresponding x_m will have the forma_mx₀+α_mg_mx₀.

For everym ∈ T₂, letI_m⁰ = [α/10, α/5]. Forβ ∈ [α/10, α/5], let x_m(β) = am(β)x0+βgmx0 wherebm(β)>0is such thatkxm(β)k= 1. For eachm ∈T2

we shall pick someβ_m ∈ [α/10, α/5]such that x_m = x_m(β_m)is a good choice for the corresponding point of type2.

We observe thatkxm(β)−xm(γ)k ≥ ^α₂|β −γ|, for all β, γ in [α/10, α/5].

Indeed if x_m(β) − x_m(γ) = (β − γ) with kk < α/2 and β 6= γ, then by definition ofx_m(β)andx_m(γ),

(b_m(β)−b_m(γ))x₀ = (γ −β)(g_mx₀−),

sog_mx₀−=± kg_mx₀−kx₀. If for example±=−in this equality, then since

|1− kg_mx₀−k |=| kg_mx₀k − kg_mx₀−k | ≤ kk, it follows that

kgmx0 +x0k=k+x0 − kgmx0−kx0k ≤ kk+kx0k kk< α,

and by definition of α, gm = −Id, a contradiction. Similarly the case ± = + would implyg_m =Id.

Now for allm ∈ T₂ divide I_m⁰ = [α/10, α/5]in three successive intervals of equal lengthα/30. Since

kx_m(β)−x_m(γ)k ≥ α

2|β−γ| ≥ α² 60

wheneverβ is in the first andγ in the last interval, it follows that there exists an intervalI_m¹ ⊂I_m⁰ of lengthα/30(which is either the first or the last subinterval), such that

β ∈I_m¹ ⇒ kx_m(β)−x₀k ≥ α² 120. We then pick β_m1 inI_m¹

1 and fix x_m1 = x_m1(β_m1). Therefore we have ensured that

kx_m1 −x₀k ≥ α² 120.

Givenn ∈N, n≥2, assume that for allm∈K2such thatm < mn, someβmand x_m =x_m(β_m)have been selected, as well as for any1≤k≤n−1andm∈K₂ such thatm≥m_k, an intervalI_m^k of length _10.3^αk, in such a way that

• I_mⁱ ⊂I_m^j ifm ≥m_i andi≥j.

• whenever0≤k < n−1, m≥m_k andβ ∈I_m^k, kx_m(β)−x_mkk ≥ α²

40.3^k+1.

Then for anym∈T₂,m≥m_n−1, dividingI_mⁿ⁻¹in three subintervals and picking the first or the last, we find by the same reasoning as aboveI_mⁿ ⊂I_mⁿ⁻¹ of length

α 10.3ⁿ with

β ∈I_mⁿ ⇒

x_m(β)−x_mn−1

≥ α² 40.3ⁿ.

We then pick β_mn inI_mⁿ_n and fix x_mn = x_mn(β_mn). So for all0 ≤ k ≤ n−1, β_mn ∈I_mⁿ

n ⊂I_m^k+1

n and therefore we have ensured that for any0≤k < n kx_mn−x_mkk ≥ α²

40.3^k+1. This easily implies that for anyk 6=n,

kx_mn−x_mkk ≥ α² 40.3^k+1, which ends up the fourth case.

Summing up the four cases we obtain the following estimates. Forx₀we have that

inf{kx₀−gxmk,(m, g)6= (0, Id)} ≥min(α/5, α/10, α²/120) =α²/120 Ifx_nis of type1, then

inf{kx_n−gx_mk,(m, g)6= (n, Id)} ≥min(α/5, α/10) =α/10 Ifx_nis of type2, andk such thatn =m_k, then

inf{kx_n−gx_mk,(m, g)6= (n, Id)} ≥min(α/10, α²

40.3^k+1) = α² 40.3^k+1. We have therefore finally proved (d), and this ends the proof of the lemma.

Theorem 8 LetXbe a separable real Banach space with an LUR-normk.kand letGbe a countable group of isometries onX such that−Id ∈ G. Assume that there exists a normalized vectorx0inXwhich separatesGand such that the orbit Gx₀ is discrete. ThenXadmits an equivalent normk|.k|such thatGis the group of isometries onX fork|.k|.

Proof : Since Gx₀ is discrete and x₀ separates G, let α ∈]0,1[ be such that kx0−gx0k ≥ α, for all g 6= Id. Then Lemma 7 applies, so let (xn)n∈K be the associated family of points. Because of Lemma 7 (d),

ck := inf{kxj−gxkk:j ∈K, g∈G,(j, g)6= (k, Id)}

is positive for each k ∈ K. Therefore (2) in Proposition 6 is satisfied; and (1) is clearly satisfied sincek.kis LUR and therefore strictly convex.

We letδ_k = c_k/3for eachk and define sequences b_k and λ_k by induction as follows. Let b₁ = 1. Let λ_0,1 be given by Proposition 6 for m = 1/2, c₁, δ₁, and b1, and pick λ1 in the interval ]λ1,1,1[. Givenbk and λk, let δk+1 = ck+1/4 and letb_k+1 satisfyingb_k+1 <min(b_k,(λ⁻¹_k −1)/2). We letλ_0,k+1 be the number given by Proposition 6 for m = 1/2andc_i, δ_i, b_i, 1 ≤ i ≤ k + 1. We fix some λk+1 >max(λ0,k+1, λk)

We then definek|.k| as theΛ, G-pimple at (x_n)_n for Λ = (λ_n)_n. Therefore Proposition 6 applies.

Observe that E = {λ⁻¹_n gx_n, g ∈ G, n ∈ K}is the set of isolated extremal points of k|.k|. Indeed for a point xof S_Λ,G either kx/kxk −gx_kk < c_k/2for some g, k, in which case by (5’) x = λ⁻¹_k gxk if it is an isolated extremal point;

or kx/kxk −gx_kk ≥ c_k/2 > δ_k for all g, kthen by (4’) k.k = k|.k| in a neigh- borhood of x and then x is not an isolated extremal point since k.k is LUR at x.

Therefore any isometryT fork|.k|mapsE onto itself. Ifn < mandg ∈ G, then T cannot map λ⁻¹_n x_n toλ⁻¹_m gx_m. Indeed ifw (resp. w⁰) is a vector so that λ⁻¹_n xnandλ⁻¹_n xn+w(resp.λ⁻¹_m gxmandλ⁻¹_m gxm+w⁰) are endpoints of a maximal line segment in the unit sphere of k|.k|, then since g is an isometry for k|.k|we may assumeg =Id, and then by (3’) and (6’),

k|wk| ≥ 1

2kwk ≥ 1

2(λ⁻¹_n −1)> bn+1 ≥bm ≥ kw⁰k ≥ k|w⁰k|.

Likewise, if n > m, we may by using T⁻¹ deduce that T cannot map λ⁻¹_n x_n to λ⁻¹_m x_m.

Finally it follows that for eachn, the orbitGx_nis preserved byT.

We finally prove thatT belongs necessarily toG. Without loss of generality we may assume thatT x₀ =x₀and then by Lemma 7 (b) it is enough to prove that T gx_n=gx_nfor allg ∈Gand anyx_nof type1or equal tox₀.

Letg ∈ G, g 6= ±Id. Let xk be the associated vector of type2of the form x_k =ax₀+βgx₀given by Lemma 7 (c). Then sinceT x_k =hx_kfor someh∈G,

T xk =ax0 +βT gx0 =h(ax0+βgx0).

So|a| kx₀−hx₀k=βkT gx₀−hgx₀kand kx₀−hx₀k ≤ α/5

1−α/5(khgx₀k+kT gx₀k))≤ α/5

1−α/5(kx₀k+ 2k|T gx₀k|)

≤ α

4(1 + 2k|x₀k|)< α,

therefore by the definition ofα, h=Id. It follows immediately from the expres- sion ofT x_kthat

T gx₀ =gx₀.

and this holds for any g ∈ G. Let now x_n be of type1, of the form a_nx₀ +z_n, according to Lemma 8 (a), and letg ∈G, then

T gx_n=T(a_ngx₀+gz_n) = a_ngx₀+T gz_n, and sinceT(gx_n)is of the formhx_nfor someh∈G, we also have

T gx_n =a_nhx₀+hz_n. Therefore

a_ngx₀+T gz_n =a_nhx₀+hz_n, soa_nkgx₀−hx₀k=kT gz_n−hz_nkand therefore,

kgx₀−hx₀k ≤ khz_nk+kT gz_nk

1−α/5 ≤ α/5 + 2k|T gz_nk|

1−α/5 ≤ 3α/5

1−α/5 < α, whenceg =hand

T gx_n =gx_n.

Finally sinceT gx_n=gx_nfor anyg ∈Gandx_nequal tox₀ or of type1, Lemma 7 (b) implies thatT =Idand therefore belongs toG.

2.2 Representable groups of linear isomorphisms

In this subsection, we give sufficient conditions for a group of isomorphisms on a Banach spaceX to be representable inX. We shall need a well-known result of Kadec about LUR-renormings, see [4] Theorem 2.6 p.48.

Theorem 9 (Kadec, 1965) Any separable real Banach space admits an equiva- lent LUR norm.

We shall also use a refinement obtained by G. Lancien for spaces with the Radon-Nikodym Property, [14] Theorem 2.1. The fact concerning the isometries in this result is not written explicitly in [14] but is obvious from the definition of the renorming. For the exact definition of the Radon-Nikodym Property we refer to [15]. In our applications we shall only use the fact that any separable dual space has the Radon-Nikodym Property.

Theorem 10 (Lancien, 1993) Any separable real Banach space with the Radon- Nikodym Property may be renormed with an equivalent LUR norm without dimin- ishing the group of isometries.

Using Theorem 9 we obtain:

Theorem 11 LetX be a separable real Banach space andGbe a finite group of isomorphisms such that−Id ∈G. ThenX admits an equivalent norm for which Gis the group of isometries onX.

Proof : By Theorem 9 we may assume that the normk.konXis LUR. Then we define an equivalent normk.k_G onXby

kxk_G= (X

g∈G

kgxk²)^1/2.

Since this is the`₂-sum of the LUR normk.kwith an equivalent norm, it is classi- cal to check that it is also LUR, see [4] Fact 2.3, and obviously anyg ∈Gbecomes an isometry for k.k_G. To apply Theorem 8 and sinceGis finite, it therefore only remains to find some x₀ such thatx₀ 6= gx₀ for all g 6= Id. But if such an x₀ didn’t exist then we would have thatX = ∪g∈GKer(Id−g), andKer(Id−g) would have non-empty interior for someg 6=Id, which would contradict the fact that no proper subspace ofXcan have non-empty interior.

We use Theorem 10 to obtain: