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arXiv:2111.10122v1 [math.FA] 19 Nov 2021
TWO NEW EXAMPLES OF BANACH SPACES WITH A
PLASTIC UNIT BALL
RAINIS HALLER, NIKITA LEO, AND OLESIA ZAVARZINA
Abstract. We prove that Banach spaces ℓ1⊕2Rand X⊕∞Y, with strictly
convex Xand Y, have plastic unit balls (we call a metric space plastic if every
non-expansive bijection from this space onto itself is an isometry).
1. Introduction
A function between two metric spaces is called an isometry if it preserves dis-
tances between points, and non-expansive if it does not increase distances between
points. We call a metric space Xplastic if every non-expansive bijection from X
onto itself is an isometry. The last notion was introduced by S. A. Naimpally, Z. Pi-
otrowski, and E. J. Wingler in [8]. It is known that every totally bounded metric
space is plastic, see [3, Satz IV] or [8, Theorem 1.1]. On the other hand, a plastic
metric space need not be totally bounded nor bounded – e.g., the set of integers with
the usual metric is plastic [8, Theorem 3.1]. There are also examples of bounded
metric spaces that are not plastic, one of our favorite examples here is a solid ellip-
soid in Hilbert space ℓ2(Z)with infinitely many semi-axes equal to 1and infinitely
many semi-axes equal to 2, see [1, Example 2.7].
It is a challenging open problem, posed by B. Cascales, V. Kadets, J. Orihuela,
and E. J. Wingler in 2016 [1], whether the unit ball of every Banach space is a plastic
metric space. The unit ball of a finite-dimensional space is compact, and therefore
plastic. So the question is really just about the infinite-dimensional spaces. So far,
the plasticity of the unit ball has successfully been proved for the following spaces
and classes of spaces:
•strictly convex spaces,
•the space ℓ1,
•ℓ1-sums of strictly convex spaces,
•spaces whose unit sphere is the union of all its finite-dimensional polyhedral
extreme subsets,
•the space c.
The first result was for strictly convex spaces and it was obtained by B. Cascales,
V. Kadets, J. Orihuela, and E. J. Wingler in 2016 [1]. The same year, V. Kadets
and O. Zavarzina proved the plasticity of the unit ball of ℓ1[5]. They generalized
this result to ℓ1-sums of strictly convex spaces in 2017 [4]. The fourth item was
obtained by C. Angosto, V. Kadets, and O. Zavarzina in 2018 [2, Theorem 4.11].
The plasticity of the unit ball of cwas proved by N. Leo in 2021 [6, Theorem 4.1].
In this paper, we present two new examples of Banach spaces whose unit balls
are plastic: the sum of ℓ1and Rby ℓ2, and the sum of any two strictly convex
spaces by ℓ∞.
1
2 RAINIS HALLER, NIKITA LEO, AND OLESIA ZAVARZINA
2. Preliminaries
We consider only real Banach paces. For a Banach space X, we denote the closed
unit ball and the unit sphere of Xby BXand SX.
Extreme points turn out to be essential to the study of plasticity of the unit ball.
Recall that for a vector space Xand for a convex subset Cof X, a point x∈C
is called an extreme point of Cif it does not belong to the interior of any non-
trivial line segment connecting two distinct points of C. We use ext Cto denote
the set of extreme points of a set C. Henceforth, we focus on extreme points of
the unit ball. Extreme points of the unit ball lie on the unit sphere. A space such
that all the points of the unit sphere are extreme is called strictly convex. Strictly
convex spaces have a property that, for any two distinct points xand yand any
non-negative scalars αand βwith α+β=kx−yk, the point β
kx−ykx+α
kx−ykyis
the only point of the space that is distance αfrom xand distance βfrom y. This
is going to be used in the proof for ℓ∞-sum of strictly convex spaces.
The next proposition describes the behaviour of a non-expansive bijection from
the unit ball of a Banach space onto itself. It provides some of the main tools for
the study of plasticity of the unit ball. The last item is the reason why are extreme
points so important to the topic at hand. This proposition is going to be used
extensively throughout both of the proofs.
Proposition 2.1 ([1, Theorem 2.3]).Let Xbe a Banach space and let F:BX→
BXbe a non-expansive bijection. Then
(1) F(0) = 0,
(2) for each x∈BX,kF(x)k ≤ kxk,
(3) if x∈SX, then F−1(x)∈SX,
(4) if x∈ext BX, then F−1(x)∈ext BXand F−1(αx) = αF −1(x)for each
α∈[−1,1].
It is also useful to know the following result.
Theorem 2.2 ([7, Theorem 2]).Let Xand Ybe normed spaces and let Ube a
subset of Xand Vbe a subset of Y. If Uand Vare convex with non-empty interior
and there exists an isometric bijection F:U→V, then Fextends to an affine
isometric bijection e
F:X→Y.
The theorem implies that if Xis a Banach space and F:BX→BXis an
isometric bijection, then Fextends to an isometric automorphism of X– a linear
isometric bijection from Xonto itself. If we want to prove that every non-expansive
bijection on the unit ball of some space Xis an isometry, it might be useful to know
what isometric bijections are there. The last result says that these are precisely the
restrictions of the isometric automorphisms of X.
There is also a sufficient condition for a non-expansive bijection F:BX→BX
to be an isometry.
Theorem 2.3 ([1, Theorem 2.6]).Let Xbe a Banach space and let F:BX→BX
be a non-expansive bijection. If F(SX) = SXand F(αx) = αF (x)for every x∈SX
and every α∈[−1,1], then Fis an isometry.
The latter will be used to finish the proof for the case of ℓ∞-sum of two strictly
convex spaces.
TWO NEW EXAMPLES OF BANACH SPACES WITH A PLASTIC UNIT BALL 3
3. The space ℓ1⊕2R
Theorem 3.1. The unit ball of ℓ1⊕2Ris a plastic metric space.
Proof. Denote the space ℓ1⊕2Rby Z. Let ekstand for the k-th element of the
canonical basis of ℓ1. Denote by hthe projection of Zonto R. For each b∈[−1,1]
denote by Lbthe set {z∈BZ:h(z) = b}. Note that
ext BZ={(aem, b): a, b ∈R, a2+b2= 1, m ∈N}.
Let (aei, b)and (cej, d)be two arbitrary extreme points. If i=j, then the
distance between these two points is p(a−c)2+ (b−d)2and we have
p(a−c)2+ (b−d)2= 2 ⇐⇒ (a−c)2+ (b−d)2= 4
⇐⇒ a2−2ac +c2+b2−2bd +d2= 4
⇐⇒ 0 = a2+ 2ac +c2+b2+ 2bd +d2
⇐⇒ 0 = (a+c)2+ (b+d)2
⇐⇒ a=−c&b=−d.
If i6=j, then the distance between (aei, b)and (cej, d)is p(|a|+|c|)2+ (b−d)2
and we have
p(|a|+|c|)2+ (b−d)2= 2 ⇐⇒ (|a|+|c|)2+ (b−d)2= 4
⇐⇒ a2+ 2|a||c|+c2+b2−2bd +d2= 4
⇐⇒ 0 = a2−2|a||c|+c2+b2+ 2bd +d2
⇐⇒ 0 = (|a| − |c|)2+ (b+d)2
⇐⇒ |a|=|c|&b=−d.
That is, the distance between points (aei, b)and (cej, d)is equal to two if and only
if either i=j,a=−c, and b=−d, or i6=j,|a|=|c|, and b=−d. In particular, if
uand vare extreme points that are distance two apart, then h(u) = −h(v). This
is going to be used in Step 1.
Step 1. Let Fbe a non-expansive bijection from BZonto itself. We need to show
that Fis an isometry. Let us first show that for each m∈Nwe have F−1(em,0) =
(θen,0), where θ∈ {−1,1}and n∈N.
Let m∈Nbe arbitrary. Choose indices iand jsuch that m,i, and jare pairwise
distinct. Note that x= (em,0),y= (ei,0), and z= (ej,0) are extreme points. Item
(4) of Proposition 2.1 implies that x′=F−1(x),y′=F−1(y), and z′=F−1(z)
are also extreme points. Note that x,y, and zare distance two apart. The non-
expansiveness of Fimplies that x′,y′, and z′are also distance two apart. Since x′,
y′, and z′are extreme points that are distance two apart, we have h(x′) = −h(y′),
h(y′) = −h(z′), and h(z′) = −h(x′), from which it follows that h(x′) = 0. As x′is
an extreme point, it has the form (aen, b), where a, b ∈R,a2+b2= 1, and n∈N.
Since h(x′) = 0, we have b= 0 and a∈ {−1,1}, which finishes the proof. For
further reference define σm=n,θm=a, and gm=θmeσ(m).
This way we obtain a function σ:N→N, which describes the permutation of
indices exerted by F. Note that this function is an injection. Indeed, let iand jbe
two indices with σ(i) = σ(j). Denote by nthe common value of σ(i)and σ(j). Then
F−1(ei,0) = (θien,0) and F−1(ej,0) = (θjen,0), where θi, θj∈ {−1,1}. Suppose
that θj=−θi. Then item (4) of Proposition 2.1 implies F−1(−ei,0) = (θjen,0),
4 RAINIS HALLER, NIKITA LEO, AND OLESIA ZAVARZINA
but the latter implies (−ei,0) = (ej,0), which is impossible. Hence θi=θj, so we
get (ei,0) = (ej,0), from which it follows that i=j. Therefore, σis indeed an
injection. This fact is going to be implicitly used in what follows. However, note
that we do not yet know whether σis also a surjection.
Now we know that F−1(em,0) = (gm,0) for every m∈N. Further, item (4) of
Proposition 2.1 implies F−1(aem,0) = (agm,0) for every a∈[−1,1]. The latter can
be restated as that F(agn,0) = (aen,0) for every n∈Nand a∈[−1,1].
Step 2. Using the same argument as in [5], we can show that for every N∈N
and real numbers a1,...,aNwith PN
n=1 |an| ≤ 1we have
F N
X
n=1
angn,0!= N
X
n=1
anen,0!,
where gnis defined as in Step 1. The proof is by induction on N. We are going to
omit the proof, since it repeats the argument from [5] almost word to word. The
proof depends on the fact that F(agn,0) = (aen,0) for every n∈Nand a∈[−1,1]
as was established in Step 1.
Step 3. The previous step and continuity of Fimply that for every real sequence
(an)∞
n=1 with P∞
n=1 |an| ≤ 1we have
F ∞
X
n=1
angn,0!= ∞
X
n=1
anen,0!.
If we denote the set {z∈BZ:h(z) = 0}by L0, the latter implies F−1(L0)⊂L0.
Step 4. We have seen that F−1(L0)⊂L0. Now, let us show that F−1(L0) = L0.
We are going to use a proof by contradiction. Suppose that F−1(L0)6=L0. Then
there is z∈L0\F−1(L0). Let sstand for F−1(0,1). Item (4) of Proposition 2.1
implies that F−1(0,−1) = −F−1(0,1). Hence F(s) = (0,1) and F(−s) = (0,−1).
Let us consider two cases.
Case 1. Suppose that s6∈ L0. Consider a continuous curve ξ: [0,1] →BZ
composed of line segments [s, z]and [z, −s]. Note that Im ξ∩L0={z}, but z6∈
F−1(L0). Hence Im ξdoes not intersect F−1(L0). Since Fis continuous, ξ′=F◦ξ
is also a continuous curve in BZ. We know that Im ξdoes not intersect F−1(L0),
so Im ξ′should not intersect L0. Since his a continuous functional, f=h◦ξ′is a
continuous function from [0,1] into R. Note that f(0) = 1 and f(1) = −1. Therefore,
there exists t∈(0,1) such that f(t) = 0, which implies that Im ξ′∩L06=∅, which
is a contradiction.
Case 2. For the second case, suppose that s∈L0. The argument is the same
except that now we consider a curve composed of line segments [s, (0,1)] and
[(0,1),−s]. Again, we see that Im ξdoes not intersect F−1(L0), thus Im ξ′should
not intersect L0. However, we have f(0) = 1 and f(1) = −1, which implies that
there exists t∈(0,1) with f(t) = 0. The latter means that Im ξ′∩L06=∅, which is
again a contradiction.
Therefore, we have F−1(L0) = L0. Note that this implies that the function σ
defined earlier is a bijection from Nonto N. This allows us to give the set of extreme
points an alternative description:
ext BZ={(agn, b): a, b ∈R, a2+b2= 1, n ∈N}.
Step 5. Let us show that F(0,1) = (0,1) or F(0,1) = (0,−1). First, item (4)
of Proposition 2.1 implies that F−1(0,1) ∈ext BZ. Hence we have F−1(0,1) =
TWO NEW EXAMPLES OF BANACH SPACES WITH A PLASTIC UNIT BALL 5
(agn, b), where a, b ∈R,a2+b2= 1, and n∈N. Consider points (agn, b)and
(agn,0). These points are distance |b|from each other. We have F(agn, b) = (0,1)
and by Step 1 we have F(agn,0) = (aen,0). The distance between these points is
√1 + a2. Since Fis non-expansive, we must have √1 + a2≤ |b|, which implies that
1 + a2≤b2. Keeping in mind the relation a2+b2= 1, we can infer that a= 0 and
b∈ {−1,1}. Therefore, either F(0,1) = (0,1) or F(0,1) = (0,−1).
For the case F(0,1) = (0,1), item (4) of Proposition 2.1 implies that F(0, b) =
(0, b)for all b∈[−1,1]. In particular, we have F(0,−1) = F(0,−1). For the case
F(0,1) = (0,−1), item (4) of Proposition 2.1 implies that F(0, b) = (0,−b)for all
b∈[−1,1]. In particular, we have F(0,−1) = F(0,1). Further we deal with the case
F(0,1) = (0,1). The case F(0,1) = (0,−1) can be handled in an analogous way.
Step 6. Note that an operator L:ℓ1→ℓ1defined by the formula
L ∞
X
n=1
angn!=
∞
X
n=1
anen
is an isometric automorphism of ℓ1. Hence the operator L′:Z→Zdefined by
L′(x, y) = (L(x), y)is an isometric automorphism of Z. Our goal is to show that F
is the restriction of the latter operator to BZ. Obviously, this will also prove that F
is an isometry. So we need to show that F(z) = L′(z)for every z∈BZ. Note that
the latter has already been established for h(z) = 0 and h(z)∈ {−1,1}. Hence, we
only need to consider the case |h(z)| ∈ (0,1). Let bbe an arbitrary real number with
|b| ∈ (0,1), let astand for √1−b2, and let us show that F(z) = L′(z)for h(z) = b.
Note that zhas the form (ax, b), where x∈Bℓ1. Also note that L′(z) = (L(ax), b).
Therefore, we need to show that F(ax, b) = (L(ax), b)for every x∈Bℓ1.
We begin with showing that F(ax, b) = (L(ax), b)holds for every x∈Sℓ1. This
is the same as to show that F−1(ay, b) = (L−1(ay), b)for every y∈Sℓ1. Hence, fix
an arbitrary y∈Sℓ1. Since F−1(ay , b)∈BZ, it has the form (cx, d), where c, d ∈R,
c2+d2= 1, and x∈Bℓ1. Further, since (ay, b)∈SZ, by item (3) of Proposition
2.1 we have (cx, d)∈SZ. Therefore, x∈Sℓ1.
Consider points (cx, d)and (0,1). The distance between these points is equal
to pc2+ (1 −d)2=√2−2d. We have F(cx, d) = (ay, b)and by Step 5 we have
F(0,1) = (0,1). The distance between these points is pa2+ (1 −b)2=√2−2b.
Since Fis non-expansive, we must have √2−2b≤√2−2d, which implies that
d≤b. Now, consider points (cx, d)and (0,−1). The distance between these points
is equal to pc2+ (1 + d)2=√2 + 2d. We have F(cx, d) = (ay, b)and by Step 5 we
have F(0,−1) = (0,−1). The distance between these points is pa2+ (1 + b)2=
√2 + 2b. Since Fis non-expansive, we must have √2 + 2b≤√2 + 2d, which implies
that b≤d. Therefore, we have b=d.
Consider now points (cx, b)and (cx, 0). The distance between these points is
equal to |b|. We have F(cx, b) = (ay, b)and by Step 2 we have F(cx, 0) = (L(cx),0).
The distance between these points is equal to pkay − L(cx)k2+b2. Since Fis non-
expansive, we must have pkay − L(cx)k2+b2≤ |b|, but this implies L(cx) = ay.
Therefore, we have F−1(ay, b) = (L−1(ay), b), which finishes the argument.
Step 7. The relation F(ax, b) = (L(ax), b)is now established for kxk= 1. Note
that the latter relation is obvious for x= 0, being a corollary of Step 5. Therefore,
let us consider the case where kxk ∈ (0,1).
6 RAINIS HALLER, NIKITA LEO, AND OLESIA ZAVARZINA
To begin with, let us consider points (ax
kxk, b),(ax, b), and (0, b). The distance
between the first two is a(1−kxk)and the distance between the second two is akxk.
Note that F(0, b) = (0, b)by Step 5 and F(ax
kxk, b) = (L(ax
kxk), b)by Step 6. Since
Fis non-expansive, we have
Lax
kxk, b−F(ax, b)
≤a(1 − kxk)
and
kF(ax, b)−(0, b)k ≤ akxk.
However, as the distance between (L(ax
kxk), b)and (0, b)is equal to a, the last two
inequalities must be in fact equalities, since otherwise we have a contradiction with
the triangle inequality. Let (P∞
n=1 ynen, d)be the expansion of F(ax, b)and let
(P∞
n=1 ˜ynen, b)be the expansion of (L(ax
kxk), b). Now we have
a=
Lax
kxk
=
∞
X
n=1 |˜yn| ≤
∞
X
n=1 |˜yn−yn|+
∞
X
n=1 |yn| ≤
v
u
u
t ∞
X
n=1 |˜yn−yn|!2
+ (b−d)2+v
u
u
t ∞
X
n=1 |yn|!2
+ (b−d)2=
Lax
kxk, b−F(ax, b)
+kF(ax, b)−(0, b)k=
a(1 − kxk) + akxk=a.
Therefore, all the inequalities in between are in fact equalities, which is only possible
when b=d.
Finally, consider points (ax, b)and (ax, 0). The distance between these two points
is |b|. By Step 2 we have F(ax, 0) = (L(ax),0). Since Fis non-expansive, it follows
that pkP∞
n=1 ynen− L(ax)k2+b2≤ |b|, hence P∞
n=1 ynen=L(ax). Therefore,
F(ax, b) = (L(ax), b)as needed.
We have shown that Fis a restriction of an isometric automorphism of Z. There-
fore, Fis an isometry as required.
TWO NEW EXAMPLES OF BANACH SPACES WITH A PLASTIC UNIT BALL 7
4. The space X⊕∞Y
In this section, we are going to prove the plasticity of the unit ball of the ℓ∞-sum
of two strictly convex Banach spaces.
Theorem 4.1. Let Xand Ybe two strictly convex Banach spaces. Then the unit
ball of X⊕∞Yis a plastic metric space.
Proof. Denote the space X⊕∞Yby Z. The unit ball of this space is the set of
all pairs (x, y)where x∈BXand y∈BY. Extreme points are the pairs (x, y )
where x∈SXand y∈SY. If one of the two spaces is trivial, then Zis itself strictly
convex, so we can limit ourselves with the case where Xand Yare both non-trivial.
Let F:BZ→BZbe an arbitrary non-expansive bijection from the unit ball of
Zonto itself. Our goal is to show that Fis an isometry. Let us fix the notation.
Denote by Gthe inverse of F. For z∈Zdenote by zxand zythe first and the
second element of z. Denote by ZXthe set
{z∈BZ:kzxk>kzyk},
denote by ZYthe set
{z∈BZ:kzxk<kzyk},
and denote by Ethe set
{z∈BZ:kzxk=kzyk}.
These three sets form a partition of the unit ball of Z– they are pairwise disjoint
and their union is BZ. The set Eis closed, the closure of ZXis ZX∪Eand the
closure of ZYis ZY∪E. Since E={αz :α∈[0,1], z ∈ext BZ}, the last item of
Proposition 2.1 implies G(E)⊂E. The proof is going to depend on the number of
dimensions of Xand Y. If Xhas more than one dimension, then ZXis a connected
set. However, if Xis R(that is, Xis one-dimensional), then the set ZXhas two
connected components: Z−
X={z∈ZX:zx<0}and Z+
X={z∈ZX:zx>0}.
Obviously, the same can be said about Yand ZY. If Xand Yare both one-
dimensional, then Zis finite-dimensional, so we can omit this case. Hence, we have
to handle two cases: 1) both spaces have more than one dimension, 2) one of the
spaces is Rand the other has more than one dimension.
1) Let us start with the first case. We know that G(E)⊂E, so it follows that
F(ZX∪ZY)⊂ZX∪ZY. The set ZX∪ZYhas two connected components, these
are ZXand ZY, so the continuity of Ftogether with the last inclusion imply that
there are four possible cases:
i) F(ZX)⊂ZXand F(ZY)⊂ZX,
ii) F(ZX)⊂ZXand F(ZY)⊂ZY,
iii) F(ZX)⊂ZYand F(ZY)⊂ZX,
iv) F(ZX)⊂ZYand F(ZY)⊂ZY.
Consider case i). For this case, the continuity of Fimplies that F(ZX)⊂ZX
and F(ZY)⊂ZX, and since the union of ZXand ZYis the unit ball, we have
F(BZ)⊂ZX, which contradicts the surjectivity of F. Case iv) is similar. Therefore,
we are left with cases ii) and iii), which we are going to refer to as cases A) and B).
2) Now, let us consider the case where one of the spaces is Rand the other
has more than one dimension. We are going to show that this reduces to case A).
Suppose that Xhas more than one dimension and Yis R. Let us introduce the
8 RAINIS HALLER, NIKITA LEO, AND OLESIA ZAVARZINA
following notations:
Z−
Y={z∈ZY:zy<0},
Z+
Y={z∈ZY:zy>0},
E−={z∈E:zy≤0},
E+={z∈E:zy≥0}.
Note that the closure of Z−
Yis Z−
Y∪E−and the closure of Z+
Yis Z+
Y∪E+. As
with the previous case, the starting point is the inclusion F(ZX∪ZY)⊂ZX∪ZY,
but the set ZX∪ZYis now comprised of three connected components: Z−
Y,Z+
Y,
and ZX. The continuity of Ftogether with the last inclusion imply that there are
27 = 33possible cases, but some of these can be excluded, because they contradict
the surjectivity of F. As a result, we are left with a total of 6 = 3! possible cases.
We can divide these cases into three groups of two, obtaining the following three
cases:
i) F(ZX)⊂ZXand F(ZY)⊂ZY,
ii) F(ZX)⊂Z−
Yand F(ZY)⊂ZX∪Z+
Y,
iii) F(ZX)⊂Z+
Yand F(ZY)⊂ZX∪Z−
Y.
The first case is the same as case A). Our goal is to exclude cases ii) and iii).
Consider case ii). The continuity of Fimplies the inclusions F(ZX)⊂Z−
Y∪E−
and F(ZY)⊂ZX∪Z+
Y∪E. Since Eis a subset of ZXand a subset of ZY,
we obtain the inclusion F(E)⊂E−, which contradicts the previously obtained
inclusion G(E)⊂E. Case iii) is similar. As we see, we are left with case A).
Thereby, we have two cases to consider: A) F(ZX)⊂ZXand F(ZY)⊂ZY, B)
F(ZX)⊂ZYand F(ZY)⊂ZX. Let us make some additional conclusions. For case
A), the continuity of Fimplies that F(ZX)⊂ZXand F(ZY)⊂ZY. As Eis a
subset of ZXand a subset of ZY, it follows that F(E)⊂ZX∩ZY. The latter is
equal to E, so we have the inclusion F(E)⊂E. We can use the same argument to
show this for case B). Now, for case A) we have F(ZX)⊂ZX,F(ZY)⊂ZYand
F(E)⊂E. Since Fis a bijection, while ZX,ZY, and Eform a partition of BZ,
we can conclude that F(ZX) = ZX,F(ZY) = ZY, and F(E) = E. Analogously, for
case B) we obtain F(ZX) = ZY,F(ZY) = ZX, and F(E) = E.
The last item of Proposition 2.1 asserts that G(ext BZ)⊂ext BZ. We can also
show that F(ext BZ)⊂ext BZ. To demonstrate this, let z∈ext BZbe arbitrary.
Since F(E) = Eand zbelongs to E,F(z)belongs to Eas well, so F(z) = αz′
where z′∈ext BZand α∈[0,1]. We need to show that α= 1. The last item of
Proposition 2.1 implies that G(αz′) = αG(z′)and G(z′)∈ext BZ, so we obtain
the equation z=αG(z′). Taking the norm of both sides, we see that α= 1. This
proves the inclusion F(ext BZ)⊂ext BZ, hence F(ext BZ) = ext BZ.
To finish the proof, we would have to consider cases A) and B) separately. How-
ever, the proofs for these two cases are almost identical, so we will only consider
case A).
First, we are going to show that
∀z, w ∈BZ(zx=wx∈SX=⇒G(z)x=G(w)x∈SX),
∀z, w ∈BZ(zy=wy∈SY=⇒G(z)y=G(w)y∈SY).
Let us demonstrate the first item. Fix z , w ∈BZand suppose zx=wx∈SX. Let
us consider four possible cases.
TWO NEW EXAMPLES OF BANACH SPACES WITH A PLASTIC UNIT BALL 9
•First, we consider the case where zy∈SYand wy6∈ SY. Since wx∈SXand
wy6∈ SY, it follows that w∈ZX. As w∈ZXand F(ZX) = ZX, we have
G(w)∈ZX. Therefore, G(w)y6∈ SY. Since zx=wx∈SX, the distance
between −zand wis two. Therefore, the distance between G(−z)and G(w)
is also two. This means that either G(−z)xand G(w)xare distance two
apart or G(−z)yand G(w)yare distance two apart. The second possibility
is excluded, because G(w)y6∈ SY. Consequently, we have the first case.
As G(−z)xand G(w)xare distance two apart, they should belong to SX.
Moreover, since Xis strictly convex, it follows that G(−z)x=−G(w)x.
Finally, since z∈ext BZ, the last item of Proposition 2.1 implies that
G(−z) = −G(z), so we have G(z)x=G(w)x∈SXas wanted.
•The same argument can be applied to the case where zy6∈ SYand wy∈SY
by swapping the roles of zand w.
•If zy∈SYand wy∈SY, then we take p∈BZsuch that py6∈ SYand px
is the same as zxand wx. Then zx=px∈SX,zy∈SY, and py6∈ SY, so
G(z)x=G(p)x∈SXby the first item of this list. Similarly, as wx=px∈
SX,wy∈SY, and py6∈ SY, we have G(w)x=G(p)x∈SX. Combining
these two yields G(z)x=G(w)x∈SXas wanted.
•An argument similar to the one used in the third item of this list can be
applied to the case where zy6∈ SYand wy6∈ SYby choosing p∈BZsuch
that py∈SYand pxis the same as zxand wx.
We have considered all four possible cases. One can use the same argument to prove
the second item. The obtained result brings us to consider functions gx:SX→SX
and gy:SY→SYthat can be defined with gx(x) = G(x, 0)xand gy(y) = G(0, y )y.
So now we have
∀z∈BZ(zx∈SX=⇒G(z)x=gx(zx)),
∀z∈BZ(zy∈SY=⇒G(z)y=gy(zy)).
We go on to prove some facts about these functions. First, let us show that these
functions are injective. Consider gx. Suppose by contrary that there are x, x′∈SX
such that x6=x′and gx(x) = gx(x′). Choose arbitrary y∈SYand consider pairs
(x, y)and (x′, y). We have G(x, y) = (gx(x), gy(y)) and G(x′, y) = (gx(x′), gy(y)).
Note that (x, y)6= (x′, y), but G(x, y ) = G(x′, y). This contradicts the injectivity
of G. Thus, gxis actually injective. We can use the same argument to show that gy
is injective too.
Let us consider surjectivity. Let x∈SXand y∈SYbe arbitrary. Then (x, y)
is extreme and hence F(x, y)is also extreme, because F(ext BZ) = ext BZ. Denote
by x′and y′the first and the second element of F(x, y). Since (x′, y′)is extreme, we
have x′∈SXand y′∈SY. Since Fsends (x, y)to (x′, y′), we have G(x′, y′) = (x, y),
thus gx(x′) = xand gy(y′) = y, which proves the surjectivity of both gxand gy.
Now that we know gxand gyare bijective, denote the inverse of gxby fxand the
inverse of gyby fy.
Let us show that gxand gyare symmetric. That is, we demonstrate that gx(−x) =
−gx(x)and gy(−y) = −gy(y)for every x∈SXand y∈SY. Pick an arbitrary
x∈SX. Let us show that gx(−x) = −gx(x). Choose an arbitrary y∈SY. Consider
pairs z= (x, y)and w= (−x, y). The distance between them is two, so the distance
between G(z)and G(w)should be also two. We know that G(z) = (gx(x), gy(y))
and G(w) = (gx(−x), gy(y)). For distance between these two pairs to be equal to
10 RAINIS HALLER, NIKITA LEO, AND OLESIA ZAVARZINA
two, elements gx(x)and gx(−x)should be distance two apart. Since Xis a strictly
convex space, it follows that gx(−x) = −gx(x). The same argument can be used to
show that gy(−y) = −gy(y)for y∈SY. Obviously, the symmetricity of gxand gy
implies that fxand fyare also symmetric.
Define a function Gx:X→Xwith Gx(0) = 0 and Gx(tx) = tgx(x)for all
x∈SXand t > 0. Similarly, define a function Gy:Y→Ywith Gy(0) = 0 and
Gy(ty) = tgy(y)for all y∈SYand t > 0. We see that Gxcoincides with gxon
SXand Gycoincides with gyon SY. Using the properties of gxand gyestablished
above, one can show that Gxand Gyare bijective and homogeneous (that is, for
every x∈BX,y∈BYand t∈Rwe have Gx(tx) = tGx(x)and Gy(ty) = tGy(y)).
Denote the inverses of Gxand Gyby Fxand Fy. It is easy to see that Fxand Fy
are also homogeneous. Moreover, we see that Fxcoincides with fxon SXand Fy
coincides with fyon SY.
Now, we are going to examine what Gdoes with a pair one element of which lies
on the sphere and the other element of which is arbitrary. Our goal is to show that
∀x∈SX∀y∈BYG(x, y) = (Gx(x), Gy(y)),
∀x∈BX∀y∈SYG(x, y) = (Gx(x), Gy(y)).
Let us consider the first item. Fix arbitrary x∈SXand arbitrary y∈BY. First,
let us consider the case y= 0. Our goal is to show that G(x, 0) = (gx(x),0). We
know that G(x, 0) = (gx(x), y0)where y0∈BY. Suppose by contrary that y06= 0.
Denote by y∗
0the element y0/ky0k ∈ SY. We see that Gsends (x, 0) to (gx(x), y0)
and (x, fy(y∗
0)) to (gx(x), y∗
0). The distance between (x, 0) and (x, fy(y∗
0)) is one,
while the distance between (gx(x), y0)and (gx(x), y∗
0)is smaller than one, which
is a contradiction. Next, consider the case y6= 0. Then y=tu where u∈SYand
t∈(0,1]. Our goal is to show that Gsends (x, tu)to (gx(x), tgy(u)). Again, we know
that G(x, tu) = (gx(x), y0)where y0∈BY. We need to show that y0=tgy(u). If
y0= 0, then Gsends both (x, 0) and (x, tu)to (gx(x),0), which contradicts the
injectivity of G. Therefore, y06= 0 and we can consider the element y∗
0=y0/ky0k ∈
SY. We see that Gsends (x, fy(y∗
0)) to (gx(x), y∗
0),(x, tu)to (gx(x), y0), and (x, 0) to
(gx(x),0). The distance between (gx(x), y∗
0)and (gx(x), y0)is 1−ky0k, the distance
between (gx(x), y0)and (gx(x),0) is ky0k. So the distance between (x, fy(y∗
0)) and
(x, tu)should be at most 1−ky0kand the distance between (x, tu)and (x, 0) should
be at most ky0k. This means that the distance between fy(y∗
0)and tu should also
be at most 1−ky0kand the distance between tu and 0should also be at most ky0k.
Since Yis a strictly convex space, we have tu =ky0kfy(y∗
0). It follows that t=ky0k
and u=fy(y∗
0). Thus we have y0=ky0ky∗
0=tgy(u)as wanted. The proof of the
second item is analogous.
From the last obtained result it follows directly that Fhas the analogous prop-
erties. That is, we have
∀x∈SX∀y∈BYF(x, y) = (Fx(x), Fy(y)),
∀x∈BX∀y∈SYF(x, y) = (Fx(x), Fy(y)).
Finally, we show that for every z∈BZwe have F(z) = (Fx(zx), Fy(zy)). Let us
first show that for every z∈BZwe have F(z)x=Fx(zx). Fix arbitrary z∈BZ.
Denote the elements of zby xand yand the elements of F(z)by x′and y′. The
element x∈BXcan be represented in the form tu where t∈[0,1] and u∈SX. As
Fx(tu) = tFx(u)by homogeneity of Fx, the equality that we need to prove takes the
form x′=tFx(u). We know that Fsends (tu, y)to (x′, y′),(u, y)to (Fx(u), Fy(y))
TWO NEW EXAMPLES OF BANACH SPACES WITH A PLASTIC UNIT BALL 11
and (−u, y)to (Fx(−u), Fy(y)). By homogeneity of Fxwe have Fx(−u) = −Fx(u).
The distance between (u, y)and (tu, y)is 1−t, the distance between (tu, y)and
(−u, y)is 1 + t. Thus the distance between (Fx(u), Fy(y)) and (x′, y′)is at most
1−tand the distance between (x′, y′)and (−Fx(u), Fy(y)) is at most 1 + t. This
implies that the distance between Fx(u)and x′is also at most 1−tand the distance
between x′and −Fx(u)is also at most 1 + t. Since Xis a strictly convex space, it
follows that x′=tFx(u)as wanted. One can use the same argument to show that
for every z∈BZwe have F(z)y=Fy(zy).
We can finish the proof by applying Theorem 2.3.
12 RAINIS HALLER, NIKITA LEO, AND OLESIA ZAVARZINA
Acknowledgements
The third author is grateful to her scientific advisor Vladimir Kadets for help with
this project. The third author was supported by the National Research Foundation
of Ukraine funded by Ukrainian State budget in frames of project 2020.02/0096
“Operators in infinite-dimensional spaces: the interplay between geometry, algebra
and topology”.
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Institute of Mathematics and Statistics, University of Tartu, Narva mnt 18, 51009
Tartu, Estonia
Email address:rainis.haller@ut.ee
Institute of Mathematics and Statistics, University of Tartu, Narva mnt 18, 51009
Tartu, Estonia
Email address:nikita.leo@ut.ee
Department of Mathematics and Informatics, V. N. Karazin Kharkiv National
University, 61022 Kharkiv, Ukraine.
Email address:olesia.zavarzina@yahoo.com
