Isometry Games in Banach Spaces

Reprint from the Bulletin of the Belgian Mathematical Society – Simon Stevin
Isometry Games in Banach Spaces
Stefano Baratella
Siu-Ah Ng
Bull. Belg. Math. Soc. Simon Stevin 15 (2008), 509–521
The Bulletin of the Belgian Mathematical Society - Simon Stevin is published by The Belgian
Mathematical Society, with financial support from the Universitaire Stichting van Belgie – Fon-
dation Universitaire de Belgique and the Fonds National de la Recherche Scientifique (FNRS).
It appears quarterly and is indexed and/or abstracted in Current Contents, Current Mathemat-
ical Publications, Mathematical Reviews, Science Citation Index Expanded and Zentralblatt f¨
ur
Mathematik.
The Bulletin of the Belgian Mathematical Society - Simon Stevin is part of Project Euclid (Cornell
University Library), an aggregation of electronic journals. It is available online to subscribers to
Project Euclid (http://projecteuclid.org).
For more informations about the Belgian Mathematical Society - Simon Stevin, see our web site
at http://bms.ulb.ac.be

---

---

Isometry Games in Banach Spaces
Stefano Baratella
Siu-Ah Ng
Abstract
We study a property of extension of partial isometries in a Banach space.
This property is formulated in game-theoretic language. It is weaker than
transitivity, self-dual for reflexive spaces and it is related to a well-known open
problem in functional analysis and to model-theoretic notions from mathemat-
ical logic.
1
Introduction
In this paper we attempt to relate and compare notions from functional analysis with
notions from model theory in mathematical logic. The latter are in turn inspired by
game theory. To be more precise, we formulate and consider a new property which
is equivalent to transitivity in separable Banach spaces but is weaker in general.
We call this property W S, for having a Winning Strategy in the game of extending
partial isometries. More specifically, suppose from the previous moves of the game,
we have two n−tuples ¯
a and ¯b of vectors in a Banach space, so that between the
subspaces they generate there is a linear isometric isomorphism T taking ai to bi
for all 1 ≤ i ≤ n. Now suppose there is given a new element an+1, we would like
to respond by an element bn+1 so that T extends to a linear isometric isomorphism
that takes an+1 to bn+1.
Property W S may provide more understanding into the following old problem
posed by Banach and Mazur (see [22]): Is every separable transitive Banach space
necessarily a Hilbert space? As far as we know, this problem remains unsolved.
The motivation behind the proposing of W S comes from similar considerations
in model theory of first order logic, as mentioned in [17], and hence it is related to
Received by the editors September 2007.
Communicated by F. Point.
1991 Mathematics Subject Classification : Primary: 46B04. Secondary: 46B10 46B08.
Key words and phrases : Banach space, transitivity, isometry, games.
Bull. Belg. Math. Soc. Simon Stevin 15 (2008), 509–521

---

510
S. Baratella – S.-A. Ng
the Banach space model theory investigated by Ward Henson and Iovino ([13], [16]),
as well as to Keisler’s BF (Back-and-Forth ) property ([19]) which we investigated
in [4] and [5] in the context of Banach spaces. Actually we will compare W S and
BF.
It is worthwhile to note that rather different notions of games in Banach spaces
are considered in [15], [23] and [10].
Despite the motivation from logic, the current paper, with the exception of the
minor result Theorem 9 (2), can be read without specific knowledge in logic. Al-
though nonstandard hulls are mentioned, they can be replaced by Banach space
ultraproducts given by [9], which are well-known among analysts. (However, for the
interested reader, nonstandard analysis and nonstandard hull constructions can be
found in [1] and [15]). The nonstandard hull construction is also briefly reviewed in
§ 5). We use [6], [20] and [18] as background in Banach space theory.
Here is an outline of the paper. In § 2 we introduce the definitions and notation
(W S and others) and make some elementary observations. We also show that W S is
weaker than transitivity. Then in § 3, we investigate whether there is any relationship
between isometry games and convexity properties. In § 4, we prove our main result
that W S is a self-dual property for reflexive spaces. Section 5 is devoted to the
study of isometry games in classical Banach spaces. Eventually in § 6 we present
some open problems.
2
Preliminaries
From now on, by isometric isomorphism we actually mean linear isometric isomor-
phism.
If X is a Banach space we denote by SX its unit sphere, namely the set of unit
vectors in X.
Recall that in a Banach space X a smooth point is a unit vector x normed by a
unique functional of unit norm, i.e. there is a unique f ∈ X′ (the dual of X) such
that f = 1 and f (x) = 1. And a space is smooth if every unit vector is smooth.
Equivalently, every point is normed uniquely, i.e. for any nonzero vector x, there is
a unique linear continuous functional f of norm 1 such that x = f (x).
A Banach space X is strictly convex if every point on SX is an extreme point,
namely it cannot be written as midpoint of two distinct unit vectors. X is uniformly
convex if for any ǫ > 0 there is δ > 0 such that, for all x, y ∈ SX, if x − y ≥ ǫ
then x+y < 1 − δ.
2
We will deal with so called Banach spaces structures of the form X = (X, ¯
x),
where ¯
x is a (possibly empty) sequence of elements from the Banach space X. Some
of the results can be generalized for structures obtained from a Banach space to-
gether with additional constants, functions and relations. Given two Banach space
structures X = (X, ¯
x) and Y = (Y, ¯
y), we write X ≡0 Y if there is an isometric
isomorphism between the subspaces spanned by {xi} and {yi}, respectively, that
maps xi to yi for all i. From a logical viewpoint (see [4]), this is the same as saying
that ¯
x and ¯
y satisfy the same quantifier-free sentences of an appropriate first order
language for Banach space structures in X and Y respectively, and hence we borrow
this notation from logic.

---

Isometry Games in Banach Spaces
511
We now give definitions of the main properties of our study.
• (Back-and-Forth) A Banach space structure X = (X, ¯
x) has BFn if whenever
(X , a1, . . . , an) ≡0 (X , b1, . . . , bn) and an+1 ∈ X, there is bn+1 ∈ X such that
(X , a1, . . . , an+1) ≡0 (X , b1, . . . , bn+1).
X has BF if it has BFn for all n.
• (Isometry Game) Given Banach space structures X and Y such that X ≡0 Y
we define a 2-player game Γ = Γ[X , Y] by describing the (n + 1)-th move in a
play of the game.
Suppose that, after n moves, (X , a1, . . . , an) ≡0 (Y, b1, . . . , bn) then player ∀
chooses some an+1 ∈ X and player ∃ responses by choosing some bn+1 ∈ Y or
∀ chooses some bn+1 ∈ Y and ∃ responses by choosing some an+1 ∈ X.
We say that player ∃ wins the play at move n + 1 if
(X , a1, . . . , an+1) ≡0 (Y, b1, . . . , bn+1).
We say that ∃ has a winning strategy if ∃ can win at every move n, no matter
how ∀ plays.
• (Winning Strategy) The Banach space structure X has W S if for all (unit)
vectors a, b satisfying (X , a) ≡0 (X , b), player ∃ has a winning strategy in
Γ[(X , a), (X , b)].
• (Transitivity) (See [7]) A Banach space X is transitive if, for all a, b ∈ SX,
there exists an isometric isomorphism of X taking a to b.
• (Almost transitivity) A Banach space X is almost-transitive if the orbits of
the action of isometric isomorphisms of X are dense in SX. Otherwise said,
X is almost-transitive if, for all ǫ > 0 and all unit vectors a, b, there exists an
isometric isomorphism T of X such that T (a) − b ≤ ǫ.
• For given rational λ ≥ 1 we define ¯a ∼ ¯
λ b if
λ−1
αiai ≤
αibi ≤ λ
αiai
for all α ∈ Qn.
So, for a Banach space X, (X, ¯
a) ≡
¯
0 (X, ¯
b) if and only if ¯
a ∼1 b.
We now remark some elementary facts about the above properties.
Remark 1.
1. Note the main difference between BF and W S : in the latter,
the partial isometry before the play of a game is not arbitrary; it must be
constructed from previous moves.
2. Let X be a Banach space with W S. Then, for all a, b ∈ SX, there exist a sepa-
rable subspace X0 ⊂ X and an isometric isomorphism T : X0 → X0 such that
T (a) = b. For, from W S we have (X, a1, . . . , an, . . . ) ≡0 (X, b1, . . . , bn, . . . ),
where a1 = a, b1 = b and {ai}i∈ω = {bi}i∈ω. Then we let X0 be the closed linear
span of {ai}, and T : X0 → X0 given by T (ai) = bi.

---

512
S. Baratella – S.-A. Ng
3. BF holds for a Banach space structure X if and only if, whenever ¯a and ¯b are
tuples of the same length such that (X , ¯a) ≡0 (X , ¯b), player ∃ has a winning
strategy in Γ[(X , ¯a), (X , ¯b)].
4. BF ⇒ W S ⇒ BF1.
5. For a Banach space of finite dimension n + 1, BFn is equivalent to BF.
6. All Hilbert spaces satisfy BF.
7. Let X be a Banach space. It is not difficult to check that, for all unit vectors
a, b ∈ X, the game Γ[(X , a), (X , b)] is closed. Hence a celebrated theorem
of game theory due to Gale and Stewart ensures the existence of a winning
strategy for one of the two players. Consequently a (possibly of little use)
criterion for W S is the non existence of a winning strategy for player ∀ in
Γ[(X , a), (X , b)] for all unit vectors a, b.
8. W S and transitivity are equivalent in separable Banach spaces.
Recall that the only finite dimensional transitive spaces are Hilbert (see [2] or
[7]). Hence a finite dimensional Banach space has W S if and only if it is Hilbert.
Also, properties W S and BF are equivalent in finite dimensional spaces and they
hold precisely in finite dimensional Hilbert spaces.
Recall the following characterization: a Banach space is smooth if and only if its
norm is Gˆateaux differentiable. We note the following:
Proposition 2. Let X be a Banach space with norm θ. Suppose that θ satisfies the
following assumption: for all u, v ∈ X such that θ(u) = θ(v) there exists a linear
and continuous mapping T : X → X such that
1. T (u) = v;
2. θT (x) = θ(x) for all x.
Let L : X → X be a linear homeomorphism and let η = θ ◦ L. Then η is a norm on
X that satisfies the same assumption as θ. Moreover, whenever θ is smooth, so is η.
Proof. First note that Gˆateaux differentiability of η follows from
η(x + ǫy) − η(x)
θ(L(x) + ǫL(y)) − θ(L(x))
η′ (y) = lim
= lim
= θ′
(L(y)).
x
L(x)
ǫ→0
ǫ
ǫ→0
ǫ
Let u and v be such that η(u) = η(v) and let T : X → X be as in the assumptions
such that T L(u) = L(v). We let S = L−1T L. Clearly, S is linear and continuous.
Moreover, S(u) = L−1T L(u) = L−1L(v) = v and, for all x,
η(S(x)) = θL(S(x)) = θT L(x) = θL(x) = η(x).

---

Isometry Games in Banach Spaces
513
Remark 3. Condition 2 in the statement of the previous proposition implies that
θ′ (y) = θ′
(T (y)) for all x and y.
x
T (x)
Notice also that if θ as above is Hilbertian, then so is η.
We prove now that W S and BF, although not preserved by equivalent renorming,
behave well with respect to certain renorming.
Corollary 4. Let L : X → X be a linear homeomorphism of a Banach space X
with norm
·
and let || · || be the norm defined by ||x|| = L(x) . Then
1. (X, · ) has W S if and only if (X, || · ||) has W S.
2. (X, · ) has BF if and only if (X, || · ||) has BF.
Proof. Follows from Proposition 2.
We remark that no finite dimensional vector space X can be equipped with a
unique norm · such that (X, · ) has BF : without loss of generality, let X = Rn
and let · be the euclidean norm. Then (Rn, · ) has BF (being a Hilbert space).
Let L be any linear isomorphism of Rn and let || · || be as in Corollary 4. Then
(Rn, || · ||) has BF.
We now prove that W S is weaker than transitivity.
Proposition 5. W S does not imply transitivity.
Proof. In the sequel [0, 1]I is always equipped with the product measure of the
Lebesgue measure on [0, 1], for all sets I.
Let
X = Lp([0, 1]ω1)
Y =
Lp([0, 1]ω2)
Z = X
Y,
α<ω1
where by
we denote p-sum and 1 ≤ p < ∞, p = 2. It is proved in [12] that Z is
not transitive. Nevertheless we claim that it has W S.
Note that each element [θ] ∈ Lp([0, 1]κ), as an equivalence class of measurable
functions, is determined by a countable set of coordinates. More precisely, for A ⊂ κ,
let πA denote the projection of [0, 1]κ onto [0, 1]A via (xi)i<κ → (xi)i∈A. Then for
all [θ] ∈ Lp([0, 1]κ), there is a countable A ⊆ κ and [ψ] ∈ Lp([0, 1]A) such that
[θ] = [ψ ◦ πA].
Note also that X ⊂ Lp([0, 1]ω2) as a subspace via [θ] → [θ ◦ πω ]. In particular,
1
as a subspace, Z ⊂ Y ≃ Lp([0, 1]ω2)
Y.
We denote by G(Y ) the group of isometric isomorphisms of Y onto itself. Again
by [12], we have that G(Y ) acts transitively on the unit sphere of Y.
In order to prove that Z has W S let a1, b1 ∈ Z (denoting both functions and
equivalence classes) with a1 ∼1 b1 and let T1 ∈ G(Y ) be such that T1(a1) = b1.
Assume now inductively that (a1, . . . , an) ∼1 (b1, . . . , bn) and T1, . . . , Tn in G(Y )
are constructed with ai, bi ∈ Z and Tj(ai) = bi, i ≤ j ≤ n. Write ai = a1 ⊕ a2 and
i
i
bi = b1 ⊕ b2, where a1, b1 ∈ X ⊂ L
, b2 ∈ Y. Now let a
i
i
i
i
p([0, 1]ω2 ) and a2
i
i
n+1 ∈ Z ⊂ Y
be given and let b = Tn(an+1). Regard b ∈ Lp([0, 1]ω2)
Y, i.e. we decompose b as
c ⊕ d, where c ∈ Lp([0, 1]ω2) and d ∈ Y. Let A ⊂ ω2 be a countable set of coordinate
indices on which c depends on.

---

514
S. Baratella – S.-A. Ng
Let B ⊂ ω1 be a countable set of coordinate indices on which a11, . . . , a1 , b1
n
1, . . . , b1
n
depend on and let ρ : ω2 → ω2 be a bijection that fixes B pointwise and ρ(A) ⊂ ω1.
Let ˆ
ρ : [0, 1]ω2 → [0, 1]ω2 be the corresponding bijection.
Let Tn+1 ∈ G(Y ) be defined as follows: for all a ∈ Y, regarding Tn(a) as an
element of Lp([0, 1]ω2)
Y, with Tn(a) = r ⊕ s, where r ∈ Lp([0, 1]ω2) and s ∈ Y,
define Tn+1(a) = r ◦ ˆ
ρ−1 ⊕ s.
Define bn+1 = c ◦ ˆ
ρ−1 ⊕ d. Since c ◦ ˆ
ρ−1 depends on ρ(A) ⊂ ω1, we have: bn+1 ∈ Z,
Tn+1(ai) = bi for i ≤ n + 1, and therefore (a1, . . . , an+1) ∼1 (b1, . . . , bn+1).
3
Isometry games, convexity and smoothness
Hilbert spaces are uniformly convex and uniformly smooth. At the same time they
have good isometry game properties, so it is natural to consider the connections
between these properties.
The following shows that convexity does not necessarily imply good behavior in
isometry games.
Remark 6. (A uniformly convex but not BF1 space.) Let R2 be normed so that it
is strictly convex (equivalently uniformly convex) but not Hilbert. Such norm can
be obtained by using a sufficiently rounded but not symmetric convex set as the unit
disc. Then BF1 cannot be satisfied in this space, for otherwise it satisfies BF and
so it has to be Hilbert (see the remarks in the previous section).
Notice that by rotating the above 2−dimensional unit disc in Rn along an axis,
we have example of an n−dimensional space which is uniformly convex but BF1
fails.
Along the lines of the proof of [8, Theorem 2.1], we can prove the following
Theorem 7. For a Banach space X with BF1 the following are equivalent:
1. X has the Radon-Nikod´ym property;
2. X is reflexive;
3. X is superreflexive;
4. X is uniformly convex.
Proof. By well-known results, every reflexive space has Radon-Nikod´ym property
and every uniformly convex space is superreflexive. Therefore it remains to prove
that 1. implies 4. For 0 ≤ ǫ ≤ 1 and x ∈ X let
∆(x, ǫ) = inf{1 − λ : there exists y ∈ X such that λx ± y ≤ 1 and y ≥ ǫ}.
It is known that if inf{∆(x, ǫ) : x ∈ SX} > 0 for all 0 < ǫ ≤ 1, then X is
uniformly convex (see [8] and [11]).
By Radon-Nikod´ym property there exists x0 ∈ SX such that ∆(x0, ǫ) > 0 for all
0 < ǫ ≤ 1 (see [8]). An easy calculation shows that BF1 implies ∆(x0, ǫ) = ∆(x, ǫ)
for all x ∈ SX and all 0 < ǫ ≤ 1. Then X is uniformly convex.

---

Isometry Games in Banach Spaces
515
Corollary 8. Properties (1) to (4) in the statement of Theorem 7 are equivalent for
Banach spaces with W S.
Proof. W S implies BF1.
The following gives dichotomy results with respect to smoothness and convexity
property.
Theorem 9. Let X be a Banach space.
1. If X has W S then either all points on SX are smooth or none is smooth;
2. If X has BF1 then either all points on SX are extreme or none is extreme.
Proof.
1. For sake of contradiction, suppose u, v are unit vectors, such that u is
smooth but v is not. Let w be a unit vector witnessing the nonsmoothness of
v. (By this we mean that there exist two unit functionals φ and θ both norming
v but φ(w) = θ(w).) Using W S we can find a closed separable subspace X0
of X including elements u, v, w, and an isometric isomorphism T : X0 → X0
such that T (u) = v. (See Remark 1(2).)
By Hahn-Banach, u is a smooth point in X0 and, by including the witness w, we
conclude that v is nonsmooth in X0. We thus get a contradiction since there
can be no isometric isomorphism mapping a smooth point to a nonsmooth
point.
2. (This is the only part of the paper where we use a little bit of logic. See [4]
for background.) Notice that BF1 is equivalent to saying that whenever two
vectors of X have the same qfpb-type (i.e. the same norm), they also satisfy
the same existential pb formulas with one existential quantifier.
Now we can prove the dichotomy by noticing that, for some ǫ > 0, either all
unit vectors satisfy the existential pb formula
∃y( x = y = 2x − y = 1 ∧ x − y ≥ ǫ)
or none does.
4
Isometry games in dual spaces
The following shows that, for reflexive Banach spaces, W S is a self-dual property.
Theorem 10. Let X be a reflexive Banach space. Then X has W S if and only if
the dual space X′ has W S.
Proof. Suppose X has W S. Then, by Theorem 7, it is strictly convex and super-
reflexive. It follows from the results in [11] (see Definition 7 and Proposition 9) that
there are smooth points on the unit sphere of X (Fr´echet differentiability is stronger
than Gˆateux differentiability). So, by Theorem 9, X is smooth.
Recall also that, for a reflexive space, strict convexity is equivalent to smoothness
of its dual.

---

516
S. Baratella – S.-A. Ng
Let φ1, θ1 be unit functionals in X′: we want to show that player ∃ has a winning
strategy in the game Γ[(X′, φ1), (X′, θ1)].
Let a1 and b1 be unit vectors in X such that a1 = φ1 = φ1(a1) = 1 and
b1 = θ1 = φ1(b1) = 1.
The main idea of the proof is that, for each play in the game Γ[(X′, φ1), (X′, θ1)],
we keep track of a simultaneous play in the game Γ[(X, a1), (X, b1)].
Suppose that, for some ¯
φ, ¯
θ ∈ (X′)n and some ¯
a, ¯b ∈ Xn, we already have the
following from the first n − 1 moves of a play in the game Γ[(X′, φ1), (X′, θ1)]:
• (X′, φ1, . . . , φn) ≡0 (X′, θ1, . . . , θn);
• (X, a1, . . . , an) ≡0 (X, b1, . . . , bn);
• (norming) ai = φi = φi(ai) = 1,
bi = θi = θi(bi) = 1,
i ≤ n.
Let φn+1 ∈ X′ with φn+1 = 1. We are going to respond by producing θn+1 ∈ X′
such that the above properties extend to some (n + 1)-tuples.
By smoothness and reflexivity of X′, let an+1 ∈ X be the unique vector so that
an+1 = φn+1(an+1) = 1. By W S in X, choose bn+1 ∈ X so that
(X, a1, . . . , an+1) ≡0 (X, b1, . . . , bn+1).
Now, by smoothness and reflexivity of X, there is θn+1 ∈ X′ so that
bn+1 = θn+1 = θn+1(bn+1) = 1.
Claim : (X′, φ1, . . . , φn+1) ≡0 (X′, θ1, . . . , θn+1).
First we construct a closed separable subspace X0 ⊂ X and an isometric isomor-
phism T : X0 → X0 so that
(i) T (ai) = bi,
i ≤ n + 1;
(ii) φi = θi ◦ T on X0, i ≤ n + 1;
(iii)
n+1
=
n+1
=
n+1
i=1 αiφi X0
i=1 αiφi X
and
n+1
i=1 αiθi X0
i=1 αiθi X , for
all α ∈ Rn+1 (i.e. the norms are the same whether one views the functionals
as functionals on X or as functionals on the subspace X0).
Define A = {α ∈ Qn+1 :
n+1 α
i=1
i = 1 }. Enumerate A as α(m), m < ω. Using
norming property on X′, for each α = α(m) ∈ A, let xm, ym ∈ X be unit vectors
such that
n+1
n+1
n+1
n+1
αiφi =
αiφi(xm),
αiθi =
αiθi(ym).
i=1
i=1
i=1
i=1
Next we define a bijection T0 whose domain is a countable subset of X as follows:
step 0: Define k0 = n + 1, ci = ai, T0(ci) = bi, i ≤ k0.
step m + 1: Suppose we have (X, c1, . . . , ck ) ≡0 (X, T0(c1), . . . , T0(ck )).
m
m
Let ck +1, . . . , ck
be a list consisting of xm, ym and images from previous
m
m+1
steps T0(ci), i ≤ km, which are distinct from cj, j ≤ km. By W S we can find vectors
T0(ci), km < i ≤ km+1 so that
(X, c1, . . . , ck
) ≡0 (X, T0(c1), . . . , T0(ck
)).
m+1
m+1

---