Isometric reflections in two dimensions and dual L1-structures

Abstract

In this manuscript we solve in the positive a question informally proposed by Enflo on the measure of the set of isometric reflection vectors in non-Hilbert 2-dimensional real Banach spaces. We also reformulate equivalently the separable quotient problem in terms of isometric reflection vectors. Finally, we give a new and easy example of a real Banach space whose dual has a non-trivial L-summand that does not come from an M-ideal in the predual.

Full text

Bull. Korean Math. Soc. 49 (2012), No. 6, pp. 1275–1289
http://dx.doi.org/10.4134/BKMS.2012.49.6.1275
ISOMETRIC REFLECTIONS IN TWO DIMENSIONS AND
DUAL L1-STRUCTURES
Francisco J. Garc
´
ıa-Pacheco
Abstract. In this manuscript we solve in the positive a question infor-
mally proposed by Enflo on the measure of the set of isometric reflection
vectors in non-Hilbert 2-dimensional real Banach spaces. We also refor-
mulate equivalently the separable quotient problem in terms of isometric
reflection vectors. Finally, we give a new and easy example of a real Ba-
nach space whose dual has a non-trivial L-summand that does not come
from an M-ideal in the predual.
1. Isometric reflection vectors in 2-dimensional spaces
The concept of isometric reflection vector seems to have been officially in-
troduced by Skorik and Zaidenberg in [13].
Definition 1.1 (Skorik and Zaidenberg, [13]).Let Xbe a real Banach space.
A vector e∈Xwith kek= 1 is said to be an isometric reflection vector if
there exists a closed, maximal subspace Mof Xsuch that X=Re⊕Mand
kλe +mk=kλe −mkfor every m∈Mand every λ∈R. We will denote by
ρe:X−→ Xthe surjective linear isometry so that ρe(λe +m) = λe −mfor
every λ∈Rand every m∈M.
Several results of [13] can be summarized into the following one.
Theorem 1.2 (Skorik and Zaidenberg, [13]).Let Xbe a real Banach space.
The following are equivalent:
(1) Xis a Hilbert space.
(2) The set of isometric reflection vectors is dense in the unit sphere.
In [3] the authors generalize the previous theorem as follows.
Theorem 1.3 (Becerra-Guerrero and Rodr´ıguez-Palacios, [3]).Let Xbe a real
Banach space. The following are equivalent:
(1) Xis a Hilbert space.
Received April 1, 2011; Revised March 7, 2012.
2010 Mathematics Subject Classification. Primary 46B20, 46C05, 46B04.
Key words and phrases. isometric reflection vector, L1-summand vector.
c
2012 The Korean Mathematical Society
1275

1276 FRANCISCO J. GARC´
IA-PACHECO
(2) The set of the isometric reflection vectors has non-empty interior in
the unit sphere.
In [2] it is provided a shorter proof of the previous result. The key of
this shortening relies on restricting to 2-dimensions. In Fall 2004, when the
author of this manuscript was a graduate student at the Math Department of
Kent State University, Per Enflo and the author had a conversation about this
shortening, moment at which Enflo asked the author the following question:
Question 1.4. Let Xbe a finite dimensional real Banach space. Consider on
the unit sphere of Xa Borel probability measure µsuch that µ({x}) = 0 for all
x∈Xsuch that kxk= 1. Assume that the set of isometric reflection vectors
has positive µ-measure. Is then Xa Hilbert space?
We would like to let the reader notice that the set of isometric reflection
vectors is closed, and hence it is a Borel subset of the unit sphere. Therefore,
the previous questions is well formulated. Before presenting any results on the
previous question we would like to remark a couple of things.
Remark 1.5.Given a real Banach space we will let BX(x, r) denote the (closed)
ball of center xand radius r. As expected, BXand SXwill denote the (closed)
unit ball and the unit sphere of X, respectively. On the other hand, let k·k0
and k·k1be two equivalent norms on X. The map
(1.1)
S(X,k·k0)→S(X,k·k1)
x7→ x
kxk1
is an homeomorphism.
Next, we will show an example, on any finite dimensional real Banach space
X, of a Borel probability measure µon SXsuch that µ({x}) = 0 for all x∈SX.
Example 1.6. Let Xbe an n-dimensional real Banach space. Choose an
isomorphism ψ:Rn→X. This induces an homeomorphism between the unit
sphere Sn−1in Rnand SXas follows:
φ:Sn−1→SX
x7→ φ(x) := ψ(x)
kψ(x)k.
This homeomorphism also defines a Borel probability measure on SXas follows:
µ(A) := µSn−1φ−1(A),
where Ais a Borel subset of SXand µSn−1is the standard measure on Sn−1.
Notice that µdepends on the choice of ψ.
As a partial positive answer to Question 1.4 we will prove the following:
Theorem 1.7 (Isometric Reflection Dichotomy Theorem).Let Xbe a 2-
dimensional real Banach space. Only one of the following two possibilities
holds:

ISOMETRIC REFLECTIONS 1277
(1) Xis a Hilbert space.
(2) The set of isometric reflection vectors is empty or finite.
Before moving into the details of the proof of the previous result we will
need the following lemma.
Lemma 1.8. Let Xbe a 2-dimensional real Banach space X. Let a, b ∈SX
with a6=±b. Denote
⌊a, b⌋:= ta + (1 −t)b
kta + (1 −t)bk:t∈[0,1].
Then
(1) ⌊a, b⌋ ⊆ BX(a, ka−bk).
(2) If f∈X∗is the unique functional such that f(a) = f(b) = 1, then
⌊a, b⌋ ⊆ f−1[1,+∞)∩SX.
(3) The equality holds above if and only if [a, b]is a maximal segment in
BX.
Assume, in addition, that bis an isometric reflection vector. Then
(4) ρb(⌊a, b⌋) = ⌊b, ρb(a)⌋.
(5) ⌊a, b⌋ ∩ ⌊b, ρb(a)⌋={b}.
(6) ⌊a, b⌋ ∪ ⌊b, ρb(a)⌋is connected.
(7) If ka−bk<1, then ⌊a, b⌋ ∪ ⌊b, ρb(a)⌋=⌊a, ρb(a)⌋.
Proof.
(1) It is a direct consequence of [7, Lemma 3.1].
(2) For any t∈[0,1] we have that kta + (1 −t)bk ≤ 1, therefore
fta + (1 −t)b
kta + (1 −t)bk=1
kta + (1 −t)bk≥1.
(3) Assume that
⌊a, b⌋=f−1[1,+∞)∩SX.
If [a, b] is not a maximal segment in BX, then there exists a number t∈
(−∞,0)∪(1,∞) such that ta+(1 −t)b∈SX. Obviously, ta+(1 −t)b∈
f−1[1,+∞)∩SX, but ta + (1 −t)b /∈ ⌊a, b⌋. Conversely, assume that
[a, b] is a maximal segment in BX. Let x∈f−1[1,+∞)∩SX\ ⌊a, b⌋.
The convexity of BXforces a,b, and xto be aligned, therefore the
segment containing them is in the unit sphere and hence [a, b] cannot
be maximal.
Assume that bis an isometric reflection vector:
(4) If t∈[0,1], then
ρbta + (1 −t)b
kta + (1 −t)bk=tρb(a) + (1 −t)b
ktρb(a) + (1 −t)bk,
since ρbis a surjective linear isometry that fixes b.

1278 FRANCISCO J. GARC´
IA-PACHECO
(5) Write ρb(a) = γb +δa with γ, δ ∈R. Notice that, since aand bare
linearly independent, we have that both δand γare different from 0.
Besides, a=γb +δρb(a) and hence
ρb(a) = −γ
δb+1
δa.
As a consequence, δ=−1. Assume that there exist t, s ∈[0,1) such
that
tb + (1 −t)a
ktb + (1 −t)ak=sb + (1 −s)ρb(a)
ksb + (1 −s)ρb(a)k,
Then
ρb(a) = ksb + (1 −s)ak(1 −t)
(1 −s)ktb + (1 −t)aka
+ksb + (1 −s)ak
(1 −s)t
ktb + (1 −t)ak−s
ksb + (1 −s)akb.
Therefore
ksb + (1 −s)ak(1 −t)
(1 −s)ktb + (1 −t)ak=−1,
which is impossible.
(6) ⌊a, b⌋ ∪ ⌊b, ρb(a)⌋is the union of two connected sets whose intersection
is non-empty.
(7) If ka−bk<1, then ka−ρb(a)k<2. Then ⌊a, b⌋ ∪ ⌊b, ρb(a)⌋is a
connected set in SXjoining aand ρb(a) and not containing −aor
−ρb(a). Thus ⌊a, b⌋ ∪ ⌊b, ρb(a)⌋=⌊a, ρb(a)⌋.
In virtue of the previous lemma we have the following short, but not less
important, remark. We spare the details of the proof to the reader.
Remark 1.9.Let Xbe a 2-dimensional real Banach space. Let e0, e1∈SXbe
isometric reflection vectors. We define the following recursive sequence:
en:= ρen−1(en−2) for all n≥2.
Notice the following:
•For every n≥2 we clearly have that enis an isometric reflection vector
of X. Indeed, simply take into consideration that ρen−1is a surjective
linear isometry on Xfor n≥2, and surjective linear isometries map
isometric reflection vectors into isometric reflection vectors.
•For every n≥1 it is clear that
ken−en−1k=
ρen−1(en−2)−ρen−1(en−1)

=ken−1−en−2k
=···
=ke1−e0k.

ISOMETRIC REFLECTIONS 1279
•If e06=±e1, then it is also clear, in virtue of the previous item and the
fifth item of Lemma 1.8, that
SX=[
n≥0
⌊en, en+1⌋ ⊆ [
n≥0
BX(en,ke1−e0k).
•The sequence (en)n∈Nmight be a periodic sequence. For instance:
(1) If e0=e1, then en=e0for all n≥0.
(2) If e0=−e1, then en=e0if nis even and en=−e0if nis odd.
(3) If e0and e1are in isometric reflection, that is, X=Re0⊕Re1and
kλe0+γe1k=kλe0−γe1kfor all λ, γ ∈R, then
e2=−e0,e3=−e1,e4=e0,....
Observe that in this case we have that
SX=⌊e0, e1⌋ ∪ ⌊e1, e2⌋ ∪ ⌊e2, e3⌋ ∪ ⌊e3, e4⌋
simply because e06=±e1.
Now we are ready to prove the Isometric Reflection Dichotomy Theorem:
Proof of Theorem 1.7. Assume that there is an infinite sequence (an)n∈N⊂SX
of isometric reflection vectors. Since SXis compact, we can assume that the
previous sequence is actually convergent. Our aim is at proving that the set
of isometric reflection vectors is dense in SX, which will be sufficient in virtue
of either [13] or [3]. Fix an arbitrary 0 < ε < 1. There are n, m ∈Nsuch
that kan−amk ≤ ε. We may rename anand amas e0and e1respectively.
According to Remark 1.9, there exists a sequence of isometric reflection vectors
(en)n≥0⊂SXsuch that ken−en−1k ≤ εfor all n∈Nand
SX⊂[
n≥0
BX(en, ε).
Since εwas arbitrarily fixed, we conclude that the set of isometric reflection
vectors is dense in SX.
Corollary 1.10. Let Xbe a 2-dimensional real Banach space. Let µbe a Borel
probability measure on SXsuch that µ({x}) = 0 for all x∈SX. If the set of
isometric reflection vectors has positive µ-measure, then Xis a Hilbert space.
Remark 1.11.The Isometric Reflection Dichotomy Theorem fails in any di-
mension strictly higher than 2. Indeed, let Xbe any real Banach space of
dimension strictly higher than 2. We will equivalently renorm Xto fail the
Isometric Reflection Dichotomy Theorem. It is easy to understand that Xis
isomorphic to a space of the form ℓ2
2⊕∞M, where Mis a non-zero real Ba-
nach space. Now observe that the unit sphere of ℓ2
2⊕∞Mhas infinitely many
isometric reflection vectors but ℓ2
2⊕∞Mis not a Hilbert space.

1280 FRANCISCO J. GARC´
IA-PACHECO
2. Applications to the separable quotient problem
The separable quotient problem is an old and famous problem originally
stated by Pelcynzki. However, it was Rosenthal who was the first one to state
it in its best known form (see [11]). We refer the reader to [10] for a recopilation
of recent results about this problem.
Problem 2.1 (Rosenthal, [11]).Let Xbe an infinite dimensional, real Banach
space. Does Xadmit an infinite dimensional, separable quotient?
The main result in this section is strongly motivated by the following equiv-
alent form of the separable quotient problem (see [12]). We first remind the
reader a couple of things:
•A barrel in a real topological vector space is a closed, absolutely convex
subset.
•A real topological vector space is said to be barrelled if every barrel
has non-empty interior, or equivalently, is a neighborhood of 0.
Theorem 2.2 (Saxon and Wilanski, [12]).Let Xbe an infinite dimensional,
real Banach space. The following conditions are equivalent:
(1) Xadmits an infinite dimensional, separable quotient.
(2) There exists a non-barrelled, dense subspace Yof X.
The idea is to reformulate the previous theorem in terms of isometric reflec-
tion vectors. Notice that the previous result states that if an infinite dimen-
sional Banach space Xadmits an infinite dimensional separable quotient, then
Xhas a non-barreled dense subspace Y, that is, a dense subspace Ywith a
barrel that has empty interior in Y. We will go a little further on this and
we will show that if Xhas an infinite dimensional separable quotient, then X
has a dense subspace Ywith a bounded complete barrel that has empty inte-
rior in Y. For this we will make use of the concept of Markushevich basis: A
pair (ei)i∈I⊂X, (e∗
i)i∈I⊂X∗is said to be a Markushevich basis when it is
biorthogonal (e∗
i(ej) = δij for all i, j ∈I), total (span {e∗
i:i∈I}is ω∗-dense
in X∗), and fundamental (span {ei:i∈I}is dense in X).
Lemma 2.3. Let Xbe an infinite dimensional, separable, real Banach space.
Let (en)n∈N⊂SX,(e∗
n)n∈N⊂X∗be a Markushevich basis for X. The linear
operator
(2.1) ℓ1−→ X
(tn)n∈N7−→ P∞
n=1 tnen
maps ω∗-closed, bounded subsets of ℓ1to sequentially ω-closed subsets of X. As
a consequence, the set
(2.2) (∞
X
n=1
tnen: (tn)n∈N∈Bℓ1)

ISOMETRIC REFLECTIONS 1281
is closed in X, and therefore it has empty interior in Xif and only if it has
empty interior in its linear span.
Proof. Let Abe a ω∗-closed, bounded subset of ℓ1. Let (xi)i∈Nbe a sequence
in the image of Aunder the operator (2.1), and assume that (xi)i∈Nis ω-
convergent to some x0∈X. For every i∈N, we can write
xi=
∞
X
n=1
ti
nen,
where ti
nn∈N∈A. For every n∈N, we will denote e∗
n(x0) by t0
n. If we fix
n∈N, then
lim
i→∞ ti
n= lim
i→∞ e∗
n(xi) = e∗
n(x0) = t0
n.
Since Ais bounded, the Banach-Steinhauss Theorem tells us that t0
nn∈N∈ℓ1
and ti
nn∈N
ω∗
−→ t0
nn∈Nas i→ ∞.
Since Ais ω∗-closed we deduce that t0
nn∈N∈A. Finally, if x06=P∞
n=1 t0
nen,
then there is m∈Nsuch that
t0
m=e∗
m(x0)6=e∗
m ∞
X
n=1
t0
nen!=t0
m,
which is impossible. 
Lemma 2.4. Let Xbe an infinite dimensional, separable, real Banach space.
Let (en)n∈N⊂SX,(e∗
n)n∈N⊂X∗be a Markushevich basis for X. The follow-
ing statements are equivalent:
(1) The basis (en)n∈Nis a Schauder basis equivalent to the ℓ1-basis.
(2) The operator
(2.3) ℓ1−→ X
(tn)n∈N7−→ P∞
n=1 tnen
is an isomorphism.
(3) The set
(2.4) (∞
X
n=1
tnen: (tn)n∈N∈Bℓ1)
has non-empty interior.
Proof. Assume that (1) holds. The operator given in (2.3) is onto, therefore it
is an isomorphism in accordance to the Open Mapping Theorem. Now, assume
that (2) holds. The set given in (2.4) has non-empty interior because it is

1282 FRANCISCO J. GARC´
IA-PACHECO
exactly the image under the isomorphism given in (2.3) of the closed unit ball
Bℓ1. Finally, assume that (3) holds. There exists r > 0 such that
BX(0, r)⊆(∞
X
n=1
tnen: (tn)n∈N∈Bℓ1).
If x∈X\ {0}, then
rx
kxk∈BX(0, r)⊆(∞
X
n=1
tnen: (tn)n∈N∈Bℓ1).
Thus, there exists a sequence (sn)n∈N∈Bℓ1so that
rx
kxk=
∞
X
n=1
snen.
In order words,
x=
∞
X
n=1
kxksn
ren
and (kxksn/r)n∈N∈ℓ1.
Lemma 2.5. Let Xbe an infinite dimensional, separable, real Banach space.
There exists a normalized Markushevich basis for Xwhich is not a Schauder
basis equivalent to the ℓ1-basis.
Proof. In virtue of Lemma 2.4, we can assume that Xis isomorphic to ℓ1.
Denote by (en)n∈Nthe canonical basis of ℓ1. Consider the sequence (un)n∈N
given by u1= e1and un= (en−en−1)/2 for n≥2. It is well known that
(un)n∈Nis a Schauder basis in ℓ1(see, for instance, [5]). However,
∞
X
n=1
1
nun=3
4e1+1
2
∞
X
n=2
1
n(n+ 1)en.
Therefore, (un)n∈Nis not equivalent to the ℓ1-basis. 
Lemma 2.6. Let Xbe an infinite dimensional, real Banach space. If Xadmits
an infinite dimensional, separable quotient, then there exists a proper, dense
subspace Eof Xand a bounded, complete barrel Mof Ewith empty interior
in E.
Proof. Let Zbe a closed subspace of Xsuch that X/Z is an infinite dimen-
sional, separable Banach space. Let p:X→X/Z be the canonical projection
of Xonto X/Z. By Lemma 2.5, we can find a bounded sequence (en)n∈N⊂X
such that (p(en))n∈Nis a Markushevich basis for X/Z contained in SX/Z which
is not a Schauder basis equivalent to the ℓ1–basis. By Lemma 2.3, the set
(∞
X
n=1
tnp(en) : (tn)n∈N∈Bℓ1)

ISOMETRIC REFLECTIONS 1283
is closed. And, by Lemma 2.4, the above set has empty interior in its linear
span. Now, consider
M:= co (∞
X
n=1
tnen: (tn)n∈N∈Bℓ1)∪BZ!
and E:= span (M). Note that
M⊆p−1 (∞
X
n=1
tnp(en) : (tn)n∈N∈Bℓ1)!.
Finally, since pis continuous and open, we deduce that Eis a dense subspace
of Xand Mis a bounded barrel of Ewith empty interior in Eand closed in
X.
Observe that, according to Theorem 2.2, the previous lemma is actually
a characterization of real Banach spaces admitting an infinite dimensional,
separable quotient. Before stating and proving the main result in this section
we will make some remarks and reminders.
Remark 2.7.Let Xbe a real Banach space.
•Assume that e∈SXis an isometric reflection vector of e. The only
functional e∗∈X∗such that e∗(e) = 1 and ker (e∗) = ker (ρe) is called
the isometric reflection functional associated to e. In [2] it is shown
that e∗∈SX∗and that it is an isometric reflection vector of X∗.
•Consider a vector e∈SXand a closed, maximal subspace Mof Xsuch
that X=Re⊕M. The following conditions are equivalent:
(1) km+λek=km−λekfor all λ∈Rand all m∈M.
(2) For all λ∈Rand all m∈M, if λe +m∈SX, then also m−λe ∈
SX.
Theorem 2.8. Let Xbe a real Banach space. Let e∈SXand e∗∈SX∗such
that e∗(e) = 1. If Xadmits an infinite dimensional, separable quotient, then
Xcan be equivalently renormed so that
•eis an isometric reflection vector and e∗is the isometric reflection
functional associated to e.
•span(((e∗)−1(1) ∩BX)−e)is a proper dense subspace of ker (e∗).
Proof. Note that ker (e∗) also has an infinite dimensional, separable quotient.
Therefore, by Lemma 2.6, there exists a proper, dense subspace Eof ker (e∗)
and a bounded, complete barrel Mof Ewith empty interior in E. Observe
that we can assume without any loss of generality that M⊂Bker(e∗). Now, X
can be endowed with the equivalent norm whose unit ball is
B=co Bker(e∗)∪(M+e)∪(M−e).
(1) The reader may notice that co Bker(e∗)∪ {e, −e}has non-empty in-
terior and so does B. Thus, Bis indeed the unit ball of an equivalent
norm on X.

1284 FRANCISCO J. GARC´
IA-PACHECO
(2) e∈bd (B) (the boundary of B) because e∗(e) = 1 = sup e∗(B).
(3) If φ∈X∗, then the linear function ψ:X→Rdefined as ψ(v+λe) =
φ(v−λe) for all v∈ker (e∗) and all λ∈Rverifies that
sup φ(B) = sup ψ(B).
Indeed, it suffices to observe that if t, s, r ∈[0,1] with t+r+s= 1,
m, m′∈M, and b∈Bker(e∗), then
tb +s(m+e) + r(m−e), tb +r(m+e) + s(m−e)∈ B
and
ψ(tb +r(m+e) + s(m′−e)) = φ(tb +s(m+e) + r(m′−e)) .
(4) eis an isometric reflection vector of Xendowed with the B-norm.
Indeed, we will rely on Remark 2.7. Let γ∈Rand z∈ker (e∗) such
that γe +z∈bd (B). There exists φ∈X∗such that 1 = sup φ(B) =
φ(γe +z). Define ψ:X→Ras ψ(v+λe) = φ(v−λe) for all v∈
ker (e∗) and all λ∈R. Simply observe that
sup ψ(B) = 1 = φ(γe +z) = ψ(z−γe).
As a consequence, z−γe ∈bd (B).
(5) span(((e∗)−1(1)∩B)−e) is a proper dense subspace of ker (e∗). Indeed,
it will be enough to show that ((e∗)−1(1) ∩B)−e=Mor, equivalently,
that (e∗)−1(1) ∩ B =M+e. On the one hand, we trivially have that
M+e⊆(e∗)−1(1) ∩ B. On the other hand, let z+e∈(e∗)−1(1) ∩ B
with z∈ker (e∗). There exist sequences (tn)n∈N,(sn)n∈N,(rn)n∈N⊂
[0,1], (mn)n∈N,(m′
n)n∈N⊂M, and (bn)n∈N⊂Bker(e∗)such that tn+
sn+rn= 1 for all n∈Nand (tnbn+sn(mn+e) + rn(m′
n−e))n∈N
converges to z+e. Because of the product topology, we can deduce
(by passing to subsequences if necessary) that (sn)n∈Nconverges to 1
and that (tn)n∈Nand (rn)n∈Nboth converge to 0. Therefore, (mn)n∈N
converges to z. Since Mis complete, we have that z∈M.
Remark 2.9.Observe that given a real Banach space Xwith an isometric
reflection vector e∈SXand isometric reflection functional e∗∈SX∗, the set
(e∗)−1(1) ∩BX−e
is a bounded, complete barrel of its linear span. Indeed, by [6, Theorem 2.2]
the above set is absolutely convex, thus it must be absorbing in its linear span.
As a consequence, by considering again Theorem 2.2, we have that Theorem
2.8 is in fact a characterization of real Banach spaces admitting an infinite
dimensional, separable quotient.
To finish this section, we present the following question (which is actually an
equivalent reformulation of the separable quotient problem in virtue of Theorem
2.8 and Remark 2.9).

ISOMETRIC REFLECTIONS 1285
Question 2.10. Let Xbe a real Banach space. Can Xbe equivalently
renormed to have an isometric reflection vector e∈SXsuch that
span (e∗)−1(1) ∩BX−e
is a proper dense subspace of ker (e∗)?
In [6] we presented the following two results that could be considered as an
approach to a positive answer to Question 2.10.
Theorem 2.11 (Garc´ıa-Pacheco, [6]).Let Xbe a real Banach space. Let
e∈SXbe an isometric reflection vector. Then
(1) span(((e∗)−1(1) ∩BX)−e)⊆ker (e∗). Conversely, if m∈ker (e∗)is
so that λ6= 0 can be found verifying that ke+λmk= 1, then m∈
span(((e∗)−1(1) ∩BX)−e).
(2) span(((e∗)−1(1) ∩BX)−e) = ker (e∗)if and only if (e∗)−1(1) ∩BXis
a smooth face of BX.
Theorem 2.12 (Garc´ıa-Pacheco, [6]).Let Xbe an infinite dimensional, real
Banach space. Then Xcan be equivalently renormed to have an isometric
reflection vector e∈SXsuch that span(((e∗)−1(1) ∩BX)−e)is a closed,
maximal subspace of ker (e∗).
3. L1-summand vectors in dual spaces
A particular case of isometric reflection vectors are the so called L1-summand
vectors (see [4]).
Definition 3.1 (Behrends, [4]).Let Xbe a real Banach space. We say that
a vector x∈SXis an Lp-summand vector of X(1 ≤p≤ ∞) if there exists a
closed maximal subspace Mof Xsuch that X=Rx⊕pM.
Combining results from [1] and [6] we have the following theorem, that
expresses the differences between L1-summand vectors and isometric reflection
vectors in 2-dimensions and in higher dimensions as well.
Theorem 3.2 (Aizpuru and Garc´ıa-Pacheco, [1]).Let Lbe an uncountably
infinite discrete topological space. Denote by b
Lthe one-point compactification
of Land by 1∈ C(b
L)the constant function on b
Lequal to 1. Then
(1) If Yis a 2-dimensional subspace of C(b
L)containing 1, then 1is an
L1-summand vector of Y, and thus it is an isometric reflection vector
of Yand a strongly exposed point of BY.
(2) 1is not an exposed point of BC(
b
L).
(3) 1is not an isometric reflection vector of C(b
L).
The aim of this section is actually at dealing with L1-summand vectors in
dual spaces. Taking a look at either [4] or [8] one can quickly realize that if

1286 FRANCISCO J. GARC´
IA-PACHECO
1< p ≤ ∞, then an Lp-summand vector of a dual space must be a norm-
attaining functional which attains its norm at an Lq-summand vector of the
predual space, where qis the conjugate exponent of p. More generally, if
1< p ≤ ∞, then an Lp-summand subspace of a dual Banach space must be
ω∗closed and thus it is the dual of an Lq-summand subspace of the predual
space (in [8] L1-summand subspaces are called L-summands and L∞-summand
subspaces are called M-summands). Precisely in [8, Example IV.1.8] it is shown
the existence of a Banach space without non-trivial M-ideals such that X∗has
a non-trivial L-summand. This example is based on the Theory of Banach
Algebras. Here we will show an easy way to find L1-summand vectors in dual
spaces that are not norm-attaining functionals at L∞-summand vectors without
using techniques of the ring theory or polynomial algebra. We will rely on the
following result (see [9]).
Theorem 3.3 (Jameson, [9]).Let Xbe a non-reflexive real Banach space. Let
x∗∗ ∈X∗∗ \X. Then
x∗∈BX∗:kx∗k+|x∗∗ (x∗)|
dist (x∗∗, X )≤1
⊆clω∗Bker(x∗∗)
⊆ {x∗∈BX∗:|x∗∗ (x∗)| ≤ 2dist (x∗∗, X )}.
In addition
(1) If X=c0, then
x∗∈BX∗:kx∗k+|x∗∗ (x∗)|
dist (x∗∗, X )≤1= clω∗Bker(x∗∗)
for all x∗∗ ∈X∗∗ \X.
(2) If X=ℓ1, then
clω∗Bker(x∗∗)={x∗∈BX∗:|x∗∗ (x∗)| ≤ 2dist (x∗∗, X )}
for all x∗∗ ∈X∗∗ \X.
Based upon the previous theorem we find the following definition.
Definition 3.4. Let Xbe a non-reflexive real Banach space. Then
(1) We say that Xis of c0dual type if
x∗∈BX∗:kx∗k+|x∗∗ (x∗)|
dist (x∗∗, X )≤1= clω∗Bker(x∗∗)
for all x∗∗ ∈X∗∗ \X.
(2) We say that Xis of ℓ1dual type if
clω∗Bker(x∗∗)={x∗∈BX∗:|x∗∗ (x∗)| ≤ 2dist (x∗∗, X )}
for all x∗∗ ∈X∗∗ \X.

ISOMETRIC REFLECTIONS 1287
Utilizing plain techniques of the linear algebra and convex geometry we will
show the existence of L1-summand vectors in dual spaces that are not norm-
attaining functionals at L∞-summand vectors in their predual.
Remark 3.5.Let Xbe a real Banach space.
•If (xi)i∈Iis a net in BXand f∈SX∗is a norm-attaining functional
such that (|f(xi)|)i∈Iis convergent to a number 0 ≤k < 1, then we
can find another net (zi)i∈I⊆BXsuch that (|f(zi)|)i∈Iis convergent
to k,|f(zi)| ≥ kfor all i∈I, and (xi−zi)i∈Iconverges to 0. Indeed,
we may assume that k > f (xi)>0 for all i∈Iand then it is sufficient
to take
zi:= k−f(xi)
1−f(xi)z+1−k
1−f(xi)xi,
for all i∈I, where zis any element in SXsuch that f(z) = 1.
•One could easily realize that the situation above still holds if fis not
norm-attaining, but it would obviously require a much more elaborate
proof.
•However, the situation above does not remain true if we ask k= 1.
Indeed, assume that Xis a real rotund Banach space whose unit sphere
contains a point zthat is not strongly exposed on the unit ball. There
are then f∈SX∗and a sequence (xn)n∈N⊂SXsuch that f(z) = 1,
(f(xn))n∈Nconverges to 1, but (xn)n∈Ndoes not converge to z. If
(zn)n∈N⊂BXis another sequence such that |f(zn)| ≥ 1 for all n∈N,
then for all n∈Nwe have that either zn=zor zn=−z, and hence
(xn−zn)n∈Ncannot converge to 0.
Remark 3.6.Let Xbe a non-reflexive real Banach space. The Bishop-Phelps
Theorem allows us to deduce that the set of norm-attaining functionals on X∗
is dense in X∗∗. Therefore, many x∗∈SX∗and x∗∗ ∈SX∗∗ can be found such
that x∗∗ (x∗) = 1 and 0 <dist (x∗∗ , X )≤1
2.
Theorem 3.7. Let Xbe a non-reflexive real Banach space of c0dual type. Let
x∗∈SX∗and x∗∗ ∈SX∗∗ such that x∗∗ (x∗) = 1 and 0<dist (x∗∗ , X )≤1
2.
The equivalent norm on X∗given for all y∗∈X∗by
ky∗k1:= |x∗∗ (y∗)|+ky∗−x∗∗ (y∗)x∗k
is a dual norm.
Proof. Let (y∗
i)i∈Ibe a net in X∗such that ky∗
ik1≤1 and assume that (y∗
i)i∈I
is ω∗convergent to some y∗∈X∗. We may assume that (|x∗∗ (y∗
i)|)i∈Iis
convergent to some number 0 ≤k≤1. Observe also that if k= 1, then
y∗=x∗and thus ky∗k1= 1. So, assume that k < 1. Notice that in virtue
of the previous remark we may assume that |x∗∗ (y∗
i)| ≥ kfor all i∈I. Since
ky∗
i−x∗∗ (y∗
i)x∗k ≤ 1− |x∗∗ (y∗
i)| ≤ 1−kfor all i∈I, we deduce that
y∗
i−x∗∗ (y∗
i)x∗
1−ki∈I
⊆Bker(x∗∗)

1288 FRANCISCO J. GARC´
IA-PACHECO
is a net ω∗convergent to y∗−kx∗
1−k.
By hypothesis, 



y∗−kx∗
1−k


+ 2 |x∗∗ (y∗)−k|
1−k≤1.
In other words,
ky∗k1=|x∗∗ (y∗)|+ky∗−x∗∗ (y∗)x∗k
≤ ky∗−kx∗k+ 2 |x∗∗ (y∗)−k|+k
≤1.
Corollary 3.8. Let Xbe a non-reflexive real Banach space of c0dual type.
Then Xcan be equivalently renormed so that its dual has an L1-summand vector
that is not a norm-attaining functional at an L∞-summand vector in X.
To finish the manuscript we will show that the hypothesis of c0dual type in
the previous result is necessary.
Remark 3.9.Let Xbe a real Banach space. Consider a6=b∈X\ {0}and
assume that kak>kbk. The triangular inequality tells us that kak − kbk ≤
ka−bk. Take a look at the following:
a
kak=kbk
kak
b
kbk+kak − kbk
kak
a−b
kak − kbk.
If kak − kbk=ka−bk, then we deduce that
a
kak∈b
kbk,a−b
ka−bk⊂SX.
Theorem 3.10. Let Xbe a non-reflexive real Banach space of ℓ1dual type. Let
x∗∈SX∗and x∗∗ ∈SX∗∗ \Xsuch that x∗∗ (x∗) = 1. Consider the equivalent
norm on X∗given by
ky∗k1:= |x∗∗ (y∗)|+ky∗−x∗∗ (y∗)x∗k
for all y∗∈X∗. If x∗is a rotund point of BX∗, i.e. x∗does not belong to any
non-trivial segment of SX∗, then k·k1is not a dual norm on X∗.
Proof. Take y∗∈SX∗such that
0< x∗∗ (y∗)<min {1,2dist (x∗∗, X )}.
By hypothesis, y∗∈clω∗Bker(x∗∗). However, according to the previous re-
mark and by assumption, we have:
ky∗k1=|x∗∗ (y∗)|+ky∗−x∗∗ (y∗)x∗k
>|x∗∗ (y∗)|+ky∗k − kx∗∗ (y∗)x∗k
= 1.
Therefore, k·k1is not a dual norm on X∗.

ISOMETRIC REFLECTIONS 1289
References
[1] A. Aizpuru and F. J. Garc´ıa-Pacheco, A short note about exposed points in real Banach
spaces, Acta Math. Sci. Ser. B Engl. Ed. 28 (2008), no. 4, 797–800.
[2] A. Aizpuru, F. J. Garc´ıa-Pacheco, and F. Rambla, Isometric reflection vectors in Banach
spaces, J. Math. Anal. Appl. 299 (2004), no. 1, 40–48.
[3] J. Becerra-Guerrero and A. Rodr´ıguez-Palacios, Isometric reflections on Banach spaces
after a paper of A. Skorik and M. Zaidenberg, Rocky Mountain J. Math. 30 (2000), no.
1, 63–83.
[4] E. Behrends et al., Lp-Structure in Real Banach Spaces, Lecture Notes in Mathematics,
613, Springer-Verlag, Berlin-Heidelberg-New York, 1977.
[5] J. Diestel, Sequences and Series in Banach Spaces, Springer-Verlag, New York, 1984.
[6] F. J. Garc´ıa-Pacheco, Geometry of isometric reflection vectors, Math. Slovaca 61 (2010),
no. 5, 807–816.
[7] ,Rotundity and connectedness in two dimensions, Bol. Soc. Mat. Mexicana (3)
15 (2009), no. 2, 165–173.
[8] P. Hadamard, D. Werner, and W. Werner, M-Ideals in Banach Spaces and Banach
Algebras, Springer-Verlag, Berlin, 1993.
[9] G. J. O. Jameson, The weak star closure of the unit ball in a hyperplane, Proc. Edinburgh
Math. Soc. (2) 18 (1972), no. 2, 7–11.
[10] J. M´ujica, Separable quotients of Banach spaces, Rev. Mat. Univ. Complut. Madrid 10
(1998), 299–330.
[11] H. P. Rosenthal, On quasi-complemented subspaces of Banach spaces, Proc. Natl. Acad.
Sci. USA 59 (1968), 361–364.
[12] S. A. Saxon and A. Wilansky, The equivalence of some Banach space problems, Colloq.
Math. 37 (1977), no. 2, 217–226.
[13] A. Skorik and M. Zaidenberg, On isometric reflections in Banach spaces, Mat. Fiz.
Anal. Geom. 4(1997), no. 1-2, 212–247.
Department of Mathematics
Texas A&M University
College Station, Texas, 77843-3368, USA
E-mail address:fgarcia@math.tamu.edu