Isomorphically Polyhedral Banach Spaces
We prove two theorems giving su±cient conditions for a Banach space to be isomorphically polyhedral.
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
2013
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| Цитувати: | Isomorphically Polyhedral Banach Spaces / V.P. Fonf, A.J. Pallarés, S. Troyanski // Журнал математической физики, анализа, геометрии. — 2013. — Т. 9, № 1. — С. 108-113. — Бібліогр.: 9 назв. — англ. |
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irk-123456789-1067402016-10-04T03:02:37Z Isomorphically Polyhedral Banach Spaces Fonf, V.P. Pallarés, A.J. Troyanski, S. We prove two theorems giving su±cient conditions for a Banach space to be isomorphically polyhedral. Доказаны две теоремы, дающие достаточные условия для того, чтобы банахово пространство было изоморфически многогранным. 2013 Article Isomorphically Polyhedral Banach Spaces / V.P. Fonf, A.J. Pallarés, S. Troyanski // Журнал математической физики, анализа, геометрии. — 2013. — Т. 9, № 1. — С. 108-113. — Бібліогр.: 9 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106740 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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We prove two theorems giving su±cient conditions for a Banach space to be isomorphically polyhedral. |
| format |
Article |
| author |
Fonf, V.P. Pallarés, A.J. Troyanski, S. |
| spellingShingle |
Fonf, V.P. Pallarés, A.J. Troyanski, S. Isomorphically Polyhedral Banach Spaces Журнал математической физики, анализа, геометрии |
| author_facet |
Fonf, V.P. Pallarés, A.J. Troyanski, S. |
| author_sort |
Fonf, V.P. |
| title |
Isomorphically Polyhedral Banach Spaces |
| title_short |
Isomorphically Polyhedral Banach Spaces |
| title_full |
Isomorphically Polyhedral Banach Spaces |
| title_fullStr |
Isomorphically Polyhedral Banach Spaces |
| title_full_unstemmed |
Isomorphically Polyhedral Banach Spaces |
| title_sort |
isomorphically polyhedral banach spaces |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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2013 |
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http://dspace.nbuv.gov.ua/handle/123456789/106740 |
| citation_txt |
Isomorphically Polyhedral Banach Spaces / V.P. Fonf, A.J. Pallarés, S. Troyanski // Журнал математической физики, анализа, геометрии. — 2013. — Т. 9, № 1. — С. 108-113. — Бібліогр.: 9 назв. — англ. |
| series |
Журнал математической физики, анализа, геометрии |
| work_keys_str_mv |
AT fonfvp isomorphicallypolyhedralbanachspaces AT pallaresaj isomorphicallypolyhedralbanachspaces AT troyanskis isomorphicallypolyhedralbanachspaces |
| first_indexed |
2025-07-07T18:55:45Z |
| last_indexed |
2025-07-07T18:55:45Z |
| _version_ |
1837015547157413888 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2013, vol. 9, No. 1, pp. 108–113
Isomorphically Polyhedral Banach Spaces
V.P. Fonf
Department of Mathematics, Ben Gurion University of the Negev
Beer-Sheva, Israel
E-mail: fonf@math.bgu.ac.il
A.J. Pallarés and S. Troyanski
Departamento de Matemáticas, Universidad de Murcia, Campus de Espinardo
30100 Murcia, Spain
E-mail: apall@um.es
stroya@um.es
Received October 9, 2012
We prove two theorems giving sufficient conditions for a Banach space
to be isomorphically polyhedral.
Key words: polyhedral norms and renormings, boundaries, polytopes,
countable covers.
Mathematics Subject Classification 2010: 46B20.
Devoted to the memory of M.I. Kadets
All Banach spaces we consider in this paper are infinite-dimensional and real.
A Banach space is called polyhedral if the unit ball of each its finite-dimensional
subspace is a polytope [7]. If a Banach space admits an equivalent norm in which
it is polyhedral then we say that it is isomorphically polyhedral. In [8] is proved
that polyhedral space cannot be isometric to a dual space.
In [9] is proved that if extBX∗ is countable then X is not reflexive. In [6]
this result is strengthen, namely it is proved that if extBX∗ can be covered by a
countable union of compact sets then X is not reflexive.
A subset B ⊂ SX∗ of the unit sphere SX∗ of a Banach space X∗ is called
a boundary (for X) if for any x ∈ X there is f ∈ B such that f(x) = ||x||.
An important example of a boundary is extBX∗ (the Krein–Milman theorem).
The first named author is supported by Israel Science Foundation, Grant 209/09. The
second author is supported by MCI MTM2011-25377 . The third named authors is supported
financially by Science Foundation Ireland under Grant Number SFI 11/RFP.1/MTH/3112, by
MCI MTM2011-22457 and the project of the Institute of Mathematics and Informatics, Bulga-
rian Academy of Science.
c© V.P. Fonf, A.J. Pallarés, and S. Troyanski, 2013
Isomorphically Polyhedral Banach Spaces
A subset B of the dual unit ball BX∗ has property (*) if, given any w∗-limit point
f0 of B (i.e. any w∗-neighborhood of f0 contains infinitely many elements of B),
we have f0(x) < 1 whenever x is in the unit sphere SX . Note that if B is a set
such that |||x||| = sup{f(x) : x ∈ B} defines a norm, and B has (*) for this norm,
then B is a boundary for the norm |||.|||.
Any separable polyhedral space has a countable boundary, and if a Banach
space has a countable boundary then it is isomorphically polyhedral space [2].
Any (isomorphically) polyhedral space saturated by spaces isomorphic to c0, that
is any (infinite-dimensional) subspace of a polyhedral space contains an isomor-
phic copy of c0 [4].
In this paper we prove two theorems giving sufficient conditions for a Banach
space to be isomorphically polyhedral.
Theorem 1. Let X be a Banach space. Then (a), (b) and (c) are equivalent
and imply (d).
(a) There are a sequence of subsets of SX , {Sk}∞k=1, SX =
⋃
Sk, and an in-
creasing sequence of norm-compact subsets of SX∗, {Fk}∞k=1, Fk = −Fk ,
with the following properties:
bk := inf
x∈Sk
max
f∈Fk
{f(x)} > 0, k = 1, 2, . . . and lim
k
bk = 1.
(b) There are a sequence {tk} ⊂ SX∗ and a sequence of positive numbers εk > 0,
k = 1, 2, . . ., such that
(i) the set B = {±(1 + εk)tk}k is a boundary with property (*) for the
equivalent norm
|||x||| = sup
k
|(1 + εk)tk(x)|, x ∈ X,
(ii) B(X,|||.|||) ⊂ intBX .
(c) There are a sequence {tn}n ⊂ SX∗ and a sequence of positive numbers {αn},
limn αn = 0, such that
SX ⊂
⋃
n
S(tn, αn),
where S(tn, αn) are the slices defined as S(tn, αn) = {x ∈ BX : tn(x) ≥
1− αn}.
(d) X is a separable isomorphically polyhedral space.
Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 1 109
V.P. Fonf, A.J. Pallarés, and S. Troyanski
P r o o f.
(a) ⇒ (b). Fix a sequence δn > 0 such that 0 < 3δn < bn and limn δn = 0.
Choose γn defined by the equation (1 + γn) = (bn − 3δn)−1 and observe that
γn > 0,
lim
n
γn = 0 and (1 + γn)(bn − 2δn) > 1, n = 1, 2, . . .
(notice that here we use that 1 ≥ bn > 0, for any n and that limn bn = 1).
Since each set Fn is compact, for each n there is a δn-net Nn = {hn
j : j =
1, 2, . . . , pn} ⊂ Fn such that ‖hn
i − hn
j ‖ ≥ δn, i 6= j and for each f ∈ Fn there is
some hn
j with ‖f − hn
j ‖ < δn.
Clearly, we can write the set ±⋃
n(1 + γn)Nn in the form B = {±(1 + εk)tk :
k = 1, 2, . . .} for suitable tk’s and εk’s, limk εk = 0. Define
|||x||| = sup{|(1 + εk)tk(x)| : k = 1, 2, . . .}.
For x ∈ SX there is some n such that x ∈ Sn and some f ∈ Fn such that
f(x) > bn − δn, and some hn
j ∈ Nn such that ‖f − hn
j ‖ < δn. Then we have
|||x||| ≥ (1 + γn)hn
j (x) ≥ (1 + γn)(f(x)− δn) > (1 + γn)(bn − 2δn) > 1 = ‖x‖.
On the other hand we have |||x||| ≤ maxk{(1 + εk)‖x‖}. Therefore
‖x‖ < |||x||| ≤ max
k
{(1 + εk)}‖x‖, x ∈ X, x 6= 0.
This proves (ii) in (b).
Finally we prove that B is a boundary of (X, |||.|||) with (*). Assume the
contrary and choose f a w∗ limit point of (1 + γn)hn
j such that there is x with
|||x||| = 1 and f(x) = 1. Since limn γn = 0, we have that ‖f‖ ≤ 1, and then
‖x‖ ≥ 1 in contradiction with (ii) we have already proof.
(b) ⇒ (a). Put
Ak = {u ∈ X : |||u||| = 1, (1 + εk)tk(u) = 1}, k = 1, 2, . . .
Since B is boundary it follows that S(X,|||.|||) =
⋃
k Ak. Define Sk = {u/||u|| : u ∈
Ak} and Fk = {±tj : j = 1, 2, . . . , k}. Clearly S(X,‖.‖) =
⋃
k Sk and
bk = inf{max{f(x) : f ∈ Fk} : x ∈ Sk}
= inf{max{ti(u/‖u‖) : i = 1, 2, . . . , k} : u ∈ Ak}
≥ inf{tk(u/‖u‖) : u ∈ Ak} =
1
(1 + εk)
inf{1/‖u‖ : u ∈ Ak} ≥ 1
(1 + εk)
> 0.
and clearly limk bk = 1.
110 Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 1
Isomorphically Polyhedral Banach Spaces
(b) ⇒ (c). Put V = B(X,|||.|||). Clearly V =
⋂
k{x ∈ X : |(1 + εk)tk(x)| ≤ 1}.
From (ii) follows that SX ⊂ ⋃
k{x ∈ X : |(1 + εk)tk(x)| > 1}. Put αk = εk
1+εk
,
k = 1, 2 . . . and finish the proof.
(c) ⇒ (b). Put εk = 2αk
1−2αk
, k = 1, 2, . . . and V =
⋂
k{x ∈ X : |(1+εk)tk(x)| ≤
1}. Since SX ⊂ ⋃
k{x ∈ X : |tk(x)| ≥ 1− αk}, it follows that
⋂
k
{x ∈ X : |tk(x)| < 1− αk} ⊂ int(BX),
and an easy verification shows that V ⊂ int(BX). Now (i) easily follows from
limk εk = 0.
(b)⇒ (d). Is clear, since X has an equivalent norm with a countable boundary
with (*).
R e m a r k. In [3] is proved that if X satisfies (c) (in the given norm) then X
is isomorphically polyhedral, and if X is isomorphically polyhedral then there is
an equivalent norm on X in which it satisfies (c). However, the following question
remains open: if X is isomorphically polyhedral then does it satisfy (c) in any
equivalent norm? There is a partial answer changing αk = α > 0 with α arbitrary
[1]. Finally, let us mention that the implication (a) ⇒ (d) is a particular case of
one of the results in [5].
Theorem 2. Let X be a Banach space. Assume that SX =
⋃
k Sk such
that each Sk has an ε-approximative countable boundary, for any ε > 0, in the
following sense
(P) for any k ∈ N and ε > 0 there is a sequence {hk
i } ⊂ (1 + ε)BX∗ such that if
V ∗
k = w∗ − cl co{BX∗ , {±hk
i }i∈N}, Vk = {x : maxx(V ∗
k ) ≤ 1},
then
(i)k Vk
⋂
Sk = ∅
(ii)k for any x ∈ ∂Vk \ SX there is hk
i with hk
i (x) = 1.
Then X is isomorphically polyhedral.
P r o o f. Fix ε ∈ (0, 1) and a sequence {εk}, εk ∈ (0, ε), limk εk = 0. Next by
using property (P) for every k ∈ N find {hk
i }i,k=1,2,... ⊂ (1 + εk)BX∗ , V ∗
k , Vk such
that (i)k and (ii)k are satisfied. Put
V ∗ = w∗ − clco{±hk
i }i,k∈N, V = {x ∈ X : maxx(V ∗) ≤ 1},
and introduce in X a new norm |||.||| with the unit ball V. By using SX = ∪kSk
and (i)k, k = 1, 2, . . . , we get
V ⊂ intBX . (1)
Journal of Mathematical Physics, Analysis, Geometry, 2013, vol. 9, No. 1 111
V.P. Fonf, A.J. Pallarés, and S. Troyanski
Indeed, let x ∈ X with ‖x‖ ≥ 1 and take z = x/‖x‖ ∈ SX , then z ∈ Sk
for some k and z 6∈ Vk. By the definition of Vk it is possible to find x∗1 ∈ BX∗ ,
x∗2 ∈ co{±hk
i }i∈N and λ ∈ [0, 1) such that λx∗1(z)+(1−λ)x∗2(z) > 1, (1−λ)x∗2(z) >
1− λ and x∗2(x) > ‖x‖ ≥ 1. Thus we have that x 6∈ V . Note that by (1) and its
definition we have that V =
⋂
k Vk.
From εk ∈ (0, ε), k = 1, 2, . . . , it follows that V ⊃ (1 + ε)−1BX . So the norm
|||.||| is (1+ε)-equivalent to the original one. We show that the set B = {±hk
i }i,k∈N
is a (countable) boundary for the space (X, |||.|||), and then by [2] we conclude
that (X, |||.|||) is isomorphically polyhedral. Fix x0 ∈ ∂V. From (1) follows that
x0 ∈ intBX . Assume to the contrary that for any h ∈ B we have h(x0) < 1. Since
x0 ∈ ∂V there is a sequence {hkn
in
}∞n=1 with limn hkn
in
(x0) = 1. We consider two
cases.
Case 1. lim supn kn = ∞. WLOG we can assume that limn kn = ∞. Let h0 ∈
X∗ be a w∗-limit point of {hkn
in
}∞n=1. Since hkn
in
∈ (1 + εkn)BX∗ and limk εk = 0,
it follows that h0 ∈ BX . However h0(x0) = 1, and hence ||x0|| ≥ 1, contradicting
x0 ∈ V ⊂ intBX .
Case 2. supn kn < ∞. WLOG we can assume that kn = l, for any n ∈ N.
Since x0 ∈ Vl (recall that x0 ∈ V = ∩kVk), and limn hl
in
(x0) = 1, it follows that
x0 ∈ ∂Vl. By (ii)l there is hl
i with hl
i(x0) = 1, which proves that B is a (countable)
boundary for (X, |||.|||).
The proof of the theorem is complete.
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