On Sets with Extremely Big Slices
A new characterization of the Radon-Nikodym property in terms of sizes of slices and equivalent norms is presented. A property opposite to the Radon-Nikodym property is studied in the context of 1-unconditional sums of Banach spaces.
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
2006
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| Cite this: | On Sets with Extremely Big Slices / Ye. Ivakhno // Журнал математической физики, анализа, геометрии. — 2006. — Т. 2, № 1. — С. 94-103. — Бібліогр.: 5 назв. — англ. |
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| citation_txt | On Sets with Extremely Big Slices / Ye. Ivakhno // Журнал математической физики, анализа, геометрии. — 2006. — Т. 2, № 1. — С. 94-103. — Бібліогр.: 5 назв. — англ. |
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| description | A new characterization of the Radon-Nikodym property in terms of sizes of slices and equivalent norms is presented. A property opposite to the Radon-Nikodym property is studied in the context of 1-unconditional sums of Banach spaces.
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Journal of Mathematical Physics, Analysis, Geometry Short Notes
2006, vol. 2, No. 1, pp. 94�103
On Sets with Extremely Big Slices
Yevgen Ivakhno
Department of Mechanics and Mathematics, V.N. Karazin Kharkov National University
4 Svobody Sq., Kharkov, 61077, Ukraine
E-mail:ivakhnoj@yandex.ru
Received March 29, 2005
A new characterization of the Radon�Nikodym property in terms of sizes
of slices and equivalent norms is presented. A property opposite to the
Radon�Nikodym property is studied in the context of 1-unconditional sums
of Banach spaces.
Key words: Radon�Nikodym property, big slice property, 1-unconditional
sums of Banach spaces.
Mathematics Subject Classi�cation 2000: 46B20, 46B22, 46G10, 46B03.
1. Introduction
In this note
stands for a set, � for a �-algebra on
, � for a probability
measure on (
;�), and �+
A for the set of all positive �-measurable subsets of
A 2 � whenever �(A) > 0. The set of all probability densities supported in
A 2 �+
is denoted by �A, i.e.,
�A = f' 2 L1(�) : ' is supported in A;' � 0; and k'k = 1g:
X is a Banach space, B(X) is the closed unit ball of X, B(x; �) = x + �B(X)
is the ball of radius � centered in x. A slice of a subset C � X determined by a
functional x� 2 X
� and � > 0 is the set
S(C; x�; �) = fx 2 C: x�(x) � sup
y2C
x
�(y)� �g: (1)
The diameter of C � X is denoted by d(C). The radius of C is de�ned as
r(C) = inffr: C � B(x; r) for some x 2 Xg:
We recall that a Banach space X is said to have the Radon�Nikodym property
(X 2 RNP) if the following equivalent conditions hold:
c
Yevgen Ivakhno, 2006
On Sets with Extremely Big Slices
(i) For every probability measurable space (
;�; �) and for every X-valued mea-
sure � on (
;�) if �(A)=�(A) 2 B(X) for all A 2 �+
then there is an
f 2 L1(�;X) (actually from L1(�;X)) such that
�(A) =
Z
A
f(!)d�(!) 8A 2 �:
(ii) Every bounded linear operator T : L1(�) ! X with T (�
) � B(X) is
representable in the sense that there exists a function f 2 L1(�;X) such
that
Tg =
Z
gfd� 8g 2 L1(�):
(iii) Every closed convex bounded subset C � X has slices of arbitrarily small
diameter.
There are many other equivalent de�nitions of this property, for more details
see, for instance, [1] and [2].
In Section 2, "Radon�Nikodym property and balls with big slices", we �nd
criteria of this property in terms of equivalent norms and radiuses of slices of
B(X) (Theorem 2) and slices of general closed convex bounded sets (Theorem 1).
An obvious inequality
r(C) � d(C) � 2r(C) (2)
implies that we may formulate the Radon�Nikodym property equivalently as the
property that every closed convex bounded set C � X has slices of arbitrarily
small radius. In the other words, the negation of the Radon�Nikodym property
just means that the radiuses of slices of some closed convex bounded set are
separated from zero. Theorem 2 shows that passing to equivalent norms in a
space X =2 RNP allows to obtain such a unit ball that the radiuses of all its slices
are separated from zero even by 1� ", where " is arbitrarily small.
The following result is used in the proof of this theorem.
De�nition 1. We say that X has the r-diminition property (X 2 rDP) if for
some positive � < 1 (the parameter of rDP) every subset C � X with r(C) <1
has a slice S satisfying
r(S) < ��r(C):
Obviously, X 2 RNP if and only if X 2 rDP with all arbitrarily small positive
values of parameter �. The result of our Theorem 1 is just the statement that a
single value of � > 0 is su�cient for RNP.
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1 95
Yevgen Ivakhno
While in the second section the negation of the Radon�Nikodym property is
considered as a property that the radiuses of slices of the unit balls (of equiva-
lent norms) uniformly tend to the radiuses of these balls, in Sect. 3, "r-big slice
property", we consider and investigate the limit case of these normed spaces.
De�nition 2. We say that X has the r-big slice property (X 2 rBSP) when
every slice of B(X) is of the radius 1.
Obviously, this property is a strengthened negation of the Radon�Nikodym
property. We investigate this notion in the context of 1-unconditional sums of
sequences of spaces (by a space with a 1-unconditional basis). Namely, we inves-
tigate the connection between the rBSP of a 1-unconditional sum of a sequence
of spaces and the rBSP of summands. It terns out that the sum has the rBSP
provided that every summand also has the rBSP. Conversely, we obtain a com-
plete characterization of such spaces with a 1-unconditional basis that the �xed
summand inherits the rBSP of the whole 1-unconditional sum.
We may speak about the d-big slice property (dBSP), i.e., the property when
every slice S of B(X) is of the diameter d(S) = d(B(X)) = 2. Obviously,
this property is stronger than the r-big slice property, due to the inequality (2).
However, we still do not know, whether these properties coincide.
In the last section (Sect. 4 "d-big slice property") we prove that every space
containing an isomorphic copy of c0 can be equivalently renormed to possess the
d-big slice property.
2. Radon�Nikodym property and balls with big slices
At �rst as announced above, we prove the following theorem:
Theorem 1. X 2 RNP if and only if X 2 rDP.
In order to prove this theorem we modify the proof ([1], ch. 5) of the result
that RNP (iii) implies RNP (ii). In this proof we use two lemmas.
Lemma 1 ([1], p. Lemma 5.6). An operator T : L1(�)! X is representable
if and only if for every " > 0 and A 2 �+
there exists a subset B 2 �+
A with
d(T (�B)) < ".
Lemma 2 ([1], p. Lemma 5.9). Let S be a slice of T�A, where A 2 �+
.
Then there is a subset B 2 �+
A such that T�B � S.
96 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1
On Sets with Extremely Big Slices
P r o o f o f T h e o r e m 1. Let X have the r-diminition property with
a parameter 1 � Æ. We �x a bounded linear operator T : L1(�) ! X satisfying
T�
� B(X) and show that it is representable using Lemma 1. Consider a set
A 2 �+
and " > 0. By the r-diminition property of X, there is a slice S of T�A
with r(S) � r(T�A)�(1 � Æ). Applying Lemma 2, we �nd a subset B 2 �+
A such
that T�B � S, hence, r(T�B) � r(T�A)�(1� Æ).
Continuing in this way, we �nd a decreasing sequence of subsets Bn 2 �+
A
(B0 = A, B1 = B, . . . ) with the property that r(T�Bn) � r(T�Bn�1)�(1 � Æ) �
r(T�B0
)�(1 � Æ)n ! 0. So, for some n the set Bn satis�es r(T�B) < ". By
arbitrariness of ", T is representable. So, X 2 RNP, as needed. The other
direction is trivial.
Now we are able to prove the main result of this section.
Theorem 2. X =2 RNP if and only if for every " > 0 there is such an equiva-
lent norm p(x) on X that every slice S of Bp(X) satis�es rp(S) > 1� ".
Lemma 3. If X =2 rDP, then for every Æ > 0 there is such a closed convex
bounded symmetric subset C � X that every slice S of C satis�es
r(S) > (1� Æ)�r(C) = 1� Æ:
P r o o f. Since the r-diminition property with any parameter is equivalent
to the Radon�Nikodym property, X does not have the rDP with any parameter,
arbitrarily close to 1. Therefore, we can �nd a bounded set V � X such that
every slice S of V is of the radius r(S) > (1 � Æ
2
)�r(V ). Besides, without loss
of generality we may assume V to be closed and convex and to be of the radius
which is greater than one. By the de�nition of radius, there is such x 2 X that
V � B(x; r(V ) + Æ
2
). Denote W = V � x. Then W � B(0; r(W ) + Æ
2
) and every
slice S of W satis�es r(S) > (1� Æ
2
)�r(W ). Take
C = co(W
[
�W ):
Since �W � B(0; r(W )+ Æ
2
), C also lies in B(0; r(W )+ Æ
2
), so, r(C) < r(W )+ Æ
2
.
Besides, r(C) � r(W ) > 1. For every slice S of C either S
T
W or S
T�W is a
slice of W or �W respectively. Therefore,
r(S) > (1� Æ
2
)�r(W ) > (1� Æ
2
)�(r(C)� Æ
2
) > r(C)�(1� Æ
2
� Æ
2r(C)
) > r(C)�(1� Æ):
Dividing C by r(C)�1 gives us the desired set.
P r o o f o f T h e o r e m 2. Applying Lemma 3 to X 2 rDP and Æ < "
2,
take the corresponding set C. Now consider a new norm p taking as its closed unit
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1 97
Yevgen Ivakhno
ball Bp the closure of C+"Bk�k: Since Bp � (1+")�Bk�k, the radius rp of every set
V � X in the sense of the new norm satis�es the inequality rp(V ) � 1
1+"
rk�k(V ).
Let S be a slice of Bp. By the construction of Bp, there is an element x 2 "Bk�k
and a slice ~S of C such that x+ ~S � S. So, rk�k(S) > (1�Æ)�rk�k(C) = 1�Æ. Then
rp(S) >
1�Æ
1+"
>
1�"2
1+"
= 1 � ", as needed. The inverse implication is also valid,
since the unit ball of a space with RNP has slices of arbitrarily small diameter.
R e m a r k. A natural question is the validity of Theorem 2 with diameters
instead of radiuses. But the same technique does not su�ce to prove the theorem.
So, this question remains open. However, Theorem 1 still holds in the case of
diameters and has the same proof.
3. r-Big slice property
In this section the unit sphere of a Banach space X is denoted by S(X). The
(open) slice of B(X) determined by a functional x� 2 S(X�) and " > 0 is the set
S(x�; ") = fx 2 B(X): x�(x) > 1� "g:
We will use this form of slices of B(X) instead of slices of the general form (1)
to simplify our arguments. Obviously, the r-big slice property in terms of open
slices coincides with the same property in the case of closed ones.
In this section E stands for a Banach space with a 1-unconditional normalized
Schauder basis. We can think of the elements of E as sequences with the property
that
k(a1; a2; : : : )kE = k(ja1j; ja2j; : : : )kE 8(aj) 2 E.
Suppose that X1;X2; : : : are Banach spaces. Their E-sum X = (X1;X2; : : : )E
consists of all sequences (xj) with xj 2 Xj and (kxjk) 2 E with the norm k(xj)k =
k(kxjk)kE . Note that E� can be represented by all sequences (a�j ) such that
sup
n
k(ja�1j; : : : ; ja�nj; 0; 0; : : : )kE� <1;
and X
� can be represented by all sequences (x�j ), x
�
j 2 X
�
j , such that
kx�k = sup
n
k(kx�1k; : : : ; kx�nk; 0; 0; : : : )kE� <1:
Theorem 3. If X1;X2; � � � 2 rBSP, then their E-sum X 2 rBSP.
P r o o f. Assume to the contrary that S(x�; Æ) � x+r�B(X) for some x 2 X,
r < 1, and some slice S(x�; Æ). Consider such positive Æk that
P
k�1 Æk < Æ=2.
98 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1
On Sets with Extremely Big Slices
Since kx�k = 1, there is such element a = (ak) 2 S(E)+ that
P
k�1 kxkkak >
1� Æ=2. Now for every k with ak 6= 0 we consider a slice of ak�B(Xk) of the form
Sk = fyk 2 Xk: kykk � ak; x
�
k(yk) > kx�kk�ak � Ækg (3)
and put Sk = f0g otherwise. Since every y = (yk) 2 X with yk 2 Sk for all k
satis�es the inequalities
x
�(y) =
X
k�1
x
�
k(yk) >
X
k�1
kx�kkak �
X
k�1
Æk > 1� Æ
and kyk = k(ky1k; ky2k; : : : )kE � k(a1; a2; : : : )kE = 1;
the following inclusion holds:
T1 + T2 + : : : � S(x�; Æ) � x � r�B(X); (4)
where Tk stands for Sk � xk. Consider any r
0 2 (r; 1) and let us show that
Tk0 � ak0 �r0�B(Xk0) (5)
for at least one value of k0 with ak0 6= 0. If such k0 does not exist, take a
y = (y1; y2 : : : ) 2 T1+T2+ : : : , such that kykk � ak�r0; then k(ky1k; ky2k; : : : )k �
k(a1r0; a2r0; : : : )k = r
0�1, which is a contradiction with (4). So, inclusion (5) holds.
Dividing it by ak0 6= 0, we obtain the following inclusion of a slice of B(Xk0):
1
ak0
�Sk0 �
xk0
ak0
+ r
0�B(Xk0);
which is a contradiction with Xk0 2 rBSP.
For every k = 1; 2; : : : consider a mapping
k: E ! R+ with
k(a) = k(a1; : : : ; ak�1; 0; ak+1; ak+2; : : : )kE :
Theorem 4. Consider a number n 2 f1; 2; : : : g and let
sup
n(S(h
�
; Æ)) = 1 (6)
for every slice S(h�; Æ) of B(E). Let X = (X1; : : : ;Xn�1;Xn;Xn+1; : : : )E. Then
X1, . . . , Xn�1, Xn+1, . . . 2 rBSP implies that X 2 rBSP, independently of
properties of X.
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1 99
Yevgen Ivakhno
P r o o f. Assume to the contrary that S(x�; Æ) � x+r�B(X) for some x 2 X,
r < 1, and some slice S(x�; Æ). Take h� = (kx�1k; kx�2k; : : : ) 2 S(E�)+ and such
Æk > 0 that
P
k�1 Æk < Æ=2. Applying (6) to S(h�; Æ=2), we get such a 2 B(E)
that h�(a) > 1 � Æ=2 and
n(a) >
p
r. By analogy with (3) for every k de�ne a
slice Sk and again deduce that
T1 + T2 + : : : � S(x�; Æ)� x � r�B(X); (7)
where Tk stands for Sk � xk. Observe that, due to positivity of
n(a), the set
K = fk: ak 6= 0; k 6= ng is nonempty. Let us show that
Tk0 � ak0 �
p
r�B(Xk0) (8)
for at least one value of k0 2 K. If such k0 does not exist, take a y = (y1; : : : ; yn�1;
0; yn+1; : : : ) 2 T1 + T2 + : : : , such that kykk � ak�
p
r (k 6= n); then
kyk �
p
r�k(a1; : : : ; an�1; 0; an+1; : : : )k =
p
r�
n(a) > r;
which is a contradiction with (7). So, inclusion (8) holds. Dividing it by ak0 6= 0,
we obtain the following inclusion of a slice of B(Xk0):
1
ak0
�Sk0 �
xk0
ak0
+
p
r�B(Xk0);
which is a contradiction with Xk0 2 rBSP.
Let us show that there is a space E satisfying the condition (6). Take E = l
2
1
and n from f1; 2g. Obviously, every slice of S(l21) includes some points from the
set f(1; 1); (1;�1); (�1; 1); (�1;�1)g and
n maps every such point to 1. There-
fore, (6) is ful�lled. So, (X1;X2)l2
1
2 rBSP if and only if one or both spaces
Xn 2 rBSP.
It is easy to understand that every other 2-dimensional space with a 1-uncon-
ditional basis does not satisfy (6). So, the class of spaces which do not satisfy the
condition (6) is not empty either.
The next Theorem 5 is the converse to Theorem 4.
Theorem 5. Consider a number n 2 f1; 2; : : : g. If there is a slice S(h�; Æ) of
S(E) with
sup
n(S(h
�
; Æ)) < 1; (9)
then (X1; X2; : : : )E 2 rBSP implies Xn 2 rBSP.
100 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1
On Sets with Extremely Big Slices
P r o o f. Assume n = 1. In every other case the proof does not di�er in
essence. Assume to the contrary that X1 =2 rBSP, i.e., S(y�1 ; Æ1) � y1 + r1�B(X1)
for some y1 2 X1, r1 < 1, and some slice S(y�1 ; Æ1). Denote C = sup
1(S(h
�
; Æ)) <
1 and observe that an arbitrarily small Æ may be chosen. Since h1 6= 0 (otherwise
C = 1), we may take Æ < Æ1�jh�1j�(1 � C). Finally, we may pass to a positive
functional h� 2 S(E)+ without any loss of generality. Now we are going to
construct a slice S(x�; Æ) and an element y 2 X such that
S(x�; Æ) � y + r�B(X) (10)
for some r. It will contradict with X 2 rBSP, so, the theorem will be proved.
Take any functional x� 2 X
� with x
�
1 = y
�
1�h�1 and satisfying kx�kk = h
�
k for all
k. We de�ne y = (y1; 0; 0; : : : ) and r = 1 � (1 � r1)�(1 � C). Observe that it is
enough to prove (10) for the set S = S(x�; Æ)
T
S(X) (a slice of S(X)) instead of
S(x�; Æ). So, let us show for any x 2 S that kx� yk < r. If x 2 S, then
x
�
1(x1) = x
�(x)�
X
k�2
x
�
k(xk) > 1� Æ �
X
k�2
kx�kkkxkk = kx�1kkx1k � Æ:
Besides, (kx1k; kx2k; : : : ) 2 S(h�; Æ), therefore, kx1k � 1�C, since kx1k � kxk�
k(0; kx2k; kx3k; : : : )k = 1�
1(kx1k; kx2k; : : : ). Consequently,
y
�
1(x1) > kx1k � Æ=h
�
1 > kx1k�(1� Æ1�(1� C)=kx1k) > kx1k�(1� Æ1):
It means that x1 lies in the slice kx1k�S(y�1 ; Æ1), hence, kx1 � y1k � r1�kx1k. So,
kx� yk � k(kx1k�r1; kx2k; kx3k; : : : )k (11)
for all x 2 S. De�ne a convex function g(r1) as the right part of (11). Then
kx�yk � g(r1) � g(1)�r1+g(0)�(1�r1) � r1+(1�r1)�C = 1�(1�r1)(1�C) = r;
as needed.
R e m a r k. Let us call the spaces possessing an equivalent norm with
the r-big slice property the erBSP-spaces (equivalent r-big slice property-spaces).
We don't know whether all the spaces without the RNP are erBSP-spaces. We
don't know even whether a space with a erBSP-subspace is a erBSP-space itself.
What can be easily deduced from the previous results is that a space with a
complemented erBSP-subspace is a erBSP-space. In fact, if X = Y � Z, and Y
is renormed to have the r-big slice property, then the l1-sum of Y and Z will be
a space with the r-big slice property isomorphic to X.
There are some other properties of Banach spaces concerning extremely large
slices of the unit ball, for example the Daugavet property ([3, 4]), which is strictly
stronger than rBSP. The question, whether a 1-unconditional sum of spaces with
the Daugavet property inherits this property, is solved [5] and the solution is far
di�erent from the result in the case of rBSP.
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1 101
Yevgen Ivakhno
4. d-Big slice property
Recall that a Banach space X is said to have the d-big slice property (X 2
dBSP) if every slice of S(X) is of diameter 2. In the other words, for every " > 0
and every slice S of S(X) there are such x and y 2 S that kx� yk > 2� ". For
example, c0 2 dBSP. Obviously, d-big slice property is stronger than the r-big
slice property, but we do not know, whether these properties coincide.
We remark that all theorems in the previous section hold true with the d-big
slice property instead of r-big slice property, but the proofs are slightly longer.
De�nition 3. A Banach space X has the equivalent d-big slice property (X 2
edBSP) if X is isomorphic to a space having the d-big slice property.
By the same reason as in the Remark at the end of the previous section if
X has a complemented subspace with the (equivalent) d-big slice property, then
X 2 edBSP. The case of noncomplemented subspaces is open with one exception:
Theorem 6. If X has a subspace isomorphic to c0, then X 2 edBSP.
P r o o f. Without loss of generality we assume that c0 is a subspace
of X. Consider the collection E of all subspaces Y � X, such that c0 � Y and
codimension of c0 in Y is �nite. Equip E with a �lter F , induced by the natural
order: the base of F is formed by collections of the form fY 2 E : Y � Y0g,
where Y0 2 E . Let U be an ultra�lter, majorating F . For each Y 2 E select
a projection PY : Y ! c0 with kPY k � 2. Such a projection exists due to
the Sobczyk theorem ([2, p. 71]). For every x 2 X and every Y 2 E with
x 2 Y denote kxkY = kPY xk
W kx � PY xk, where
W
stands for maximum of
two numbers. The equivalent norm on X which we need, we de�ne as follows:
kxk0 = limU kxkY . The expression under the limit is de�ned for Y 2 E big enough,
it is bounded, so the limit exists. It is evident, that k � k0 is an equivalent norm
on X, 1
2
kxk � kxk0 � 3kxk. Let us prove, that (X; k � k0) possesses the d-big slice
property. Fix a slice S of the unit ball B (X; k � k0), generated by a functional f
and an " > 0. Let e 2 S,
f(e) > 1� "
2
: (12)
Denote by en, e
�
n, n 2 N the canonical basic of c0 and the corresponding coordinate
functionals. Since feng tends weakly to 0, there is an m 2 N with
jf(em)j <
"
4
: (13)
Consider elements of the form e+ tem. For every Y 2 E , e 2 Y we have
ke+temkY = kPY (e+tem)k
_
ke+tem�PY (e+tem)k = kPY (e)+temk
_
ke�PY ek
102 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1
On Sets with Extremely Big Slices
= jt+ e
�
m(PY e)j
_
kPY e� e
�
m(PY e)emk
_
ke� PY ek: (14)
Denote a = limU e
�
m(PY e), q = limU kPY e�e�m(PY e)emk
W ke�PY ek and remark,
that a and q do not depend on t. According to (14)
ke+ temk0 = maxfjt+ aj; qg: (15)
Since e 2 S, we know that jaj � 1 and q � 1. By (15) the elements x1 =
e+ (1� a)em, x2 = e� (1 + a)em belong to the unit ball B (X; k � k0), and due to
(12) and (13), f(xj) � 1� "; j = 1; 2. Hence x1; x2 2 S. To complete the proof it
is su�cient to notice that kx1 � x2k0 = 2.
References
[1] Y. Benyamini and J. Lindenstrauss, Geometric nonlinear functional analysis. V. 1.
AMS, Coll. Publ.-48, Providens, RI, 2000.
[2] J. Diestel, Sequences and series in Banach spaces. Graduate texts in Math. 92.
Springer�Verlag, New York, 1984.
[3] I.K. Daugavet, On a property of completely continuous operators in the space C.
� Usp. Mat. Nauk 18 (1963), No. 5, 157�158 (Russian).
[4] V.M. Kadets, R.V. Shvidkoy, G.G. Sirotkin, and D. Werner, Banach spaces with
the Daugavet property. � Trans. Amer. Math. Soc. 352 (2000), 855�873.
[5] D. Bilik, V. Kadets, R. Shvidkoy, and D. Werner, Narrow operators and the Dau-
gavet property for ultraproducts. � Positivity 9 (2005), 45�62.
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 1 103
|
| id | nasplib_isofts_kiev_ua-123456789-106584 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1812-9471 |
| language | English |
| last_indexed | 2025-12-01T11:56:29Z |
| publishDate | 2006 |
| publisher | Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
| record_format | dspace |
| spelling | Ivakhno, Ye. 2016-09-30T19:25:27Z 2016-09-30T19:25:27Z 2006 On Sets with Extremely Big Slices / Ye. Ivakhno // Журнал математической физики, анализа, геометрии. — 2006. — Т. 2, № 1. — С. 94-103. — Бібліогр.: 5 назв. — англ. 1812-9471 https://nasplib.isofts.kiev.ua/handle/123456789/106584 A new characterization of the Radon-Nikodym property in terms of sizes of slices and equivalent norms is presented. A property opposite to the Radon-Nikodym property is studied in the context of 1-unconditional sums of Banach spaces. en Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України Журнал математической физики, анализа, геометрии On Sets with Extremely Big Slices Article published earlier |
| spellingShingle | On Sets with Extremely Big Slices Ivakhno, Ye. |
| title | On Sets with Extremely Big Slices |
| title_full | On Sets with Extremely Big Slices |
| title_fullStr | On Sets with Extremely Big Slices |
| title_full_unstemmed | On Sets with Extremely Big Slices |
| title_short | On Sets with Extremely Big Slices |
| title_sort | on sets with extremely big slices |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/106584 |
| work_keys_str_mv | AT ivakhnoye onsetswithextremelybigslices |