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ISSN   1078-6236   International Institute for General Systems Studies, Inc. 
 

Dentability and Convexity 

Shang Shaoqiang1 , Suyalatu Wulede2  
1Department of Mathematics, Harbin Institute of Technology , Harbin 150001, China 

2College of Mathematics Science, Inner Mongolia Normal University, Huhhot 010022, 

China 

Email: yizhitu@163.com, suyila@imnu.edu.cn 

Abstract  In this paper, we introduced the notion of weak denting point of ( )U X  
(respectively, uniformly dentable) and described the characterization of reflexive very convex 
(respectively, uniformly convex) spaces by using the notion of weak denting point (respectively, 
uniformly dentable), and studied the properties of them. 

Keywords  Dentability  Convexity  Denting point  Banach Space 

1. Introduction and Preliminaries 

Throughout this paper, X  will denote a real Banach space and X will denote its conjugate 

space . Set ( ) { : , 1}, ( ) { : , 1}, { : ( ), ( ) 1 }.xS X x x X x U X x x X x S f f S X f x x           

For a convex set ,C X ext C will denote the extreme point of C . For ( )f S X and 

0 ,  set ( , )F f   will denote the slice{ : ( ) : ( ) 1 }.x x U X f x    The weak topology of X is 

denoted by ( , )X X   and the weak  topology of X is denoted by ( , ).X X   Let D be a 

subset of .X D is said to be dentable if for any 0  there is a x D   such that 

( \ ( )) ,x co D B x   where ( ) { : , }.B x x x X x x        A point x D  is said to be 

denting point of D if for any 0 ,   we have ( \ ( )) .x co D B x  

M.A.Rieffel [1] first introduced the notion of dentability and proved that X has the 

Radon-Nikodym property whenever every bounded subset of X is dentable. H.B.Maynard[2] 
improved the result of M.A.Rieffel and proved that X has the Radon-Nikodym property if and 
only if X is dentable. It is known that there is a close connection between the extreme point and 
the denting point. For example, if M is a compact convex set in Banach space ,X  then x is 

extreme point of M if and only if x is denting point of .M In 1993, Congxin Wu and Yongjin 

Li [3] introduced the notion of strongly convex Banach spaces and proved that if X is reflexive 
Banach space, then X is dentable if and only if every point of ( )S X  is denting point of ( ).U X  

This is only a result about describing the straight relations between dentability and convexity. 
                                                 

* This research has been financially supported by the National Natural Science Foundation of China 

(No.11061022). 

∗

110-122 Advances in Systems Science and Applications (2011), Vol. 11,No.1-2



 

 

Definition 1 Let B be a bounded subset of .X  A point 0x B is said to be weakly 

exposed point of B if there is a function ( )x S X   such that 0( ) sup{ ( ): }x x x x x B    and 

for any sequence 0{ } , ( ) ( ) , ( )n nx B x x x x n     imply that 0 , ( ).w
nx x n   

Definition 2 A point 0 ( )x U X  is said to be very extreme point of ( )U X  if for 

0  there exists a weak neighborhood V of 0 in weak topology ( , )X X  such that for any  

( )f S X does not exist , ( )a b U X V   satisfying that ( ) ,{ (1 ) } ( ) ,f a b ta t b U X V       

0

1
( ) ,

2
x a b  where [ 0 ,1 ].t   

Definition 3  Let D be a subset of .X D is said to be weak dentable if for any weak neig 

-hborhood V of 0 in weak topology ( , ),X X  there is a Vx D such that ( \( )).w
v vx co D x V   

A point x D is said to be weak denting point of D if for any weak neighborhood V of 0  

in weak topology ( , ),X X   we have ( \ ( )) .wx co D x V   

Definition 4  Let D be a subset of .X  A point x D   is said to be weak  denting 

point of D if for any weak  neighborhood V of 0 in weak  topology ( , ),X X   we have 

( \ ( )) .wx co D x V


   

Definition5 ( )U X is said to be uniformly dentable. If for any 0, ( )x S X    there 

exists 0  such that inf{ ( ) ( ( ( ) \ ( ))) , } .xx x x co U X B x x S        

Definition 6 [4] A space X  is said to be uniformly convex if and only if for any 0  there 

is a 0  such that for , ( ) ,x y S X if 
1

,
2

x y   then .x y    

Definition 7 [5]  A space X  is said to be very convex if and only if for any ( ) ,x S X  

{ } ( )nx S X and for some xx S  there holds ( ) 1, ( ),nx x n   then 0,( ).w
nx x n   

Definition 8 [3]  A space X  is said to be strongly convex if and only if for any 
( ) ,x S X  

{ } ( )nx S X and for some xx S  there holds ( ) 1, ( ),nx x n   then 0 , ( ).nx x n   

Lemma 1 [6] X  is uniformly convex if and only if for any 0, ( ),f S X    there is a 

0  and some compact set C with dim 1C  such that ( , ) { : ( , ) }.F f y X d y C     

Lemma 2  Let X be a strictly convex space. If 
01 0{ } ( ), ( ) , ,n n xx U X x S X x S   

     

0 0( ) ( ) , ( ),nx x x x n   then there exists a net 1{ } { }n nx x
  

  such that .wx x

   

Proof  If
01 0 0 0{ } ( ), ( ) , , ( ) ( ) , ( ),n n x nx U X x S X x S x x x x n     

      we may assume 

that n mx x  for all .n m ( )U X is weak  compact set, so there exists 0 ( )x U X  such that 

0x
  is accumulation point of 1{ }n nx 

  about  weak  topology ( , ).X X   We construct a 

family of sets 
0 0

{ :
x x

U U  is weak  neighborhood of 0x
  in weak  topology ( , )}X X  and 

Advances in Systems Science and Applications (2011), Vol.11, No.1-2                  111 



 

Define a order by inclusive relation, i.e., 
0 0x x

U U  if and only if 
0 0

,
x x

U U  thus we obtain 

a ordered set . We also construct another family of sets 
0 0

1{ { } :n nx x
U x U 

 
  is weak  

neighborhood of 0x
  in weak   topology ( , )},X X   then by Zermelo axiom, there is a 

mapping f  such that 
0 0

1 1( { } ) { } .n n n nx x
f U x U x 

   
   Put 

0
1( { } ) ,n nx

x f U x 
  

  then  

1{ } { }n nx x 
  

   is a net in X  and .wx x

    

Denoting 0 0| ( ) | ,r x x then 1.r  Otherwise, 0 1 ,r   we consider a neighborhood 

0 0 0

1
{ :| ( ) ( )| (1 )}

2
z z x x x r     of point 0 .x Because 0| ( )| 1,nx x  there is an integer N  such that 

for all n N  there holds inequality 0

1
| ( )| ,

2n

r
x x 

 hence 0

1
: | ( ) |

2n n

r
x x x     

 
is finite set, 

furthermore, there exists a weak  neighborhood V  in weak  topology ( , )X X  such that 

0

1
:| ( ) |

2n n

r
x x x V      
 

  because of ( , )X X  is Hausdorff  topology. By 0 ,wx x

   

we know that there is a 0 such that for 0  there holds 0 0 0

1
{ :| ( ) ( )| (1 )} .

2
x z z x x x r V
         

One hand, by 0 0 0

1
{ :| ( ) ( )| (1 )},

2
x z z x x x r
       we have 0

1
| ( )| .

2

r
x x
 

 On the other hand, 

by 0 0 0

1
{ } { }, { :| ( ) ( )| (1 )}

2nx x x z z x x x r V 
           and 0

1
:| ( ) | ,

2n n

r
x x x V      
 

 we 

have 0

1
| ( )| ,

2

r
x x
 

 which leads to a contradiction, this shows that 0 0( ) 1,x x  Noticing that 

0( ) 1x x   

and the hypothesis that Xis strictly convex, we have 0 ,x x  but 0x x  is impossible because  

of 0 0 0, ( ) ( ), ( )w
nx x x x x x n

      and 1{ } { } ,n nx x 
  

  which leads to 0 .wx x

   

2. Main Results and Proof 

Theorem 1 Let X be a reflexive Banach space. Then X is very convex if and only if every 
point of ( )S X is weak denting point of ( ).U X  

Proof  We divide the proof into three parts. Firstly, we will prove that if every point of 
( )S X is weak denting point of ( ),U X then X is strictly convex. 

 Suppose that X is not strictly convex, then there exist three different point 0 1 2, ,x x x in 

)(XS such that 0 1 2

1 1
.

2 2
x x x  We choose a scalar 0l   and define an linear functional 0f  

on a subspace 0 1 2{ : , }X x x R      of X such that 0 1 2( ) ( ) ( )f x x l l           

112                          Shang: Dentability and Convexity



 

 

for some 0.  Since 0X  is finite dimensional subspace of ,X  then there exists a real number 

0M   such that 0|| || .f M  By Hahn-Banach theorem, there exists ( )f S X  such that  

 || ||f M and 0( ) ( )f y f y  whenever 0 .y X  Therefore we have 

1 0 1 0 1 2( ) ( ) ( 0 ) ( ) 0( ) ,f x f x f x x l l l            

2 0 2 0 1 2( ) ( ) (0 ) 0( ) ( ) ,f x f x f x x l l l            

0 1 2 0 1 2

1 1 1 1 1 1
( ) ( ) ( ) .

2 2 2 2 2 2
f x f x x f x x l l l               

   
 

We consider a weak neighborhood 0 0: ( ) ( ) |
2

x V x f x f x
     

 
of 0x in weak topology 

( , ),X X   where V denotes the weak neighborhood of 0 in weak topology ( , ),X X   then 

1 2 0, ,x x x V  hence 1 2 0, ( ) \ ( ) ,x x U X x V  it follows that  0 0( \ ( )).wx co U X x V   

This contradicts that 0x is weak denting point of ( ).U X  

     Secondly, we will prove the necessity of theorem 1. 

  By Hahn-Banach theorem, we know that for ( )x S X  there exists a ( )f S X such 

that ( ) 1.f x  If ( ) , ( ) 1, ( ) ,n nx U X f x n   then || || 1, ( ) .nx n  Let ,
|| ||

n
n

n

x
y

x
 then 

( ) 1.nf y  Because X is very convex (which implies strictly convex), we ( ) ( ( ) \{ })f x f U X x  

and , ( ).w
ny x n  On the other hand, | ( ) | ,n n n nf x x f x y f y x       hence 

| ( ) ( )| 0, ( ). nf x f x n   This shows that x is weakly exposed point of ( )U X and f  is 

corresponding weakly exposing function. If ( ) \ ( )y U X x V  (where V is the weak 

neighborhood of 0 in weak topology ( , ),X X  then there exists scalar 0m  such that 

( ) ( ) ,f x f y m   hence    ( ) sup ( ): ( )\( ) sup ( ): ( ( )\( )) ,wf x m f y y U X x V f y y co U X x V        

This shows that ( ( ) \ ( )) ,wx co U X x V   hence x is weak denting point of ( ).U X  

Thirdly, we will prove the sufficiency of theorem 1. 

Suppose that 1( ),{ } ( )n nx S X x S X
   and ( ) 1, ( )nf x n  for some .xf S  

By the reflexivity of ,X we know that ( )U X is weak sequential compact, hence there exists a 

subsequence 1 1{ } { }
kn k n nx x 

   such that , ( ) .
k

w
nx x k   Because ( )U X is closed  

convex set, we have ( ) ( ) ( ) ,
w

U X U X U X  hence 1.x  On the other hand, ( ) 1.x f x    

This shows that 1.x  By ( ) ( ) 1f x f x  and the fact that X is strictly convex, we have .x x  

Furthermore, we can deduce that , ( ) ,w
nx x n  this shows that X  is very convex. 

Assume the contrary,i.e., there exist weak neighborhood x V of x in weak topology ( , ),X X   

and a subsequence 1 1{ } { }
in i n nx x 

   such that .
inx x V   By the assumption that every point 

of ( )S X is weak denting point of ( ) ,U X we have ( ( ) \ ( )).wx co U X x V  Hence there is 

a function ( , )g X X X    which separates x and ( ( ) \ ( )) ,wco U X x V i.e., there is 

Advances in Systems Science and Applications (2011), Vol.11, No.1-2                  113 



 

a scalar 0r  such 

that ( ) sup ( ( ( ) \ ( ))).wg x r g co U X x V   Evidently, ( ( )\( )) ,
i

w
nx co U X x V   

thus ( ) ( ) .
ing x g x r   On the other hand, By the reflexivity of ,X we know that there exist a 

subsequence 1 1{ } { }
i il

n l n ix x 
   such that 1{ }

il
n lx 

 converges weakly to x  . Which contradicts 

that ( ) ( ) .
ing x g x r  .                

Theorem 2 X is uniformly convex if and only if ( )U X is uniformly dentable and 

reflexive. 
Proof  We divide the proof into three parts. Firstly, we will prove that X is uniformly 

convex if and only if for any 0 , ( ) ,x S X    there exists 00 , ( )x S X    such that 

0( ) 1,x x  then  0( , ) : , .F x x x X x x       

  Suppose that X is uniformly convex. By lemma 1 we know that for any 0, ( ),x S X     

there exist a scalar 0 and a compact setC with dim 1C   such that ( , ) :F x x x   

( , ) .d x C  Let ( , ) ,x F x  then ( ) 1 .x x    We select a 00 , ( )x S X   such that 

0( ) 1x x   because of uniform convexity implies reflexivity, then 0( ) ( ) ,x x x x    it 

follows that 0
0( ) 1 .

2 2 2

x x
x x

 
    By the assumption that X is uniformly convex, we 

have 0 .x x   This shows that  0( , ) : , .F x x x X x x       

Conversely, suppose that for any 0 , ( ) ,x S X    there exists 00, ( )x S X    

such that 0( ) 1,x x  then  0( , ) : , .F x x x X x x      By lemma 1 we know 

that X is uniformly convex. 

    Secondly, we will prove the necessity of theorem 2. 
    Suppose that X is uniformly convex. Evidently, X is reflexive. 

Let ( ) , .xx S X x S   

By we have proved above, there exist 0  such that  ( , ) : , ,F x y y X y x      i.e., 

if ( ) ( ) ,x x x y    then .y x   Hence, for ( ) \ ( ) ,y U X B x we have .y x    

(where ( ) { : , }B x y y X y x     ). We take ,
2

  then ( ) ( ) .x x x y       

Therefore,  

inf{ ( ) ( ( ( ) \ ( )))} .x x x co U X B x      This shows that ( )U X is uniformly dentable. 

    Thirdly, we will prove the sufficiency of theorem 2. 

By the reflexivity of ,X we know that for ( )x S X  there exists 0 ( )x S X such that 

0( ) 1.x x  Hence, there exists 0  such that 0 0{ ( ) ( ( ( ) \ ( )))}x x x co U X B x    because 

of ( )U X is uniformly dentable. We take   and let ( , ) ,x F x   then ( ) 1 ,x x     

114                          Shang: Dentability and Convexity



 

 

i.e., 0( ) ( ) .x x x x      By the inequality 0 0{ ( ) ( ( ( ) \ ( )))} ,x x x co U X B x    we have 

0 0( ) { : , }.x B x x x X x x      This shows that  0( , ) : , .F x y y X y x      By we 

have proved above, we know that X is uniformly convex. 

Theorem 3 If X  is strictly convex space, then weak  denting points of ( )U X are dense 

in ( ).S X   

 Proof  Firstly, we will prove that if for ( )x S X  there exists 0 ( )x S X such that 

0( ) 1,x x  then x is weak  denting points of ( ).U X For any weak  neighborhood V of 0 

in weak  topology ( , ),X X   there exists a scalar 0r  such that for any ( ) \ ( )y U X x V     

there holds inequality 0 0( ) ( ) .x x y x r   Assume the contrary,i.e., 0 0( ) sup ( ( )\( )),x x x U X x V     

it is obvious that there exists ( ) \ ( )ny U X x V     such that 0 0( ) ( ) 1,( ).ny x x x n     

By Lemma 2, we know that there is a 1{ } { }n nx y 
  

   such that ,wx x

   which 

contradicts to that ( ) \ ( ).ny U X x V    Thus 

         0 0( ) sup ( ) : ( ) \ ( )x x r y x y U X x V         

                 
 0sup ( ) : ( ( ) \ ( ) )y x y co U X x V       

                 
 0sup ( ) : ( ( ) \ ( ) ) ,wy x y co U X x V

       

hence 0 ( ( ) \ ( )).wx co U X x V
    This shows that x is weak denting points of ( ).U X  

Secondly, we will prove that for any ( )x S X  and 0  there is a ( )y S X   

which attains its norm on ( )S X such that .y x     

 Let ( ).x S X  For 0,  take a  0,1 such that
 
 

25 1
0 .

4 5 1







 


We consider 

the norm neighborhood 
1

:
4

z z x
      

 
of x  in norm topology ( , ) ,X   then 

5 1 3 1
.

4 4
z

  
   Thus 

      
z z x x x x

x x
z z z z z z

      
 

     
        

                   1 4 4 4
1 1

4 5 1 5 1 5 1

 
  

 
         

 
 

25 1
.

4 5 1







 


 

By Bishp-Phelp theorem, we know that there exist 0 0, ( )z X z S X   with  0 0 0z z z   such  

that 0

1
: .

4
z z z x

        
 

Taking 0

0

,
z

y
z





 then  0 1y z  and ,y x     

this shows that weak  denting points of ( )U X are dense in ( ).S X  

Advances in Systems Science and Applications (2011), Vol.11, No.1-2                  115 



 

Theorem4  Let D  be a bounded closed convex set of X .Then D  is weak dentable if 

and only if for any weak neighborhood V of 0 in weak topology ( , ),X X   there is a slice 

 ( , , ) : , ( ) sup ( )S x D y y D x y x D       and 0 ( , , )x S x D such that 0( , , ) .S x D x V    

Where 0  . 

Proof  Sufficiency. For any 1 2, , , nf f f X  and 0 ,  let  
1

: ( ) | .
n

i
i

V x f x 


   

then, by the conditions given here, we know that there exist a slice 0 ( , , )x S x D and 

0 ( , , )x S x D such that 0( , , ) ,S x D x V    it follows that 0( ) sup ( ) .x x x D     

Noticing that 

 0\ ( ) \ ( , , ). : , ( ) sup ( )D V x D S x D y y D x y x D         and

 : , ( ) sup ( )y y D x y x D     is weak closed convex set of weak topology ( , ),X X  we 

know that  0( \ ( )) : , ( ) sup ( ) ,wco D x V y y D x y x D       it follows that 

0 0( \ ( )) ,wx co D x V  this shows that D is weak dentable. 

 Necessity. Suppose that D is weak dentable. For any 1 2, , , nf f f X  and 0 ,  let 

1
: ( ) | .

2

n

i
i

V x f x




   
 

 Then, there a is 0x X such that 0 0( \ ( )) .wx co D x V  Hence 

there is a function x X  which separates 0x and 0( \ ( )) ,wco D x V i.e., there is a 

scalar 0r  such that 0 0( ) sup ( ( \ ( ))).wx x r x co D x V     

Let 0sup ( ) ( ) ,x D x x r     then 0( ) sup ( ) sup ( ) ,x x x D r x D         

this shows that 0 ( , , ) .x S x D  Furthermore, for any ( , , ) ,y S x D  we can deduce 

0 0( ) sup ( ) sup ( ) sup ( ) ( ) ( ) ,x y x D x D x D x x r x x r             which leads 

to 0 .y x V   Otherwise, 0 ,y x V  furthermore 0 0\( ) ( \( )).wy D x V co D x V    By the 

inequality 0 0( ) sup ( ( \ ( ))),wx x r x co D x V    we have 0( ) ( ).x x r x y  
 

A contradiction. 

Theorem5  Let X  be a reflexive Banach space. If every point of ( )S X is weak denting 

point of ( ) ,U X then every point of ( )S X is weakly exposed point of ( ).U X  

Proof  Suppose that 0 ( )x S X is weak denting point of ( ) ,U X by Hahn-Banach 

Theorem we know that there is a function 0 ( )x S X   such that 0 0( ) 1.x x  We will prove 

that if 0x attains its norm at another point 0 ( ) ,y S X  then 0 0 .x y Otherwise, 0 0 ,x y it 

is obvious that    0 0 0 0 0

1 1
1 1,

2 2
x y x y x     thus  0 0

1
1.

2
y x  By Hahn-Banach  

Theorem we know that there is a function 0 ( )y S X  such that    0 0 0 0 .y y y x  We consider 

a weak neighborhood 0 0 0 0
0 0 0

| ( ) ( ) |1
: ( ( ))

2 4

y y y x
V x y x y x

 
 

    
 

of point 0 0

1
( ) ,

2
y x  

116                          Shang: Dentability and Convexity



 

 

it is clear that 0 0, ,x y V thus  0 0, \ ,x y U X V  furthermore  0 0

1
( ) ( \ ),

2
wy x co U X V   

this shows that  0 0

1

2
y x is not weak denting point of ( ) ,U X which contradicts to the 

hypothesis that every point of ( )S X is weak denting point of ( ).U X  

 If 1 0 0 0{ } ( ) , ( ) ( ) 1, ( ) ,n n nx U X x x x x n  
     then, by the hypothesis that X is reflexive 

space we know that there exist 1 1{ } { }
kn k n nx x 

  and ( )x U X such that ,( ).
k

w
nx x k   

Hence 0 0 0 0( ) ( ) ( ) 1, ( ) ,
knx x x x x x k      it follows that 1,x  this shows that ( ).x S X  

By the proof above, we have 0 ,x x  thus 0 , ( )
k

w
nx x k    and we can deduce that 

0 , ( ).w
nx x n  If 1{ }n nx 

 does not converges weakly to 0 ,x then there exist a weak 

neighborhood 1V of 0 in weak topology ( , )X X  and a subsequence 1{ }
in ix 

 of 1{ }n nx 
  such 

that 1 0 1{ } ( ) .
in ix x V

    Noticing that 0x is weak denting point of ( )U X , we know that 

0 0 1( ( ) \ ( )).wx co U X x V  Hence there is a function x X  which separates 0 1( ( ) \ ( ))wco U X x V  

and 0 ,x i.e., there is a scalar 0r  such that  0 0 1sup ( ( ( ) \ ( ))).wx x r x co U X x V     

 Noticing that 1 0 1{ } ( ) ,
in ix x V

    we have 0 1 0 1( ) \ ( ) ( ( ) \ ( )).
i

w
nx U X x V co U X x V     

By the inequality 0 0 1( ) sup ( ( ( ) \ ( ))),wx x r x co U X x V    we have 0( ) ( ).
inx x r x x   On the 

other hand, by we have proved above, there exists a subsequence 1{ }
jn jx 

  of 1{ }
in ix 

  such that 

1{ }
jn jx 

  converges weakly to 0 ,x which leads to    0 0 .x x r x x  
 

A contradiction. Hence 

0 ( )x S X is weakly exposed point of ( ).U X  

Theorem6  Let X be a separable reflexive Banach space. If 0 ( )x U X is extreme 

point of ( ) ,U X  then 0 ( )x S X is weak denting point of ( ).U X  

Proof  Suppose that 0 ( )x U X is extreme point of ( ).U X If 0 ( )x S X is not weak 

denting point of ( ),U X then there exists a weak neighborhood V of 0 in weak topology ( , )X X   

such that 0 0( ( ) \ ( )).wx co U X x V  Noticing that 0( ( )\( )) ( )wco U X x V U X  and the assumption  

that X is reflexive Banach space, we know that 0( ( ) \ ( ))wco U X x V is weak compact set. By 

Krein-Milman theorem, we have 0 0( ( ( ) \ ( ))) ( ( ) \ ( )),w w wco ext co U X x V co U X x V   hence 

0 0( ( ) \ ( )) ( ) \ ( ) .
wwext co U X x V U X x V    In fact, by the assumption that X is separable  

Banach space, we know that weak topology ( , )X X  is metrizable space, hence, for 

any 0( ( ) \ ( )) ,wx ext co U X x V  there exists a sequence 1{ (1 ) }n n n n nt x t y 
  such 

that 

 (1 ) , .w
n n n nt x t y x n    Where   00,1 , , ( ) \ ( ).n n nt x y U X x V    By the 

reflexivity of ,X  we know that there exists a sequence    in n such that 0 0, ,
i i

w w
n nx x y y   

0, ( ).
i

w
nt t i  It is obvious that  0 0 0 00,1 , , ( )\( )

w
t x y U X x V   and (1 ) ,

i i i i

w
n n n nt x t y x    

Advances in Systems Science and Applications (2011), Vol.11, No.1-2                  117 



 

thus  0 0 0 01 .x t x t y    

 Case (I):  If 0 0,t   then 0 0( ) \ ( ) ;
w

x y U X x V    

 Case (II):  If 0 1,t   then 0 0( ) \ ( ) ;
w

x x U X x V    

 Case (III): If  0 0,1 ,t  then, by 0 0 0, ( ) \ ( )
w

x y U X x U  0( ( ) \ ( ))wco U X x V   

and 0( ( ) \ ( )) ,wx ext co U X x V  we have 0 0 0( ) \ ( ) .
w

x x y U X x V    Combining 

case (I),(II) and (III), we have 0 0( ( ) \ ( )) ( ) \ ( ) .
wwext co U X x V U X x V   Because 

  0 0( ( ( ) \ ( ))) ( ( ) \ ( )) ,w wext U X co U X x V ext co U X x V   we have  

0 0 0( ( ) \ ( )) ( ) \ ( ) .
wwx ext co U X x V U X x V    This is a contradiction. 

Theorem7  If 0 ( )x S X is weak denting point of ( )U X , then 0x is very extreme 

point of ( )U X . 

Proof  Suppose that 0 ( )x S X is not very extreme point of ( ),U X then there exist 0 0 ,   

0 ( )f S X  such that for any neighborhood V of 0 in weak topology ( , ),X X  there exists 

, ( )a b U X V  satisfying that     0 0

1
( ),| | , 1 ( ) ,

2
x a b f a b ta t b U X V        where  0,1 .t   

We consider a weak neighborhood 0
0: | ( ) |

6
V x f x

   
 

of 0 in weak topology ( , )X X   

and a family of sets  V (where V  is balanced convex neighborhood of 0 in weak topology 

( , ),X X   then weak topology ( , )X X  has a neighborhood base  V V  of 0. For any 

,V V  there exist , ( ) ( )a b U X U V     satisfying that 0 0

1
( ), ( ) ,

2
x a b f a b         

{ (1 ) } ( ) ,ta t b U X V V       where [0,1].t   

Let ,a a a b b b             (where , ( ) , ,a b U X a b V V          ), then 

0
0 0 0 0

1 1 1
( ) ( ) , ( ) | ( ( )) | ,

2 2 2 2
x a b a b f a x f a b      

            

0 0 0
0 0 0 0 0( ) ( ) | ( ) | .

2 6 3
f a x f a x f a a   

            

Noticing that ( )U X is convex set and V V  is balanced convex set, we know that 

1 1 1
( ) ( ) , ( ) , ( ) .

2 2 2
a b U X a b V V a b V V                        

 Let 0
0 0: ( ) .

3
V x f x

   
 

It is obvious that 0 0 .a x V   Similarly, 0 0 .b x V   Hence  

118                          Shang: Dentability and Convexity



 

 

0, ( ) \ ( ) ,a b U X x V     furthermore, 0 0

1
( ) ( ( ) \ ( )).

2
a b co U X x V    

 

On the other hand, 0

1 1
( ) ( )

2 2
a b x a b          and

1
(

2
a  ) ,b V V    so we have 

0

1
( ) .

2
a b x V V       Because  V V  is neighborhood base of 0 in weak 

topology ( , ),X X  from the definition of weak closure about weak topology ( , ),X X  we 

know that 0 0 0( ( ) \ ( )).wx co U X x V  Which contradicts that 0 ( )x S X is weak denting point of 

( ).U X  

Theorem8  Let X be a reflexive separable Banach space. X X is a Banach space with 

the norm  ( , ) max , .x y x y If 1 2,x x are weak denting points 

of ( ) ,U X then 1 2( , )x x is weak denting point of ( ) ( ) ( ).U X U X U X X   . 

Proof  Suppose that 1 2,x x are weak denting points of ( ) .U X By theorem 7 we know 

that 1 2,x x  are very extreme points of ( ).U X It is obvious that 1 2,x x are extreme points 

of ( ).U X We will prove that 1 2( , )x x is extreme point of ( ) ( ).U X U X If 

1 2 1 2( , ), ( , ) ( ) ( )y y z z U X U X 
 

and 1 2 1 2 1 2 1 1 2 2

1 1 1 1 1 1
( , ) ( , ) ( , ) ( , ),

2 2 2 2 2 2
x x y y z z y z y z     then  1 1 1 2 2 2

1 1
( ), .

2 2
x y z x y z     

Hence 1 1 1 2 2 2, .x y z x y z    This shows 

that 1 2 1 2 1 2( , ) ( , ) ( , ).x x y y z z  Therefore 1 2( , )x x is extreme point of ( ) ( )U X U X . By 

Theorem 6, 1 2( , )x x is weak denting point of ( ) ( ) ( ).U X U X U X X    

Theorem9  Let X be a very convex space and X X  is a Banach space with the norm 

 ( , ) max , .x y x y If 1 1, ( ) ,x y S X then 1 1( , )x y is weak denting point 

of ( ).U X X  If 1 11, 1,x y 
 

then 1 1( , )x y is not weak denting point of ( ).U X X  

Proof  Firstly, we will prove that ( , ( ) ) ( , ) ( , ).X X X X X X X X         

For 0, ( ) ,f X X      we consider a eighborhood{( , ) : ( , ) }x y f x y  of (0,0) in 

weak topology ( ,( ) )X X X X   and define 1 2( ) ( ,0), ( ) (0, )f x f x f x f x   for any ,x X  

then ( )if x f x  and ( 1,2)if i  are linear functions. This shows that 

*, ( 1,2).if X i  We construct a set 1 2: ( ) : ( ) ,
2 2

G x f x x f x
          

   
then G is open set 

in ( , ) ( , ).X X X X    

For any ( , ) ,x y G we have 1 2( , ) ( ,0) (0, ) ( ) ( ) ,
2 2

f x y f x f y f x f y
         hence 

{( , ) : | ( , ) | }.G x y f x y   This shows that topology ( , ) ( , )X X X X   is smaller than 

topology ( , ( ) ).X X X X    

 For *
1 20, , ,f f X   we also consider the above open setG  in topology and define 

Advances in Systems Science and Applications (2011), Vol.11, No.1-2                  119 



 

1 2( , ) ( ) , ( , ) ( )f x y f x g x y f x  for ( , ) ,x y X X  then 1 1( , ) . ( , ) ,f x y f x f x y 

2 2( , ) . ( , )g x y f x f x y  and ( , ) , ( , )f x y g x y are linear functions. This shows 

that , ( ) ,f g X X   we consider a 

neighborhood ( , ): ( , ) ( , ): ( , )
2 2

x y f x y x y g x y
        

   
 of (0,0)  

in weak topology ( ,( ) ),X X X X   then ( , ) : ( , ) ( , ) : ( , ) .
2 2

x y f x y x y g x y G
         

   
  

This shows that topology ( ,( ) )X X X X   is smaller than topology ( , ) ( , ).X X X X   This 

completes the proof that ( ,( ) )X X X X   ( , ) ( , )X X X X    . 

Secondly, we will prove that if 1 1, ( ),x y S X then 1 1( , )x y is weak denting point of ( ).U X X  

If 1 1, ( ) ,x y S X  then, by Hahn Banach Theorem we know that there 

exists 1 ( )f S X such that 1 1( ) 1.f x  We choose ( )nx U X with 1,nx  such 

that  1 1 1( ) ( ) 1, .nf x f x n    

By the assumption that X is very convex space, we have 1, ( ) .w
nx x n  It follows 

that for any weak neighborhood V of 0 in weak topology ( , ),X X 
 there exists 1 0r  such  

that for any 1( ) \ ( )x U X x V  there holds 1 1 1 1( ) 1 ( ) .f x f x r   Otherwise, there 

exists 1( ) \ ( )nx U X x V  such that 1( ) 1,( ),nf x n  which leads 

to 1,( ).w
nx x n  This contradicts that 1( )\( )nx U X x V  .Similarly, there 

exists 2 ( )f S X such that 2 1( ) 1f y  and for any weak neighborhood V of 0 in weak 

topology ( , ),X X  there exists 2 0r  such that for any   1\ ( )y U X y V  there 

holds 2 1 2 2( ) 1 ( ) .f y f y r   For any ( , ) ,x y X X   let 1 2( , ) ( ) ( ) ,f x y f x f y   

then f is linear function on ,X X  and 

1 2 1 2( , ) ( ) ( ) . .f x y f x f y f x f y    2 2

1 22 f f  2 2
x y  

 2 2 2 2

1 2 1 24 max , 4 ( , )f f x y f f x y    . 

This shows that ( ) .f X X    

We consider any weak neighborhood W of (0,0) in weak topology 

( ,( ) )X X X X   and notice that 1 1, ( ) ,x y S X  then, 

by ( ,( ) )X X X X   ( , ) ( , ),X X X X     we know that there exists a weak 

neighborhood V of 0 in weak topology ( , )X X  such 

that 1 1 1 1( ) ( ) ( , ) .x V y V x y W     For 1 1( , ) ( ( ) ( ) ) \ (( ) ( )),x y U X U X x V y V     we 

have 1( ) \ ( )x U X x V  or 1( ) \ ( ).x U X y V  Without loss of generality, we may assume 

120                          Shang: Dentability and Convexity



 

 

that  1( ) \ ,x U X x V   then 

1 1 1 1 1 2 1 2 1( , ) ( , ) ( ( ) ( )) ( ( ) ( )) .f x y f x y f x f x f y f y k      Thus  

1 1 1( , )f x y k   1 1sup ( , ) : ( , ) ( ) ( ) \ ( ) ( )f x y x y U X U X x V y V      

                   1 1sup ( , ) : ( , ) ( ) ( ) \ (( , ) )f x y x y U X U X x y W     

                   1 1sup ( , ) : ( , ) ( ( ) ( ) \ (( , ) ))f x y x y co U X U X x y W     

                   1 1sup ( , ) : ( , ) ( ( ) ( ) \ (( , ) )) .wf x y x y co U X U X x y W     

Hence 1 1 1 1( , ) ( ( ) ( ) \ (( , ) )).wx y co U X U X x y W   Obviously,

( ) ( ) ( ),U X X U X U X    

which leads to 1 1 1 1( , ) ( ( ) \ (( , ) )) ,wx y co U X X x y W   this shows that 1 1( , )x y is weak 

denting point of ( )U X X . 

Thirdly, we will prove that if 1 11, 1,x y  then 1 1( , )x y is not weak denting point of ( ).U X X  

Case (i). If 1 1( ), , (0,1),x S X y a a    then, there exist 1 1

2 2 2 2
1 , 1

3 3

a a
y y y y

a a

             
   

 

such that 1 1 1 1

2 2
(1 ) 1, (1 ) 1.

3 3
y y y y a a y y y y a a                 By  

Hahn-Banach Theorem, there exists *f X such that 1

2 2 2 2
( ) 1, ( ) 1 , ( ) 1 ,

3 3

a a
f y f y f y

a a

        

hence 1 1

2 2 2 2
( ) , ( )

3 3

a a
f y y f y y

a a

      and 0

2 2
.

6

a

a
 
 We consider a weak neighborhood 

1 0{ : | ( )| }x f x y   of 1y in weak topology ( , ) ,X X   then 1 0{ : | ( )| }x f x y   can be written by 

1 0 1{ : | ( )| } ,x f x y y V     where V  is a weak neighborhood of 0 in weak topology ( , ).X X   

Evidently, 1, ,y y y V     it follows that 1 1 1 1( , ),( , ) ( ( ) ( ) \ ( )( )).x y x y U X U X x V y V        

By ( ,( ) ) ( , ) ( , ),X X X X X X X X        we know that there exists a weak neighborhood 

V of 0 in weak topology ( , )X X  such that 1 1 1 1( , ) ( ) ( ) .x y V x V y V      Noticing  

that 1 1 1 1

1 1
( , ) ( , ) ( , )

2 2
x y x y x y   and the equality ( ) ( ) ( ) ,U X X U X U X    we have 

1 1 1 1 1 1( , ) ( ( ) ( ) \ ( ) ( )) ( ( ) \ (( , ) )) ,x y U X U X x V y V co U X X x y V         furthermore, 

1 1 1 1( , ) ( ( ) \ (( , ) )).wx y co U X X x y V  
 

Hence, 1 1( , )x y is not weak denting point o ( ).U X X  

Case (ii). If 1 10, ( ),y x S X  let
1

( ) , .
2

z U X z  By Hahn-Banach Theorem, there 

exists *f X  such that ( ) 1.f z  We consider a neighborhood
1

: ( )
2

V x f x
    
 

of 0 in weak 

Advances in Systems Science and Applications (2011), Vol.11, No.1-2                  121 



 

topology ( , ),X X  then , .z z V   Which leads 

to 1 1 1( , ) , ( , ) ( ( ) ( ) \ ( ) (0 )).x z x z U X U X x V V        

Similar to case (i), there exists a V such that 1 1( ,0) ( ) (0 ).x x V V    
 

Noticing that 1( ,0)x   

1 1

1 1
( , ) ( , ),

2 2
x z x z   we have 1 1 1( ,0) ( ( )\( ) (0 )) ( ( )\(( ,0) )).w wx co U X X x V V co U X X x V          

This shows that 1( ,0)x is not weak denting point of ( ).U X X . 

Theorem10  Let X be a strongly convex space. X X is a Banach space with the 

norm  ( , ) max , .x y x y If 1 1, ( ) ,x y S X then 1 1( , )x y is denting point 

of ( ).U X X If 1 11, 1,x y   then 1 1( , )x y is not denting point of ( ).U X X  

Proof  Case (i). 1 1, ( ).x y S X Using a similar method to that used in the proof of 

theorem 9, We can prove that 1 1( , )x y is denting point of ( ).U X X  

Case (ii).
 1 11, 1.x y   By the assumption that X is strongly convex space, we know 

that X is very convex space. By theorem 9, we know that 1 1( , )x y is not weak denting point 

of ( ) ,U X X so there exists a weak neighborhood V of (0,0) in weak 

topology ( , ),X X  such that 1 1 1 1( , ) ( ( ) \ ( , ) ).wx y co U X X x y V   It is obvious that there 

exists a scalar 0  such that 1 1 1 1( , ) ( , ) .B x y x y V    

Hence 1 1 1 1 1 1( , ) ( ( )\ ( , )) ( ( )\ ( , )).wx y co U X X B x y co U X X B x y      

This shows that 1 1( , )x y is not denting point of ( ).U X X  

References 

[1] M.A.Rieffel. Dentable subsets of Banach spaces with applications to a Radon-Nikodym     
theorem, Funct.Anal.Proc.(Conf. Irvine, Calif., 1966), Acad.Press.London. Thompson.Wash   
-ington, D.C. 71-77. 
[2] H.B.Maynard. A geometrical characterization of Banach spaces having the Radon-Nikodym       
   property. Trans.Amer.Math.Soc, 1973, 185: 493-500. 
[3] C.X.Wu, Y.J.Li. Strong convexity in Banachspaces. Chin.J.Math, 1993, 13(1): 105-108. 
[4] J.A.Clarkson. Uniformly convex spaces. Trans.Amer.Math.Soc, 1936, 40: 396-414. 
[5] Tegusi, Suyalatu, Y.J.Li. Very convex Banach spaces. Northeast.Math.J, 1997, 13(1): 1-4. 
[6] X.N.Fang. Slice and convexity, smoothness of Banach spaces. Chin.J.Math, 1999, 19(3): 

293-298. 
[7]Xu Xu,Tai Kuang. Q-operator for multi-attribute decision-making with incomplete 

informati-on. Advances in systems science and applications, 2006, 1: 148-154. 
[8] Long Zhou, Mianyun Chen. Investigation on fussy degree criterion of electric power 

equipm-ent. Advances in systems science and applications, 2004, 2: 319-324. 
[9] Jinghong Pan, Xuemou Wu. A pansystem topology approach grayscale image processing.  

Advances in systems science and applications, 2003, 2: 151-156. 
 

122                          Shang: Dentability and Convexity