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Chaper 3
Weak Topologies. Reflexive
Space .Separabe Space. Uniform
Convex Spaces.
III.1
The weakest Topology
Recall on the weakest topology which
renders a family of mapping
continuous
i : X  Yi
iI
arbitary set
topological space
To define the weakest topology on X
such that
i
for each
Let
Yi
is continuous from X to
iI
1
wi  F (Yi ),  ( wi )
must be open in X
For any finite set
(*)
1
 ( wi ),
iF
FI
wi
: open in
The family of the sets of the form (*)
form a base of a topology F of X
The topology is the weakest topology
that renders all
i
continuous
Yi
Proposition III.1
Let
xn 
F
be a sequence in X, then
F(Yi )
xn 
 x  i ( xn ) 
i ( x) i  I
" " For each i  I
 i ( x )  w and w is open in Yi
1
1
 x   i ( w) and  i ( w) is open in X
 N s.t.
1
n  N  xn   i ( w )
  i ( xn )  w
Hence  i ( xn )   i ( x )
" " It is sufficient to consider U 
1
i ( wi )
iF
1
where x   i ( wi )  i  F ,
For each i,
 i ( x )  wi
 N i s.t.
n  N i   i ( xn )  wi
Let N  max N i i  F  , then
n  N   i ( xn )  wi
1
i  F
 xn   i ( wi )
iF
Proposition III.2
Let Z be a topological space and

i
Z
 X Yi i  I
Then

is continuous
is continuous from Z to
 i 
Yi  i  I
"  " For each i  I
If w is open in Yi , then
i 1 ( w) is open in X ,
1
1
 ( i ( w)) is open in Z sin ce  is continuous
Hence  i  is continuous
" " Let U 
1

 i ( wi )
iF

1
U  
1 
  i
 iF
 
iF

1
1

( wi ) 

1
  i ( wi )
1
(



)
( wi ) is open in Z
 i
iF
III.2
Definition and properties of
the weak topology σ(E,E´)
Definition σ(E,E´)
E: Banach space
E´: topological dual of E
 f ( x)  f , x  x  E, f  E
see next page
 ( E , E )
Definition : The weak topology
is the weakest topology on E such that
f :E
 R
is continuous for each
f  E
Proposition III.3
The topology
is Hausdorff
 ( E , E ) on E
Let x  y  E
By sec ond geometric form of Hahn  Banach Thm
 f  E  and   R s.t.
f , x   and
f ,y 
Let O1  z  E : f , z   , and
O2  z  E : f , z   ,
( , ) is  ( E , E ) open in E ,
1
O2   f ( , )  is  ( E , E ) open in E ,
then O1   f
1
x O1 , y O2 , and O1  O2  
Hence the topo log y  ( E , E ) on E is Hausderff .
Proposition III.4
Let
x0  E
neighborhood of
; we obtain a base of
x0
by consider
sets of the form
V  x  E : fi , x  x0   , i  F 
  0, fi  E
, and F is finite
where
Proposition III.5
Let
(i)
(ii) if
xn 
be a sequence in E. Then
xn  x weakly  f , xn  f , x  f  E 
xn  x
xn  x
strongly, then
weakly.
(iii) if
xn  x
 xn 
weakly, then
is bounded and
x  lim inf xn
n 
(iv) if
xn  x
weakly and
fn  f
strongly in E´, then
f n , xn  f , x
(i ) By Pr oposition III .1
(ii) Assume xn  x strongly.
Since
f , xn  f , x  f , xn  x  f xn  x
f , xn  f , x
By Pr oposition III .1, xn  x weakly
(iii) If xn  x weakly, then by (i )
f , xn  f , x
 f  E
By Corollary II .2, sup xn  
n
and
x  lim inf xn
n 
f n , xn  f , x
(iv)
 f n , xn  f , xn  f , xn  f , x
 f n  f , xn  f , xn  f , x
 f n  f , xn  f , xn  f , x
 f n  f xn  f , x n  f , x
By (i )
By (iii)
f , xn  f , x  0
 xn is
Since f n  f ,
Hence
i.e.
bound
fn  f  0
f n , xn  f , x  0
f n , xn  f , x
Exercise
Let E , F be real normed vector space
consider on E and F the topologies
 ( E , E )
and
 ( F , F )
respectively.
Then the product topology on E X F is
 ( E  F , E   F )
Proposition III.6
If
dim E  
,then
 ( E , E ) is strong topology on E.
To show that a strong open set is  ( E , E )  open.
It is sufficient to show that for x0  E and
for an open nhd . U of x,
there is a  ( E , E )  open nhd . V of s.t. V  U
Suppose B ( x0, r )  U .
Choose a basis e1 , e2 ,, en of E
s.t ei  1  i  1,, n
n
For x  E , x   f i ( x )ei , f i  E   i  1,, n
i 1
n
x  x0   f i , x  x0 ei
i 1
n
  f i , x  x0 ei
i 1
n
  f i , x  x0
i 1
 n
if
f i , x  x0    i  1,, n
Hence if we let
V  x  E ; f i , x  x0    i  1,, n
then V  B ( x0 , n )
Choose  s.t. n  r, then
V  B ( x0 , r )  U
Remark
If
dim E  
,then
 ( E , E )
is strictly weaker then the
strong topology.
Let S  x  E ; x  1
S is strongly closed .
Claim : S  ( E , E )  B (0,1)
pf : Let x0  B (0,1) and
let V  x  E ; f i , x  x0   i  1,, n
where   0 and f1 ,, f n  E 
There is y0  E s.t. y0  0 and
f i , y0  0  i  1,, n
because co dim  x  E ; f i , x  0 
n
i 1
[ If there is no such y0 , then the map
 : E  R n defined by
x   f1 ( x ),, f ( xn )  is 1  1
then dim R ( )  n
 dim E  n
]
Consider the function g (t )  x0  ty0
g is continuous fun. of x, g (0)  x0  1
and lim g (t )   , then there is t0  0 s.t
t 
x0  t0 y0  1
i.e x0  t0 y0  S
But
f i , x0  t0 y0  x0  0  i  1,, n
then x0  t0 y0 V
 x0  t0 y0 V  S
 x0  S
 ( E , E )
III.3
Weak topology, convex set
and linear operators
Theorem III.7
Let
CE
be convex, then
C is weakly closed
if and only if
C is strongly closed.
To show : If C is strongly closed , then
C is weakly closed .
Let x0  C , there is 0  f  E  and   R
c
f , x0    f , x
s.t.
 x C
Let V  x  E ; f , x   , then
V is a nhd . of x0 and V  C  
then V  C
c
 C is weakly open
 C is weakly closed
c
Remark
The proof actually show that every
every strongly closed convex set
is an intersection of closed
half spaces
H f ,  x  E f , x    C
Corollary III.8
If
 : E  ( , ]
is convex l.s.c. w.r.t. strongly topology
then

In particular, if
then
is l.s.c. w.r.t.
 ( E , E )
xn  x weakly ( ( E, E))
 ( x )  lim inf  ( xn )
n 
Let
then
then
then
A  x  E ; ( x )    ,   R
A is strongly closed and convex.
A is  ( E , E ) closed
 is  ( E , E )  l.s.c
Observation :
Let  ( x )  x , then
 is strongly continuous and convex
 is strongly l.s.c and convex
then  is  ( E , E )  l.s.c
Hence if xn  x weakly ( ( E , E )), then
x  lim inf xn
n 
Theorem III.9
Let E and F be Banach spaces and let
T :EF
be linear continuous (strongly) , then
T is linear continuous on E with
to F with
 ( F , F )
And conversely.
 ( E , E )
By Pr oposition III .2
it is sufficient to show that
x  f , Tx is continuous from ( E , ( E , E ))
x  f , Tx is strongly continuous
 x  f , Tx is  ( E , E )  l.s.c
 x   f , Tx is  ( E , E )  l.s.c
 x  f , Tx is  ( E , E )  u.s.c
 x  f , Tx is  ( E , E )  continuous
Conversely, sup pose that T is linear continuous
from ( E , ( E , E )) to ( F , ( F , F ))
then G (T ) is closed in E  F w.r.t
product topo log y  ( E , E )   ( F , F )
which is  ( E  F , E   F )
i.e. G (T ) is closed w.r.t.  ( E  F , E   F )
G (T ) is convex in E  F
then G (T ) is closed in E  F w.r.t strong topo log y
Closed Graph Theorem 
T is continuou from ( E ,
E
) to ( F ,
F
)
On
Remark
E,  ( E, E) is weak topology
j : E  E 
by
j ( x ), f  f , x  f  E 
j( x)  x
In genernal j is not surjective
If
j ( E )  E 
E is called reflexive
III.4
The weak* topology
σ(E′,E)
The weak* topology
 ( E , E )
is the weakest topology on E´
such that
f  f ,x
is continuous for all
xE
Proposition III.10
The weak* topology on E´
is Hausdorff
Proposition III.11
One obtains a base of a nhds for a
f0  E
by considering sets of the form
V  f  E ; f  f 0 , xi   , i  1,, n
  0 , x1, x2 ,, xn in E
Proposition III.12
Let
(i)
 f n  be a sequence in E´, then
f

 f
*
n
for  ( E , E )

  fn , x  f , x  x  E 
(ii) If
fn  f
fn  f
strongly, then
for  ( E, E)
(iii) If
then
fn  f
fn  f
for  ( E, E)
for  ( E, E )
fn 
 f
*
(iv) If
fn
then
is bounded and
f  lim inf f n
n 
(v) If
fn 
 f
*
and
xn  x
strongly, then
f n , xn  f , x
Lemma III.2
Let X be a v.s. and
 ,1,,n
are linear functionals´on X such that
i (v)  0
i  1,, n   (v)  0
n
  1 ,, n  R s.t.    i i
i 1
Let F : X  R n 1 be defined by
F (u )   (u ),1 (u ),, n (u )   u  X
Then F ( X )  R( F ) is closed convex set in R
Since (1,0,,0)  F ( X ) ,
  , 1 ,, n  R not all zero and   R s.t.
n
     (u )   ii (u )  u  X
i 1
n
  (u )   i i (u )  0  u  X
i 1
 0
i
  (u )     i (u )
i 1 
n
n 1
Proposition III.13
If
 : E  R
is linear continuous´w.r.t
 ( E , E )
then there is
x  E s.t.  ( f )  f , x  f  E 
There is a nhd . V of 0 for  ( E , E )
s.t.  ( f )  1
may take V as follows
V  f  E ; f , xi   i  1,, n
In particular, if
f , xi  0  i  1,, n
then  ( f )  0
n
Lemma III .2   ( f )   i f , xi
i 1
n
 f ,  i xi
i 1
n
Take x   i xi
i 1
then  ( f )  f , x
 f  E
 f  E
Corollary III.14
If H is a hyperplane in E´ closed w.r.t
 ( E , E )
Then H is of the form
H   f  E ; f , x   ,0  x  E ,  R
H is of the form
*


H   f  E ; ( f )   , where 0    E ,   R
Suppose f 0  H , then there is V  H
c
where V  f  E ; f  f 0 , xi    i  1,, n
x1 , x2 ,, xn  E ,   0
V is convex
(13) Either  ( f )    f V
(13) or  ( f )    f V
From (13)  ( g )     ( f 0 )  g W  V  f 0
From(13)  ( g )     ( f 0 )  g W  V  f 0
W is a  ( E , E ) nhd . of 0
then  is  ( E , E ) continuous
By Theorem III .13
 x  E s.t.  ( f )  f , x  f  E 
then H   f  E ; f , x   
III.5 Reflexive spaces
On
Remark
E,  ( E, E) is weak topology
j : E  E 
by
j ( x ), f  f , x  f  E 
j( x)  x
j is isometry
j(E) is closed vector subspace of
E 
In genernal j is not surjective
If
j ( E )  E 
E is called reflexive
Lemma 1 (Helly)
p.1
Let E be a Banach space,
f1, f 2 ,, f n  E
1,2 ,,n  R
and
are fixed.
Then following statements are
equivalent
Lemma 1 (Helly)
(i)
  0,  x  BE s.t.
fi , x  i  
(ii)
p.2
n
n
i 1
i 1
 ii   i fi
where
 i  1,2,, n
1 ,  2 ,,  n  R
BE  x  E ; x  1
n
(i )  (ii) Let 1 ,  2 ,,  n  R and s    i
i 1
n
n
i 1
i 1
(i )    i f i , x    i i  s
n
n
i 1
i 1
   i i    i f i , x  s
n
   i f i  s
i 1
By letting   0, we have
n
n
i 1
i 1
 ii   i f i
(ii)  (i ) Let   1 , 2 ,, n   R
define the map  : E  R
n
n
and
by
 ( x )   f1 , x ,  f 2 , x ,,  f n , x  
Then (i )     ( BE )
Suppose that (i ) doesn' t hold .
then    ( BE )
Since  ( BE ) is a closed convex set in R ,
n
by strictly separation principle
there is a   1 ,  2 ,,  n   R
 ( x)        
n
i 1
i 1
   i fi , x     i i
n
i 1
i 1
s.t.
 x  BE
n
n
n
 x  BE
   i f i      i  i which contradicts (ii)
Lemma 2 (Goldstine)
Let E be a Banach space. Then
JB
E
is dense in
w.r.t the weak* topology
 ( E, E) on E
BE 
Let   BE  and V be a nhd of  in the form
V    E ;    , f i   i  1,, n
Let  i   , f i
 1 ,  2 ,,  n  R
i  1,, n and note that
we have
n
n
n
i 1
i 1
i 1
 ii   ,  i f i   i f i
Helly Lemma    0,  x  BE s.t.
f i , x   i   i  1,, n
 Jx , f i   , f i   i  1,, n
 Jx   , f i   i  1,, n
 Jx V
Theorem
(Banach Alaoglu-Bornbaki)
BE    f  E ; f  1
is compact w.r.t.
 ( E , E )
Theorem
A Banach space E is reflexive
if and only if
BE
is compact w.r.t weak topology
 ( E , E )
" " Assume that E is reflexive.
Then JBE  BE  , sin ce J is isometry
Claim : J : E ,  ( E , E   E ,  ( E , E 
For any f  E 
1
( f  J 1 )( )  f , J 1 ( )   , f
  E 
then f  J 1 is continuous
from E , ( E , E  to R
By Pr oposition III .2,
J 1 is continuous
from E ,  ( E , E  to E ,  ( E , E 
Since BE  is compact w.r.t  ( E , E ) and by Claim,
BE  J 1 ( BE  ) is compact w.r.t  ( E , E )
" " Suppose that BE is compact w.r.t  ( E , E )
J is continuous with norm
By Thm III .9, J is continuous from
( E , ( E , E )) to ( E , ( E , E ))
 J is continuous from
( E , ( E , E )) to ( E , ( E , E )),
because  ( E , E ) is wea ker than  ( E , E )
 JBE is compact w.r.t  ( E , E )
By Goldstine Lemma , JBE  BE 
then JE  E 
Hence E is reflexive.
Exercise
Suppose that E is a reflexive
Banach space . Show that
evere closed vector subspace M
of E is reflexive.
Since E is reflexive, BE is compact w.r.t.  ( E , E )
M is strongly closed
then M is closed w.r.t.  ( E , E )
 BM  BE  M is compact w.r.t.  ( E , E )
For any x0  M and a  ( M , M )  nhb. V of x0
in the form V  x  M ; f i , x  x0   ,  i  1,, n
where f i  M   i  1,, n
Claim : V  M c is a  ( E , E )  nhb. of x0
 fi ( x)
ˆ
[ Let f i ( x )  
 f ( x0 )
if x  M
if x  M
Since M is strongly closed , fˆi  E 

V  M c  x  E ; fˆ , x  x0   ,  i  1,, n
is a  ( E , E )  nhb. of x0 ]

If
V I
is a family of  ( M , M )  open sets
such that V  BM


V  M 
I

then  V  M c  BM and by Claim
I
c
I
is a family of  ( E , E )  open sets
Since BM is compact w.r.t.  ( E , E ),
n


1 ,,  n  I s.t.  V  M c  BM
i 1
i
n
 V  BM
i 1
i
Hence BM is compact w.r.t  ( M , M )
Therefore M is reflexive
Corollary 1
Let E be a Banach space. Then
E is reflexive if and only if
E
is reflexive
(1) Suppose that E is reflexive.
Then  ( E , E )   ( E , E )
By Alaoglu Bornbaki Thm
BE  is compact w.r.t  ( E , E )
 BE  is compact w.r.t  ( E , E )
Hence E  is reflexive.
( 2) Suppose that E  is reflexive.
then by (1), E  is reflexive.
Since JE is closed subspace of E ,
by Exercise, JE is reflexive
Since J is isometry between E and JE ,
E is reflexive.
Corollary 2
Let E be a reflexive Banach space.
Suppose that if K is closed convex
and bounded subset of E . Then
K is compact w.r.t
 ( E , E )
Since K is strongly closed and convex ,
by Thm III .7 K is closed w.r.t  ( E , E )
Since K is bounded , K  mBE for some m  0
Since E is reflexive, mBE is compact w.r.t  ( E , E )
K is compact w.r.t  ( E , E ).
Uniformly Convex
A Banach space is called
uniform convex if for all ε>0 ,
there is δ>0 such that if
x, y  BE with x  y  
x y
then
 1
2
x y
2
x
y
Counter Example
for Uniformly Convex
Consider
(R ,  )
2
( x1 , x2 )  x1  x2
is not uniform convex.
see next page
y
(0,1)
x y
2
(-1,0)
x
(1,0)
(0,-1)
Example
for Uniformly Convex
Consider
(R ,  )
( x1,, xn ) 
n
2
x1

2
x2
 
2
xn
is uniform convex.
see next page
For any   0
For any x, y  B(0,1) and
2

2
2
x y  x y 2 x  y

2
2
 x y 2 x  y
2
x y 
2
 ( Parallelog ram
 x  y
2
 42
x y

2
2
1
2
4
2

x y
2


 1
 1  1  1 

2
4
4

Take   1  1 
x y
 1
2
2
4
, we have




Thm)
Theorem
A uniformly convex Banach space E
is reflexive.
Let   E ,   1
To show that   JBE
Since JBE is closed in E  strongly,
it is sufficient to show that for   0,
 x  BE s.t.   Jx  
Given   0, let   0 be as in the definition of
uniform convexity
 f  E  with
f  1 s.t.
, f 1

2


Let V    E ;    , f  ,
2

V is a nhb. of  w.r.t  E , E 
Since JBE is dense in BE  w.r.t  E , E ,
 x  BE s.t. Jx V
Claim :   Jx  BE 
[ Suppose not. Then  W  Jx  BE  c
Since BE  is compact w.r.t  E , E ,
BE  is closed w.r.t  E , E 
 Jx  BE  is closed w.r.t  E , E 
 W is open w.r.t  E , E 
  xˆ  BE s.t. Jxˆ V  W
We have then (sin ce Jx, Jxˆ V )
f , x  , f 

and
f , xˆ   , f 

2
 2  , f  f , x  xˆ    x  xˆ  
2
x  xˆ

 
  , f   (1  )   1  
2
2
2 2
 x  xˆ  

But
x  xˆ  J ( x  xˆ )  Jx  Jxˆ   , sin ce Jxˆ W
a contradiction.