Download Mazm-Orlicz Theorem

Mazm-Orlicz Theorem
Linearly open
Let E be a real vector space.
C  E is called linearly open if
each po int of C is an int erior po int of C
i.e for any x  C , y  E
x  ty  C if
t is sufficient small
Lemma 1
Let E be a real vector space.
Let C be the convex set in E with IntC  
let x0  E , x0  IntC. Then
there is nonzero linear function f s.t
f ( x0 )  f ( x )
 x C
May assume 0  IntC.
If
y0  IntC , let Cˆ  C  y0
xˆ0  x0  y0
f ( xˆ0 )  f ( yˆ )
 yˆ  Cˆ
 f ( x0  y0 )  f ( y  y0 )
 f ( x0 )  f ( y )
 y C
 y C
Let PC be the min kowski gauge function of C
Proof
of
Lemma 1
Let G  Rx 0 , g (x0 )  
Then g  G
  R
*
Claim : g (x0 )  PC (x0 )
  R
pf :
Since x0  IntC , PC ( x0 )  1
(1) For   0,
g (x0 )      1   PC ( x0 )  PC (x0 )
p.1
of
(2) ForProof
  0,
Lemma 1
p.2
g (x0 )    0  PC (x0 ) , sin ce PC ( x )  0
x  E
Apply Hahn  Banach Theorem
 f  E extending g such that
*
f ( x )  PC ( x )
x  E
Then for x  C
f ( x )  PC ( x )  1  g ( x0 )
 f ( x0 )
Hence f ( x0 )  f ( x )
, sin ce f extends g
x  C
Remark
Let E be a normed real vector space.
Let C be an open convex subset of E
let x0  E , x0  C. Then
 f  E s.t
f ( x0 )  f ( x )
 x C
May
assume
0C
From the proof
of
Lemma 1
Proof of Remark
p.1
*
 f E
s.t 1  f ( x0 )
and
f ( x )  PC ( x )
Then
for
x  C
x C
f ( x0 )  1  PC ( x )
, sin ce
 f ( x)
f ( x0 )  f ( x )
 r0  0
s.t
r0 x  C
 PC ( x )  r0 1
C
is
open
by (*)
Hence
if
 PC ( x )  r0 1 x
x  C
if
x 1
x 1

 f ( x )  PC ( x )  r0 1 x
 f  E
(*)
x C

x C
Corollary To Lemma 1
Let E be a real vector space.
Let C be a linearly open convex set in E
with 0 C. Then
 f E
*
f ( x)  0
s.t
 x C
From the proof
 0
*
Proof
 f E
of
sof
.t
Corollary
0  f ( 0)  f ( x )
x  C
f ( x)  0
Suppose
Lemma 1
for some x  C
Choose y  C with
f ( y)  0
Since C is linearly open ,
x  ty  C
if
t
is sufficiently small
 0  f ( x  ty)  tf ( y )  0, if
which is impossible
Hence
f ( x)  0
x  C
t0
Convex cone
Let E be a real vector space.
P is a convex cone if
for any x1 ,, xk  P
 x  P
k
i
i 1
i
i  0 , but not all zero
Define P
Let E be a real vector space.
Let p : E  R be sublinear .
let P  x  E p( x)  0, then
?
P is a linearly open convex cone
(1) To show that P is a convex cone.
of
For any xProof
,

x

P
1
k
p ( xi )  0
?
P.1
i  1,, k
 p (   i xi )   p ( i xi )
k
k
i 1
i 1
   i p ( xi )
k
i  0
i 1
0
   i xi  P
k
  i  0 but not all zero
  i  0 but not all zero
i 1
Hence P is a convex cone.
(1) To show that P is linearly open.
of
For any x  P, y  E
Proof
?
P.2
p( x  ty)  p( x )  p(ty)
 tp( y ) , t  0
 p( x )  
(  t ) p (  y ) , t  0
0
, if t is sufficiently small
 p( x  ty)  0
 x  ty  P
, if t is sufficiently small
, if t is sufficiently small
Hence P is linearly open.
Lemma 2(証明很重要)
Let E be a real vector space.
Let p : E  R be sublinear .
Let P  x  E p( x)  0
If P   , l  E , and l  0, then
The following statements are equivalent :
*
( I ) l ( x)  0  x  P
( II )   0 s.t l  p on E
( II )  ( I ) obvious
Proof of Lemma 2
( I )  ( II )
Define t : E  R
2
P.1
by
x  ( p ( x ), l ( x ))
Let Q be the convex hull of tE
i.e Q is the set of all finite
convex combinations of
Let R  ( r1 , r2 ) r1 , r2  0,
2
Claim : R  Q  
2
po int s in tE.
v Q
For
v 
  tx
n
  0,
where
Proof of Lemma 2
i
i 1
i
i

n
i 1
   i ( p ( xi ),  l ( xi ))
n
i 1
 (  p ( i xi ),   l ( i xi ) )
n
n
i 1
i 1
 (   i p ( xi ),  l (   i xi ) )
n
n
i 1
i 1

n
If
i
p ( xi )  0,
then
i 1
p (   i xi ) 
n
i 1

n
i
xi 

n
p ( xi )  0, then
i
i 1
P
i 1
 l (   i xi )  0
n
by
i 1
  l (   i xi )  0
n
i 1
 v  R2
Hence
R2  Q 

(I )
i
 1,
P.2
xi  E
Consider C  R2  Q   ( R 2  v )
vQ
Proof of Lemma
2
P.3
C is linearly open and convex.
By Corollary to Lemma 1,
there is a nonzero linear functional f on R 2
s.t f (u  v )  0
u  R2 , v  Q
i.e f (u )  f ( v )
u  R2 , v  Q
i.e (1 ,  2 )  R
2
with
2
1
 2  0
(*) 1r1   2 r2  1 p ( x )   2l ( x )
2
s.t
r1  0, r2  0, and x  E
 r   r is bunded from above for all r  0, r  0, x  E
   0, Proof
0
of Lemma 2
P.4
Let ( r , r )  (0,0), we have
 l ( x )   p( x )
x  E
Claim :   0,   0
If   0, then   0
  l ( x)  0
x  E
 l ( x)  0
x  E
 l  0 , which is impossible
Hence   0
1 1
2 2
1
1
2
1
2
2
1
1
2
1
2
2
1
2
If  2  0, then 1  0
Proof of Lemma 2
 1 p( x )  0,
x  E
 p( x )  0,
x  E
 P 
P.5
, which is impossible
Hence  2  0
then 1  2l ( x )  p( x )
1
x  E
Main Lemma
p. 1
Let E be a real vector space.
Let p : E  R be sublinear .
Let S be an arbitary nonempty set.
Let  : S  R be a map.
The following statements are equi var ent :
Main Lemma
(I )  l  E
*
p. 2
with l  p s.t
l ( ( s))  0
s  S
( II ) p(  i ( si ))  0
k
i 1
for all finite subset
s ,, s 
1
and all 1  0,, n  0
n
of S
( I )  ( II )
P( 
n

Proof
of
main
Lemma
  ( s ))  l (   ( s )) by
n
i
i
i 1
i
i
(I )
P.1
i 1
  i l ( ( si ))
n
i 1
0
by ( I )
( II )  ( I )
If
P   , then p( x )  0
x  E
take l  0
Suppose P  
Let Q be the convex cone generated by S .
( II )  P  Q  
Consider C  Q  P  x  y x  Q, y  P
Proof of main Lemma
P.2
C is linearly open convex cone with 0  C.
By Corollary to Lemma 1,
 l  E , l  0 s.t
*
l( z)  0
z  C
i.e for all x  Q and y  P
l ( rx )  l ( y ), l ( x )  l ( ry )
 l( y)  0
r0 
l ( x )  0
 r0
 yP
(*)
 x Q
(**)
  0 s.t l  p on E
  Proof of main Lemma
l ( ( s ))  0 s  S
, by Lemma 2
P.3
Re name l by l , we have
l  p on E and l ( ( s ))  0
s  S
Theorem
p. 1
Let S be an arbitary set.
For i  1,2
Let pi be a sublinear functional on
a real vector space Ei and
 be a map from S to E .
i
i
The following statements are equi var ent :
Theorem
p. 2
( I )  l1  E1 , l2  E2 with l1  p1 , l2  p2 s.t
*
*
l1 ( 1 ( s))  l2 ( 2 ( s))
s  S
( II )  p1 (  i 1 ( si ))  p2 (  i 2 ( si ))
n
n
i 1
i 1
for all finite subset
s ,, s 
1
and all 1  0,, n  0
n
of S
Let E  E1  E2
of Theorem
Let p : EProof
 R by
P.1
u, v  p (u)  p (v )
1
2
Let  : S  E by
s    1 ( s ),  2 ( s )
Then apply main Lemma and observe that
(1) Every linear functional l on E can
be represente d by a pair l1 , l2 
l1  E1 , l2  E2 through
*
*
l ( u, v)  l1 (u )  l2 ( v )
( 2)
l , l  
1
2
p  l1  p1 , l2  p2
Theorem
(3) l ( (Proof
s ))  of
0
l (  1 ( s ),  2 ( sP.2
)  0
 l1 (   1 ( s ))  l2 (  2 ( s ))  0
 l1 (  1 ( s ))  l2 (  2 ( s ))
p (  i ( si ))  0
n
(3)
i 1
0
 p( 



(
s
)
,


(
s
)
  i 1 i  i 2 i 
i 1
 i 1

n
n
 p1 (   i 1 ( si ))  p2 (  i 2 ( si ))  0
n
n
i 1
i 1
  p1 (   i 1 ( si ))  p2 (  i 2 ( si ))
n
n
i 1
i 1
Corollary (Mazm-Orlicz)
p. 1
Let E be a real vector space.
Let p : E  R be sublinear .
Let S be an arbitary nonempty set.
Let  : S  R be a map.
Let  : S  R be a map.
The following statements are equi var ent :
Theorem
(I )  l  E
*
p. 2
with l  p s.t
 ( s)  l ( ( s))
   ( s )  p(  ( s ))
n
( II )
s  S
n
i
i 1
i
i
i
i 1
for all finite subset
s ,, s 
1
and all 1  0,, n  0
n
of S
Proof of Corollary
Let E1  R and E2  E
Let p1 (t )  t and p2  p
Let  1   and  2  
Then apply Theorem

1
p1
E1
l1
S
R
p2

2
 
1
E2
E1  R
l2
p1  id
l1  id
p2  p
S
 
2
E2  E
l2  l
R
Let E be a real vector space.
Mazm-Orlicz Thm implies HahnLet p : E 
R
be
sublinear
.
Banach Thm p.1
Let G  E be v.s.s
Let g  G
*
with g  p on G.
In Mazm  Orlicz Thm,
take S  G ,   g ,
 is identification map of G int o E
i.e.  ( s )  s  s  G
For any
finite subset
s1 , s2 ,, sk of G
and 1  0,  2  0,,  k  0
k
k
k
Thm
implies
Hahng
(
s
)

g
(

s
)

p
(

Mazm-Orlicz


i
i
i i
i si )
i 1
i 1
i 1
Banach
Thm p.2
 ( II ) in Mazm  Orlicz Thm hold
 (I )
i.e.  l  E with l  p on E s.t
*
g ( s )  l ( ( s ))  l ( s )  s  G
 g ( s)  l ( s)  s  G
 g ( s)  l ( s)  s  G
 g ( s)  l ( s)  s  G