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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 0C 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 t0 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 R2 Hence R2 Q (I ) i 1, P.2 xi E Consider C R2 Q ( R 2 v ) vQ 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 R2 , v Q i.e f (u ) f ( v ) u R2 , 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 r0 l ( x ) 0 r0 yP (*) 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