A Short Proof of the Schröder-Simpson Theorem
A Short Proof of the Schr¨ oder-Simpson Theorem
A Short Proof of the Schr¨oder-Simpson Theorem Jean Goubault-Larrecq
Summer Topology Conference — July 23-26, 2013
A Short Proof of the Schr¨ oder-Simpson Theorem Continuous Valuations
Continuous Valuations Definition + Continuous valuation ν = map ν : O(X ) → R such that: ν(∅) U⊆V
ν(U ∪ V ) + ν(U ∩ V ) [↑ ν( Ui ) i∈I
=
0
(strictness)
⇒ ν(U) ≤ ν(V ) =
ν(U) + ν(V )
=
sup↑i∈I ν(Ui )
(monotonicity) (modularity) (continuity)
Similar to measures, except give weights to opens. Needed in semantics of programming languages [JonesPlotkin89] Under assumptions on X , cont. valuation=regular measure [KeimelLawson05].
A Short Proof of the Schr¨ oder-Simpson Theorem Continuous Valuations
The Weak Topology +
+
Let Rσ = R with Scott topology (opens = (t, +∞]). Definition Let V(X ) = set of continuous valuations of X , with the weak topology, R coarsest making ν 7→ x h(x)dν continuous, for every continuous h. Subbasic opens: Z [h > r ] = {ν ∈ V(X ) | h(x)dν > r } x
As the usual weak topology, except test functions h are + continuous to Rσ (=lsc).
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
A Riesz Representation Theorem Write
+
C(Y ) = space of continuous maps [Y → R ] (=lsc) Y ∗ = subspace of linear continuous maps
Theorem (Kirch93, Tix95) The following is a homeomorphism: C(X )∗ o
G 7−→ λU∈O(X )·G (χU ) λh∈C(X )·
R x
h(x)dν ←−[ ν
No condition on X at all Easy proof
/
V(X )
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
The Schr¨oder-Simpson Representation Theorem Write
+
C(Y ) = space of continuous maps [Y → R ] (=lsc) Y ∗ = subspace of linear continous maps
Theorem (Schr¨oderSimpson05) The following is an isomorphism of cones: C(X ) o
h 7−→ λν∈V(X )·
R x∈X
h(x)dν
/
V(X )∗
No condition on X at all Schr¨oder and Simpson’s proof [Simpson09] was elaborate
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
Topological Cones
Definition A cone (C , +, .) is as a vector space, except scalar product r .c is with non-negative reals r . Topological cone = cone with + : C × C → C , . : R+ σ × C → C continuous. Example: V(X ), weak topology Non-example: C(X ), Scott topology (unless X core-compact) . . . product not jointly continuous P P Linear maps: ψ( i ri νi ) = i ri ψ(νi ) for ri ≥ 0
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
The “obvious” way to a proof
C(X ) o
h 7−→ λν∈V(X )· ?
R x∈X
h(x)dν
/
V(X )∗ +
Main task: given ψ : V(X ) → R R linear continuous, find h ∈ C(X ) such that ψ(ν) = x∈X h(x)dν for every ν R If h exists, then ψ(δx ) = x h(x)dδx = h(x)
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
The “obvious” way to a proof
C(X ) o
h 7−→ λν∈V(X )·
R x∈X
h(x)dν
?
/
V(X )∗ +
Main task: given ψ : V(X ) → R R linear continuous, find h ∈ C(X ) such that ψ(ν) = x∈X h(x)dν for every ν R If h exists, then ψ(δx ) = x h(x)dδx = h(x)
So only one possible choice: h(x) = ψ(δx ) Need to show
Z
ψ(δx )dν = ψ(ν)
(1)
x
for every ν ∈ V(X )
. . . really hard.
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
The “obvious” way to a proof (2)
Let’s try and show (1): Z
ψ(δx )dν = ψ(ν)
x
Obvious if ν =
Pm
i=1 ai δxi
(simple)
Easy for ν quasi-simple (directed sup of simple valuations), by continuity Problem: not all continuous valuations are quasi-simple.
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
Previous proofs [Schr¨oderSimpson 05], [Simpson 09]: elaborate series of deep results and finely bounding inequalities [Keimel 12]: imitates classical proof of similar, measure-theoretic theorem; develops nice theory of quasi-uniform separation in (quasi-uniform) cones.
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
Previous proofs [Schr¨oderSimpson 05], [Simpson 09]: elaborate series of deep results and finely bounding inequalities [Keimel 12]: imitates classical proof of similar, measure-theoretic theorem; develops nice theory of quasi-uniform separation in (quasi-uniform) cones. Our proof: look at the S weakTopen ψ −1 (1, +∞] . . . must be of the form i∈I nj=1 [hij > rij ] . . . we can take rij = 1; . . . use Lemma 1 to eliminate unions . . . use Lemma 2 to eliminate intersections . . . so ψ −1 (1, +∞] = [h > 1]: this is the right h.
A Short Proof of the Schr¨ oder-Simpson Theorem The proof
Proof, Lemma 1 A surprising observation (mostly due to [Keimel12]) R R Lemma 1 (Linear cont. ψ cannot tell between sup and sup)
Let hi ∈ C(X ), i ∈R I . TFAE: (1) ψ(ν) ≥ Rsupi x hi (x)dν for every ν (2) ψ(ν) ≥ x supi hi (x)dν for every ν R R Note: supi not directed, supi x hi (x)dν 6= x supi hi (x)dν.
Proof (1/2). Only 1 ⇒ 2 is non-trivial.
Trick 1. Every hi directed sup of step Pfunctions so can assume each hi step, = k aik χUik Trick 2. supi∈I = sup↑J finite ⊆I supi∈J R by continuity of , can assume I finite.
A Short Proof of the Schr¨ oder-Simpson Theorem The proof
Proof, Lemma 1 A surprising observation (mostly due to [Keimel12]) R R Lemma 1 (Linear cont. ψ cannot tell between sup and sup)
Let hi ∈ C(X ), i ∈R I . TFAE: (1) ψ(ν) ≥ Rsupi x hi (x)dν for every ν (2) ψ(ν) ≥ x supi hi (x)dν for every ν Proof (2/2). Assume hi =
P
k
aik χUik , i ∈ I finite.
Let Cm be atoms of Boolean algebra of sets generated by Uik On eachR Cm , hi constant = aim so x supi hi (x)dν|Cm = max R i aim ν(Cm ) = supi x hi (x)dν|Cm R By 1, ψ(ν|Cm ) ≥ x supi hi (x)dν|Cm P Since ψ additive and ν = m ν|Cm , 2 follows.
U11 U12 U21 C1 C2 C3 C4 C5
U11 U12 U21 C1 C2 C3 C4 C5 U11 U12 U21 C1 C2 C3 C4 C5
U11 U12 U21 C1 C2 C3 C4 C5 U11 U12 U21 C1 C2 C3 C4 C5
U11 U12 U21 C1 C2 C3 C4 C5
U11 U12 U21 C1 C2 C3 C4 C5 U11 U12 U21 C1 C2 C3 C4 C5
A Short Proof of the Schr¨ oder-Simpson Theorem The proof
Proof, Lemma 1 A surprising observation (mostly due to [Keimel12]) R R Lemma 1 (Linear cont. ψ cannot tell between sup and sup)
Let hi ∈ C(X ), i ∈R I . TFAE: (1) ψ(ν) ≥ Rsupi x hi (x)dν for every ν (2) ψ(ν) ≥ x supi hi (x)dν for every ν Lemma 1, Corollary ψ −1 (1, +∞] =
S
i∈I [hi
> 1] ⇒ ψ −1 (1, +∞] = [h > 1] with h = supi hi .
Proof. ⊆ obvious. For ⊇:
R [hi > 1] ⊆ ψ −1 (1, +∞] implies ψ(ν) ≥ x hi (x)dν for all ν, i R So 1 holds, hence 2: ψ(ν) ≥ x h(x)dν. R If ν ∈ [h > 1], ψ(ν) ≥ x h(x)dν > 1, so ν ∈ ψ −1 (1, +∞].
A Short Proof of the Schr¨ oder-Simpson Theorem The proof
Proof, Lemma 2 Lemma 2 T If nj=1 [hj > 1] ⊆ ψ −1 (1, +∞], Proof, Lemma 2 T there is h ∈ C(X ) such that nj=1 [hj > 1]⊆ [h > 1] ⊆ ψ −1 (1,Lemma +∞] 2
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
Proof. R R Let F (ν) = ( x h1 (x)dν, · · · , x hn (x)dν)
A Short Proof of the Schr¨ oder-Simpson Theorem T Representation Theorems If nj=1 [hj
A Short Proof of the Schr¨ oder-Simpson Theorem
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
U
U21 C1 Lemma C2 C3 C42 Proof,
U
Representation 11 Theorems 12
in R
+n
1 (1, +1], > 1] ✓ T there is h 2 ¬X such that nj=1
CProof. 5
Let F (⌫) = (
Proof, Lemma 2 ⇤ = 12 ⇤ 1 ⇤ > 1 Proof, Lemma
R
x
h1 (x)d⌫, · ·
1
A = F (complement of convex non-empty
Lemma 2 U = (1, +1]n open T 1 (1, By separation If 2nj=1 [hj > 1] ✓ +1], Ptheorem (e.g Lemma 2 Lemma Tc )n = Tnthere is h 2 ¬X1 such that find ⇤(~ i ci linear T P [hij > 1] ✓ > +1], 1] ✓ (1, +1], j=1 If n [h >If1] ✓ [h1j(1,
P j=1 Separate by Λ(~c ) = j γj cj linear continuous h TnLet h = T is hthat 2 ¬X such that [h1] [h +1] > 1] j > there is h 2 there ¬X such [h > 1] ✓ j=1 [h > ✓1] ✓ (1, Proof. U RU U C CR C C (e.g. [Keimel 2006]) Let F (⌫) = ( x h1 (x)d⌫, · · · , x hn (x) Proof. Proof. R R R R P Let F (⌫) = (LethFA (x)d⌫, · ·x,hU1 (x)d⌫, h (x)d⌫) in (⌫) · · 1C ·of , ⇤R h1n1(1, (x)d⌫) U⇤ = C⇤+1]) C ==FU·((complement >C1in R xC Let h = j γj hj . Note 1 (1, +1]) A = F (complement of (1, +1]) A = Fconvex (complement of non-empty R P R convex non-empty non-empty ⇤n = 1 ⇤ 1 ⇤ > 1 U = (1, +1] open h(x)dν = j γj x hj (x)dν = Λ(F (ν)) U = (1, +1]convex x U open = (1, +1]n open By separation (e.g. [Keimel 2 Ptheorem By separation (e.g.theorem [Keimel 2006]),[Keimel 2006]), Inequalities follows. Bytheorem separation P find ⇤(~ ccontinuous ) = i (e.g. i ci linear continuous P find ⇤(~c ) = c linear j=1
j
n j=1
x
1
i i
i
j 11
11 1
x12 n
12
21
21
1
1
1
U11 U12 U21 =1
2
+n C 2 C 1
1
2
3
3
C3 C4 C5 U11
4
4
U12 U21
>1
=1
U
n
U11 U12 U21 C1 C2
A U11 U12 U21= 1C1 C2 1 C3 =1
Let h =
i i i⇤(~ c) = P i ci linear continuous P find LetP h = i i i hi i i hi
Let h =
i
i hi
1
1
>
A Short Proof of the Schr¨ oder-Simpson Theorem The proof
End of proof Let ψ ∈ V(X )∗ ,
write ψ −1 (1, +∞] as
Lemma 2 (recap) There is hi ∈ C(X ) s.t. So ψ −1 (1, +∞] =
Tn i
j=1 [hij
S
i∈I [hi
S
i∈I
Tni
j=1 [hij
> 1].
> 1] ⊆ [hi > 1] ⊆ ψ −1 (1, +∞]
> 1]
A Short Proof of the Schr¨ oder-Simpson Theorem The proof
End of proof Let ψ ∈ V(X )∗ ,
write ψ −1 (1, +∞] as
Lemma 2 (recap) There is hi ∈ C(X ) s.t. So ψ −1 (1, +∞] = Lemma 1, Corollary
Tn i
j=1 [hij
S
i∈I [hi
S
i∈I
Tni
j=1 [hij
> 1].
> 1] ⊆ [hi > 1] ⊆ ψ −1 (1, +∞]
> 1]
Then ψ −1 (1, +∞] = [h > 1] with h = supi hi . For all ν, t, ψ(ν) > t So ψ(ν) =
R
x h(x)dν.
iff ν/t ∈ ψ −1 (1, +∞] iff ν/t ∈ [h > 1]
iff
R
x
h(x)dν > t.
A Short Proof of the Schr¨ oder-Simpson Theorem The proof
Recap
Given ψ ∈ V(X )∗ , R there is an h ∈ C(X ) such that ψ(ν) = x h(x)dν (1) S T This h derived from the shape of weak opens i∈I nj=1 [hij > rij ] . . . and taking ν = δx in (1) implies h(x) = ψ(δx ).
A Short Proof of the Schr¨ oder-Simpson Theorem The proof
Recap We have proved: Theorem (Schr¨oderSimpson05) The following is an isomorphism of cones: C(X ) o
h 7−→ λν∈V(X )·
R x∈X
h(x)dν
λx·ψ(δx ) ←−[ ψ
/
V(X )∗
Notes: ∗ V(X )∗ ∼ = C(X ) (here) + V(X ) ∼ = C(X ) (Riesz-Kirch-Tix) ⇒ V(X ) and C(X ) are dual cones (C(X ) with weak∗ topology)
No assumption needed on X . Short proof Questions?
A Short Proof of the Schr¨oder-Simpson Theorem Jean Goubault-Larrecq
Summer Topology Conference — July 23-26, 2013
A Short Proof of the Schr¨ oder-Simpson Theorem Continuous Valuations
Continuous Valuations Definition + Continuous valuation ν = map ν : O(X ) → R such that: ν(∅) U⊆V
ν(U ∪ V ) + ν(U ∩ V ) [↑ ν( Ui ) i∈I
=
0
(strictness)
⇒ ν(U) ≤ ν(V ) =
ν(U) + ν(V )
=
sup↑i∈I ν(Ui )
(monotonicity) (modularity) (continuity)
Similar to measures, except give weights to opens. Needed in semantics of programming languages [JonesPlotkin89] Under assumptions on X , cont. valuation=regular measure [KeimelLawson05].
A Short Proof of the Schr¨ oder-Simpson Theorem Continuous Valuations
The Weak Topology +
+
Let Rσ = R with Scott topology (opens = (t, +∞]). Definition Let V(X ) = set of continuous valuations of X , with the weak topology, R coarsest making ν 7→ x h(x)dν continuous, for every continuous h. Subbasic opens: Z [h > r ] = {ν ∈ V(X ) | h(x)dν > r } x
As the usual weak topology, except test functions h are + continuous to Rσ (=lsc).
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
A Riesz Representation Theorem Write
+
C(Y ) = space of continuous maps [Y → R ] (=lsc) Y ∗ = subspace of linear continuous maps
Theorem (Kirch93, Tix95) The following is a homeomorphism: C(X )∗ o
G 7−→ λU∈O(X )·G (χU ) λh∈C(X )·
R x
h(x)dν ←−[ ν
No condition on X at all Easy proof
/
V(X )
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
The Schr¨oder-Simpson Representation Theorem Write
+
C(Y ) = space of continuous maps [Y → R ] (=lsc) Y ∗ = subspace of linear continous maps
Theorem (Schr¨oderSimpson05) The following is an isomorphism of cones: C(X ) o
h 7−→ λν∈V(X )·
R x∈X
h(x)dν
/
V(X )∗
No condition on X at all Schr¨oder and Simpson’s proof [Simpson09] was elaborate
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
Topological Cones
Definition A cone (C , +, .) is as a vector space, except scalar product r .c is with non-negative reals r . Topological cone = cone with + : C × C → C , . : R+ σ × C → C continuous. Example: V(X ), weak topology Non-example: C(X ), Scott topology (unless X core-compact) . . . product not jointly continuous P P Linear maps: ψ( i ri νi ) = i ri ψ(νi ) for ri ≥ 0
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
The “obvious” way to a proof
C(X ) o
h 7−→ λν∈V(X )· ?
R x∈X
h(x)dν
/
V(X )∗ +
Main task: given ψ : V(X ) → R R linear continuous, find h ∈ C(X ) such that ψ(ν) = x∈X h(x)dν for every ν R If h exists, then ψ(δx ) = x h(x)dδx = h(x)
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
The “obvious” way to a proof
C(X ) o
h 7−→ λν∈V(X )·
R x∈X
h(x)dν
?
/
V(X )∗ +
Main task: given ψ : V(X ) → R R linear continuous, find h ∈ C(X ) such that ψ(ν) = x∈X h(x)dν for every ν R If h exists, then ψ(δx ) = x h(x)dδx = h(x)
So only one possible choice: h(x) = ψ(δx ) Need to show
Z
ψ(δx )dν = ψ(ν)
(1)
x
for every ν ∈ V(X )
. . . really hard.
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
The “obvious” way to a proof (2)
Let’s try and show (1): Z
ψ(δx )dν = ψ(ν)
x
Obvious if ν =
Pm
i=1 ai δxi
(simple)
Easy for ν quasi-simple (directed sup of simple valuations), by continuity Problem: not all continuous valuations are quasi-simple.
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
Previous proofs [Schr¨oderSimpson 05], [Simpson 09]: elaborate series of deep results and finely bounding inequalities [Keimel 12]: imitates classical proof of similar, measure-theoretic theorem; develops nice theory of quasi-uniform separation in (quasi-uniform) cones.
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
Previous proofs [Schr¨oderSimpson 05], [Simpson 09]: elaborate series of deep results and finely bounding inequalities [Keimel 12]: imitates classical proof of similar, measure-theoretic theorem; develops nice theory of quasi-uniform separation in (quasi-uniform) cones. Our proof: look at the S weakTopen ψ −1 (1, +∞] . . . must be of the form i∈I nj=1 [hij > rij ] . . . we can take rij = 1; . . . use Lemma 1 to eliminate unions . . . use Lemma 2 to eliminate intersections . . . so ψ −1 (1, +∞] = [h > 1]: this is the right h.
A Short Proof of the Schr¨ oder-Simpson Theorem The proof
Proof, Lemma 1 A surprising observation (mostly due to [Keimel12]) R R Lemma 1 (Linear cont. ψ cannot tell between sup and sup)
Let hi ∈ C(X ), i ∈R I . TFAE: (1) ψ(ν) ≥ Rsupi x hi (x)dν for every ν (2) ψ(ν) ≥ x supi hi (x)dν for every ν R R Note: supi not directed, supi x hi (x)dν 6= x supi hi (x)dν.
Proof (1/2). Only 1 ⇒ 2 is non-trivial.
Trick 1. Every hi directed sup of step Pfunctions so can assume each hi step, = k aik χUik Trick 2. supi∈I = sup↑J finite ⊆I supi∈J R by continuity of , can assume I finite.
A Short Proof of the Schr¨ oder-Simpson Theorem The proof
Proof, Lemma 1 A surprising observation (mostly due to [Keimel12]) R R Lemma 1 (Linear cont. ψ cannot tell between sup and sup)
Let hi ∈ C(X ), i ∈R I . TFAE: (1) ψ(ν) ≥ Rsupi x hi (x)dν for every ν (2) ψ(ν) ≥ x supi hi (x)dν for every ν Proof (2/2). Assume hi =
P
k
aik χUik , i ∈ I finite.
Let Cm be atoms of Boolean algebra of sets generated by Uik On eachR Cm , hi constant = aim so x supi hi (x)dν|Cm = max R i aim ν(Cm ) = supi x hi (x)dν|Cm R By 1, ψ(ν|Cm ) ≥ x supi hi (x)dν|Cm P Since ψ additive and ν = m ν|Cm , 2 follows.
U11 U12 U21 C1 C2 C3 C4 C5
U11 U12 U21 C1 C2 C3 C4 C5 U11 U12 U21 C1 C2 C3 C4 C5
U11 U12 U21 C1 C2 C3 C4 C5 U11 U12 U21 C1 C2 C3 C4 C5
U11 U12 U21 C1 C2 C3 C4 C5
U11 U12 U21 C1 C2 C3 C4 C5 U11 U12 U21 C1 C2 C3 C4 C5
A Short Proof of the Schr¨ oder-Simpson Theorem The proof
Proof, Lemma 1 A surprising observation (mostly due to [Keimel12]) R R Lemma 1 (Linear cont. ψ cannot tell between sup and sup)
Let hi ∈ C(X ), i ∈R I . TFAE: (1) ψ(ν) ≥ Rsupi x hi (x)dν for every ν (2) ψ(ν) ≥ x supi hi (x)dν for every ν Lemma 1, Corollary ψ −1 (1, +∞] =
S
i∈I [hi
> 1] ⇒ ψ −1 (1, +∞] = [h > 1] with h = supi hi .
Proof. ⊆ obvious. For ⊇:
R [hi > 1] ⊆ ψ −1 (1, +∞] implies ψ(ν) ≥ x hi (x)dν for all ν, i R So 1 holds, hence 2: ψ(ν) ≥ x h(x)dν. R If ν ∈ [h > 1], ψ(ν) ≥ x h(x)dν > 1, so ν ∈ ψ −1 (1, +∞].
A Short Proof of the Schr¨ oder-Simpson Theorem The proof
Proof, Lemma 2 Lemma 2 T If nj=1 [hj > 1] ⊆ ψ −1 (1, +∞], Proof, Lemma 2 T there is h ∈ C(X ) such that nj=1 [hj > 1]⊆ [h > 1] ⊆ ψ −1 (1,Lemma +∞] 2
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
Proof. R R Let F (ν) = ( x h1 (x)dν, · · · , x hn (x)dν)
A Short Proof of the Schr¨ oder-Simpson Theorem T Representation Theorems If nj=1 [hj
A Short Proof of the Schr¨ oder-Simpson Theorem
A Short Proof of the Schr¨ oder-Simpson Theorem Representation Theorems
U
U21 C1 Lemma C2 C3 C42 Proof,
U
Representation 11 Theorems 12
in R
+n
1 (1, +1], > 1] ✓ T there is h 2 ¬X such that nj=1
CProof. 5
Let F (⌫) = (
Proof, Lemma 2 ⇤ = 12 ⇤ 1 ⇤ > 1 Proof, Lemma
R
x
h1 (x)d⌫, · ·
1
A = F (complement of convex non-empty
Lemma 2 U = (1, +1]n open T 1 (1, By separation If 2nj=1 [hj > 1] ✓ +1], Ptheorem (e.g Lemma 2 Lemma Tc )n = Tnthere is h 2 ¬X1 such that find ⇤(~ i ci linear T P [hij > 1] ✓ > +1], 1] ✓ (1, +1], j=1 If n [h >If1] ✓ [h1j(1,
P j=1 Separate by Λ(~c ) = j γj cj linear continuous h TnLet h = T is hthat 2 ¬X such that [h1] [h +1] > 1] j > there is h 2 there ¬X such [h > 1] ✓ j=1 [h > ✓1] ✓ (1, Proof. U RU U C CR C C (e.g. [Keimel 2006]) Let F (⌫) = ( x h1 (x)d⌫, · · · , x hn (x) Proof. Proof. R R R R P Let F (⌫) = (LethFA (x)d⌫, · ·x,hU1 (x)d⌫, h (x)d⌫) in (⌫) · · 1C ·of , ⇤R h1n1(1, (x)d⌫) U⇤ = C⇤+1]) C ==FU·((complement >C1in R xC Let h = j γj hj . Note 1 (1, +1]) A = F (complement of (1, +1]) A = Fconvex (complement of non-empty R P R convex non-empty non-empty ⇤n = 1 ⇤ 1 ⇤ > 1 U = (1, +1] open h(x)dν = j γj x hj (x)dν = Λ(F (ν)) U = (1, +1]convex x U open = (1, +1]n open By separation (e.g. [Keimel 2 Ptheorem By separation (e.g.theorem [Keimel 2006]),[Keimel 2006]), Inequalities follows. Bytheorem separation P find ⇤(~ ccontinuous ) = i (e.g. i ci linear continuous P find ⇤(~c ) = c linear j=1
j
n j=1
x
1
i i
i
j 11
11 1
x12 n
12
21
21
1
1
1
U11 U12 U21 =1
2
+n C 2 C 1
1
2
3
3
C3 C4 C5 U11
4
4
U12 U21
>1
=1
U
n
U11 U12 U21 C1 C2
A U11 U12 U21= 1C1 C2 1 C3 =1
Let h =
i i i⇤(~ c) = P i ci linear continuous P find LetP h = i i i hi i i hi
Let h =
i
i hi
1
1
>
A Short Proof of the Schr¨ oder-Simpson Theorem The proof
End of proof Let ψ ∈ V(X )∗ ,
write ψ −1 (1, +∞] as
Lemma 2 (recap) There is hi ∈ C(X ) s.t. So ψ −1 (1, +∞] =
Tn i
j=1 [hij
S
i∈I [hi
S
i∈I
Tni
j=1 [hij
> 1].
> 1] ⊆ [hi > 1] ⊆ ψ −1 (1, +∞]
> 1]
A Short Proof of the Schr¨ oder-Simpson Theorem The proof
End of proof Let ψ ∈ V(X )∗ ,
write ψ −1 (1, +∞] as
Lemma 2 (recap) There is hi ∈ C(X ) s.t. So ψ −1 (1, +∞] = Lemma 1, Corollary
Tn i
j=1 [hij
S
i∈I [hi
S
i∈I
Tni
j=1 [hij
> 1].
> 1] ⊆ [hi > 1] ⊆ ψ −1 (1, +∞]
> 1]
Then ψ −1 (1, +∞] = [h > 1] with h = supi hi . For all ν, t, ψ(ν) > t So ψ(ν) =
R
x h(x)dν.
iff ν/t ∈ ψ −1 (1, +∞] iff ν/t ∈ [h > 1]
iff
R
x
h(x)dν > t.
A Short Proof of the Schr¨ oder-Simpson Theorem The proof
Recap
Given ψ ∈ V(X )∗ , R there is an h ∈ C(X ) such that ψ(ν) = x h(x)dν (1) S T This h derived from the shape of weak opens i∈I nj=1 [hij > rij ] . . . and taking ν = δx in (1) implies h(x) = ψ(δx ).
A Short Proof of the Schr¨ oder-Simpson Theorem The proof
Recap We have proved: Theorem (Schr¨oderSimpson05) The following is an isomorphism of cones: C(X ) o
h 7−→ λν∈V(X )·
R x∈X
h(x)dν
λx·ψ(δx ) ←−[ ψ
/
V(X )∗
Notes: ∗ V(X )∗ ∼ = C(X ) (here) + V(X ) ∼ = C(X ) (Riesz-Kirch-Tix) ⇒ V(X ) and C(X ) are dual cones (C(X ) with weak∗ topology)
No assumption needed on X . Short proof Questions?
