The equivalence between a standard H-cone of functions on a saturated set and Exc U

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1 The equivalence between a standard H-cone of functions on a saturated set and Exc U D. Mărginean, "Petru Maior" University of Tg. Mureş Abstract In this paper we show that if S is a standard H-cone of functions on a saturated set X then there exists a proper, sub-markovian resolvent of kernels U = (U α ) α>0 on (X, B) such that S Exc U. Keywords: H-cone, standard H-cone, excessive measure. 1 Introduction We know that if V = (V α ) α>0 is a proper, sub-markovian resolvent on (E, B), absolutely continous with respect to a positive, finite measure such that initial kernel V is proper then the convex cone E V pb of measurable, V-excessive functions is a standard H-cone and for any standard H-cone of functions on a saturated set X and p S a nearly continous it exists a finite measure m and a sub-markovian resolvent absolutely continous with respect to m, V = (V α ) α>0 on (X, B) (B is Borel sets of natural topology on X) such that S = E V pb and V 1 = p, where V is initial kernel of V. An ordered convex cone S is called H-cone if the following axioms are satisfied: 1. For any non-empty family A S there exists A and we have s + A = (s + A), for any s S; 2. For any increasing and dominated family A S there exists A and we have (A + u) = A + u, for any u S; 3. S satisfies the Riesz decomposition property i.e. for any s, s 1, s 2 S such that s s 1 + s 2, there exist t 1, t 2 S satisfying s = t 1 + t 2, t 1 s 1, t 2 s 2. 1

2 Definition 1. A convex cone S of positive numerical functions on a set X is called H-cone of functions (on X) if the following conditions are satisfied: 1. For any s, t, u S such that s + u t + u we have s t. 2. For any increasing family F S dominated by an element of S the function F belongs to S. 3. For any non-empty family F S there exists the greatest lower bound F in S and we have (s + F ) = s + F for any s S. 4. The Riesz decomposition property holds in s. 5. inf(x, t + α) S for any s, t S and for any α R S separates the points of X and for any x X, there exists s S such that s(x) > 0. Definition 2. An element u S is called weak unit (or strictly positive) if we have s = n N(s nu) for any s S. Definition 3. Let u S be a weak unit. An element s S is called u-continuous if for any ε > 0 and for any family F S increasing to s there exists t F such that s t + εu. An element s S is called universally continuous if it is v-continuous for any weak unit v S. Definition 4. A H-cone S is called standard H-cone if the following conditions are fulfilled: S possesses a weak unit There exists a countable subset of S 0 which is increasingly dense 2

3 We know that any standard H-cone is isomorphic to a H-cone of functions. Let (E, B) be a Lusin measurable space and U = (U α ) α>0 be a proper sub- Markovian resolvent of kernels on (E, B) such that the set of all B-measurable, U-excessive functions is min-stable, contains the positive constant functions and generates B. We denote by Exc U the set of all σ-finite U-excessive measures. It remind that a U-excessive measure on (E, B) is a positive measure ξ on (E, B) provided ξ(αv α f) ξ(f) for all f pb and α > 0. If µ is a positive measure on (E, B) then µ U is a U-excessive measure. A measure ξ Exc U is called potential provided there exists a measure µ on (E, B) such that ξ = µ U. We know that for any family (ξ i ) U-excessive measures ξ i is excessive and if family is an increasing and dominated by an excessive measure then ξ i is a U-excessive. For every σ-finite measure µ on (E, B) such that there exists ξ Exc U with µ ξ, we put R(µ) = {ξ Exc U ξ µ}. If ξ 1, ξ 2 Exc U and ξ 2 ξ 1 then exists ξ Exc U such that R(ξ 1 ξ 2 ) + ξ = ξ 1. Therefore Exc U is H-cone and particularly cone of potentials. 2 S Exc U Teorema 1. Let S a standard H-cone of functions on a saturated set X such that S = S. Then it exists a sub-markovian, proper resolvent U = (U α ) α>0 absolutely continuous with respect to a finite measure m, such that S Exc U. Proof. We know from (T [2]), that exists a finite measure m and two sub- Markovian resolvents on (X, B) V, W, which are in duality and absolutely continuous with respect to m, such that initial kernels V and W are bounded, V 1 > 0 and S = E V pb. If ξ Exc W ξ m. Let m(f) = 0 W f = 0 L(ξ, W f) = 0 ξ(f) = 0. Hence ξ m. It follows that (as in. P [1]) there exists s E V pb unique and finite m-a.e., such that ξ = s m. We have that L(s m, W g) = (s m)(g) = m(s g), for all g pb. But [s, W g] = m(s g) (as in T [2]). It follows that L(s m, W g) = [s, W g], for all 3

4 g pb, s E V pb. Therefore L(s m, t) = [s, t], for all t E W pb, s E V pb. Hence L(ξ, t) = [s, t], for all t E W pb, ξ Exc W. Let ξ Exc W fixed. We know that the map E W pb ϕ R +, ϕ(t) = L(ξ, t) is from (E W pb) and (as in P [2]) it follows that there exists uniquely map s E V pb, such that L(ξ, t) = [s, t], for all t E V. From above it follows that [s, t] = [s, t], for all t E W pb. Hence (as in T [2]) s = s. Therefore for everyξ Exc W pb it exists and is uniquely s E V pb, such that ξ = s m. Conversely for every s E V pb, we have that s m Exc W. We can define a map We show that: γ : Exc W E V pb, γ(ξ) = s. 1. γ(ξ 1 + ξ 2 ) = γ(ξ 1 ) + γ(ξ 2 ), ξ 1, ξ 2 Exc W ; 2. ξ 1 ξ 2 γ(ξ 1 ) γ(ξ 2 ), ξ 1, ξ 2 Exc W ; 3. For any increasing family (ξ i ) Exc W to a measure ξ Exc W we have that γ(ξ i ) γ(ξ). 1. Let ξ 1, ξ 2 Exc W. From above it results that there exists unique s 1, s 2 E V pb such that ξ 1 = s 1 m andξ 2 = s 2 m. Then γ(ξ 1 +ξ 2 ) = s 1 +s 2 = γ(ξ 1 )+γ(ξ 2 ). 2. Assume ξ 1, ξ 2 Exc W, such that ξ 1 ξ 2. It results that there exists s 1, s 2 E V pb unique such that s 1 m s 2 m s 1 s 2 m-a.e. Hence s 1, s 2 E V pb it results (as in T b [2]), s 1 s 2, i.e. γ(ξ 1 ) γ(ξ 2 ). Conversely assume γ(ξ 1 ) γ(ξ 2 ), i.e. s 1 s 2 s 1 m s 2 m ξ 1 ξ Let (ξ i ) Exc W an increasing family to a measure ξ Exc W. Then ( ) (s i ) E V pb, s E V pb unique such that ξ i = s i m, for all i I and ξ = s m. From s i m s m αv α s i αv α s, for all α > 0. It follows that s i = α>0 αv α s i = α>0 αv α s i = αv α s = s. α Therefore γ(ξ i ) = γ(ξ). From above it results that S Exc W. We take U = W = (W α ) α>0 S Exc U. 4

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