Dedicated to Professor Philippe G. Ciarlet on his 70th birthday VECTOR SEMINORMS, SPACES WITH VECTOR NORM, AND REGULAR OPERATORS ROMULUS CRISTESCU The rst section of this paper deals with the properties of some type of vector seminorms. In the second section, vector spaces with vector norm are considered while in the third section some theorems about the regular operators are given. Operators dened on a vector space with vector norm are considered in the fourth section. The terminology is that of [3]. AMS 2000 Subject Classication: 47860. Key words: vector seminorm, regular operators. 1. VECTOR SEMINORMS If X is a vector space over R and Y an ordered vector space, a mapping P of X into Y is said to be a vector seminorm if P (x1 + x2) ≤ P (x1) + P (x2), ∀x1, x2 ∈ X, P (αx) = |α| P (x), ∀α ∈ R, ∀x ∈ X. If X and Y are ordered vector spaces, a vector seminorm P : X → Y is said to be monotone if 0 ≤ a ≤ b ∈ X ⇒ P (a) ≤ P (b) while it is said to be absolute monotone if ± x ≤ c ∈ X ⇒ P (x) ≤ P (c). An ordered vector space is said to be (o)-complete if every majorized subset has a supremum. The following theorem generalize some results given in [7] for real semi- norms. Theorem 1.1. Let X be a directed vector space, G a majorized vector subspace of X, and Y an (o)-complete ordered vector space. If P0 : G → Y is REV. ROUMAINE MATH. PURES APPL., 53 (2008), 56, 407418 408 Romulus Cristescu 2 a vector seminorm, then by putting (1.1) P (x) = inf {P0(a) + P0(z) | a, z ∈ G; ± (x − z) ≤ a} , ∀x ∈ X, the following statements hold. (i) The operator P : X → Y given by (1.1) is an absolute monotone vector seminorm and P (z) ≤ P0(z), ∀z ∈ G. (ii) If Q : X → Y is an absolute monotone vector seminorm and if Q(z) ≤ P0(z), ∀z ∈ G, then Q(x) ≤ P (x), ∀x ∈ X. (iii) We have P | G = P0 if and only if the vector seminorm P0 is absolute monotone. Proof. (i) For any x ∈ X there exists the element P (x) given by (1.1). Indeed, since G is a majorizing vector subspace of the directed vector space X, for any elements x ∈ X and z ∈ G there exists an element a ∈ G such that ± (x − z) ≤ a. On the other hand, we have P0 (c) ≥ 0, ∀c ∈ G, and the space Y is (o)-complete. As in the case of real seminorms, it is easily veried that the operator P given by (1.1) is a vector seminorm. We also have (1.2) P (x) = inf {P0 (b) | x ≤ b ∈ G} , ∀x ∈ X+. Indeed, denoting by P 0(x) the right-hand side in (1.2), we obviously have 0 P (x) ≤ P (x). Let now x ∈ X+ and ± (x − z) ≤ a with a, z ∈ G. Then 0 ≤ x ≤ a + z ∈ G, whence 0 P (x) ≤ P0 (a + z) ≤ P0 (a) + P0(z). So, P 0(x) ≤ P (x). Therefore (1.2) holds. The vector seminorm P is absolute monotone. Indeed, if ± x ≤ v in X then, taking a ∈ G such that v ≤ a, it follows from ± x ≤ a that P (x) ≤ P0(a) by (1.1). Then, by (1.2), we get P (x) ≤ P (v). Obviously P (z) ≤ P0(z), ∀z ∈ G. (ii) Let Q : X → Y be an absolute monotone vector seminorm such that Q(z) ≤ P0(z), ∀z ∈ G. If x ∈ X and ± (x − z) ≤ a with a, z ∈ G, then Q(x) ≤ Q(x − z) + Q(z) ≤ Q(a) + Q(z) ≤ P0(a) + P0(z) whence Q(x) ≤ P (x). (iii) Suppose that P0 is absolute monotone and let x ∈ G. Let also, z, a ∈ G such that ± (x − z) ≤ a. We have P0(x − z) ≤ P0(a) whence, as in the preceding case regarding the vector seminorm Q, we obtain P0(x) ≤ P (x). We also have P (x) ≤ P0(x) by statement (i). Conversely, if then is obviously absolute monotone. P | G = P0 P0 3 Vector seminorms, spaces with vector norm, and regular operators 409 Denition 1.1. If X is a directed vector space and Y an (o)-complete ordered vector space, a vector seminorm P : X → Y is said to be solid if P (x) = inf {P (a) | ± x ≤ a ∈ X} , ∀x ∈ X. Remark 1.1. Let X be a directed vector space and Y an (o)-complete ordered vector space. If G is a majorizing vector subspace of X and P0 : G → Y a solid vector seminorm, then the formula P (x) = inf{P0(a) | ± x ≤ a ∈ G}, ∀x ∈ X, yields a solid vector seminorm P : X → Y such that P | G = P0. Remark 1.2. Let X be a vector lattice and Y an (o)-complete ordered vector space. A vector seminorm P : X → Y is solid if and only if |x1| ≤ |x2| in X implies P (x1) ≤ P (x2). This assertion is veried as in the case Y = R. Remark 1.3. Let X be a vector lattice, Y an (o)-complete ordered vector space and P : X → Y a solid vector seminorm. Then P is additive on the positive cone X+, if and only if P is of the form (1.3) P (x) = U(|x|), ∀x ∈ X, where U : X → Y is a positive linear operator. Indeed, if P is additive on X+ then, by putting, U(x) = P (x+) − P (x−), ∀x ∈ X, we obtain a positive linear operator U : X → Y for which (1.3) holds. Con- versely, if U : X → Y is a positive linear operator, then by (1.3) we obviously obtain a solid vector seminorm. We recall now that if X and Y are vector lattices, an operator U : X → Y is said [3] to be disjunctive if x1 ⊥ x2 in X implies U(x1) ⊥ U(x2). Theorem 1.2. If X and Y are vector lattices and P : X → Y is a monotone and disjunctive vector seminorm, then (1.4) P (x1 ∧ x2) = P (x1) ∧ P (x2), ∀x1, x2 ∈ X+, and if x1 ∧ x2 = 0 in X, then (1.5) P (x1 ∨ x2) = P (x1) ∨ P (x2). Proof. Let x1, x2 ∈ X+. We have (x1 − (x1 ∧ x2)) ∧ (x2 − (x1 ∧ x2)) = 0, whence (1.6) P (x1 − (x1 ∧ x2)) ∧ P (x2 − (x1 ∧ x2)) = 0. 410 Romulus Cristescu 4 On the other hand 0 ≤ P (xi) − P (x1 ∧ x2) ≤ P (xi − (x1 ∧ x2)), i = 1, 2, and, by (1.6), (P (x1) − P (x1 ∧ x2)) ∧ (P (x2) − P (x1 ∧ x2)) = 0. Therefore, (1.4) holds. If x1 ∧ x2 = 0 then x1 ∨ x2 = x1 + x2, hence P (x1 ∨ x2) = P (x1 + x2) ≤ P (x1) + P (x2) = P (x1) ∨ P (x2) since P (x1) ⊥ P (x2). On the other hand, we have P (x1) ∨ P (x2) ≤ P (x1 ∨ x2), so that (1.5) holds. Remark 1.4. Let X and Y be vector lattices and let P : X → Y be a solid vector seminorm. Then P is disjunctive if and only if x1 ∧ x2 = 0 in X implies P (x1) ∧ P (x2) = 0. Theorem 1.3. Let X be a vector lattice and Y an (o)-complete ordered vector space. If P : X → Y is a (non-zero) solid vector seminorm, then there exists a (non-zero) positive linear operator U : X → Y such that (1.7) |U(x)| ≤ P (x), ∀x ∈ X. Proof. The proof is similar to that given in the case of functionals [4]. By putting Q(x) = P (x+), ∀x ∈ X, we obtain a sublinear operator Q : X → Y . Let us consider an element a > 0 of X such that P (a) > 0. By a corollary of the Hahn-Banach-Kantorovich theorem [5] there exists a linear operator U : X → Y such that U(a) = Q(a) and U(x) ≤ Q(x), ∀x ∈ X. It is easy to see that U is positive and that inequality (1.7) holds. Theorem 1.4. Let X be a vector lattice, Y an (o)-complete ordered vector space and P (X, Y ) the set of all solid vector seminorms mapping X into Y . If Pi ∈ P(X, Y ), i = 1, 2, then, with respect to the pointwise order in the set P(X, Y ), there exists P1 ∧ P2 in this set, and we have (P1 ∧ P2)(x) = inf {P1(a) + P2(b) | a, b ∈ X+; a + b = |x|} , ∀x ∈ X. The proof is similar to that given in the case Y = R (see [2]). 5 Vector seminorms, spaces with vector norm, and regular operators 411 2. SPACES WITH VECTOR NORM Let F be a real vector space and X a vector lattice. A vector seminorm P : F → X is said to be a vector norm if P (x) = 0 implies x = 0. If the vector space F is endowed with a vector norm P : F → X, then F is called a v-normed vector space and is denoted by . We shall also write FX kfkX = P (f). Let FX be a v-normed vector space. A subset A of FX is said to be (v)-bounded if there exists x0 ∈ X such that kfkX ≤ x0, ∀f ∈ A. A sequence of elements of the space is said to be a - (fn)n∈N FX (v) Cauchy sequence if there exists a sequence of elements of such that (an)n∈N X an ↓ 0 and n∈N (2.1) kfi − fjkX ≤ an, ∀i, j ≥ n. We say that a sequence -converges to an element (fn)n∈N (v) f ∈ FX if the sequence (kfn − fk ) (o)-converges (i.e.