Vector Seminorms, Spaces with Vector Norm, and Regular Operators

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 seminorms. 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. converges with respect to X n∈N the order) to 0 in the space X. The element f is called the (v)-limit of the sequence (fn)n∈ , and we write f = (v)-lim fn. N n The space FX is said to be (v)-complete if any (v)-Cauchy sequence of elements of FX (v)-converges to an element of the space. The space FX is called a strictly v-normed vector space if whenever with , there exists , kfkX = a1 + a2 ai ∈ X+, i = 1, 2 fi ∈ FX , i = 1, 2 such that , and kfikX = ai, i = 1, 2 f = f1 + f2. A strictly v-normed vector space which is also (v)-complete is said to be a space of type (BK ) . The notion of a (v)-convergent series in a v-normed vector space can be introduced in a natural manner: ∞ n X X (v)- fn = (v)- lim fi n n=1 i=1 if the right-hand member (v)-limit exists. If FX is a v-normed vector space with respect to an Archimedean vector lattice , a sequence of elements of is said to -converge to an X (fn)n∈N FX (vρ) element f ∈ FX if the sequence (kfn − fk ) (ρ)-converges (i.e. converges X n∈N with regulator [3]) to 0 in the space X. Remark 2.1. If a -Cauchy sequence of elements of a -normed (v) (fn)n∈N v vector space has a -convergent subsequence, then the sequence is (v) (fn)n∈N (v)-convergent. In the next theorem, by regular space we mean (as in [4]) an Archimedean vector lattice in which any (o)-convergent sequence is also (ρ)-convergent. 412 Romulus Cristescu 6

Theorem 2.1. Let FX be a v-normed vector space and consider the following two conditions:

(i) the space FX is (v)-complete; ∞ (ii) if a series of the form P , , is -convergent, then kfnkX fn ∈ FX (o) n=1 ∞ P the series fn is (v)-convergent. n=1 Condition (i) implies (ii), and if X is a σ-complete vector lattice which is also a regular space, then conditions (i) and (ii) are equivalent. Proof. Suppose that condition (i) is satised and let be a sequence (fn)n∈N of elements of such that the series P be -convergent. Denoting FX kfnkX (o) n n X sn = fi, n ∈ N, i=1 there exists a sequence of elements of such that and (wn)n∈N X wn ↓ 0 n∈N

ksn − smkX ≤ wm, m, n ∈ N; m ≤ n. Therefore, is a -Cauchy sequence and since is -complete, (sn)n∈N (v) FX (v) the sequence is -convergent. Consequently, (i) implies (ii). (sn)n∈N (v) Suppose now that X is a σ-complete vector lattice which also is a regular space and suppose that condition (ii) is satised. Let be a -Cauchy (fn)n∈N (v) sequence of elements of Let be a sequence of elements of such FX . (an)n∈N X that an ↓ 0 and n∈N

(2.1) kfi − fjkX ≤ an, i, j ≥ n. Since is a regular space, there exists and a sequence X w ∈ X+ (εn)n∈N of real numbers, such that εn ↓ 0 and an ≤ εnw, ∀n ∈ N. According to (2.1) n∈ there exists a strictly increasing sequenceN of natural numbers such that (jn)n∈N 1 (2.2) fj − fjn ≤ w, ∀n ∈ . n+1 X 2n N Since the vector lattice X is σ-complete, it follows from (2.2) that the series corresponding to the sequence is -convergent. (kfjn+1 − fjn kX )n∈N (o) By hypothesis, the series P is -convergent. It follows from (fjn+1 − fjn ) (v) n the equation n X fjn+1 = fj1 + fjk+1 − fjk , ∀n ∈ N, k=1 7 Vector seminorms, spaces with vector norm, and regular operators 413 that the sequence is -convergent while, by Remark 2.1, the se- (fjn )n∈N (v) quence is -convergent. Consequently, if is a -complete vector (fn)n∈N (v) X σ lattice which also is a regular space, then conditions (i) and (ii) are equivalent. Theorem 2.2. Let X and Y be complete vector lattices and G a vector sublattice of X endowed with a solid vector norm with values in Y . If X is a Dedekind extension of G, then the vector norm given on G can be extended to a solid vector norm on X by the formula

(2.3) kxkY = inf {kakY | |x| ≤ a ∈ G} , ∀x ∈ X.

Proof. If x1, x2 ∈ X then, denoting

A = {kakY | a = b + c, |x1| ≤ b ∈ G, |x2| ≤ c ∈ G} ,

B = {kbkY | |x1| ≤ b ∈ G} ,C = {kckY | |x2| ≤ c ∈ G} , we have

kx1 + x2kY ≤ inf A ≤ inf (B + C) = inf B + inf C = kx1kY + kx2kY . On the other hand, one can easily verify that

kαxkY = |α| kxkY , ∀α ∈ R, ∀x ∈ X. If 0 6= x ∈ X then there exists b ∈ G such that 0 < b ≤ |x|. If |x| ≤ a ∈ G then 0 < b ≤ a, hence

0 < kbkY ≤ kakY . It follows is immediately that . Therefore, by (2.3), we obtain a kxkY 6= 0 vector norm on X. It is easily veried that this vector norm is solid and that it is an extension of the vector norm given on G.

3. REGULAR OPERATORS

Let X be a vector lattice and Y a complete vector lattice. We shall denote by R(X,Y ) the set of all regular operators which map X into Y . It is known that R(X,Y ) is a complete vector lattice with respect to the usual operations and the order given by the cone of positive operators. We shall also denote r X = R (X, R) . As in the rst section (see Theorem 1.4) we denote by P(X,Y ) the set of all solid vector seminorms on X into Y and shall consider the pointwise order in this set.

Theorem 3.1. Let Ui ∈ R(X,Y ) and Pi ∈ P(X,Y ), i = 1, 2, such that

(3.1) |Ui(x)| ≤ Pi(x), ∀x ∈ X, i = 1, 2. 414 Romulus Cristescu 8

Then the inequality

(3.2) |(|U1| ∧ |U2|)(x)| ≤ (P1 ∧ P2)(x), ∀x ∈ X. holds. Proof. It follows from (3.1) that

| |Ui| (x)| ≤ Pi(x), ∀x ∈ X, i = 1, 2, and by Theorem 1.4 we have

|(|U1| ∧ |U2|)(x)| ≤ (|U1| ∧ |U2|)(|x|) =

= inf {|U1| (a) + |U2| (b) | a, b ∈ X+; a + b = |x|} ≤

≤ inf {P1(a) + P2(b) | a, b ∈ X+; a + b = |x|} = (P1 ∧ P2)(x). Therefore, (3.2) holds. Denition 3.1. Let p : X → R be a seminorm. An operator U : X → Y is said to be p-bounded if there exists y0 ∈ Y such that |U(x)| ≤ p(x)y0, ∀x ∈ X. Remark 3.1. Theorem 3.1 implies the following proposition given in [2]. If

U1,U2 ∈ R (X,Y ), p1, p2 are solid real seminorms on X and Ui is pi-bounded, i = 1, 2, then |U1| ∧ |U2| is p1 ∧ p2-bounded. Theorem 3.2. Let P : X → Y be a solid vector seminorm, G a normal subspace of X and V0 : G → Y a positive linear operator such that

(3.3) |V0(a)| ≤ P (a), ∀a ∈ G.

Then there exists a positive linear operator V : X → Y such that V | G = V0 and

(3.4) |V (x)| ≤ P (x), ∀x ∈ X.

Proof. By the Hahn-Banach-Kantorovich theorem [5], there exists a linear operator U : X → Y such that U | G = V0 and U(x) ≤ P (x), ∀x ∈ X. It follows that U is an (o)-bounded linear operator, hence U ∈ R(X,Y ). Let us now consider the positive part U+ of U. If x ∈ X+ then it follows from 0 ≤ z ≤ x that U(z) ≤ P (z) ≤ P (x), hence

U+(x) = sup {U(z) | 0 ≤ z ≤ x} ≤ P (x). For any x ∈ X we now have

|U+(x)| ≤ U+ (|x|) ≤ P (|x|) = P (x). 9 Vector seminorms, spaces with vector norm, and regular operators 415

Therefore, putting V = U+, inequality (3.4) holds. On the other hand, since G is a normal subspace of X, if 0 ≤ a ∈ G then [0, a] ⊂ G. It follows that

V (a) = sup U([0, a]) = V0(a) and, being a vector sublattice of , we have , G X V (x) = V0(x) ∀x ∈ G. Remark 3.2. It follows by Theorems 3.2 and 3.1 that if G is a normal subspace of the space X, if 0 ≤ Vi ∈ R (G, Y ) ,Pi ∈ P (X,Y ) , i = 1, 2, and

|Vi(a)| ≤ Pi(a), ∀a ∈ G, i = 1, 2, then there exists a positive V ∈ R(X,Y ) such that V | G = V1 ∧ V2 and

|V (x)| ≤ (P1 ∧ P2)(x), ∀x ∈ X.

4. LINEAR OPERATORS ON v-NORMED VECTOR SPACES

Let X,Y be vector lattices and FX ,GY v-normed vector spaces. We recall that an operator U : FX → GY is said to be a (v)-regular operator [3] if it is additive and if there exists a positive linear operator V : X → Y such that

(4.1) kU(f)kY ≤ V (kfkX ) , ∀f ∈ FX . v We shall denote by R (FX ,GY ) the set of all (v)-regular operators which map FX into GY . If the space Y is Archimedean, then a (v)-regular operator which maps FX into GY is a linear operator. Remark 4.1. If Y is a complete vector lattice and in the spaces X,Y we consider the modulus as vector norm, then Rv(X,Y ) = R(X,Y ).

Remark 4.2. By a proposition given in [3], if FX is a strictly v-normed v vector space, Y is a complete vector lattice and U ∈ R (FX ,GY ), then there exists the smallest positive linear operator V which satises inequality (4.1). In this case we shall denote such an operator by . The map kUkR v given by the formula is a vector P : R (FX ,GY ) → R(X,Y ) P (U) = kUkR v norm on the vector space R (FX ,GY ). We recall now that a subset A of an Archimedean vector lattice is said to be -annihilating [3] if for any sequence of elements of and (o) (xn)n∈N A any sequence of real numbers which converges to zero, one has - (αn)n∈N (o) lim αnxn = 0. An Archimedean vector lattice Z is said to be of (o)-boundedness n type if any (o)-annihilating subset of Z is (o)-bounded. The following theorem generalizes a result given in [8]. 416 Romulus Cristescu 10

Theorem 4.1. Let X,Y be Archimedean vector lattices and FX ,GY v- normed vector spaces. Let be a sequence of linear operators which map (Un)n∈N FX into GY . Consider the following statements. S (i) For any (v)-bounded subset A of FX , the set Un(A) is (v)-bounded. n∈N (ii) If (vρ)-lim fn = 0 in FX , then (vρ)-lim Uj (fn) = 0 for any sequence n n n of natural numbers. (jn)n∈N Statement (i) implies (ii) and if Y is of (o)-boundedness type, then (i) and (ii) are equivalent.

Proof. Suppose (i) holds and let (vρ)-lim fn = 0 in FX . Then [5] there n exists a sequence of natural numbers such that and - (λn)n∈N λn ↑ ∞ (ρ) n∈N lim λn kfnk = 0. It follows that the sequence (λnfn)n∈ is (v)-bounded. By n X N hypothesis, there exists y0 ∈ Y such that

(4.2) kUk (λnfn)kY ≤ y0, ∀k, n ∈ N. If is an arbitrary sequence of natural numbers, then by (4.2) we have (jn)n∈N 1 kUjn (fn)kY ≤ y0, ∀n ∈ N, λn hence (vρ)-lim Uj fn = 0. Consequently, (i) ⇒ (ii). n n Let us now assume that Y is of (o)-boundedness type and suppose (ii) holds. Let A be a (v)-bounded subset of the space FX . Denote [ B = Uk(A) k∈N and let us consider a sequence of elements of . Let and (gn)n∈N B jn ∈ N with . Consider now a sequence of real num- fn ∈ A gn = Ujn (fn) (αn)n∈N bers convergent to zero. Since the sequence (kfnk ) is (o)-bounded, the X n∈N sequence (αn kfnk )n∈ is ρ-convergent to 0, hence (vρ)-lim αnfn = 0. By X N n hypothesis,

(vρ) -lim Uj (αnfn) = 0 n n hence (ρ)-lim αn kgnk = 0. Consequently, the set {kgk | g ∈ B} is (o)- n Y Y annihilating, therefore it is (o)-bounded. In other words, the set B is (v)- bounded.

Remark 4.3. Let FX be a strictly v-normed vector space, GY a v-normed v vector space with Y complete vector lattice and Un ∈ R (FX ,GY ) , n ∈ N. If the sequence is -bounded in then statement (i) of (kUnkR)n∈ (o) R (X,Y ) Theorem 4.1 holds. N 11 Vector seminorms, spaces with vector norm, and regular operators 417

In the next theorem we consider the modulus as vector norm in a complete vector lattice Y . Theorem 4.2. Let F be a vector lattice and E a vector sublattice of F . Consider a solid vector norm on F with values in a vector lattice X such that FX and EX are strictly v-normed vector spaces. If Y is a complete vector lattice and U0 : EX → Y is a positive (v)-regular operator, then there exists a positive -regular operator such that and (v) U : FX → Y U | E = U0 kUkR = kU0kR . Proof. Denoting by F + the positive cone of the space F and putting + Q(f) = inf kU0kR (khkX ) | f ≤ h ∈ F , ∀f ∈ F, we obtain a sublinear operator Q : F → Y . It is easily veried that

(4.3) Q(f) ≤ kU0kR (kfkX ) , ∀f ∈ F. The inequality

(4.4) U0(f) ≤ Q(f), ∀f ∈ E, + also holds. Indeed, if f ∈ E and f ≤ h ∈ F , then f ≤ f+ ≤ h with f+ ∈ E and we have

U0(f) ≤ U0 (f+) ≤ kU0kR (kf+kX ) ≤ kU0kR (khkX ) whence (4.4) follows. By the Hahn-Banach-Kantorovich theorem [5], there exists a linear operator U : F → X such that U | E = U0 and (4.5) U(f) ≤ Q(f), ∀f ∈ F, whence, by (4.3), |U(f)| ≤ kU0kR (kfkX ) , ∀f ∈ F. Therefore, v and . On the other hand, if U ∈ R (FX ,Y ) kUkR = kU0kR f ∈ F and f ≤ 0, then Q(f) = 0 and it follows by (4.5) that U(f) ≤ 0, that is, U is a positive operator.

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Received 16 July 2008 University of Bucharest Faculty of Mathematics and Computer Science Str. Academiei 14 010014 Bucharest, Romania