ABSTRACT

Order unit normed linear spaces are a special type of regularly ordered normed linear spaces and therefore the first section is a short collection of the fundamental results on this type of normed linear spaces. The connection between order unit normed linear spaces and base normed linear spaces within the category of regularly ordered normed linear spaces is described in Section 2, and Section 3 at last, contains the results on Banach limits in an arbitrary order unit normed linear space. It is shown that the original results on Banach limits are valid for a greater range.

Keywords:

Order Unit Normed Spaces, Base Normed Spaces, Banach Limits

1. Introduction

Most, if not all, publications where Banach limits are investigated take place in an order unit normed real linear space. Order unit normed linear spaces are a special type of regularly ordered normed linear spaces and therefore the first section is a short collection of the fundamental results on this type of normed linear spaces, for the reader's convenience. The connection between order unit normed linear spaces and base normed linear spaces within the category of regularly ordered normed linear spaces is described in Section 2, and Section 3 at last, contains the results on Banach limits in an arbitrary order unit normed linear space. It is shown that the original results on Banach limits are valid in a for greater range. For a further generalisation of vector valued Banach limits in a different direction we refer to a recent paper of R.Armario, F. Kh. Garsiya-Pacheko and F. Kh Peres-Fernandes [1] .

2. Regularly Ordered Normed Linear Spaces

An ordered normed linear space with order norm and order cone is called regularly ordered iff the cone is -closed and proper and is a Riesz norm, i.e. if

(Ri 1) For implies i.e. is absolutely mo- notone, and

(Ri 2) For x Î E with there exists a with and

hold. (see [2] [3] ).

Lemma 1. Let, for an ordered linear space with proper and -closed cone (Ri 1) hold. Then each of the following two conditions is equivalent to (Ri 2)

(Ri 3) For and there exists a such that and hold.

(Ri 4) For any

holds.

Proof. The proof is straightforward. Condition (Ri 2) implies that generates If (Ri 2) holds, then for and there is with

hence (Ri 3) is proved for

(Ri 3) implies that generates and (Ri 1) implies

Because of (Ri 3), for and there is a such that and for any and hence proving (Ri 4). Moreover, (Ri 4) obviously implies (Ri 2) which completes the proof. □

In [3] K. Ch. Min introduced regularly ordered normed spaces as a natural and canonical generalization of Riesz spaces. A crucial point in this generalization was the definition of the corresponding homomorphisms compatible and most closely related to the structure of these spaces, such that, in addition, the set of these special homomorphisms is again a regularly ordered normed linear space in a canonical way. This is done by

Definition 1. If are regularly ordered linear spaces a bounded linear mapping is called positive iff holds. A bounded linear mapping is called regular iff it can be expressed as the difference of two positive linear mappings [3] .

The set

is a linear space by the obvious operations. One introduces the cone

which is obviously proper and generates One often writes as abbreviation for and consequently calls an positive and writes if for in i.e. The positive part of the unit ball in a regularly ordered space with norm is denoted by

Lemma 2. Let be regularly ordered normed linear spaces with norm and cone If and denotes the usual supremum norm, then

holds.

Proof. For with there is with which implies

and

hence

□

Now, we proceed to define the norm in the space by

(*)

Proposition 1. For regularly ordered normed spaces is a Riesz norm on and makes a regularly ordered normed linear space. For holds and in general

Proof. The proof that is a seminorm is straightforward. In order to show that one starts with and Let and with Then follows and

(i)

Using and one obtains in the same way

and, multiplying by −1

(ii)

Adding (i) and (ii) yields

hence

and

Now yields i.e. is a norm. If then

hence and is a Riesz norm because of Lemma 1, (Ri 4) and the definition of. is obviously -closed and therefore also -closed because of

□

In the following will always denote this norm of regular linear mappings. Note that Reg-Ord is a symmetric, complete and cocomplete monoidal closed category and the inner hom-functor has as an adjoint, the tensor product [3] .

3. Order Unit and Base Ordered Normed Linear Spaces

The order unit normed linear spaces are a special type of regularly ordered normed linear spaces , as are the base normed linear spaces [3] [4] . For investigating a special type of mathematical objects, however, it is always best to use the type of mappings most closely related to the special structure of the objects (the Bourbaki Principle). Hence, for investigating order unit normed spaces we do not look at the full subcategory of Reg-Ord generated by the order unit normed spaces but introduce a more special type of regular linear mappings. The same method, by the way, has been successful for another type of regularly ordered spaces, namely the base normed (Banach) spaces (cp. [3] [5] [6] ).

Definition 2. For two order unit normed linear spaces with order unit define

and

Proposition 2. Let be order unit spaces with order units Then

i) is a -closed convex base of and the -closed unit ball of in the supremum norm

ii) is a -closed proper subcone of

Proof. (1) Let and with Then, for follows which implies and i.e. showing that is -closed. Now, and imply i.e. and even because Hence follows, even

Let be a convex combination of

Then

follows, i.e.

which proves that is convex.

Now with and implies and i.e. is a -closed base of and

(ii) This follows from (i) (see [7] , 3.9 p. 128). □

Corollary 1. For order unit normed linear spaces

is a base-normed ordered linear space with base and base norm denoted by and are closed in the base norm

Proof. That is a base normed space follows from Proposition 2 and the definition. That base and cone are base normed closed follows from the fact that they are -closed (see Proposition 2) and because the -topology is weaker than the -topology (see Proposition 2 and [7] , 3.8.3, p. 121).

□

Remark 1. If is a Banach space, with the norm because are Banach spaces, then is superconvex (see [3] [6] ) and is a base normed Banach space (see [3] [4] [7] ).

Definition 3. The order unit normed linear spaces together with the linear mappings with constitute the category Ord-Unit of order- unit normed linear spaces which is a not full subcategory of Reg-Ord.

There is an equally important subcategory of Reg-Ord, the category of based normed linear spaces.

Definition 4. A base normed ordered linear space “base normed linear space” for short, is a regular ordered linear space with proper closed cone and norm which is induced by a base of (see [4] [7] ). If are base normed linear spaces, put

The elements of are monotone mappings, is a base set in and it is -closed. Let denote the proper closed cone generated by.

is a base normed space of special mappings from to The base normed linear spaces and these linear mappings form the not full subcategory BN-Ord of Reg-Ord (see [6] [8] [9] ), which is therefore a closed category.

What remains in this connection is to investigate special morphisms particularly adapted to these subcategories between spaces belonging to two different of these subcategories Ord-Unit and BN-Ord. We start this with investigating the intersection of these subcategories.

Proposition 3. Let be a regular ordered normed linear space. Then is a base and order unit norm iff is isomorphic to by a regular positive isomorphism.

Proof. If is the order unit and if we omit the index at the norm, then trivially and hold. Let and assume As (see [7] ), holds and follows or which implies because is additive on This implies and hence which gives a contradiction. Therefore and the assertion follows as and

Hence, the isomorphism is defined by □

It should be noted that this isomorphism is an isomorphism in the category Ord- Unit of order unit normed spaces and also in BN-Ord. So, loosely speaking,

Now the “general connection” between Ord-Unit and BN-Ord is investigated via the morphisms:

Proposition 4. If is a base normed and an order unit normed linear space, then is an order unit normed linear space.

Proof. Define by and extend positive linearly by for to which can be uniquely extended to a monotone, linear mapping in in the usual way. Obviously, with the Reg-Ord norm, as is a positive mapping. Take a with i.e.

and hence for or

whence for For arbi- trary and

follows implying for or This means, for arbitrary that This shows that is an order unit in Denoting the order unit norm by follows. □

This is a slightly different version of the proof of Theorem 1 in Ellis [7] .

Surprisingly a corresponding result also holds if Ord-Uni and BN-Ord

Proposition 5. If is an order unit and is a base normed ordered linear space, then Reg-Ord. is a base normed ordered linear space.

Proof. Define

where denotes the order unit of One shows first that is a base set. For this, let i.e. and implying for

that is and hence this implies that

For and obviously and is convex. Besides, the above proof shows, that any can be written as with

Obviously and if with and implying

from which follows because of and finally

□

It is interesting that by defining the subspaces and

of for order unit or base normed spaces

respectively, one gets a number of results which for the bigger space have either not yet been proved or were more difficult to prove because the assumptions for are weaker (see [10] [11] ). The Pro- positions 4 and 5 are an exception because here the general space has the special structure of an order unit or base normed spaces, respectively.

There are different ways to generalize the structure of in many fields of mathematics. In analysis one is primarly interested in aspects of order, norm and convergence. Now, essentially, with 1, the usual order and the absolute value (considered as a norm) forms the intersection which both generalize in different (dual) directions. The above results seem to indicate that the order unit spaces are at least as important as generalizations of as the base normed spaces while in many publications the latter type seems to play the dominant role. Propositions 4 and 5 are particularly interesting because the hom-spaces have a special structure if the arguments do not belong to the same of the two subcategories and

4. Banach Limits

For the introduction of Banach Limits we first prove, following a proof method of W. Roth in [12] , Theorem 2.1, a special variant of the Hahn-Banach Theorem for order unit normed linear spaces.

Theorem 6. (Hahn-Banach Theorem for Order Unit Spaces) Let be an order unit normed space with order unit ordering cone and norm and let the following conditions be satisfied:

i) is a sublinear monotone function with

ii) is a surjective positive linear mapping.

iii) For any the set mapping is a right inverse of with

is monotone and

iv) is a muliplicative group of positive automorphisms of

v) For any and for every, and hold.

Then there exists a positive linear functional with:

a) and

b)

c)

d)

for and

Proof. Define

Obviously holds, hence A partial order “” is defined in by putting, for

Let be a non-empty, with respect to “” totally ordered subset and define

As for all is well defined and finite and

holds.

If then for all and hence for all follows, i.e. is monotone.

Let then for all and hence

As obviously for it follows that is sublinear. Also trivially satisfies the conditions a)-d) as well as Consequently, and is a lower bound of in Zorn’s Lemma now implies the existence of (at least) one minimal element in with respect to which will be denoted by

Define for

As, for

(1)

Taking and in the defining equation of yields

(2)

implying

(2a)

Now, the remaining equations in the assertion will be proved for Take the inequality from the defining equation of then

contributing

to Conversely, leads to

contributing

to the definition of and one gets

(3a)

To show the invariance of under start with from Then

contributing

to

An inequality of leads to

and

as contribution to. Hence

(3b)

Verbatim, this proof carries over to the equation

(3c)

A new function is now introduced by

(4)

If then is and therefore

For, implies and yields

because of the definition of. Hence

and

(5)

follows which implies in, particular, for as is positive.

Taking and one has

(6)

in particular If and in (4) then

(7)

follows i.e. monotonicity.

Consider now, for

then

i.e.

Now for

and

The mapping

is, for fixed bijective, therefore

(8)

holds because for one has So is sublinear.

We now show that also satisfies the equations of the assertions of the theorem. Take from the defining set of Then

contributing

to Conversely, contributing to yields by applying

Hence

This implies

(9)

The proof of the remaining two equations of the assertions follows almost verbatim this pattern of proof and one gets:

and (6) implies Hence is proved which implies

(10)

because of (6) and the minimality of

Now, looking again at the definition (4) of and putting and one gets

(11)

which together with (1) yields and in combination with (1) and (10) gives

(12)

Now, for

and

and since and are superlinear, one has

which implies that is is superadditive and because of (12) positively homogeneous, i.e. superlinear. This implies that is linear because of (12) and satisfies all the equations in the assertion, which completes the proof. □

Banach limits are almost always defined as continuous extensions of a continuous linear functional in an order unit normed space. Hence, for the introduction of Banach limits we need Theorem 6 in a continuous form. Surprisingly Theorem 6 already contains all the necessary continuity conditions as the following Corollary shows:

Corollary 2. Let the assertions (i)-(v) of Theorem 6 be satisfied and put and for Then

i) and is an isometrical order unit normed subspace of which is closed.

ii) is continuous and is a positive, continuous linear functional with

iii) Any satisfying Theorem 6 is a positive, continuous linear extension of

Proof. i): Obviously, and, hence, For holds and this is an inequality in and which proves i).

ii): and are both monotone, sublinear and superlinear. As for holds, is linear and positive on. If then and s. th. norm of is and is in 0 continuous hence also for any and Hence, we get for any in Theorem 6

implying the continuity of and even because In particular, this holds also for □

It is remarkable that with respect to the continuity properties, the continuity of and do not play any role.

Definition 5. With the notations of Corollary 2 any such is called a Banach limit of

One defines

of course, depends on the parameters but in order to make the notation for the following not too cumbersome we will mostly omit them and write simply

Proposition 7. For the following statements hold:

i) is a convex subset of of the base normed Banach space the dual space of

ii) is weakly-*-closed, weakly closed and also -closed, where denotes the usual dual norm of of

iii) is a superconvex base set contained in.

Proof. i): Let be the set of

all abstract convex combinations, then, for obvi-

ously as for

and obviously all other equations in Theorem 6 are

satisfied, too.

ii): One first proves that is -closed, because from this , the other two assertions of ii) then follow at once. If and there is a with for then i.e. follows for which implies for and for Analogously, one shows the other equations in Theorem 6 for because they hold for the

iii): Obviously, holds, hence is bounded and because of ii) -closed. Then, it is a general result that is superconvex (see [13] , Theorem 2.5).

□

Because is as a subset of trivially a base set

is a proper cone and

is a base normed ordered linear space. To simplify notation, we will write instead of in the following.

Theorem 8. If the norm induced by in is denoted by and the topology induced by the weak-*-topology in by then

is a compact, base normed Saks space (see [13] , Theorem 3.1) and an isometrical subspace of

Proof. As is a weakly-*-closed base set and a subset of which is weakly-*-compact because of Alaoglu-Bourbaki it is also weakly-*-compact and the space generated by the closed cone (see [7] , Theorem 3.8.3, [13] , Theorem 3.2, [14] ) is a compact, base normed Saks space. The last assertion is obvious. □

The result of Theorem 8 is essentially the definition of a functor from any category with objects satisfying the assertions of Theorem 6 to the category of compact, base normed Saks spaces ( [13] , Theorem 3.1). This functor will be investigated by the authors in a forthcoming paper.

5. Summary

The main result of the paper offers a Hahn-Banach theorem for order unit normed spaces (Theorem 6) from which novel conclusions on Banach limits are drawn. The result of Theorem 8 gives rise to the definition of a functor which goes from any category with objects satisfying the assertions of Theorem 6 into the category of compact, base normed Saks spaces.

Cite this paper

Pallaschke, D. and Pumplün, D. (2016) Banach Limits Revisited. Advances in Pure Mathematics, 6, 1022-1036. http://dx.doi.org/10.4236/apm.2016.613075

References

NOTES

*Dedicated to Reinhard Börger, a brilliant and enthusiastic mathematician full of new ideas.