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 ![](//html.scirp.org/file/5-5301219x16.png)
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.
NOTES
![]()
*Dedicated to Reinhard Börger, a brilliant and enthusiastic mathematician full of new ideas.