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http://www.scirp.org/journal/apm ISSN Online: 2160-0384 ISSN Print: 2160-0368

DOI: 10.4236/apm.2016.613075 December 23, 2016

Banach Limits Revisited

*

Diethard Pallaschke1, Dieter Pumplün2

1Institute of Operations Research, Karlsruhe Institute of Technology (KIT), Karlsruhe, Germany 2Faculty of Mathematics and Computer Science, Fern Universität Hagen, Hagen, Germany

Abstract

Order unit normed linear spaces are a special type of regularly ordered normed li-near 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 E with order '′ ′′≤ , norm □E and order cone

*Dedicated to Reinhard Börger, a brilliant and enthusiastic mathematician full of new ideas. How to cite this paper: Pallaschke, D. and

Pumplün, D. (2016) Banach Limits Revi-sited. Advances in Pure Mathematics, 6, 1022-1036.

http://dx.doi.org/10.4236/apm.2016.613075

Received: November 13, 2016 Accepted: December 20, 2016 Published: December 23, 2016

Copyright © 2016 by authors and Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

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( )

C E is called regularly ordered iff the cone C E

( )

is □E-closed and proper and

E

□ is a Riesz norm, i.e. if

(Ri 1) For x y, ∈ − ≤ ≤E, y x y implies xEy E, i.e. □E is absolutely mo-

notone, and

(Ri 2) For x ∈ E with xE <1 there exists a yE with yE <1 and − ≤ ≤y x y

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

Lemma 1. Let, for an ordered linear space E with proper and

E

□ -closed cone

( )

C E (Ri 1) hold. Then each of the following two conditions is equivalent to (Ri 2)

(Ri 3) For xE and ε> 0 there exists a yE such that − ≤ ≤y x y and

<

E E

y x +ε hold.

(Ri 4) For any xE

{

}

inf | ,

E E

x = y − ≤ ≤y x y

holds.

Proof. The proof is straightforward. Condition (Ri 2) implies that C E

( )

generates

E If (Ri 2) holds, then for xE and ε > 0 there is yE with

,

E

x

y y

x ε

− ≤ ≤

+

hence (Ri 3) is proved for y0:=

(

xE

)

y.

(Ri 3) implies that C E

( )

generates E and (Ri 1) implies

{

}

0

inf : .

E E

xy − ≤ ≤y x y = x

Because of (Ri 3), for xE and ε> 0 there is a yE such that − ≤ ≤y x y

and y E < xE +ε, for any ε >0 and hence x0− ≤ε xEx0 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 E ii, =1, 2 are regularly ordered linear spaces a bounded linear

mapping f E: 1→E2 is called positive iff f C E

(

( )

1

)

C E

( )

2 holds. A bounded

linear mapping is called regular iff it can be expressed as the difference of two positive linear mappings [3].

The set

(

1 2

)

{

1 2

}

Reg−Ord E E, := f f E: →E regular linear mapping

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

(

)

(

Reg Ord 1, 2

)

:

{

Reg Ord

(

1, 2

)

,

(

( )

1

)

( )

2

}

CE E = f f ∈ − E E f C EC E

(3)

as abbreviation for x1∈C E

( )

1 and consequently calls an f ∈Reg Ord−

(

E E1, 2

)

positive and writes f ≥0, if f x

( )

1 ≥0, for x1≥0 in E1, i.e. x1∈C E

( )

1 . The positive part of the unit ball in a regularly ordered space E with norm

E

is

denoted by

( )

( )

( )

,

( )

{

and 1 .

}

E

E C E E E x x E

∆ = ∩ = ∈ □ ≤

Lemma 2. Let Ei be regularly ordered normed linear spaces with norm □i and cone C E

( )

i , 1, 2.i= If gC

(

Reg−Ord

(

E E1, 2

)

)

and g ∞ denotes the usual supremum norm, then

( )

( )

{

1 2 1 1

}

: sup

g = g x x ∈ ∆ E

holds.

Proof. For x1∈E1 with x1 1<1 there is y1∈E1, 1,y1 < with − ≤y1 x1≤ y1 which implies

( )

1

( )

1

( )

1

g y g x g y

− ≤ ≤

and

( )

1 2

( )

1 2,

g xg y

hence

( )

{

}

( )

( )

{

}

( )

( )

{

}

( )

{

}

1 2 1 1

1 2 1 1 1 1 1 2 1 1

1 2 1 1

sup 1

sup 1 ,

sup

sup 1 .

g g x x

g y y y C E

g y y E

g y y g

= ≤

≤ ≤ ∈

= ∈ ∆

≤ ≤ =

□ Now, we proceed to define the norm □∗ in the space Reg Ord−

(

E E1, 2

)

by

{

}

: inf .

f ∗ = g − ≤ ≤g f g (*)

Proposition 1. For regularly ordered normed spaces E E1, 2,

□ is a Riesz norm on Reg Ord−

(

E E1, 2

)

and makes Reg Ord−

(

E E1, 2

)

a regularly ordered normed linear space. For f ≥0 f ∗= f holds and in general

.

f f

Proof. The proof that ∗

is a seminorm is straightforward. In order to show that

∞ ≤

□ □ one starts with f g, ∈Reg Ord−

(

E E1, 2

)

and − ≤ ≤g f g. Let x1 1<1

and − ≤y1 x1≤y1 with y11<1, ,x y1 1∈E1. Then x1+y1≥0 follows and

(

1 1

)

(

1 1

)

(

1 1

)

.

g x y f x y g x y

− + ≤ + ≤ + (i) Using g− ≥f 0 and y1− ≥x1 0 one obtains in the same way

(

1 1

)

(

1 1

)

(

1 1

)

g y x f y x g y x

− − ≤ − ≤ −

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(

1 1

)

(

1 1

)

(

1 1

)

.

g y x f x y g y x

− − ≤ − ≤ − (ii) Adding (i) and (ii) yields

( )

1

( )

1

( )

1

g y f x g y

− ≤ ≤

hence

( )

1 2

( )

1 2

f xg y

and

( )

{

}

( )

{

}

1 2 1 1

1 2 1 1 1 1 1

sup 1

sup 1 and .

f f x x

g y y y x y g

= ≤

≤ ≤ − ≤ ≤ ≤

Now

( )

∗ yields f f ∗, i.e. □∗ is a norm. If f ≥0 then

{

}

: inf | 0 ,

f ∗ = g ≤ ≤f gf

hence f ∗ = f and □∗ is a Riesz norm because of Lemma 1, (Ri 4) and the definition of

( )

∗ . C

(

Reg−Ord

(

E E1, 2

)

)

is obviously □∞-closed and therefore also

□ -closed because of f f ∗.

□ 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 Reg Ord−

(

□ □,

)

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 Ei with order unit , 1, 2,

i

e i= define

(

)

{

(

)

( )

}

0 1, 2 : Reg Ord 1, 2 , 0 and 1 2

Bs E E = f f ∈ − E E ff e =e

and

(

)

(

)

0 1, 2 : 0 1, 2 .

C E E =R Bs+ E E

Proposition 2. Let E E1, 2 be order unit spaces with order units e e1, 2. Then i) Bs0

(

E E1, 2

)

is a □∞-closed convex base of C0

(

E E1, 2

)

and

(

)

(

)

0 1, 2 1, 2 ,

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supremum norm □∞

ii) C0

(

E E1, 2

)

is a □∞-closed proper subcone of C

(

Reg−Ord

(

E E1, 2

)

)

.

Proof. (1) Let fnBs0

(

E E1, 2

)

, nN, and f ∈Reg Ord−

(

E E1, 2

)

with

lim n = 0.

n→∞ ff ∞ Then, for x1∈E1, limf x

( )

1 =n→∞fn

( )

x1 follows which implies f ≥0 and f e

( )

1 =e2, i.e. fBs0

(

E E1, 2

)

showing that Bs0

(

E E1, 2

)

is □∞ -closed. Now, fBs0

(

E E1, 2

)

and − ≤e1 x1≤e1 imply − ≤e2 f x

( )

1 ≤e2 i.e. f ∞≤1 and even f =1 because f e

( )

1 =e2, Hence Bs0

(

E E1, 2

)

⊂○∞

(

E E1, 2

)

follows, even

(

)

(

(

)

)

0 1, 2 1, 2 .

Bs E E ⊂ ∂ ○ E E

Let

1 1

, 0, 1 , 1,

n n

i i i i

i i

f i n

α α α

= =

≥ ≤ ≤ =

be a convex combination of fiBs0

(

E E1, 2

)

. Then

( )

1 2 2

1 1 1

0 and ,

n n n

i i i i i

i i i

f f e e e

α α α

= = =

 

= =

 

follows, i.e.

(

)

0 1 2

1

, , n

i i i

f Bs E E

α

=

which proves that Bs0

(

E E1, 2

)

is convex.

Now αfg with α β >, 0 and f g, ∈Bs0

(

E E1, 2

)

implies

( )

( )

2 1 1 2,

e f e g e e

α =α =β =β and α β= , i.e. Bs0

(

E E1, 2

)

is a □∞-closed base of C E E

(

1, 2

)

and 0∉Bs0

(

E E1, 2

)

.

(ii) This follows from (i) (see [7], 3.9 p. 128). □ Corollary 1. For order unit normed linear spaces E ii, 1, 2, =

(

1 2

)

(

1 2

)

(

1 2

)

Ord−Unit E E, =C E E, −C E E,

is a base-normed ordered linear space with base Bs0

(

E E1, 2

)

and base norm denoted

by □0. C E E

(

1, 2

)

and Bs0

(

E E1, 2

)

are closed in the base norm □0.

Proof. That Ord−Unit

(

E E1, 2

)

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 □0.-closed (see Proposition 2) and because the □0.-topology is weaker than the □0.-topology (see Proposition 2 and [7], 3.8.3, p. 121).

□ Remark 1. If Reg Ord−

(

E E1, 2

)

is a Banach space, with the norm □0. because

, 1, 2, i

E i= are Banach spaces, then Bs0

(

E E1, 2

)

is superconvex (see [3] [6]) and

(

1 2

)

Reg Ord− E E, is a base normed Banach space (see [3][4][7]).

Definition 3. The order unit normed linear spaces together with the linear mappings

1 2

:

f EE with f∈Ord−Unit

(

E E1, 2

)

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 E with proper closed cone C E

( )

and norm

E

(6)

base normed linear spaces, put

(

1, 2

)

:

{

: 1 2 linear and

(

( )

1

)

( )

2 .

}

Bs E E = f f EE f Bs EBs E

The elements of Bs E E

(

1, 2

)

are monotone mappings, Bs E E

(

1, 2

)

is a base set in

(

1 2

)

Reg Ord− E E, and it is □-closed. Let C E E

(

1, 2

)

denote the proper closed cone generated by Bs E E

(

1, 2

)

.

(

1 2

)

(

1 2

)

(

1 2

)

BN Ord− E E, :=C E E, − ,C E E

is a base normed space of special mappings from E1 to E2. 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

(

,

( )

,

)

E

E C E □ be a regular ordered normed linear space.

Then □E is a base and order unit norm iff

(

E C E,

( )

, □E

)

is isomorphic to

[

]

(

R, 0,∞ ,□

)

by a regular positive isomorphism.

Proof. If eE is the order unit and if we omit the index E at the norm, then

trivially e =1 and e>0 hold. Let bBs E

( )

and assume b=/e. As

( )

( )

[

,

]

Bs E ⊂○E = −e e (see [7]), 0< <b e holds and d:= − >e b 0 follows or

,

e= +b d which implies 1= e = +b d = b + d , because □ is additive on

( )

.

C E This implies d =0 and hence e=b which gives a contradiction. Therefore

( ) { }

Bs E = e and the assertion follows as C E

( )

=R Bs E+

( )

=R e+ and

( )

( )

.

E=C EC E =Re

Hence, the isomorphism is i E: →R defined by i

( )

αe =α α, .∈R □ 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,

{ }

Ord−Unit∩BN Ord− = R .

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

Proposition 4. If E1 is a base normed and E2 an order unit normed linear space, then Reg Ord−

(

E E1, 2

)

is an order unit normed linear space.

Proof. Define ε:C E

( )

1 →C E

( )

2 by ε

(

Bs E

( )

1

)

=

{ }

e2 and extend ε positive

linearly by ε α

( )

x1 =αe2, for α ≥0, ,x1∈Bs E

( )

1 to ε:C E

( )

1 →C E

( )

2 which can be uniquely extended to ε:E1→E2, a monotone, linear mapping in Reg−Ord in the usual way. Obviously, ε = ε , with □ the Reg-Ord norm, as ε is a

positive mapping. Take a g∈Reg Ord−

(

E E1, 2

)

with g ≤1, i.e.

( )

(

1

)

( )

2

[

2, 2

]

gE ⊂○ E = −e e and hence − ≤e2 g b

( )

1e2 for b1Bs E

( )

1 or

( )

b1 g b

( )

1

( )

b1

ε ε

(7)

( )

d1 g d

( )

1

( )

d1

ε ε

− ≤ − ≤ follows implying −ε

( )

x1 ≤g x

( )

1 ≤ε

( )

x1 for x1∈E1 or .

g

ε ε

− ≤ ≤ This means, for arbitrary g=/0, that − g ε≤ ≤g g ε. This shows

that ε is an order unit in Reg Ord−

(

E E1, 2

)

. Denoting the order unit norm by

0

0

gg follows. □

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

Surprisingly a corresponding result also holds if E1∈ Ord-Uni and E2 BN-Ord Proposition 5. If E1 is an order unit and E2 is a base normed ordered linear space, then Reg-Ord

(

E E1, 2

)

. is a base normed ordered linear space.

Proof. Define

(

)

{

(

(

)

)

( )

( )

}

0 1, 2 : Reg Ord 1, 2 and 1 2

Bs E E = f fCE E f eBs E

where e1 denotes the order unit of E1 One shows first that Bs0

(

E E1, 2

)

is a base set. For this, let gC

(

Reg−Ord

(

E E1, 2

)

)

g=/0, i.e. g>0 and g e

( )

1 >0, implying for

( )

1 2

1 :

f g

g e

=

that is f >0, and f e

( )

1 2=1, hence f e

( )

1 ∈Bs E

( )

2 . this implies that

(

)

0 1, 2 .

Bs E E = ∅/

For f g, ∈Bs0

(

E E1, 2

)

and 0≤ ≤α 1, obviously αf + −

(

1 α

)

gBs0

(

E E1, 2

)

and Bs0

(

E E1, 2

)

is convex. Besides, the above proof shows, that any

(

)

(

Reg Ord 1, 2

)

, 0,

gCE E g > can be written as gf with

(

)

0 1 2

0, ,f Bs E E .

α ≥ ∈

Obviously 0∉Bs0

(

E E1, 2

)

and if αfg with α β ≥, 0 and f g, ∈Bs E E

(

1, 2

)

implying

( )

1

( )

1

f e g e

α =β

from which α β= follows because of f e

( ) ( )

1 ,g e1Bs E E

(

1, 2

)

and finally f =g.

□ It is interesting that by defining the subspaces Ord−Unit

(

E E1, 2

)

and

(

1 2

)

BN Ord− E E, of Reg Ord−

(

E E1, 2

)

for order unit or base normed spaces 1, ,2

E E respectively, one gets a number of results which for the bigger space

(

1 2

)

Reg Ord− E E, have either not yet been proved or were more difficult to prove

because the assumptions for Reg Ord−

(

E E1, 2

)

are weaker (see [10] [11]). The Pro- positions 4 and 5 are an exception because here the general space Reg Ord−

(

E E1, 2

)

has the special structure of an order unit or base normed spaces, respectively.

There are different ways to generalize the structure of R in many fields of

mathematics. In analysis one is primarly interested in aspects of order, norm and convergence. Now, essentially, R with 1, the usual order and the absolute value

(considered as a norm) forms the intersection Ord−Unit∩BN Ord− =

{ }

R , which

both generalize R in different (dual) directions. The above results seem to indicate

that the order unit spaces are at least as important as generalizations of R as the base

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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

Reg−Ord and BN−Ord

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 E be an order

unit normed space with order unit e, ordering cone C E

( )

and norm

E

□ and let

the following conditions be satisfied:

i) p E: →R is a sublinear monotone function with p e

( )

=1,

ii) S E: →E is a surjective positive linear mapping.

iii) For any xE, the set mapping Tx is a right inverse of S E: →E with

( )

(

)

.

x

T S x =x

x

T is monotone and −Tx

( )

y =Tx

( )

y , Tx

(

y+z

)

=Tx

( )

y +T0

( )

z , , ,x y zE.

iv) G is a muliplicative group G of positive automorphisms of E.

v) For any x y, ∈E and for every γ ∈G, p S x

(

( )

)

=p x

( )

, p T

(

x

( )

y

)

= p y

( )

and

( )

(

)

( )

p γ x = p x hold.

Then there exists a positive linear functional µ:ER with:

a) µ

( )

xp x

( )

and 1,µ

( )

e =

b) µ

(

S x

( )

)

( )

x , c) µ

(

Tz

( )

x

)

( )

x , d) µ γ

(

( )

x

)

( )

x , for x z, ∈E and γ ∈G.

Proof. Define

( ) ( )

( )

{

( )

(

)

(

( )

)

(

( )

)

( )

}

: : , sublinear and monotone with ,

, for all , and . p

z

M s s E R p x s x p x

s S x s T x s γ x s x x z E γ G

= → − − ≤ ≤

= = = ∈ ∈

Obviously pMp holds, hence Mp = ∅/ . A partial order “≤” is defined in Mp by putting, for s s1, 2∈Mp,

( )

( )

1 2 if and only if 1 2 for all .

ss s xs x xE

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

( )

{

( )

}

0 inf , .

s x = s x sO xE

As − − ≤p

( ) ( )

x s xp x

( )

for all sO s0 is well defined and finite and

( )

0

( )

( )

p x s x p x

− − ≤ ≤

holds.

If xy then s x

( ) ( )

s y for all sO and hence for all xy s, 0

( )

xs0

( )

y follows, i.e. s0 is monotone.

(9)

(

)

( )

( )

0 0 0 .

s x+ys x +s y

As obviously s0

( )

αxs0

( )

x for α ≥0 it follows that s0 is sublinear. Also

( )

0

s x trivially satisfies the conditions a)-d) as well as s e0

( )

=1. Consequently,

0 p

sM and s0 is a lower bound of O in Mp. Zorn’s Lemma now implies the existence of (at least) one minimal element in Mp with respect to ≤ which will be denoted by µ.

Define for x0∈E:

( )

{

( )

( )

}

0 x0 sup p x1 x2 x x1, 2 E and x1 x0 x2 .

α = − − −µ ∈ ≤ +

As, for xE, − − ≤p

( )

x µ

( )

xp x

( )

( )

1

( )

1

(

0 2

)

( )

0

( )

2 ,

p x µ x µ x x µ x µ x

− − ≤ ≤ + ≤ +

( )

1

( )

2

( )

0

p x µ x µ x

− − ≤

( )

( )

( )

0 0 0 0

and α x ≤µ xp x follows. (1)

Taking x1=x0 and x2 =0 in the defining equation of α0 yields

( )

0 0

( )

0

( )

0

p x α x p x

− − ≤ ≤ (2) implying

( )

0 e 1.

α = (2a) Now, the remaining equations in the assertion will be proved for α0

( )

x . Take the inequality x1≤S x

( )

0 +x2 from the defining equation of α0

(

S x

( )

0

)

, then

( )

(

( )

)

( )

0 1 0 0 2 0 0 2

x x

T xT S x +x =x +T x

contributing

( )

(

x0 1

)

(

0

( )

2

)

(

x0

( )

1

)

( )

2

( )

1

( )

2

p T x µ T x p Tx µ x p x µ x

− − − = − − − = − − −

to α0

( )

x0 . Conversely, x1 ≤x0+x2 leads to

( )

1

(

0 2

)

( ) ( )

0 2

S xS x +x =S x +S x

contributing

( )

(

1

)

(

( )

2

)

( )

1

( )

2

p S x µ S x p x µ x

− − − = − − −

to the definition of α0

(

S x

( )

0

)

and one gets

( )

(

)

( )

0 S x0 0 x0 .

α =α (3a) To show the invariance of α0

( )

x0 under T zz, ,∈E start with x1 ≤x0+x2 from

( )

0 x0 .

α Then

( )

1

(

0 2

)

( )

0 0

( )

2

z z z

T xT x +x =T x +T x

contributing

( )

(

z 1

)

(

0

( )

2

)

( )

1

( )

2

p T x µ T x p x µ x

− − − = − − −

(10)

An inequality x1≤Tz

( )

x0 +x2 of α0

(

Tz

( )

x0

)

. leads to

( )

1

(

z

( )

0 2

)

0

( )

2

S xS T x +x =x +S x

and

( )

(

1

)

(

( )

2

)

( )

1

( )

2

p S x µ S x p x µ x

− − − = − − −

as contribution to α0

( )

x0 . Hence

( )

(

)

( )

0 Tz x0 0 x0 , . z E

α =α ∈ (3b) Verbatim, this proof carries over to the equation

( )

(

)

( )

0 x0 0 x0 , , . z E G

α γ =α ∈ γ∈ (3c) A new function is now introduced by

( )

x : inf

{

( )

y 0

( )

x0 0 and x y x0 ,x y0, E

}

.

µ = µ +λα λ≥ ≤ +λ ∈ (4)

If λ=0 then x≤ +y λx0 is xy and therefore

( )

( )

( )

( )

0 0

( )

0 .

p x µ x µ y µ y α x

− − ≤ ≤ = +

For λ>0, x≤ +y λx0 implies 0

x y

x

λ ≤ +λ and yields

( )

0 0

x y

p µ α x

λ λ

   

 

    because of the definition of α0

( )

x0 . Hence

( )

( )

0

( )

0 for 0

p x µ y λα x λ

− − ≤ + >

and

( )

( )

p x µ x

− − ≤  (5)

follows which implies in, particular, µ

( )

x ≥0 for x≥0, as − −p

( )

x is positive.

Taking y=x and λ =0 one has

( )

( )

( )

( )

p x µ x µ x p x

− − ≤  ≤ ≤ (6)

in particular − − ≤p

( )

x µ

( )

xp x

( )

. If x1≤x2 and x2≤y2+λx0 in (4) then

( )

x1

( )

x2

µ ≤µ (7) follows i.e. monotonicity.

Consider now, for xiE i, 1, 2,=

( )

xi inf

{

( )

yi i 0

( )

x0 yi E, i 0 and xi yi ix0

}

,

µ = µ +λ α ∈ λ ≥ ≤ +λ

then

(

x1 x2

)

(

y1 y2

) (

1 2

) ( )

0 x0

( )

y1 1 0

( )

x0

( )

y2 2 0

( )

x0

µ + ≤µ + + λ λ α+ ≤µ +λ α +µ +λ α

i.e.

(

x1 x2

)

( )

x1

( )

x2 .

µ + ≤µ +µ

Now for σ >0,

( )

x : inf

{

( )

yσ σ 0

( )

x0 yσ E, σ 0 and x yσ σx0

}

(11)

and

( )

x : inf

{

( )

y 0

( )

x0 y E, 0 and x y x0

}

.

σµ = µ σ +σλα ∈ λ≥ σ ≤σ +λσ

The mapping

(

y

) (

 σ σλ σy,

)

), 0,>

is, for fixed σ >0, bijective, therefore

( )

x

( )

x , 0,

µ σ =σµ σ ≥ (8) holds because for σ=0, one has µ

( )

0x =µ

( )

0 =0. So µ is sublinear.

We now show that µ also satisfies the equations of the assertions of the theorem. Take x≤ +y λx0 from the defining set of µ

( )

x . Then

( )

(

0

)

( )

( )

0 follows,

S xS yx =S yS x

contributing

( )

(

S y

)

0

(

S x

( )

0

)

( )

y 0

( )

x0

µ +λα =µ +λα

to µ

(

S x

( )

)

. Conversely, S x

( )

≤ +y λx0 contributing µ

( )

y +λα0

( )

x0 to µ

(

S x

( )

)

yields by applying Tx

(

0

)

( )

0

(

0

)

.

x x

xT yx =T y +T λx

Hence

( )

(

Tx y

)

1 0

(

T0

(

x0

)

)

( )

y 1 0

(

x0

)

( )

y

( )

x0 .

µ + ⋅α λ =µ + ⋅α λ =µ +λα

This implies

( )

(

S x

)

( )

x .

µ =µ (9) The proof of the remaining two equations of the assertions follows almost verbatim this pattern of proof and one gets:

( )

(

Tz x

)

( )

x , ,x z E

µ =µ ∈

( )

(

x

)

( )

x , x E, G

µ γ =µ ∈ γ∈

and (6) implies µ

( )

e =1. Hence µ∈Mp is proved which implies

µ µ= (10) because of (6) and the minimality of µ∈Mp.

Now, looking again at the definition (4) of µ and putting y=0, 1λ= and x=x0 one gets

( )

x0 0

( )

x0

µ ≤α (11) which together with (1) yields µ

( )

x0 ≤α0

( )

x0 ≤µ

( )

x0 and in combination with (1) and (10) gives

( )

( )

0 x x , x E.

α =µ ∈ (12)

(12)

( )

{

( )

( )

}

0 x0 sup p x1 x2 ,x x1 2 E and x1 x0 x2 .

α = − − −µ ∈ ≤ +

and

( )

{

( )

( )

}

0 y0 sup p y1 y2 y y1, 2 E and y1 y0 y2 .

α = − − −µ ∈ ≤ +

and since − −p

( )

x and −µ

( )

x are superlinear, one has

( )

1

( )

2

( )

1

( )

2

(

1 1

)

(

2 2

)

0

(

0 0

)

p x µ x p y µ y p x y µ x y α x y

− − − − − − ≤ − − − − + ≤ +

which implies α0

( )

x0 +α0

( )

y0 ≤α0

(

x0+y0

)

that is α0

( )

x is superadditive and because of (12) positively homogeneous, i.e. superlinear. This implies that µ

( )

x 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

( )

( )

{

}

:

p

U = x p x = − −p x and λ0

( )

x := p x

( )

for xUp. Then

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

ii) p E: →R is continuous and λ0:UpR is a positive, continuous linear functional with λ0

( )

e =1.

iii) Any µ:ER satisfying Theorem 6 is a positive, continuous linear extension

of λ0.

Proof. i): Obviously, eUp and, hence, xUp =/

{ }

0 . For ,

p E E

xUEx e≤ ≤x x e holds and this is an inequality in E and Up which

proves i).

ii): p x

( )

and − −p

( )

x are both monotone, p x

( )

sublinear and

( )

( )

p x p x

− − ≤ superlinear. As for xUp, p x

( )

= − −p

( )

x holds, p x

( )

is linear

and positive on Up. If − ≤ ≤e x e then − ≤1 p x

( )

≤1, and p x

( )

≤1 s. th. norm of p is

( )

{

}

sup 1

p = p x − ≤ ≤e x e = and p is in 0 continuous hence also for any xE

and

( )

, .

E

p xx xE Hence, we get for any µ in Theorem 6

( )

( )

( )

E E

x p x µ x p x x

− ≤ − − ≤ ≤ ≤

implying the continuity of µ and µ ≤1, even µ =1 because µ

( )

e =1. In

particular, this holds also for λ0. □ It is remarkable that with respect to the continuity properties, the continuity of

, x, ,i

S T xE and γ ∈G, do not play any role.

Definition 5. With the notations of Corollary 2 any such µ is called a Banach limit of λ0.

One defines

(

)

{

0

}

(13)

(

)

Ban−Lim E e p S T G, , , , □, , of course, depends on the parameters S T, □, ,G but in

order to make the notation for the following not too cumbersome we will mostly omit them and write simply Ban−Lim

(

E e p, ,

)

.

Proposition 7. For Ban−Lim

(

E e p, ,

)

. the following statements hold:

i) Ban−Lim

(

E e p, ,

)

. is a convex subset of Bs E

( )

′ of the base normed Banach

space E′, the dual space of E.

ii) Ban−Lim

(

E e p, ,

)

. is weakly-*-closed, weakly closed and also □′-closed,

where E

denotes the usual dual norm of

E

of E′.

iii) Ban−Lim

(

E e p, ,

)

. is a superconvex base set contained in Bs E

( )

′ .

Proof. i): Let

(

1

)

(

1

)

1

, , : , , 0, 1 , ,

n

n n i i

i

n N

α α α α α α

=

 

∈ Ω = ≥ = ∈ 

  be the set of

all abstract convex combinations, then, for λ ∈i Ban−Lim

(

E e p, ,

)

, 1≤ ≤i n, obvi-

ously

(

)

1

Ban Lim , , , n

i i i

E e p

α λ

=

∈ −

as

( )

1 0 n i i i x α λ = ≥

for x≥0,

( )

1 1 1

, 1

n n n

i i i i i

i i i

p e

α λ α λ α

= = =

≤ = =

and obviously all other equations in Theorem 6 are satisfied, too.

ii): One first proves that Ban−Lim

(

E e p, ,

)

. is □′-closed, because from this , the

other two assertions of ii) then follow at once. If λ ∈n Ban−Lim

(

E e p, ,

)

, nN, and

there is a λE′ with λ λn− ∗′→0 for n→ ∞, then λn

( )

x −λ∗

( )

x →0, i.e.

( )

( )

lim n

n→∞λ x =λ∗ x follows for xE, which implies λ∗

( )

x ≥0, for

( )

( )

( )

0, 1, , ,

x≥ λ∗ e = λ∗ xp x xX and λ∗

( )

x =λ0

( )

x for xUp. Analogously, one shows the other equations in Theorem 6 for λ because they hold for the

( )

, , . n x x E n N

λ ∈ ∈

iii): Obviously, Ban−Lim

(

E e p, ,

)

Bs E

( )

′ holds, hence Ban−Lim

(

E e p, ,

)

. is

bounded and because of ii) □′-closed. Then, it is a general result that

(

)

Ban−Lim E e p, , . is superconvex (see [13], Theorem 2.5).

□ Because Ban−Lim

(

E e p, ,

)

. is as a subset of Bs E

( )

′ trivially a base set

(

)

(

Ban Lim , ,

)

: Ban Lim

(

, ,

)

CE e p =R+E e p

is a proper cone and

(

, ,

)

:

(

(

, ,

)

)

(

(

(

, ,

)

)

anim E e p =C BanLim E e pC BanLim E e p

B L

is a base normed ordered linear space. To simplify notation, we will write

(

E p,

)

instead of

(

E e p, ,

)

in the following.

Theorem 8. If the norm induced by Ban−Lim

(

E p,

)

in Ban−Lim

(

E p,

)

is

denoted by □Ban and the topology induced by the weak-*-topology σ

(

E E′,

)

in

(

)

an− im E p,

B L by τB∗, then

(

)

(

(

)

)

(

an im E p, , Ban,C Ban Lim E p, ,τB

)

− −

B L □

(14)

( ) (

)

(

E′, □′,C EE E′,

)

.

Proof. As Ban−Lim

(

E p,

)

is a weakly-*-closed base set and a subset of Bs E

( )

which is weakly-*-compact because of Alaoglu-Bourbaki it is also weakly-*-compact and the space Ban−Lim

(

E p,

)

generated by the closed cone C

(

Ban−Lim

(

E p,

)

)

(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.

References

[1] Armario, R., Garcia-Pacheco, F.J. and Pérez-Fernández, F.J. (2013) On Vector-Valued Ba-nach Limits. Functional Analysis and Its Applications, 47, 82-86. (In Russian) (Transl. (2013) Functional Analysis and Its Applications, 47, 315-318.)

[2] Davis, R.B. (1968) The Structure and Ideal Theory of the Predual of a Banach Lattices.

Transactions of the AMS, 131, 544-555.

https://doi.org/10.1090/S0002-9947-1968-0222604-8

[3] Min, K.Ch. (1983) An Exponential Law for Regular Ordered Banach Spaces. Cahiers de Topologie et Géometrie Differentielle Catégoriques, 24, 279-298.

[4] Wong, Y.C. and Ng, K.F. (1973)Partially Ordered Topological Vector Spaces. Oxford Ma-thematical Monographs, Clarendon Press, Oxford.

[5] Pumplün, D. (1999) Elemente der Kategorientheorie. Spektrum Akademischer Verlag, Heidelberg, Berlin.

[6] Pumplün, D. (2002) The Metric Completion of Convex Sets and Modules. Results in Ma-thematics, 41, 346-360.https://doi.org/10.1007/BF03322777

[7] Jameson, C. (1971) Ordered Linear Spaces. Lecture Notes in Mathematics (Volume 141), Springer, Berlin.

[8] Pumplün, D. (1995) Banach Spaces and Superconvex Modules. In: Behara, M., et al. (Eds.),

Symposia Gaussiana, de Gruyter, Berlin, 323-338.

https://doi.org/10.1515/9783110886726.323

[9] Pumplün, D. and Röhrl, H. (1989) The Eilenberg-Moore Algebras of Base Normed Spaces. In: Banaschewski, Gilmour, Herrlich, Eds., Proceedings of the Symposium on Category Theory and its Applications to Topology, University of Cape Town, Cape Town, 187-200. [10] Ellis, E.J. (1966) Linear Operators in Partially Ordered Normed Vector Spaces. Journal of

(15)

[11] Klee Jr., V.L. (1954) Invariant Extensions of Linear Functionals. Pacific Journal of Mathe-matics, 4, 37-46.

[12] Roth, W. (2000) Hahn-Banach Type Theorems for Locally Convex Cones. Journal of the Australian Mathematical Society Series A, 68, 104-125.

https://doi.org/10.1017/S1446788700001609

[13] Pumplün, D. (2011) A Universal Compactification of Topological Positively Convex Sets and Modules. Journal of Convex Analysis, 8, 255-267.

[14] Pumplün, D. (2003) Positively Convex Modules and Ordered Normed Linear Spaces. Jour-nal of Convex AJour-nalysis, 41, 109-127.

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