Fréchet AM-Spaces and AL-Spaces

Mustapha Laayouni

doi:10.34257/GJSFRF258652

1. Introduction

In this section, we provide a brief overview of definitions and known results in Fréchet spaces. (For more details, see ^{[5, 7, 8]}).

In applications, the topology on the spaces studied is not normable but is often defined in a more appropriate way by means of a system of seminorms. We are interested in the case where these systems are countable and more specifically in Fréchet cas. Recall that a Fréchet space is a complete metrizable topological linear space E having a neighbourhood basis $\left( V _ { n } \right) _ { n }$ of zero consisting of convex sets $V _ { n }$ such that $V _ { n + 1 } \subset V _ { n }$ for all $n \in$ N. The topology of E can be generated by an increasing sequence $( \left| \right| \left| \right| _ { n } ) _ { n }$ of separating (ie ∩ker $( | | \mathbf { \partial } | | _ { n } ) = \{ 0 \} )$ seminorms which are gauges of $\left( V _ { n } \right) _ { n }$ , that is to say, $\left| \left| x \right| \right| _ { n } ^ { n } = \operatorname* { i n f } \left\{ \alpha > 0 : \alpha ^ { - 1 } x \in V _ { n } \right\}$ for all $n \in \mathbb N$ and $x \in E$ . The null space of the seminorm $| | \mathbf { \Psi } | | _ { n }$ is the closed subspace of E given by ker $( | | | | | _ { n } ) =$ $\{ x \in E : | | x | | _ { n } = 0 \}$ $E _ { n }$ denotes the normed space obtained by equipping $E / \ker \left( | | \mathbf { \mathbf { \Omega } } | | | _ { n } \right)$ with the norm $\lvert \lvert x +$ ker $( \rfloor | \left. \right. _ { n } ) \rvert | _ { n } = | | x | | _ { n }$ for all $x \in E$ $\overline { { E _ { n } } }$ denotes the completion of $E _ { n } . \ \pi _ { n }$ (respectively, $\overline { { \pi _ { n } } } )$ is the natural homomorphism from E to $E _ { n }$ (respectively, to ${ \overline { { E _ { n } } } } )$ , which is obviously continuous. Let $m \geqslant n$ define the $\pi _ { n m } : E _ { m } \to E _ { n }$ by $\pi _ { n m } ( x + \ker ( | | | _ { m } ) ) = x + \ker ( | | | | _ { n } )$

wich is clearly a continuous homomorphism onto $E _ { n }$ . Consequenctly, $\pi _ { n m }$ can be uniquely extended to a continuous homomorphism $\overline { { \pi _ { n m } } }$ from $\overline { { E _ { m } } }$ into ${ \overline { { E _ { n } } } }$

It is an important tool in work to represent each Fréchet space as the projective limit of a sequence of Banach spaces.

Let $X _ { k }$ be a sequence of Banach spaces and assume that for each $k \in \mathbb N$ a continuous map $\theta _ { k } : X _ { k + 1 } \to X _ { k }$ is given. We say that this constitutes a projective system of Banach spaces

Definition 1.1 The subset

endowed with the relative product topology is called the projective limit of the projective system.

$\prod_{k \in \mathbb{N}} X_k$ is a space under coordinatewise defined algebraic operations.

Since the product of complete spaces is complete (^[7], p. 194) and the topology on $\prod_{k \in \mathbb{N}} X_k$ is generated by the sequence of seminorms

$\prod_{k \in \mathbb{N}} X_k$ is a Fréchet space.

It is not hard to prove that $\varprojlim X_k$ is a closed subspace of $\prod_{k \in \mathbb{N}} X_k$ , so $\varprojlim X_k$ is a Fréchet space, too. We might define another topology besides the $(q_n)$ -topology on $\varprojlim X_k$ by

With these notations, we have:

Lemma 1.1 The two systems of seminorms $(q_n)_n$ and $(p_n)_n$ define equivalent topologies on $\varprojlim X_k$ .

Proof. Let $( x _ { k } ) _ { k } \in \varprojlim { X _ { k } }$ . For $n \in \mathbb N$ there is $c _ { n } > 0$ such that $| | x _ { n } | | _ { n } \leqslant$ $q _ { n } \left( ( x _ { k } ) _ { k } \right) \leqslant c _ { n } | | x _ { n } | | _ { n }$ since

Therefore the seminormsystem $p _ { n } \left( ( x _ { k } ) _ { k } \right) ~ = ~ | | x _ { n } | | _ { n }$ defines an equivalent topology on $\varprojlim X _ { k }$

We now consider projective systems

where each $\overline { { E _ { n } } }$ is a Banach space with norm $| | { \big | } | _ { n }$ and each $\overline { { \pi _ { n ( n + 1 ) } } }$ is the continuous linear map as described above. Then we realize that

is a Fréchet space. Moreover, it is well known that the canonical map $\psi : E \to \prod_n \overline{E_n}$ defined by $\psi(x) = \left( \overline{\pi_n}(x) \right)$ is an isomorphism onto (^[8], p. 230). Henceforth we will freely identify E with $\varprojlim \overline{E_n}$ :

Now, we will equip E with an order $\leqslant ,$ compatible with the algebraic structure of $E .$ Recall that the partially ordered space $( E , \leqslant )$ is a lattice if each pair of elements $x , y \in E$ has a supremum (or least upper bound) and an infimum (or greatest lower bound). We denote the supremum and infimum of two elements $x , y \in E$ by $x \vee y ,$ and $x \wedge y$ respectively. For a vector $x$ in the Riesz space $( E , \leqslant )$ , the positive part $x ^ { + }$ , the negative part $x ^ { - }$ , and the absolute value x are defined by $x ^ { + } = x \vee 0 , x ^ { - } = - x \vee 0$ and $| x | = x ^ { + } \vee x ^ { - }$ . The reader can find several details concerning Riesz spaces in ^[1]. A seminorm $| | \mathbf { \Psi } | | _ { n }$ on a Riesz space $( E , \leqslant )$ is a lattice seminorm (or a Riesz seminorm) $\mathrm { i f } \ | x | \leqslant | y |$ implies $\vert \vert x \vert \vert _ { n } \leqslant \vert \vert y \vert \vert _ { n }$ or, equivalently,

$| | \mathbf { \Psi } | | _ { n }$ is absolute, which means that $\| x \| _ { n } = \| \left| x \right| \| _ { n }$ for all $x ;$ and
$| | \mathbf { \Psi } | | _ { n }$ is monotone on the positive vectors: $0 \leqslant x \leqslant y$ implies $\vert \vert x \vert \vert _ { n } \leqslant \vert \vert y \vert \vert _ { n }$

A subset $S$ of the Riesz space $E$ is solid if $\left| y \right| \leqslant \left| x \right|$ and $x \in S$ imply $y \in S$ . It follows from ^[1] that the gauge $| | \bigstar | | _ { n }$ of $V _ { n }$ is a lattice seminorm, if and only if $V _ { n }$ is solid. Now consider the case where all $V _ { n }$ are convex-solid, that is, for all n the seminorm $| | \bigstar | | _ { n }$ is lattice. $( E , \leqslant , | | \mathbf { \omega } | | _ { n } )$ is called locally convex-solid Riesz spaces. If E is in addition complete, then $( E , \leqslant , | | \mathbf { \omega } | | _ { n } )$ is called a Fréchet lattice. In the sequel, we shall use the previous notations.

2. Am and al context in Fréchet lattice

Recall that a vector subspace F of a E is a Riesz subspace if it is closed under the lattice operations on E. That is $x \in F$ implies $| x | \in F$ . A solid vector subspace $F$ of $E$ is called an ideal. A subset S of $E$ is order closed if $\{ x _ { \alpha } \} \subset S$ and $x _ { \alpha } \stackrel { o } { \to } x$ imply $x \in S$ . An order closed ideal is called a band. an ideal F is a band if and only if $\{ x _ { \alpha } \} \subset F$ and $0 \leqslant x _ { \alpha } \uparrow x$ imply $x \in F$

Lemma 2.1 Let $(E, \leqslant, (\lVert ~ \rVert_n))$ be a Fréchet lattice. Then for each $n$ , $\ker(\lVert ~ \rVert_n)$ is an ideal.

Proof. Fix an integer $n$ , since the seminorms are lattice then $\vert\vert\vert x \vert\vert\vert_n = \lVert x \rVert_n = 0$ when $x \in \ker(\lVert ~ \rVert_n)$ . The Riesz subspace $\ker(\lVert ~ \rVert_n)$ satisfies $|y| \leqslant |x|$ and $x \in \ker(\lVert ~ \rVert_n)$ imply $\lVert y \rVert_n = \vert\vert\vert y \vert\vert\vert_n \leqslant \vert\vert\vert x \vert\vert\vert_n = \lVert x \rVert_n = 0$ . Then $y \in \ker(\lVert ~ \rVert_n)$ and $\ker(\lVert ~ \rVert_n)$ is an ideal.

Definition 2.1 Let $\pi_n(x), \pi_n(y) \in E_n$ . Then we write $\pi_n(x) \leqslant \pi_n(y)$ whenever there exist elements $u \in \pi_n(x)$ and $v \in \pi_n(y)$ such that $u \leqslant v$ .

We can easily verify that the following statements are equivalent, and therefore, this definition is well justified:

$\pi _ { n } \left( x \right) \leqslant \pi _ { n } \left( y \right)$
For all $u \in \pi _ { n } \left( x \right)$ ther exists a $v \in \pi _ { n } \left( y \right)$ satisfying $u \leqslant v$
For all $u \in \pi _ { n } \left( x \right)$ and for all $v \in \pi _ { n } \left( y \right)$ there exists $w \in$ ker $( | | \ | | | _ { n } )$ satisfying $w \leqslant v - u$

Proposition 2.1 For each $n$ , $E_n$ endowed with the partial ordering $\leqslant$ is a Riesz space. In particular, $\pi_n(x) \vee \pi_n(y) = \pi_n(x \vee y)$ and $\pi_n(x) \wedge \pi_n(y) = \pi_n(x \wedge y)$ , for all $x, y \in E$ .

Proof. See ^[6].

In Riesz space theory, two special types of Riesz seminorms, namely abstract L and M-seminorms, play an important role.

Definition 2.2 A lattice seminorm $\lVert\rVert_n$ on a Riesz space is:

an M-seminorm if $x, y \geqslant 0$ implies $\lVert x \vee y \rVert_n = \max \left\{ \lVert x \rVert_n, \lVert y \rVert_n \right\}$
an L-seminorm if $x, y \geqslant 0$ implies $\lVert x + y \rVert_n = \lVert x \rVert_n + \lVert y \rVert_n$ .

Since $( E , | | | _ { n } )$ is a seminormed Riesz space and that ker $( \left| \left| \mathbf { \alpha } \right| \right| \mathbf { \alpha } _ { n } )$ is an ideal of $E$ , then it is known that the quotient space $E _ { n }$ is also a seminormed Riesz space when equipped with its quotient Seminorm. It is enough to see that if $| | \mathbf { \Psi } | | _ { n }$ is an A-seminorm (resp. L-seminorm), then the same is true for the quotient seminorm. Specifically, we have the following

Lemma 2.2 If $(\alpha_k)$ and $(\beta_k)$ are two convergent sequences of positive real numbers, then

Proof. Assume at first that $\alpha _ { k } \leqslant \beta _ { k }$ for all sufficiently large $k \in \mathbb N$ , then $\alpha \leqslant \beta$ an hence the lemma holds. Now assume that for each $s \in \mathbb N$ there are tow integers $k _ { s } \geqslant s$ and $k _ { s } ^ { \prime } \geqslant s$ for which $\alpha _ { k _ { s } } < \beta _ { k _ { s } }$ and $\beta _ { k _ { s } ^ { \prime } < \alpha _ { k _ { c } ^ { \prime } } }$ . From this it follows that $\alpha = \operatorname* { l i m } _ { s \infty } \alpha _ { k _ { s } } \leqslant \operatorname* { l i m } _ { s \infty } \beta _ { k _ { s } } = \beta$ and $\beta = \operatorname* { l i m } _ { s \infty } \beta _ { k _ { s } ^ { \prime } } \leqslant \operatorname* { l i m } _ { s \infty } \alpha _ { k _ { s } ^ { \prime } } = \alpha$ Then $\alpha = \beta$ and the lemma follows straightforwardly.

Which allows us to state the following.

Proposition 2.2 Let $\lVert ~ \rVert_n$ be a seminorm on a Riesz space E. We have:

(i) $\lVert ~ \rVert_n$ is an M-seminorm implies that $(\overline{E_n}, \lVert ~ \rVert_n)$ is an AM-space.

(ii) $\lVert ~ \rVert_n$ is an L-seminorm implies that $(\overline{E_n}, \lVert ~ \rVert_n)$ is an AL-space.

Proof. Let x, y be in $\overline { { E _ { n } } }$ There exist $( x _ { k } ) _ { k }$ and $( y _ { k } ) _ { k }$ in E such that
lim $| | \overline { { x } } - \overline { { \pi _ { n } } } ( x _ { k } ) | | _ { n } = 0$ and lim $| | \overline { { y } } - \overline { { \pi _ { n } } } ( y _ { k } ) | | _ { n } = 0$ \

(i) It follows from the continuity of $\overline { { \pi _ { n } } }$ that,

and this leads to the desired conclusion.

(ii) Likewise, we have,

Thus the proposition has been proved.

Definition 2.3 A Fréchet lattice $(E, \leqslant, \lVert ~ \rVert_n)$ is an AMF-space (respectively, ALF-space) if for each $n \geqslant 1$ , $\lVert ~ \rVert_n$ is an M-seminorm (respectively, L-seminorm).

Example 2.1 Consider the universal sequence space $\mathbb { R } ^ { \mathbb { N } }$ with the usual operations on the coordinates. The partially ordered is the pointwise ordering, $( x _ { k } ) _ { k } \ \leqslant \ ( y _ { k } ) _ { k }$ if $x _ { k } ~ \leqslant ~ y _ { k }$ for each $k . \mathrm { ~ \mathbb ~ { ~ R ^ { N } ~ } ~ }$ becomes a Riesz space satisfying $( x _ { k } ) _ { k } \lor ( y _ { k } ) _ { k } \ : = \ : ( x _ { k } \lor y _ { k } ) _ { k }$ and $\left| ( x _ { k } ) _ { k } \right| = ( \left| x _ { k } \right| ) _ { k } .$ For $n \in \mathbb { N } .$ define the seminorm $q _ { n }$ by $q _ { n } \left( ( x _ { k } ) \right) = \operatorname* { m a x } \left\{ | x _ { k } | ; k = 1 , 2 , . . . , n \right\}$ We see that $\left| ( x _ { k } ) _ { k } \right| \leqslant \left| ( y _ { k } ) _ { k } \right|$ implies $q _ { n } \left( ( x _ { k } ) _ { k } \right) \leqslant q _ { n } \left( ( y _ { k } ) _ { k } \right)$ for all $n \geqslant 1$ . Then $\left( \mathbb { R } ^ { \mathbb { N } } ; ( q _ { n } ) _ { n } \right)$ is a Fréchet lattice. More precisely, $\left( \mathbb { R } ^ { \mathbb { N } } ; ( q _ { n } ) _ { n } \right)$ is an AMF-space. Indeed, for $n \in \mathbb { N } , \mathrm { i f ~ } ( x _ { k } ) _ { k }$ and $( y _ { k } ) _ { k }$ are in $\mathbb { R } ^ { \mathbb { N } }$ then

Thus $q _ { n }$ is an M-seminorm .

Furthermore, we have the following.

Theorem 2.1 If $(E_k, \lVert ~ \rVert_k)_k$ is a sequence of AM-spaces then the cartesian product $E = \prod_n E_n$ is an AMF-space.

Proof. For each non-negative $n _ { \mathrm { : } }$ the mapping $p _ { n } : E \to [ 0 , + \infty [$ defined by

becomes a seminorm satisfying ker $( 0 _ { k }$ denote the zero of $E _ { k } )$ . Obviously, we have the following assertions:

If ${ \overline { { E _ { n } } } } = E { \big / } \ker ( p _ { n } )$ denotes the quotient space of E by the subspace ker $\left( p _ { n } \right)$ then ${ \overline { { E _ { n } } } } \equiv E _ { 1 } \times E _ { 2 } \times \ldots \times E _ { n }$
$\overline { { p _ { n } } } \left( \overline { { ( x _ { k } ) _ { k } } } \right) = p _ { n } \left( ( x _ { k } ) _ { k } \right) = \operatorname* { m a x } \left\{ | | x _ { k } | | _ { k } ; k = 1 , 2 , . . . , n \right\}$ . E is a Riesz space under the usual ordering where $( ( x _ { k } ) _ { k } ) \leqslant ( ( y _ { k } ) _ { k } )$ whenever $x _ { k } \leqslant y _ { k }$ . The infimum and supremum of two vectors x and y are given by

Then, $( ( x _ { k } ) _ { k } ) ^ { + } \ = \ ( ( x _ { k } ) _ { k } \lor ( 0 _ { k } ) _ { k } ) \ = \ ( x _ { k } ^ { + } ) _ { k }$ and $\left( \left( x _ { k } \right) _ { k } \right) ^ { - } \ = \ \left( x _ { k } ^ { - } \right) _ { k }$ So, $| ( ( x _ { k } ) _ { k } ) | = ( | x _ { k } | ) _ { k }$ and,

Moreover,

We therefore conclude that the $\left( { p _ { n } } \right) _ { n }$ are indeed AM-seminorms, as claimed.

The next result gives a certain way of looking at all possible AMF-spaces.

Theorem 2.2 Every AMF-space E is the projective limit of a sequence of AM-spaces.

Proof. Let $(\lVert ~ \rVert_n)_n$ be a sequence of seminorms on E which defines its topology. With the notation above, it follows from Proposition 2.2 that $(\overline{E_n}, \lVert ~ \rVert_n)$ is an AM-space. Furthermore, according to (1.1) $E = \varprojlim \overline{E_n}$ . Obviously, the topological isomorphism $\psi : x \to (\overline{\pi_n}(x))_n$ that allowed us to make this identification is lattice-like. Therefore, the proof is complete.

On the other hand, using the notations above, we have the following.

Theorem 2.3 If $(E_k, \lVert ~ \rVert_k)_k$ is a sequence of AL-spaces then the cartesian product $E = \prod_n E_n$ is an ALF-space.

Proof. For each non-negative n, the mapping $p _ { n } : E \to [ 0 , + \infty [$ defined by

becomes a seminorm satisfying ker $( 0 _ { k }$ denote the zero of $E _ { k } )$ . Obviously, we have the following assertions:

$\bigcap \ker ( p _ { n } ) = \left\{ 0 _ { E } \right\}$
If ${ \overline { { E _ { n } } } } = E { \big / } \ker ( p _ { n } )$ denotes the quotient space of E by the subspace ker $\left( p _ { n } \right)$ then ${ \overline { { E _ { n } } } } \equiv E _ { 1 } \times E _ { 2 } \times \ldots \times E _ { n }$
${ \overline { { p _ { n } } } } \left( { \overline { { ( x _ { k } ) _ { k } } } } \right) = p _ { n } \left( ( x _ { k } ) _ { k } \right) = \sum _ { k = 1 } ^ { n ^ { \prime } } | | x _ { k } | | _ { k } .$ E is a Riesz space under the

usual ordering where $( ( x _ { k } ) _ { k } ) \leqslant ( ( y _ { k } ) _ { k } )$ whenever $x _ { k } \leqslant y _ { k }$ for each $k = 1$ The infimum and supremum of two vectors x and y are given by

Then, $( ( x _ { k } ) _ { k } ) ^ { + } \ = \ ( ( x _ { k } ) _ { k } \lor ( 0 _ { k } ) _ { k } ) \ = \ \left( x _ { k } ^ { + } \right) _ { k }$ and $\left( ( x _ { k } ) _ { k } \right) ^ { - } \ = \ \left( x _ { k } ^ { - } \right) _ { k }$ So,

$| ( ( x _ { k } ) _ { k } ) | = ( | x _ { k } | ) _ { k }$ and,

Moreover,

Remark 2.1 (a) A linear functional $\varphi : E \to \mathbb{R}$ on a Riesz space E is strictly positive if $x > 0$ implies $\varphi(x) > 0$ . In functional analysis, several Fréchet spaces are not normable. It is important to notice that not every nonormable Fréchet lattice admits a strictly positive linear form. Indeed, it is enough to consider the mapping $x \mapsto \lVert x \rVert_\varphi = \varphi(|x|)$ from $E$ to $[0, +\infty[$ to realize that $E$ becomes normable whenever $\varphi$ is strictly positive linear functional on $E$ .

(b) A not normable Fréchet lattice must not admit an order unit. Otherwise, if e is an order unit in $E_n$ then the mapping $x \to \lVert x \rVert_e = \inf \{ \lambda > 0 : |x| \leqslant \lambda e \} = \min \{ \lambda \geqslant 0 : |x| \leqslant \lambda e \}$ from $E$ to $[0, +\infty[$ , will be a norm on $E$ .

(c) For a not normable Fréchet lattice its positive cone has empty interior in any linear topology.

Recall that a topological space X is completely regular if for each member x of X and each neighborhood $U$ of $x$ there is a continuous function $f$ on $X$ to the closed unit interval such that $f ( x ) = 0$ and $f$ is identically one on $X - U$ . A Hausdorff space is called a k-space if every subset intersecting each compact subset in a closed set is itself closed. Examples of k-spaces are locally compact and first countable spaces (^[7], p. 231) . A Hausdorff space X is called hemicompact if there is a countable compact exhaustion $K _ { 1 } \subset K _ { 2 } \subset \ldots \subset K _ { n } \subset K _ { n + 1 } \subset \ldots$ . of X such that for each compact subset $K \subset X$ there is $n \in \mathbb { N }$ so that $K \subset K _ { n }$ (for instance, $X = \mathbb { R } , K _ { n } = [ - n , n ]$ 2 $n = 1 , 2 , 3 ; \ldots )$ . We denote by $\mathcal { C } \left( \boldsymbol { X } \right)$ the space of all continuous real-valued functions on X (with pointwise operations). The following determine a class of spaces for which this one becomes a Fréchet space (^[5], p. 69).

Theorem 2.4 Let X be a completely regular space. Then $\mathcal { C } \left( X \right)$ is a Fréchet space iff X is a hemicompact k-space.

X denote a completely regular hemicompact k-space and let $\left( K _ { n } \right) _ { n }$ stand for a countable compact exhaustion. The seminorms on $\mathcal { C } \left( X \right)$ are given by: $| | f | | _ { n } = \operatorname* { s u p } { \big \{ } | f ( x ) | : x \in K _ { n } { \big \} }$ . The ordering is defined pointwise. That is, $f \leqslant g$ whenever $f ( x ) \leqslant g ( x )$ for each $x \in X$ . As in Banach lattices, we will show that FAM-spaces are the abstract versions of the $\mathcal { C } \left( X \right)$ -spaces (X completely regular hemicompact k-space).

In algebras where the unit is lacking, the notion of an approximate identity was introduced to fill this gap (see ^[4]) . With this idea in mind, we will introduce the notion of an approximate order unit in Riesz spaces.

3. Approximate order unit

Recall that a directed set is a partially ordered set Λ such that, given $\lambda _ { 1 }$ and $\lambda _ { 2 }$ in Λ, there exists $\lambda \in \Lambda$ with $\lambda _ { k } \leqslant \lambda \left( k = 1 , 2 \right)$ . A net in E is a mapping of a directed set into E.

The principal ideal generated by {e} in E is denoted $E _ { e }$ . Clearly,

It is well known that, if E is either a Banach lattice or an order complete Riesz space, then for each $e \in E _ { + }$ . the principal ideal $E _ { e } ,$ equipped with the norm:

is an AM-space, with unit e ^[1].

Definition 3.1 Let E be a Riesz space. An approximate order unit in E is a net $\{ e _ { \lambda } \} _ { \lambda \in \Lambda }$ of strictly positive elements such that:

$\alpha \leqslant \beta$ implies $e _ { \alpha } \leqslant e _ { \beta }$
$\pi _ { \alpha } \left( e _ { \alpha } \right)$ is an order unit of the completion ${ \overline { { E _ { \alpha } } } } \ o f \ ( E { \ ' } { \Big / } \ker \left( \left| \right| \ \left| \alpha \right) \right)$ , for some increasing sememinorms $\big ( \big | \big | \big | _ { \alpha } \big ) _ { \alpha \in \Lambda }$ on $E$
E is the projective limit of the $\overline { { E _ { \alpha } } }$ , with respect to the homomorphisms $\pi _ { \alpha \beta } : E _ { \beta } = E { \big / } \ker \left( | | \mathbf { \Gamma } | | _ { e _ { \beta } } \right) \to E _ { \alpha } = E { \big / }$ ker $( | | \mathbf { \epsilon } | | _ { e _ { \alpha } } )$ , for each $\alpha \leqslant \beta ,$ that $i s ,$ $E = \varprojlim \overline { { E _ { \alpha } } }$

Example 3.1 Let ${ \mathbb K } _ { n } = \{ 0 , 1 , 2 , . . . , n \}$ endowed with the indiscrete topology. $\mathcal { C } \left( \mathbb { K } _ { n } \right)$ becomes an AM-space having unit the constant function 1. Let us set $e _ { n } ( x ) = 1$ if $x \in \mathbb { K } _ { n }$ and $e _ { n } ( x ) = 0$ otherwise, so that $e _ { n } \in \mathcal { C } \left( \mathbb { N } \right)$ . C (N) will be seen without notice as a Riesz space with countable order approximate identity $\left( e _ { n } \right) _ { n }$

Another example is obteined wehn an order complet Riesz space $E$ is the increasing union of bands $( B _ { e _ { \alpha } } ) _ { \alpha \in \Lambda }$ . A detailed analysis of this case will also be given.

We begin by extend the norm $| | \bigstar \bigstar | | _ { e }$ given on the band $\boldsymbol { B } _ { e _ { \alpha } }$ to the entire space E while preserving the monotonicity (for inclusion) of the kernels. This is why we use bands, which satisfy Riesz’s decomposition theorem, (see also ^[1]).

Theorem 3.1 (F. Riesz) Every band B in an order complete Riesz space E is a projection band. That is, $E = B \oplus B ^ { \perp }$

We retain the notation $| | \bigstar \bigstar | | _ { e }$ for the mapping from E to $\mathbb { R } _ { + }$ defined by $| | x | | _ { e } = | | u + v | | _ { e } = | | u | | _ { e }$ whenever $x = u + v$ is given in the Riesz decomposition $E = B _ { e } \oplus B _ { e } ^ { \scriptscriptstyle \perp }$

Lemma 3.1 Let E be an order complete Riesz space and and let e and f be in E such that $0 < e \leqslant f$ . Then we have:

$| | \big | | _ { f } \leqslant | | \big | | _ { e }$
$B _ { f } ^ { \perp } = \ker \left( | | \mathbf { \Gamma } | | _ { f } \right) \subset \ker \left( | | \mathbf { \Gamma } | | _ { e } \right) = B _ { e } ^ { \perp }$
The mapping $\pi_{ef}: E / \ker(\lVert~\rVert_f) \to E / \ker(\lVert~\rVert_e)$ defined by $\pi_{ef}(x + \ker(\lVert~\rVert_f)) = x + \ker(\lVert~\rVert_e)$ is a continuous injective lattice homomorphism.

Proof. (i) Let $x \in E$ . Since $B _ { e } \subset B _ { f } , { \mathrm { i f ~ } } x = u + v$ in the Riesz decompostion $E = B _ { e } \oplus B _ { e } ^ { \perp }$ then $x = ( u + u ^ { \prime } ) + v ^ { \prime }$ with $u + u ^ { \prime } \in B _ { f }$ and $v ^ { \prime } \in B _ { f } ^ { \bot }$ . Obviously, we have $u ^ { \prime } \in B _ { e } ^ { \bot }$ so that $| u + u ^ { \prime } | = | u | \vee | u ^ { \prime } |$ , which implies

Thus the assetion (i) has been proved.

(ii) and (iii) follow directly from (i).

Property (iii) allows us to identify $E / \ker(\lVert~\rVert_f)$ with a subspace of $E / \ker(\lVert~\rVert_e)$ .

Lemma 3.2 Let F be a dense ideal in a Banach lattice E. If F admits an order unit e, then e is an order unit in E.

Proof. Let $x \in E$ and $\left( a _ { n } \right) _ { n }$ a sequence in F such that $| | x | - | a _ { n } | | \mid \to 0$ when $n + \infty$ . Since the mapping $u | x | \vee u$ is norm continuous, then we can assume (by replacing $\left( \left| a _ { n } \right| \right) _ { n }$ by $( | x | \vee | a _ { n } | ) _ { n } )$ that $| x | \leqslant | a _ { n } |$ holds for each n. For each n there is $\lambda _ { n } > 0$ satisfying $| a _ { n } | \leqslant \lambda _ { n } e$ and so, $| x | \leqslant \lambda _ { n } e$ This means that e is an order unit in E.

Combining lemmas 3.1 and 3.2 leads to the conclusion that $\{ e _ { \lambda } \} _ { \lambda \in \Lambda }$ is an approximate order unit in E.

Theorem 3.2 Let X be a completely regular hemicompact k-space. Then $\mathcal { C } \left( \boldsymbol { X } \right)$ is an AMF-space having a countable order approximate unit.

Proof. Let $\left( K _ { n } \right) _ { n }$ be an admissible exhaustion of X. Define the seminorm $| | \mathbf { \Psi } | | _ { n } { \mathrm { ~ b y ~ } } | | f | | _ { n } = \operatorname* { s u p } { \big \{ } | f ( x ) | : x \in \mathcal { K } _ { n } { \big \} }$ . In view of Theorem 3.1, $\mathcal { C } \left( X \right)$ is a Fréchet space. For each $n \in \mathbb { N }$ , It is not hard to see that $( | | | | _ { n } ) _ { r }$ are lattice seminorms satisfying $f , g \geqslant 0$ implies $| | f \vee g | | _ { n } = \operatorname* { m a x } \left\{ | | f | | _ { n } , | | g | | _ { n } \right\}$ for all $f , g \in { \mathcal { C } } \left( X \right)$ . So it only remains to show the existence of a countable order approximate unit.

Urysohn’s theorem ^[1] shows that for each positive integer $n$ , there exists a continuous function $e_n: X \to [0, 1]$ such that $e_n(x) = 1$ for all $x \in K_n$ and $e_n(x) = 0$ for all $x \in X - K_{n+1}$ . Since the lattice isomorphism $\pi_n$ preserves order units, it follows that $\pi_n(e_n)$ is an order unit in $E_n = \mathcal{C}(X)/\ker(\lVert~\rVert_n)$ . Define a mapping $i_n: E_n \to \mathcal{C}(K_n)$ by $i_n(\pi_n(f)) = f|_{K_n}$ (restriction of $f$ to $K_n$ ). Obviously, $i_n$ is a lattice homomorphism and satisfies $\pi_n(f) = \pi_n(g) \Leftrightarrow f|_{K_n} = g|_{K_n}$ for all $f, g \in \mathcal{C}(X)$ . Since the constant function $1 = i_n(\pi_n(e_n))$ is an order unit in $\mathcal{C}(K_n)$ , then $\pi_n(e_n)$ is an order unit in $\mathcal{C}(X)/\ker(\lVert~\rVert_n)$ . Using Lemma 3.2, we get $\overline{\pi}_n(e_n) = e_n + \ker(\lVert~\rVert_n)$ is an order unit in $\overline{E}_n$ , with $E = \varprojlim \overline{E}_n$ . We see that the statement is fulfilled.

In the classes of Banach lattices, AM-spaces are the abstract versions of the C(K)-spaces (K compact Hausdorff). We will establish a type of Kakutani-Bohnenblust-M. Krein-S. Kerin theorem for Fréchet lattices.

Theorem 3.3 Let E be a Fréchet lattice order complete. E is an AMF-space with a countable approximate order unit if and only if it is lattice isometric to $\mathcal { C } \left( \boldsymbol { X } \right)$ for some completely regular hemicompact k-space X. The space X is unique up to homeomorphism.

Proof. Assume that E is an AMF-space with a countable approximate order unit. Then, by Definition 3.1, $E = \varprojlim B _ { e _ { r } }$ projective limit of a sequence of AM-spaces $\boldsymbol { B _ { e _ { n } } }$ with order unit $\dot { e } _ { n } ~ \left( B _ { e _ { n } } \right.$ is lattice isomorphic to $E / B _ { e _ { n } } ^ { \perp } )$ . It follows from Kakutani-Bohnenblust-M. Krein’s Theorem (see ^[1]) that each $\boldsymbol { B } _ { \boldsymbol { e } _ { n } }$ is lattice isometric to $\mathcal { C } \left( K _ { n } \right)$ for some compact Hausdorff space $K _ { n }$ . The space $K _ { n }$ is unique up to homeomorphism. Since $B_{e_n} \subset B_{e_{n+1}}$ , then up to homeomorphism, $K_n \subset K_{n+1}$ . Thus, $X = \bigcup K_n$ satisfies the desired conclusions.

Conversely, if E is lattice isometric to $\mathcal { C } \left( X \right)$ for some completely regular hemicompact k-space X, Theorem 3.1 asserts that E is an AMF-space with a countable approximate order unit.

It is well known that when replacing a normed vector space with a metric vector space, there is a risk of losing the convexity of the balls, which is a very useful tool in functional analysis. Order is no exception, since one of the nice results of AL and AM-spaces, which is as follows:

Theorem 3.4 ^[2] A Banach lattice E is an AL-space (resp. an AM-space) if and only if $E'$ is an AM-space (resp. an AL-space).

There will be nothing new to add to Fréchet lattices, since we have:

Theorem 3.5 ^[8] If E is locally convex and metrizable, E' is metrizable if and only if E is normable.