§5.5. Combining vertical and lateral extensions
Chapter
5
As we have seen, the lateral extension differs from the vertical extension in the sense that the vertical extensions of Γ and ϕ can always be made, but for lateral extension we had to assume the spaceΓto be stable andϕto be laterally extendable (see 5.4.11). In this section we investigate when one can make a lateral extension of another (say vertical) extension. Furthermore we will compare different extensions and combinations of extensions.
Instead of (ΓL)V and((ΓL)V)L we write ΓLV andΓLV L; similarlyϕLV = (ϕL)V etc.
5.5.1. By Theorem 5.4.18 the following holds for a stable directed linear subspace
∆ofFX and a laterally extendable order preserving linear mapω: ∆→E: If∆ L is stable, thenωL is laterally extendable (and so∆LL exists). If∆V is stable, thenωV is laterally extendable (and so ∆V L exists). We will use these facts without explicit mention.
5.5.2. The following statements follow from the definitions and theorems we have: (a) ΓV ⊂ΓLV andϕLV =ϕV onΓV.
(b) ΓL⊂ΓLV andϕLV =ϕL onΓL. (c) ϕV =ϕL onΓL∩ΓV.
For (d), (e) and (f) letΓV be stable.
(d) ΓLV ⊂ΓV LV andϕV LV =ϕLV onΓLV. (e) ΓV L⊂ΓV LV andϕV LV =ϕV L onΓV L. (f) ϕLV =ϕV L onΓLV ∩ΓV L.
Observe that as a consequence of (a) and (b): If f ∈ΓL and g ∈ΓV andf ≤g (or
f ≥ g), then ϕL(f) ≤ ϕV(g) (or ϕL(f) ≥ ϕV(g)). Moreover, as a consequence of (c) and (d); ifΓV is stable: If f ∈ΓLV andg ∈ ΓV LV and f ≤g (or f ≥g), then
ϕLV(f)≤ϕV LV(g)(orϕLV(f)≥ϕV LV(g)).
5.5.3. Note that if Γ is stable andϕis laterally extendable, then we can extendΓ
to ΓV,ΓL and ΓLV. If, moreover, ΓV is stable, then we can also extend Γ to ΓV L and ΓV LV. However, “more stability” will not give us larger extensions than ΓV LV. Indeed, if ΓLV is stable then ΓLV ⊂ΓLV L = ΓV L (see Theorem 5.5.8). If moreover
ΓV LV is stable, then evenΓV LV L= ΓV LV = ΓV L.
5.5.4 Lemma.
(a) If f ∈ Γ+LV, then there exists a countable Λ ⊂Γ with Λ ≤f and ϕLV(f) =
supϕ(Λ).
(b) If ΓV is stable and f ∈ΓV L+ , then there exists a countableΛ ⊂Γ with Λ≤f andϕV L(f) = supϕ(Λ).
Proof. (a) There existσ1, σ2, . . . inΓLwithσn≤f for alln∈Nandsupn∈NϕL(σn) =
ϕLV(f). Hence, we are done if for everyσin ΓL withσ≤f there is a countable set
Λσ ⊂ {ρ∈Γ :ρ≤f}such that every upper bound forϕ(Λσ)majorizesϕL(σ). But that is not hard to prove. For such a σ, by Lemma 5.4.17 there exists a partition
Chapter
5
(Bm)m∈Nfor which (5.32) holds. Now let Λσ be{PMm=1σ1Bm:M ∈N}.
(b) SupposeΓV is stable. Let(An)n∈N be aϕV-partition forf. Then the setΛf =
{PNn=1f1An : N ∈ N} is a countable subset of ΓV and supϕV(Λf) = ϕV L(f).
Moreover, for everyN ∈Nthere is a countable set ΛN ⊂ {σ∈Γ :σ≤PNn=1f1An}
for whichsupϕ(ΛN) =ϕV(PNn=1f1An). TakeΛ =
S
N∈NΛN.
5.5.5 Theorem. For (b),(c),(d) and (e) letΓV be stable andf be partially in ΓV. (a) If f ∈ΓLV, then
f ∈ΓV ⇐⇒ there exist π, ρ∈Γ withπ≤f ≤ρ6. (b) If f ∈ΓV L, then
f ∈ΓV ⇐⇒ there exist π, ρ∈Γ withπ≤f ≤ρ6. (c) f ∈ΓLV ⇐⇒ f ∈ΓV L and there existπ, ρ∈ΓL withπ≤f ≤ρ. (d) If ϕV(ΓV)is splitting inE, then
f ∈ΓV L ⇐⇒ there existπ, ρ∈ΓV L with π≤f ≤ρ. (e) If ϕV(ΓV)is splitting in E, then
f ∈ΓV L∩ΓLV ⇐⇒ there existπ, ρ∈ΓL withπ≤f ≤ρ.
Proof. The proofs of (a) and (b) are similar to the proof of (c) and therefore omitted. (c) ⇐: By Lemma 5.5.4 (b) there exist countable sets Λ,Υ⊂Γwith Λ≤f−π
and Υ ≤ ρ−f for which supϕ(Λ) = ϕV L(f −π) and supϕ(Υ) = ϕV L(ρ−f). Then Λ +π and ρ−Υare countable subsets of ΓL with Λ +π ≤f ≤ρ−Υ and
supϕL(Λ +π) =ϕV L(f) = infϕL(ρ−Υ). Hence f ∈ΓLV.
⇒: Let f ∈ ΓLV and be partially in ΓV. There exists a π ∈ ΓL for which
f −π ∈ Γ+LV, hence we may assume f ≥ 0. Let (An)n∈N be a ΓV-partition for
f, i.e., f1An ∈ ΓV and thus ϕLV(f1An) =ϕV(f1An) for all n ∈ N (see 5.5.2(a)).
Then ϕLV(f) ≥ PNn=1ϕV(f1An) for all N ∈ N. Let h ∈ E be such that h ≥
PN
n=1ϕV(f1An) for all N ∈ N. From Lemma 5.4.17 we infer that h ≥ ϕL(σ) for
every σ∈ΓL withσ≤f. We conclude thatPnϕV(f1An) =ϕLV(f), i.e., f ∈ΓV L.
(d)⇐: We may assumeπ= 0. Let(An)n∈Nbe aϕV-partition forρwithf1An∈
ΓV for alln∈N. Then 0≤ϕV(f1An)≤ϕV(ρ1An)for alln∈Nand
P
nϕV(ρ1An)
exists inE. Hence, so doesPnϕV(f1An), i.e., f ∈ΓV L.
(e) is a consequence of (c) and (d).
In the following example all functions inΓLV are partially inΓV.
5.5.6 Example. Consider X = N,I = P(N), E =F; let D be a linear subspace of E and let DV be the vertical extension of D with respect to the inclusion map
D→E. LetΓ =c00[D]andϕ: Γ→E beϕ(f) =Pn∈Nf(n). ThenΓV =c00[DV].
6By the definition of ideal in [11] or [40] (note thatΓ
V is directed) this means thatΓV is the smallest ideal inΓLV (and for (b); inΓV L) that containsΓ.
§5.5. Combining vertical and lateral extensions
Chapter
5
Let f ∈ ΓLV. We will show that f(k) ∈ DV and thus that f is partially in ΓV. Letσn, τn ∈ΓL be such thatσn ≤f ≤τn and infn∈Nϕ(τn) = supn∈Nϕ(σn). Then
infn∈N(τn(k)−σn(k))≤infn∈Nϕ(τn−σn) = 0. Sinceσn(k), τn(k)∈Dfor all n∈N, we havef(k)∈DV.
Thus every f ∈ΓLV is partially inΓV. SinceΓV is stable, by Theorem 5.5.5(c) we conclude thatΓLV ⊂ΓV L.
5.5.7 Lemma. Suppose thatΓLV is stable. Then every f ∈ΓLV is partially inΓV. Proof. Let f ∈ ΓLV and letπ, ρ ∈ΓL be such that π ≤f ≤ρ. Let (An)n∈N be a
ϕ-partition for bothπ andρ. Thenf1An ∈ΓLV and π1An ≤f1An ≤ρ1An for all
n∈N. By Theorem 5.5.5(a) we conclude thatf1An∈ΓV.
5.5.8 Theorem. Suppose that ΓV and ΓLV are stable. Then ΓLV ⊂ΓV L= ΓLV L. Write Γ = ΓV L and ϕ= ϕV L. If Γ is stable, then ΓL = Γ and ϕL =ϕ. If ΓV is stable, thenΓV = Γ andϕV =ϕ.
In particular, if ϕL(ΓL) is mediated in E and ϕV(ΓV) is splitting in E, then ΓV,
ΓLV and ΓV L are stable (see Theorem 5.4.25) and thus ΓLV ⊂ Γ, Γ = ΓV = ΓL,
ϕ=ϕV =ϕL, soΓ = Γ(andϕ=ϕ).
Proof. The inclusion ΓLV ⊂ ΓV L follows by Theorem 5.5.5(c) and Lemma 5.5.7. We prove ΓLV L ⊂ ΓV L. For f ∈ Γ+LV L there is a ϕLV-partition for f and since
ΓLV ⊂ΓV L this is also a ϕV L-partition forf, hence there exists a ϕV-partition for
f, i.e.,f ∈ΓV L.
Suppose Γ is stable. Then ΓL = (ΓV L)L = ΓV L = Γ and ϕL =ϕ by Theorem 5.4.18(a).
SupposeΓV to be stable. AsΓV is stable we can apply the first part of the theorem to ΓV instead ofΓ. Indeed, (ΓV)V and (ΓV)LV are stable, since (ΓV)V = ΓV and
(ΓV)LV = ΓV. Hence,(ΓV)LV ⊂(ΓV)V L= ΓV L, i.e.,ΓV ⊂Γ(andϕV =ϕ). SupposeϕL(ΓL)is mediated inEandϕV(ΓV)is splitting inE. ThenΓL,ΓV and
ΓLV are stable by Theorem 5.4.25(a),(b) and (c). Consequently, again by Theorem 5.4.25(b)ΓV L is stable.
5.5.9 Corollary. SupposeE is mediated (and thus splitting), Γ = ΓV L. ThenΓ =
ΓV = ΓL, soΓ = Γ(andϕ=ϕ).
At the end of Section 5.5 we will show that sometimes ΓV L ( ΓLV (Example 5.5.14) and sometimes ΓLV ( ΓV L (Example 5.5.15). Note that this implies that
ΓV LV can be strictly larger then eitherΓV LorΓLV.
Theorem 5.5.8 raises the question whether stability of ΓV entails ΓV L ⊂ ΓLV. In general the answer is negative; see Example 5.5.15. In Theorem 5.5.10 we give conditions sufficient for the inclusion.
5.5.10 Theorem. SupposeΓV is stable. Consider these two statements. (a) For every f ∈Γ+V L there is aρ inΓ+L with f ≤ρ.
Chapter
5
(b) E satisfies:
If Y1, Y2,· · · ⊂E are nonempty countable with infYn= 0 for alln∈N, then there exist y1∈Y1, y2∈Y2, . . . such that
X
n
yn exists in E. (5.43) If (a) is satisfied, then ΓV L⊂ΓLV. (b) implies (a).
Proof. If (a) is satisfied, then by Theorem 5.5.5(c) follows thatΓV L⊂ΓLV.
Suppose (b). Let f ∈ Γ+V L. Let (An)n∈N be a ϕV-partition for f. For n ∈ N, let
Υn⊂Γbe a countable set withf1An≤Υn and
ϕV(f1An) = infϕ(Υn). (5.44)
We may assume σ1An = σ for all σ ∈ Υn. Choose σn ∈ Υn for n ∈ N such that
P n(ϕ(σn)−ϕV(f1An))and thus P nϕ(σn)exist inE. Thenρ:= P n∈Nσn is inΓ+L withf ≤ρ.
5.5.11. We will discuss examples of spacesE for which (5.43) holds.
(I) IfEis a Banach lattice withσ-order continuous norm, thenEsatisfies (5.43) (one can findyn∈Yn withkynk ≤2−n).
(II) Let (X,A, µ) be a complete σ-finite measure space and assume there exists a
g∈L1(µ)with g >0 µ-a.e.. Then the spaceE of equivalence classes of measurable functions X →Rsatisfies (5.43): It is sufficient to prove that if Z1, Z2,· · · ⊂E are nonempty countable with infZn = 0 for all n ∈ N, then there exists z1 ∈ Z1, z2 ∈
Z2, . . . and az∈E such thatzn≤zfor alln∈N(forZn take2nYn). One can prove that such az exists by mapping the equivalence classes of measurable functions into
L1(µ)by the order isomorphismf 7→(arctan◦f)g. (III)RN is a special case of (II), therefore satisfies (5.43).
5.5.12 Theorem. Let E be mediated and splitting and satisfy (5.43) (e.g. E be a Banach lattice with σ-order continuous norm (Theorem 5.4.24), or E is the space mentioned in 5.5.11(II)). ThenΓV is stable and ΓV L= ΓLV,ϕV L=ϕLV.
Proof. This is a consequence of Theorem 5.5.8 and Theorem 5.5.10.
For a Riesz space F and a Riesz subspace ΓofFX we will now investigate under which conditions on ϕ(Γ), ϕL(ΓL) and ϕV(ΓV) the spaces ΓLV and ΓV L are Riesz subspaces ofFX.
5.5.13 Theorem. SupposeF is a Riesz space andΓ is a Riesz subspace of FX. If
ϕ(Γ) is splitting in E and ϕL(ΓL) is mediated in E, then ΓLV is a Riesz subspace of FX. Ifϕ(Γ) is mediated inE andϕ
V(ΓV)is splitting in E, thenΓV L is a Riesz subspace of FX.
In particular, ifE is mediated (and thus splitting), then bothΓLV andΓV L are Riesz subspaces of FX.
Proof. Note first that ifϕ(Γ)is mediated inE, thenΓV is stable by Theorem 5.4.25(b). For a proof, combine Theorem 5.4.32 and Corollary 5.3.10.
§5.5. Combining vertical and lateral extensions
Chapter
5
The next example illustrates thatΓLV is not always included inΓV L(given that
ΓV is stable) even ifE and F are Riesz spaces and Γ,ΓLV,ΓV L Riesz subspaces of
FX.
5.5.14 Example. [ΓV L(ΓLV = ΓV LV]
For an elementb= (β1, β2, . . .)ofRNwe write b=Pn∈Nβnen.
Consider X ={0,1,2, . . .} andI =P(X). Let E=c, F =RN,Ω = FX. We view the elements ofΩas sequences(a, b1, b2, . . .)witha, b1, b2,· · · ∈RN.
Define setsΓ⊂Θ⊂Ωand a mapΦ : Θ→RNby
Θ ={(a, β1e1, β2e2, . . .) :a∈c, β1, β2,· · · ∈R}, (5.45) Φ(a, β1e1, β2e2, . . .) =a+ X n∈N βnen (a∈c, β1, β2,· · · ∈R), (5.46) Γ ={(a, β1e1, β2e2, . . .) :a∈c, (β1, β2, . . .)∈c00}. (5.47) ThenΦ(Γ) =c=E; letϕ= Φ|Γ. From the definition it is easy to see thatΓis stable andϕis laterally extendable. We leave it to the reader to verify thatΓV = Γ,
ΓL={(a, β1e1, β2e2, . . .) :a∈c, (β1, β2, . . .)∈c} (5.48) andϕL= ΦonΓL.
It follows thatΓV is stable andΓV L= ΓL⊂ΓLV = ΓV LV. We proveΓV L6= ΓLV. To this end, defineh∈Ωby
( h(n) = (−1)ne n (n= 1,2, . . .), h(0) =−Pn∈Nh(n) =− P n∈N(−1)nen. (5.49)
Ash(0)∈/cwe haveh1{0}∈/ Γ; in particular,his not partially inΓ, soh /∈ΓL= ΓV L. It remains to prove h∈ΓLV. Fork∈N, defineτk, σk:X→RN: τk(0) =−Pkn=1(−1)nen+P∞n=k+1en, τk(n) =h(n) = (−1)nen (n= 1, . . . , k), τk(n) =en (n=k+ 1, k+ 2, . . .), (5.50) σk(0) =−Pkn=1(−1)nen−P∞n=k+1en, σk(n) =h(n) = (−1)nen (n= 1, . . . , k), σk(n) =−en (n=k+ 1, k+ 2, . . .). (5.51) Thenτk, σk∈ΓL,τk≥h≥σk,ϕL(τk) = Φ(τk) = 2Pn>ken,ϕL(σk) =−2Pn>ken, soinfk∈NϕL(τk) = supk∈NϕL(σk) = 0, andh∈ΓLV.
The next example illustrates that ΓV L is not always included inΓLV; it provides an example of anf ∈Γ+V L for which there exist noρ∈Γ+L withf ≤ρ(see Theorem 5.5.5(c)).
5.5.15 Example. [ΓLV (ΓV L]
Chapter
5
D is order dense7 in C[0,1](see [42, Example 4.4]). Hence, for allf
∈E there exist
(gn)n∈N,(hn)n∈N in D with f = infn∈Ngn = supn∈Nhn. Therefore E is the vertical extension ofD with respect to the inclusion mapD→E.
Take X = N, I = P(N), F = E = C[0,1],Γ = c00[D] ⊂ FN = EN and let
ϕ : Γ → E be given by ϕ(f) = Pn∈Nf(n). Since this situation is the same as in Example 5.5.6 withDV =E, we haveΓV =c00[E]andΓLV ⊂ΓV L.
Furthermore (see 5.4.6) Γ+L ={f ∈(D+)N:X n f(n)exists inE}, (5.52) Γ+V L={f ∈(E+)N:X n f(n)exists inE}. (5.53)
We construct anf ∈Γ+V Lthat is not inΓLV. Forn∈Nletfn be the ‘tent’ function defined by
fn(0) = 0; fn(n1) = 1; fn(1i) = 0 ifi∈N, i6=n;
fn is affine on the interval [1+i1 ,1i]for alli∈N. (5.54)
1 0 f1 1 2 1 1 0 fn (n >1) 1 n+1 1 n 1 n−1 1 Figure 5.1: Graph offn.
ThenP∞n=1fn=1(0,1]pointwise, soPnfn=1inC[0,1]. Hencef = (f1, f2, f3, . . .)
∈Γ+V L.
We will prove thatf /∈ΓLV; by showing there exists noρ∈ΓL for whichf ≤ρ. Supposeρ∈ΓLandf ≤ρ. Thenρ= (ρ1, ρ2, . . .)whereρ1, ρ2, . . . are elements of
D+ andj =P
nρn exists inE=C[0,1]. LetM be the largest value ofj. Every ρn is a quadratic function that maps[0,1]into[0, M]. Consequently (see the postscript)
|ρn(x)−ρn(y)| ≤4M|x−y| (x, y∈[0,1], n∈N). (5.55) In particular, ρn(0)≥ρn(n1)−4M1n ≥fn(n1)−4Mn1 = 1−4Mn1 ≥ 12 forn≥8M. Asj(0)≥Pn≥Nρn(0)for allN ∈N, this is a contradiction.
Postscript. Leth:x7→ax2+bx+cbe a quadratic function on[0,1]and0≤h(x)≤M
7A subspaceDof a partially ordered vector spaceE is calledorder denseinE ifx= sup{d∈