dominación y extensiones óptimas de operadores con rango esencial compacto en espacios de banach...

Dominacion y extensiones optimas de operadores conrango esencial compacto en espacios de Banach de

funciones

P. Rueda, Enrique A. Sanchez Perez

Seminario de Medidas Vectoriales en IntegracionI.U.M.P.A.-U. Politecnica de Valencia

Diciembre 2012

E. Sanchez Extensiones optimas de operadores

Let 1 < p,p′ < ∞ such that 1/p + 1/p′ = 1.We say that an operator T from X(µ) into a Banach space E satisfies a 1/p-th powerapproximation if for each ε > 0 there is a function hε ∈ X(µ)[1/p′] such that for everyfunction f ∈ BX(µ) there is a function g ∈ BX(µ)[1/p]

such that ‖T (f −ghε )‖E < ε.

problem: Let X(µ) be an order continuous Banach function space over a finitemeasure µ and let T : X(µ)→ E be a Banach space valued operator with compactessential range. When is the maximal extension of T compact?

Motivation: Compactness. Grothendieck’s result.

Theorem

Let E be a Banach space and let K ⊆ E be a weakly closed subset. If for every ε > 0there exists a weakly compact subset K (ε) of X such that K ⊆ K (ε) + εBE , then K isweakly compact.

Theorem

Let 1≤ p < ∞ and X an order continuous Banach function space over a finite measurespace. Let T : X → (E ,τ) be a continuous operator and let U be a basis of absolutelyconvex open neighborhoods of 0 in (E ,τ). Consider the following assertions:

(a) The operator T : X → (E ,τ) is compact.

(b) For each g ∈ BX[1/p′ ]the image T (gBX[1/p]

) is relatively τ-compact in E and forevery U ∈U there exists gU ∈ X[1/p′] such that T (BX )⊂ T (gUBX[1/p]

) + U.

(c) For each g ∈ BX[1/p′ ]the image T (gBX[1/p]

) is relatively τ-compact in E and forevery U ∈U there exists KU > 0 such that T (BX )⊂ T (KUBX[1/p]

) + U.

Then (a) implies (b) and (c). If besides τ is metrizable then (b) or (c) implies (a).

Proof.

(a)⇒(b). If T : X → (E ,τ) is compact then T (gBX[1/p]) is relatively τ-compact in E for all

g ∈ BX[1/p′ ].

Take U ∈U . Then T (BX )τ ⊂ T (BX ) + U

⊂ T (∪g∈BX[1/p′ ]gBX[1/p]

) + U = ∪g∈BX[1/p′ ](T (gBX[1/p]

) + U).

By (a) T (BX )τ

is τ-compact. Since each set of the form T (gBX[1/p]) +U is τ-open, there

exist finitely many g1, . . . ,gl ∈ BX[1/p′ ]such that

T (BX )τ ⊂ ∪l

i=1T (gi BX[1/p]) + U.

By the lattice properties of the norm of X[1/p], for any h1,h2 ∈ X[1/p′] satisfying|h1| ≤ |h2| a.e., we have that h1BX[1/p]

⊆ |h2|BX[1/p].

Define gU = ∑li=1 |gi |. Since |gi | ≤ gU , we obtain that gi BX[1/p]

⊂ gUBX[1/p], and so

T (gi BX[1/p])⊂ T (gUBX[1/p]

) for all i = 1, . . . , l . Therefore,

∪li=1T (gi BX[1/p]

) + U ⊂ T (gUBX[1/p]) + U

and (b) is proved.E. Sanchez Extensiones optimas de operadores

Proof.

g ∈ BX[1/p′ ].

⊂ T (∪g∈BX[1/p′ ]gBX[1/p]

) + U = ∪g∈BX[1/p′ ](T (gBX[1/p]

) + U).

By (a) T (BX )τ

T (BX )τ ⊂ ∪l

i=1T (gi BX[1/p]) + U.

⊆ |h2|BX[1/p].

) + U ⊂ T (gUBX[1/p]) + U

Proof.

g ∈ BX[1/p′ ].

⊂ T (∪g∈BX[1/p′ ]gBX[1/p]

) + U = ∪g∈BX[1/p′ ](T (gBX[1/p]

) + U).

By (a) T (BX )τ

T (BX )τ ⊂ ∪l

i=1T (gi BX[1/p]) + U.

⊆ |h2|BX[1/p].

) + U ⊂ T (gUBX[1/p]) + U

Proof.

g ∈ BX[1/p′ ].

⊂ T (∪g∈BX[1/p′ ]gBX[1/p]

) + U = ∪g∈BX[1/p′ ](T (gBX[1/p]

) + U).

By (a) T (BX )τ

T (BX )τ ⊂ ∪l

i=1T (gi BX[1/p]) + U.

⊆ |h2|BX[1/p].

) + U ⊂ T (gUBX[1/p]) + U

Proof.

g ∈ BX[1/p′ ].

⊂ T (∪g∈BX[1/p′ ]gBX[1/p]

) + U = ∪g∈BX[1/p′ ](T (gBX[1/p]

) + U).

By (a) T (BX )τ

T (BX )τ ⊂ ∪l

i=1T (gi BX[1/p]) + U.

⊆ |h2|BX[1/p].

) + U ⊂ T (gUBX[1/p]) + U

Proof.

g ∈ BX[1/p′ ].

⊂ T (∪g∈BX[1/p′ ]gBX[1/p]

) + U = ∪g∈BX[1/p′ ](T (gBX[1/p]

) + U).

By (a) T (BX )τ

T (BX )τ ⊂ ∪l

i=1T (gi BX[1/p]) + U.

⊆ |h2|BX[1/p].

) + U ⊂ T (gUBX[1/p]) + U

(b)⇒(c) Let ε > 0 and U ∈U . Take V ∈U such that V + V ⊂ U.

Consider gV ∈ X[1/p′] given by (b). By continuity there exists δ > 0 such thatT (δBX )⊂ V . Since simple functions are dense in X[1/p′], we find a simple function ssuch that

‖(gV −s)f‖X ≤ ‖gV −s‖X[1/p′ ]< δ

for every f ∈ BX[1/p], and so T ((gV −s)BX[1/p]

)⊆ V .

Take KU := ‖s‖L∞(µ). Then, again by the ideal property of X[1/p], we obtain

T (BX )⊆ T (gV BX[1/p]) + V ⊆ T (sBX[1/p]

) + T ((gV −s)BX[1/p]) + V

⊆ T (sBX[1/p]) + V + V ⊆ T (KUBX[1/p]

) + U. The proof is finished

Let us assume now that τ is metrizable. We prove (b)⇒(a) (the proof for (c)⇒(a) is thesame).

Take an arbitrary U ∈U and chose gU ∈ X[1/p′] such that T (BX )⊂ T (gUBX[1/p]) + U.

By (b) the set T (gUBX[1/p]) is relatively τ-compact in E .

Then we have shown that for any U ∈U , there is a relatively τ-compact setT (gUBX[1/p]

) such that T (BX )⊂ T (gUBX[1/p]) + U. As τ is metrizable this implies that

T (BX ) is relatively τ-compact.

‖(gV −s)f‖X ≤ ‖gV −s‖X[1/p′ ]< δ

)⊆ V .

) + T ((gV −s)BX[1/p]) + V

⊆ T (sBX[1/p]) + V + V ⊆ T (KUBX[1/p]

‖(gV −s)f‖X ≤ ‖gV −s‖X[1/p′ ]< δ

)⊆ V .

) + T ((gV −s)BX[1/p]) + V

⊆ T (sBX[1/p]) + V + V ⊆ T (KUBX[1/p]

‖(gV −s)f‖X ≤ ‖gV −s‖X[1/p′ ]< δ

)⊆ V .

) + T ((gV −s)BX[1/p]) + V

⊆ T (sBX[1/p]) + V + V ⊆ T (KUBX[1/p]

‖(gV −s)f‖X ≤ ‖gV −s‖X[1/p′ ]< δ

)⊆ V .

) + T ((gV −s)BX[1/p]) + V

⊆ T (sBX[1/p]) + V + V ⊆ T (KUBX[1/p]

‖(gV −s)f‖X ≤ ‖gV −s‖X[1/p′ ]< δ

)⊆ V .

) + T ((gV −s)BX[1/p]) + V

⊆ T (sBX[1/p]) + V + V ⊆ T (KUBX[1/p]

Let 1 < p < ∞. Let X(µ) be an order continuous Banach function space and E be aBanach space. A continuous operator T : X(µ)→ E is essentially compact if and only iffor every h ∈ X[1/p′] the map Th : X[1/p]→ E given by Th(·) := T (h ·), is compact.

Corollary

Let 1 < p < ∞. Let X(µ) be an order continuous Banach function space and E be aBanach space. If T : X(µ)→ E is a continuous essentially compact operator then therestriction T |X[1/p]

: X[1/p]→ E is compact.

Let 1 < p < ∞. Let X(µ) be an order continuous Banach function space and E be aBanach space. A continuous operator T : X(µ)→ E is essentially compact if and only iffor every h ∈ X[1/p′] the map Th : X[1/p]→ E given by Th(·) := T (h ·), is compact.

Corollary

Let 1 < p < ∞. Let X(µ) be an order continuous Banach function space and E be aBanach space. If T : X(µ)→ E is a continuous essentially compact operator then therestriction T |X[1/p]

: X[1/p]→ E is compact.

Theorem

Let 1≤ p < ∞ and let X(µ) be an order continuous Banach function space. Thefollowing statements for a continuous operator T : X → E are equivalent:

(i) T is compact.

(ii) T is essentially compact and for every ε > 0 there exists hε ∈ X[1/p′] such thatT (BX )⊂ T (hε BX[1/p]

) + εBE .

(iii) T is essentially compact and for every ε > 0 there exists Kε > 0 such thatT (BX )⊂ T (Kε BX[1/p]

) + εBE .

Examples

(a) Take X(µ) = L1[0,1] and 1 < p < ∞. Then X[1/p] = Lp[0,1] isometrically. Consider apartition Ai∞

i=1 of [0,1], where µ(Ai ) > 0 for all i ∈ N. Take a sequence of non-nullmeasurable sets Bi∞

i=1 such that Bi ⊆ Ai for all i and ri := µ(Bi )/µ(Ai ) ↓ 0 and theoperator T : L1(µ)→ `1 given by T (f ) := ∑

i=1(∫

Bif dµ)ei ∈ `1, f ∈ L1[0,1]. Let us show

that the requirement on T (χA) : A ∈Σ of being relatively compact is fulfilled.Consider the sequence ri ei∞

i=1. For every A ∈Σ,

T (χA) =∞

∑i=1

µ(Bi ∩A)ei ≤∞

∑i=1

µ(Ai )(ri ei )

in the order of `1, and so each T (χA) is in the closure of the convex hull of ri ei∞

i=1.Thus, it is relatively compact. In fact, any `1-valued continuous operator defined on anorder continuous Banach function space is essentially compact. However, since forevery i we can find a norm one function fi of L1(µ) with support in Bi , the setT (BL1[0,1]) includes all ei∞

i=1, and so T is not compact. Consequently, there is ε > 0such that there is no K such that T (BL1[0,1])⊆ KT (BLp [0,1]) + εB`p .

(b) Although operators of compact range with domain in an L1-space allow sometimesgood characterizations —due mainly to 1-concavity of L1—, this fact is not crucial inthe example above. Consider p ≥ q ≥ 1 and the class of spacesEp,q = (⊕p)∞

i=1Lq(Ai ,µ|Ai), where the Ai ′s are chosen as in (a), with the norm

|‖f‖|p,q :=( ∞

∑i=1‖f |Ai

‖pLq [0,1]

)1/p, f ∈ Ep,q .

It is a Banach function space of Lebesgue measurable functions on [0,1] over ameasure µa given by µa(A) := ∑

i=1 ai µ(Ai ∩A), where a is any strictly positive elementof B

`p′ . Notice also that it contains L1[0,1]. The same calculations that in the abovecase prove that T defined as in (a) is continuous also if defined from Ep,1 to `p , and itsessential range T is relatively compact, but the operator is not compact. Then, since(Ep,1)[1/p] = Ep2,p , we get that there is an ε > 0 such that T (BEp,1 ) * KT (BEp2 ,p

) + εB`1

for all constants K > 0.

Corollary

Let 1 < p < ∞. Let X(µ), E and T as above, and assume that T is essentially compact.Let Φ : BX → BX[1/p]

be a function and suppose that

lımK→∞

supf∈BX

∥∥T (f χ|f |≥K |Φ(f )|)‖= 0.

Then T is compact.

Corollary

Let T be an essentially compact (positive) kernel operator T : X(µ)→ Y (ν) such thatthe kernel k satisfies that

lımµ(A)→0

∥∥∥‖χAk(x ,y)‖X ′ (y)∥∥∥

Then T is compact.

Corollary

Let 1 < p < ∞. Let X(µ), E and T as above, and assume that T is essentially compact.Let Φ : BX → BX[1/p]

be a function and suppose that

lımK→∞

supf∈BX

∥∥T (f χ|f |≥K |Φ(f )|)‖= 0.

Then T is compact.

Corollary

Let T be an essentially compact (positive) kernel operator T : X(µ)→ Y (ν) such thatthe kernel k satisfies that

lımµ(A)→0

∥∥∥‖χAk(x ,y)‖X ′ (y)∥∥∥

Then T is compact.

Proposition. Let 1≤ p < ∞. The following are equivalent:

(a) The integration operator Im : L1(m)→ (E ,‖ ‖E ) is compact.

(b) R(m) is relatively compact and for every ε > 0 there exists gε ∈ Lp′ (m) such thatIm(BL1(m))⊂ Im(gε BLp(m)) + εBE .

(c) R(m) is relatively compact and for every ε > 0 there exists Kε > 0 such thatIm(BL1(m))⊂ Im(Kε BLp(m)) + εBE .

Corollary

Let m : Σ→ E be a vector measure with relatively compact range. Suppose that theintegration map Im : L1(m)→ E satisfies an 1/p-th power approximation. ThenL1(m) = L1(|m|).

Corollary

Let m : Σ→ E be a vector measure with relatively compact range. Suppose that theintegration map Im : L1(m)→ E satisfies an 1/p-th power approximation. ThenL1(m) = L1(|m|).

Theorem

Let 1 < p < ∞, let X(µ) be an order continuous Banach function space, E a Banachspace and T : X(µ)→ E a continuous operator. The following statements areequivalent for T .

(i) T is essentially compact, and for each ε > 0 there is hε ∈ X[1/p′] such that

supf∈BL1(mT )

∩X(µ)

g∈BLp (mT )∩X[1/p]

(∥∥T (f −hε g)∥∥

))< ε.

(ii) T is essentially compact, and for each ε > 0 there is a constant Kε > 0 such that

supf∈BL1(mT )

∩X(µ)

g∈BLp (mT )∩X[1/p]

(∥∥T (f −Kε g)∥∥

))< ε.

(iii) The optimal domain of T is L1(|mT |) and the extension ImT is compact, i.e. Tfactorizes compactly as

X(µ)T // r

[i] $$HHHHH

HHHHE .

L1(|mT |)

;;xxxxxxxxx

Corollary

Let 1 < p < ∞, let X(µ) be an order continuous Banach function space, E a Banachspace and T : X(µ)→ E a µ-determined essentially compact continuous operator. Thefollowing statements are equivalent.

(i) Each extension of T to an order continuous Banach function space satisfies a1/p-th power approximation.

(ii) ImT satisfies a 1/p-th power approximation.

(iii) T admits a maximal extension that satisfies a 1/p-th power approximation.

We show a factorization theorem for homogeneous maps. The proof is based in aseparation argument that has shown to be the key for proving factorization theoremsfor operators between Banach function spaces.

Proposition. Let E be a Banach space, and let Y (ν) and Z (ν) be Banach functionspaces such that Y (ν)⊆ Z (ν). Let U ⊆ E be an homogeneous set, and φ : U→ Y (ν)and P : U→ Z (ν) be bounded homogeneous maps. Assume also that Y Z has theFatou property and (Y Z )′ is order continuous. Then the following statements areequivalent.

(i) There is a constant K > 0 such that for every x1, ...,xn ∈ U and A1, ...,An ∈Σ,

∑i=1|P(xi )|χAi

‖L1(ν) ≤ K‖n

∑i=1|φ(xi )|χAi

‖(Y Z )′ .

(ii) There is a function g ∈ Y Z such that for every x ∈ U there is a function hx ∈ BL∞(ν)

depending only on x/‖x‖ such that P(x) = g ·hx ·φ(x). In other words, Pfactorizes through the homogeneous map φ given by φ(x) := hx ·φ(x) as

UP // p

φ !!BBB

B Z (ν)

Y (ν)

<<yyyyyyyy

We show a factorization theorem for homogeneous maps. The proof is based in aseparation argument that has shown to be the key for proving factorization theoremsfor operators between Banach function spaces.

Proposition. Let E be a Banach space, and let Y (ν) and Z (ν) be Banach functionspaces such that Y (ν)⊆ Z (ν). Let U ⊆ E be an homogeneous set, and φ : U→ Y (ν)and P : U→ Z (ν) be bounded homogeneous maps. Assume also that Y Z has theFatou property and (Y Z )′ is order continuous. Then the following statements areequivalent.

(i) There is a constant K > 0 such that for every x1, ...,xn ∈ U and A1, ...,An ∈Σ,

∑i=1|P(xi )|χAi

‖L1(ν) ≤ K‖n

∑i=1|φ(xi )|χAi

‖(Y Z )′ .

(ii) There is a function g ∈ Y Z such that for every x ∈ U there is a function hx ∈ BL∞(ν)

depending only on x/‖x‖ such that P(x) = g ·hx ·φ(x). In other words, Pfactorizes through the homogeneous map φ given by φ(x) := hx ·φ(x) as

UP // p

φ !!BBB

B Z (ν)

Y (ν)

<<yyyyyyyy

DEFINITIONS: Banach function spaces

(Ω,Σ,µ) be a finite measure space.

L0(µ) space of all (classes of) measurable real functions on Ω.

A Banach function space (briefly B.f.s.) is a Banach space X ⊂ L0(µ) with norm‖ · ‖X such that if f ∈ L0(µ), g ∈ X and |f | ≤ |g| µ-a.e. then f ∈ X and ‖f‖X ≤ ‖g‖X .

A B.f.s. X has the Fatou property if for every sequence (fn)⊂ X such that 0≤ fn ↑ fµ-a.e. and supn ‖fn‖X < ∞, it follows that f ∈ X and ‖fn‖X ↑ ‖f‖X .

We will say that X is order continuous if for every f , fn ∈ X such that 0≤ fn ↑ fµ-a.e., we have that fn→ f in norm.

APPENDIX: DEFINITIONS. Vector measures and integration

Let m : Σ→ E be a vector measure, that is, a countably additive set function,where E is a real Banach space.

A set A ∈Σ is m-null if m(B) = 0 for every B ∈Σ with B ⊂ A. For each x∗ in thetopological dual E∗ of E , we denote by |x∗m| the variation of the real measurex∗m given by the composition of m with x∗. There exists x∗0 ∈ E∗ such that |x∗0m|has the same null sets as m. We will call |x∗0m| a Rybakov control measure for m.

A measurable function f : Ω→ R is integrable with respect to m if(i)

∫|f |d |x∗m|< ∞ for all x∗ ∈ E∗.

(ii) For each A ∈Σ, there exists xA ∈ E such that

x∗(xA) =∫

Af dx∗m, for all x∗ ∈ E .

The element xA will be written as∫

A f dm.

∫|f |d |x∗m|< ∞ for all x∗ ∈ E∗.

x∗(xA) =∫

A f dm.

∫|f |d |x∗m|< ∞ for all x∗ ∈ E∗.

x∗(xA) =∫

A f dm.

∫|f |d |x∗m|< ∞ for all x∗ ∈ E∗.

x∗(xA) =∫

A f dm.

∫|f |d |x∗m|< ∞ for all x∗ ∈ E∗.

x∗(xA) =∫

A f dm.

DEFINITIONS: Spaces of integrable functions

Denote by L1(m) the space of integrable functions with respect to m, wherefunctions which are equal m-a.e. are identified.

The space L1(m) is a Banach space endowed with the norm

‖f‖m = supx∗∈BE∗

∫|f |d |x∗m|.

Note that L∞(|x∗0m|)⊂ L1(m). In particular every measure of the type |x∗m| is finiteas |x∗m|(Ω)≤ ‖x∗‖ · ‖χΩ‖m.

Given f ∈ L1(m), the set function mf : Σ→ E given by mf (A) =∫

A f dm for all A ∈Σis a vector measure. Moreover, g ∈ L1(mf ) if and only if gf ∈ L1(m) and in thiscase

∫g dmf =

∫gf dm.

∫|f |d |x∗m|.

∫g dmf =

∫gf dm.

∫|f |d |x∗m|.

∫g dmf =

∫gf dm.

∫|f |d |x∗m|.

∫g dmf =

∫gf dm.

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