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LIPSCHITZ TENSOR PRODUCT

M. G. CABRERA-PADILLA, J. A. CHAVEZ-DOMINGUEZ, A. JIMENEZ-VARGAS, AND MOISES VILLEGAS-VALLECILLOS

Abstract. Inspired by ideas of R. Schatten in his celebrated monograph [22] on a theory of cross-spaces, we introducethe notion of a Lipschitz tensor productX ⊠ E of a pointed metric spaceX and a Banach spaceE as a certain linearsubspace of the algebraic dual of Lip0(X,E∗). We prove thatX⊠E is linearly isomorphic to the linear space of all finite-rank continuous linear operators from (X#, τp) into E, whereX# denotes the space Lip0(X,K) andτp is the topology ofpointwise convergence ofX#. The concept of Lipschitz tensor product of elements ofX# andE∗ yields the spaceX#

iE∗

as a certain linear subspace of the algebraic dual ofX⊠E. To ensure the good behavior of a norm onX⊠E with respect tothe Lipschitz tensor product of Lipschitz functionals (mappings) and bounded linear functionals (operators), the conceptof dualizable (respectively, uniform) Lipschitz cross-norm onX⊠E is defined. We show that the Lipschitz injective normε, the Lipschitz projective normπ and the Lipschitzp-nuclear normdp (1 ≤ p ≤ ∞) are uniform dualizable Lipschitzcross-norms onX⊠E. In fact,ε is the least dualizable Lipschitz cross-norm andπ is the greatest Lipschitz cross-norm onX⊠ E. Moreover, dualizable Lipschitz cross-normsα on X⊠ E are characterized by satisfying the relationε ≤ α ≤ π. Inaddition, the Lipschitz injective (projective) norm onX ⊠ E can be identified with the injective (respectively, projective)tensor norm on the Banach-space tensor product between the Lipschitz-free space overX andE. In terms of the spaceX#i E∗, we describe the spaces of Lipschitz compact (finite-rank, approximable) operators fromX to E∗.

Introduction

The Lipschitz space Lip0(X,E) is the Banach space of all Lipschitz mapsf from a pointed metric spaceX to aBanach spaceE that vanish at the base point ofX, under the Lipschitz norm given by

Lip( f ) = sup

{‖ f (x) − f (y)‖

d(x, y): x, y ∈ X, x , y

}.

The elements of Lip0(X,E) are referred to as Lipschitz operators. IfK is the field of real or complex numbers, thespace Lip0(X,K), denoted byX#, is called the Lipschitz dual ofX. A comprehensive reference for the basic theoryof the spaces of Lipschitz functions is the book [23] by N. Weaver.

The use of techniques of the theory of algebraic tensor product of Banach spaces to tackle the problem of theduality for Lipschitz operators fromX to E goes back to the seventies with the works [17, 18] of J. A. Johnson.Recently, the second-named author [4] has adopted this approach to describe the duals of spaces of Lipschitzp-summing operators fromX to E∗ for 1 ≤ p ≤ ∞. The notion of Lipschitzp-summing operators between metricspaces, a nonlinear generalization ofp-summing operators, was introduced by J. D. Farmer and W. B. Johnson in[10], where a nonlinear version of the Pietsch factorization theorem was established. The article [10] has motivatedthe study of Lipschitz versions of different types of bounded linear operators in the works [4, 5, 6,7, 16].

The reading of the paper [4] invites to give a definition for the tensor product ofX andE. In [17], J. A. Johnsonproved that the dual of the closed linear subspace of Lip0(X,E∗)∗ spanned by the functionalsδx ⊠ eon Lip0(X,E∗)with x ∈ X ande ∈ E, defined by (δx ⊠ e)( f ) = 〈 f (x), e〉, is isometrically isomorphic to Lip0(X,E∗). It is wellknown that the dual of the projective tensor product of Banach spacesE andF can be identified with the spaceof all bounded linear operators fromE to F∗, so the predual of Lip0(X,E∗) provided by Johnson’s result plays the

Date: August 11, 2014.2010Mathematics Subject Classification.26A16, 46B28, 46E15, 47L20.Key words and phrases.Lipschitz map; tensor product;p-summing operator; Lipschitz compact operator.The second author’s research was partially supported by NSFgrants DMS-1001321 and DMS-1400588, and the third author’sone by Junta

of Andalucıa grant FQM-194.

1

http://arxiv.org/abs/1408.1874v1

2 M. G. CABRERA-PADILLA, J. A. CHAVEZ-DOMINGUEZ, A. JIMENEZ-VARGAS, AND MOISES VILLEGAS-VALLECILLOS

role of the projective tensor product in the linear theory and this fact suggests to call Lipschitz tensor product ofXandE the linear subspace of the algebraic dual of Lip0(X,E∗) spanned by the functionalsδx ⊠ e. In [4], this spaceis called the space ofE-valued molecules onX, a generalization of the Arens–Eells space Æ(X) of scalar-valuedmolecules onX (see [1, 23]).

Our purpose is to develop a theory of the Lipschitz tensor product X ⊠ E of a pointed metric spaceX and aBanach spaceE by following the original ideas of R. Schatten [22] used to construct the algebraic tensor productof two Banach spaces. We are also motivated by the problem of researching the spaces of Lipschitz compact(finite-rank, approximable) operators fromX to E∗ introduced in [16].

We now describe the contents of this paper. In Section 1, we introduce and study the Lipschitz tensor productX ⊠ E. We show that

⟨X ⊠ E, Lip0(X,E∗)

⟩forms a dual pair and identify linearly the space Lip0(X,E∗) with a

linear subspace of the algebraic dual ofX ⊠ E, and the spaceX ⊠ E with the spaceF ((X#, τp); E) of all finite-rankcontinuous linear operators from (X#, τp) to E, whereτp denotes the topology of pointwise convergence ofX#.

In Section 2, we define the concept of a Lipschitz tensor product of a Lipschitz functionalg ∈ X# and a boundedlinear functionalφ ∈ E∗ as a certain linear functionalg⊠ φ on X ⊠ E, and consider the associated Lipschitz tensorproduct ofX⊠E, denoted byX#

iE∗, as the linear subspace of the algebraic dual ofX⊠E spanned by the elementsg⊠φ. It is showed that the spaceX#

iE∗ is linearly isomorphic to the space Lip0F (X,E∗) of all Lipschitz finite-rankoperators fromX to E∗.

In Section 3, we give the notion of a Lipschitz tensor productof a base-point preserving Lipschitz map betweenpointed metric spacesh: X → Y and a bounded linear operator between Banach spacesT : E → F as a certainlinear operatorh⊠ T from X ⊠ E to Y⊠ F.

In Section 4, we consider a normα on X ⊠ E and obtain a normed spaceX ⊠α E and its completionX⊠αE. Weare interested in the called Lipschitz cross-norms which are those satisfying the conditionα(δ(x,y) ⊠ e) = d(x, y) ‖e‖for all x, y ∈ X ande ∈ E. Desirable attributes for Lipschitz cross-norms onX ⊠ E is that they behave well withrespect to the formation of Lipschitz tensor products of functionals and operators. In this line, we study dualizableLipschitz cross-norms and uniform Lipschitz cross-norms on X ⊠ E.

In Section 5, given a dualizable Lipschitz cross-normα on X ⊠ E, we construct a normα′ on X#i E∗ called

the associated Lipschitz norm ofα. The spaceX#i E∗ with the normα′ will be denoted byX#

iα′ E∗ and itscompletion byX#

iα′E∗.In Sections 6, 7, 8 and 9 we investigate, respectively, the dual normL induced onX ⊠ E by the norm Lip of

Lip0(X,E∗), the Lipschitz injective normε, the Lipschitz projective normπ and the Lipschitzp-nuclear normdp

with p ∈ [1,∞]. All these norms onX ⊠ E are uniform and dualizable Lipschitz cross-norms. In fact,ε is theleast dualizable Lipschitz cross-norm andπ is the greatest Lipschitz cross-norm onX⊠E. Furthermore, dualizableLipschitz cross-normsα on X ⊠ E are characterized as those which satisfy the relationε ≤ α ≤ π. We also provethat L agrees withπ and justify the terminologies “injective” and “projective” for the Lipschitz normsε andπ,respectively. We identifyX⊠ε E and its completionX⊠εE with F ((X#, τp); E) and its closure in the operator normtopology, respectively. We also show that the Lipschitz injective (projective) norm onX⊠ E can be identified withthe injective (respectively, projective) tensor norm on the Banach-space tensor product between the Lipschitz-freespace overX andE.

In Section 10, we deal with the space Lip0F(X,E∗) of all Lipschitz finite-rank operators fromX to E∗ and itsclosure in the Lipschitz norm topology, the space of Lipschitz approximable operators fromX to E∗. It is provedthat the former space is isometrically isomorphic toX#

iπ′ E∗, and the latter toX#iπ′E∗. The approximation

property for Banach spaces was introduced by Grothendieck in his famous memory [15]. We show that ifX#

has the approximation property, then the space of all Lipschitz compact operators fromX to E∗ is isometricallyisomorphic toX#

iπ′E∗ for any Banach spaceE. We close the paper giving a new expression of the normπ′.

Notation. Given two metric spaces (X, dX) and (Y, dY), let us recall that a mapf : X→ Y is said to be Lipschitzif there is a real constantC ≥ 0 such thatdY( f (x), f (y)) ≤ CdX(x, y) for all x, y ∈ X. The least constantC for which

LIPSCHITZ TENSOR PRODUCT 3

the preceding inequality holds will be denoted by Lip(f ), that is,

Lip( f ) = sup

{dY( f (x), f (y))

dX(x, y): x, y ∈ X, x , y

}.

A pointed metric spaceX is a metric space with a base point inX which we always will denote by 0. We willconsider a Banach spaceE overK as a pointed metric space with the zero vector as the base point. As is customary,BE andSE stand for the closed unit ball ofE and the unit sphere ofE, respectively. Given two pointed metric spacesX andY, Lip0(X,Y) denotes the set of all base-point preserving Lipschitz maps fromX to Y.

For two linear spacesE andF, L(E, F) stands for the linear space of all linear operators fromE into F. In thecase thatE andF are Banach spaces, we denote byL(E, F) the Banach space of all bounded linear operators fromE into F with the usual norm, and byF (E, F) its subspace of finite-rank bounded linear operators. In particular,the algebraic dualL(E,K) and the topological dualL(E,K) are denoted byE′ andE∗, respectively. For eache ∈ Eandφ ∈ E′, we frequently will write〈φ, e〉 instead ofφ(e).

1. Lipschitz tensor products

The Lipschitz tensor product of a pointed metric spaceX and a Banach spaceE, which we will denote fromnow on byX ⊠ E, can be constructed as a space of linear functionals on Lip0(X,E∗).

Definition 1.1. Let X be a pointed metric space and E a Banach space. For each x∈ X, letδ(x,0) : Lip0(X,E∗)→ E∗

be the linear map defined by

δ(x,0)( f ) = f (x)(f ∈ Lip0(X,E∗)

).

For each(x, y) ∈ X2, let δ(x,y) : Lip0(X,E∗)→ E∗ be the linear map given by

δ(x,y) = δ(x,0) − δ(y,0).

Let ∆(X,E∗) denote the linear subspace of L(Lip0(X,E∗),E∗) spanned by the set{δ(x,y) : (x, y) ∈ X2

}. For any

γ ∈ ∆(X,E∗) and e∈ E, letγ ⊠ e: Lip0(X,E∗)→ K be the linear functional given by

(γ ⊠ e)( f ) = 〈γ( f ), e〉(f ∈ Lip0(X,E∗)

).

In particular, for any(x, y) ∈ X2 and e∈ E, letδ(x,y) ⊠ e be the element ofLip0(X,E∗)′ defined by(δ(x,y) ⊠ e

)( f ) =

⟨δ(x,y)( f ), e

⟩=

⟨δ(x,0)( f ) − δ(y,0)( f ), e

⟩= 〈 f (x) − f (y), e〉 , ∀ f ∈ Lip0(X,E∗).

The Lipschitz tensor product X⊠ E is defined as the vector subspace ofLip0(X,E∗)′ spanned by the set{δ(x,y) ⊠ e: (x, y) ∈ X2, e ∈ E

}.

The following properties of the Lipschitz tensor product can be checked easily.

Lemma 1.1. Letλ ∈ K, (x, y), (x1, y1), (x2, y2) ∈ X2 and e, e1, e2 ∈ E.

(i) λ(δ(x,y) ⊠ e

)= (λδ(x,y)) ⊠ e= δ(x,y) ⊠ (λe).

(ii)(δ(x1,y1) + δ(x2,y2)

)⊠ e= δ(x1,y1) ⊠ e+ δ(x2,y2) ⊠ e.

(iii) δ(x,y) ⊠ (e1 + e2) = δ(x,y) ⊠ e1 + δ(x,y) ⊠ e2.(iv) δ(x,x) ⊠ e= δ(x,y) ⊠ 0 = 0.

We say thatδ(x,y) ⊠ e is an elementary Lipschitz tensor. Note that each elementu in X ⊠ E is of the formu =

∑ni=1 λi(δ(xi ,yi) ⊠ ei), wheren ∈ N, λi ∈ K, (xi , yi) ∈ X2 andei ∈ E. This representation ofu is not unique. It is

worth noting that each elementu of X⊠E can be represented asu =∑n

i=1 δ(xi ,yi)⊠ei sinceλ(δ(x,y)⊠e) = δ(x,y)⊠ (λe).This representation ofu admits the following refinement.


Lemma 1.2. Every nonzero Lipschitz tensor u∈ X ⊠ E has a representation in the form∑m

i=1 δ(zi ,0) ⊠ di , where

m= min

k ∈ N : ∃z1, . . . , zk ∈ X, d1, . . . , dk ∈ E | u =k∑

i=1

δ(zi ,0) ⊠ di

and the points z1, . . . , zm in X are distinct from the base point0 of X and pairwise distinct.

Proof. Let u ∈ X ⊠ E and let∑n

i=1 δ(xi ,yi) ⊠ ei be a representation ofu. We have

u =n∑

i=1

δ(xi ,yi) ⊠ ei

=

n∑

i=1

(δ(xi ,0) − δ(yi ,0)

)⊠ ei

=

n∑

i=1

δ(xi ,0) ⊠ ei +

n∑

i=1

δ(yi ,0) ⊠ (−ei)

=

n∑

i=1

δ(xi ,0) ⊠ ei +

2n∑

i=n+1

δ(yi−n,0) ⊠ (−ei−n)

=

2n∑

i=1

δ(zi ,0) ⊠ di,

where

δ(zi ,0) ⊠ di =

{δ(xi ,0) ⊠ ei if i = 1, . . . , n,δ(yi−n,0) ⊠ (−ei−n) if i = n+ 1, . . . , 2n.

Then, by the well-ordering principle ofN, there exists a smallest natural numberm for which there is a represen-tation ofu in the form

∑mi=1 δ(zi ,0) ⊠ di . Sinceu , 0, it is clear thatzi , 0 for somei ∈ {1, . . . ,m}. This implies that

zj , 0 for all j ∈ {1, . . . ,m}. Otherwise, observe that∑m

i=1,i, j δ(zi ,0) ⊠ di would be a representation ofu containingm− 1 terms and this contradicts the definition ofm. Moreover, ifzj = zk for somej, k ∈ {1, . . . ,m} with j , k, wewould have

u =m∑

i=1, j,i,k

δ(zi ,0) ⊠ di +(δ(zj ,0) ⊠ (d j + dk)

),

and this is impossible. Hence the pointszi are pairwise distinct. �

We can concatenate the representations of two elements ofX ⊠ E to get a representation of their sum.

Lemma 1.3. Let u, v ∈ X⊠E and let∑n

i=1 δ(xi ,yi)⊠ei and∑m

i=1 δ(x′i ,y′i )⊠e′i be representations of u and v, respectively.

Then∑n+m

i=1 δ(x′′i ,y′′i ) ⊠ e′′i , where

(x′′i , y′′i ) =

{(xi , yi) if i = 1, . . . , n,(x′i−n, y

′i−n) if i = n+ 1, . . . , n+m,

e′′i =

{ei if i = 1, . . . , n,e′i−n if i = n+ 1, . . . , n+m,

is a representation of u+ v.

We now describe the action of a Lipschitz tensoru ∈ X ⊠ E on a functionf ∈ Lip0(X,E∗).

Lemma 1.4. Let u=∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E and f ∈ Lip0(X,E∗). Then

u( f ) =n∑

i=1

〈 f (xi) − f (yi), ei〉 .

Our next aim is to characterize the zero Lipschitz tensor. For it we need the following Lipschitz operators.


Lemma 1.5. Let g∈ X# andφ ∈ E∗. The map g· φ : X → E∗, given by(g · φ)(x) = g(x)φ, belongs toLip0(X,E∗)andLip(g · φ) = Lip(g) ‖φ‖.

Proof. Clearly,g · φ is well defined. Letx, y ∈ X. For anye ∈ E, we obtain

|〈(g · φ)(x) − (g · φ)(y), e〉| = |〈(g(x) − g(y))φ, e〉| = |g(x) − g(y)| |〈φ, e〉| ≤ Lip(g)d(x, y) ‖φ‖ ‖e‖ ,

and so‖(g · φ)(x) − (g · φ)(y)‖ ≤ Lip(g) ‖φ‖d(x, y). Theng · φ ∈ Lip0(X,E∗) and Lip(g · φ) ≤ Lip(g) ‖φ‖. For theconverse inequality, note that

|g(x) − g(y)| ‖φ‖ = ‖(g · φ)(x) − (g · φ)(y)‖ ≤ Lip(g · φ)d(x, y)

for all x, y ∈ X, and therefore Lip(g) ‖φ‖ ≤ Lip(g · φ). �

Proposition 1.6. If u =∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, then the following assertions are equivalent:

(i) u = 0.(ii)

∑ni=1 (g(xi) − g(yi)) 〈φ, ei〉 = 0 for every g∈ BX# andφ ∈ BE∗ .

(iii)∑n

i=1 (g(xi) − g(yi)) ei = 0 for every g∈ BX#.

Proof. (i) implies (ii): If u = 0, thenu( f ) = 0 for all f ∈ Lip0(X,E∗). Sinceu =∑n

i=1 δ(xi ,yi) ⊠ ei , it follows that∑ni=1〈 f (xi) − f (yi), ei〉 = 0 for all f ∈ Lip0(X,E∗) by Lemma 1.4. For anyg ∈ BX# andφ ∈ BE∗ , the functiong · φ is

in Lip0(X,E∗) by Lemma 1.5, and therefore we haven∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉 =

n∑

i=1

〈(g(xi) − g(yi)) φ, ei〉 =

n∑

i=1

〈(g · φ)(xi) − (g · φ)(yi), ei〉 = 0.

(ii) implies (iii): If (ii) holds, then 〈φ,∑n

i=1(g(xi) − g(yi))ei〉 = 0 for everyg ∈ BX# andφ ∈ BE∗ . SinceBE∗

separates the points ofE, it follows that∑n

i=1(g(xi) − g(yi))ei = 0 for all g ∈ BX#.(iii) implies (i): By Lemma 1.2, we can writeu =

∑mi=1 δ(zi ,0) ⊠ di , where the pointszi in X are pairwise distinct

and different from the base point 0. It follows thatn∑

i=1

δ(xi ,yi) ⊠ ei +

m∑

i=1

δ(zi ,0) ⊠ (−di) = u− u = 0,

and, by using the fact proved above that (i) implies (iii), wehaven∑

i=1

(g(xi) − g(yi)) ei +

m∑

i=1

(g(zi) − g(0)) (−di) = 0

for all g ∈ BX#. If (iii) holds, we get thatm∑

i=1

g(zi)di =

n∑

i=1

(g(xi) − g(yi)) ei = 0

for all g ∈ BX#. Set

r = min({

d(zi , zj) : i, j ∈ {1, . . . ,m}, i , j}∪ {d(zi , 0) : i ∈ {1, . . . ,m}}

).

Clearly,r > 0. Given j ∈ {1, . . . ,m}, defineg j : X→ R by

g j(x) = max{0, r − d(x, zj)

}.

It is easy to check thatg j ∈ BX#, g j(zj) = r andg j(zi) = 0 for all i ∈ {1, . . . ,m} \ { j}. Hence 0=∑m

i=1 g j(zi)di = rd j ,therefored1 = d2 = · · · = dm = 0 and thusu = 0. �

According to Definition 1.1,X ⊠ E is a linear subspace of Lip0(X,E∗)′. Furthermore, we have the next fact.


Theorem 1.7.⟨X ⊠ E, Lip0(X,E∗)

⟩forms a dual pair, where the bilinear form〈·, ·〉 associated to the dual pair is

given, for u=∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E and f ∈ Lip0(X,E∗), by

〈u, f 〉 =n∑

i=1

〈 f (xi) − f (yi), ei〉 .

Proof. Note that〈u, f 〉 = u( f ) by Lemma 1.4. It is plain that〈·, ·〉 : (X ⊠ E) × Lip0(X,E∗) → K is a well-definedbilinear map and that Lip0(X,E∗) separates points ofX ⊠ E. Moreover, if f ∈ Lip0(X,E∗) and〈u, f 〉 = 0 for allu ∈ X ⊠ E, then〈 f (x), e〉 =

⟨δ(x,0) ⊠ e, f

⟩= 0 for all x ∈ X ande ∈ E. This implies thatf = 0 and thusX ⊠ E

separates points of Lip0(X,E∗). This completes the proof of the theorem. �

Since⟨X ⊠ E, Lip0(X,E∗)

⟩is a dual pair, Lip0(X,E∗) can be identified with a linear subspace of (X ⊠ E)′ as

follows.

Corollary 1.8. For every map f∈ Lip0(X,E∗), the functionalΛ( f ) : X ⊠ E→ K, given by

Λ( f )(u) =n∑

i=1

〈 f (xi) − f (yi), ei〉

for u =∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, is linear. We say thatΛ( f ) is the linear functional on X⊠ E associated to f . Themap f 7→ Λ( f ) is a linear monomorphism fromLip0(X,E∗) into (X ⊠ E)′.

Proof. Let f ∈ Lip0(X,E∗). By Theorem 1.7, note thatΛ( f )(u) = 〈u, f 〉 for all u ∈ X⊠E. It is immediate thatΛ( f )is a well-defined linear functional onX ⊠ E and thatf 7→ Λ( f ) from Lip0(X,E∗) into (X ⊠ E)′ is a well-definedlinear map. Finally, letf ∈ Lip0(X,E∗) and assume thatΛ( f ) = 0. Then〈u, f 〉 = 0 for all u ∈ X ⊠ E. SinceX ⊠ Eseparates points of Lip0(X,E∗), it follows that f = 0 and this proves that the mapΛ is one-to-one. �

We next show thatX ⊠ E is linearly isomorphic to the linear spaceF ((X#, τp); E) of all finite-rank linearoperators fromX# into E which are continuous from the topology of pointwise convergenceτp of X# to the normtopology ofE.

Definition 1.2. Let X be a pointed metric space. For f∈ X#, x ∈ X andε ∈ R+, we put

B( f , x, ε) ={g ∈ X# : |g(x) − f (x)| < ε

}.

LetS be the family of sets{B( f , x, ε) : f ∈ X#, x ∈ X, ε ∈ R+

}. Then the topology of pointwise convergenceτp on

X# is the topology generated byS.

We can check that (X#, τp) is a locally convex space. Next we describe its dual space.

Lemma 1.9. Let X be a pointed metric space. Then(X#, τp)∗ = lin({δx : x ∈ X})(⊂ (X#)′), whereδx is the functionalon X# defined byδx(g) = g(x).

Proof. Define the linear functionalT : X# → K by T(g) =∑n

i=1 λig(xi) for all g ∈ X#, wheren ∈ N, λ1, . . . λn ∈ K

and x1, . . . , xn ∈ X. Put r = 1 +∑n

i=1 |λi | and letε > 0 be arbitrary. Ifg ∈⋂n

i=1 B(0, xi, ε/r), then |T(g)| ≤∑ni=1 |λi | |g(xi)| < ε. This proves thatT is continuous onX# when it is equipped with the topologyτp. Conversely,

we need to show that every elementS in (X#, τp)∗ is of that form. SinceS is continuous in theτp-topology, thereis an open neighborhoodV of 0 such that|S(g)| < 1 for all g ∈ V. We can suppose thatV =

⋂ni=1 B(0, xi, ε) for

suitablen ∈ N, x1, . . . , xn ∈ X andε > 0. Take f ∈⋂n

i=1 kerδxi . Thenm f ∈ V for eachm ∈ N. By the linearity ofS, it follows that|S( f )| < 1/m for all m ∈ N and soS( f ) = 0. This shows that

⋂ni=1 kerδxi ⊂ kerS and the lemma

follows from a known fact of linear algebra. �

Theorem 1.10. The map J: X ⊠ E→ F ((X#, τp); E), given by

J(u)(g) =n∑

i=1

(g(xi) − g(yi)) ei


for u =∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E and g∈ X#, is a linear isomorphism.

Proof. Let u =∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E. It is immediate thatJ(u) : X# → E is well defined and linear. Note thatJ(u)(X#) ⊂ lin{e1, . . . , en} and henceJ(u) has finite-dimensional range. In order to prove thatJ(u) is continuousfrom (X#, τp) to E, it is sufficient to see thatJ(u) is continuous at 0. Letε > 0. Denoter = 2(1+

∑ni=1 ||ei ||) and put

zi = xi for i = 1, . . . , n andzi = yi−n for i = n+ 1, . . . , 2n. TakeU = ∩2ni=1B(0, zi, ε/r). For anyg ∈ U, we have

‖J(u)(g)‖ ≤n∑

i=1

(|g(xi)| + |g(yi)|) ‖ei‖ ≤

n∑

i=1

2ε

r‖ei‖ < ε,

as required. Moreover, by Proposition 1.6, the mapu 7→ J(u) from X ⊠ E to F ((X#, τp); E) is well defined. Wenow show thatJ is linear. Ifλ ∈ K, thenλu =

∑ni=1 δ(xi ,yi) ⊠ (λei) and so

J(λu)(g) =n∑

i=1

(g(xi) − g(yi)) (λei) = λn∑

i=1

(g(xi) − g(yi)) ei = λJ(u)(g)

for all g ∈ X#. Now, letv =∑m

i=1 δ(x′i ,y′i ) ⊠ e′i ∈ X⊠ E and take the representation

∑n+mi=1 δ(x′′i ,y

′′i ) ⊠ e′′i of u+ v given in

Lemma 1.3. Then we have

J(u+ v)(g) =n+m∑

i=1

(g(x′′i ) − g(y′′i )

)e′′i

=

n∑

i=1

(g(x′′i ) − g(y′′i )

)e′′i +

n+m∑

i=n+1

(g(x′′i ) − g(y′′i )

)e′′i

=

n∑

i=1


n+m∑

i=n+1

(g(x′i−n) − g(y′i−n)

)e′i−n

=

n∑

i=1


m∑

i=1

(g(x′i ) − g(y′i )

)e′i

= J(u)(g) + J(v)(g)

for all g ∈ X#. It remains to show thatJ is bijective. On one hand, assume thatJ(u) = 0. ThenJ(u)(g) =∑ni=1(g(xi) − g(yi))ei = 0 for all g ∈ X#, this implies thatu = 0 by Proposition 1.6 and soJ is one-to-one. On

the other hand, ifT ∈ F ((X#, τp); E), take a basis{e1, . . . , en} of T(X#). For eachg ∈ X#, there are uniqueλ

(g)1 , . . . , λ

(g)n ∈ K such thatT(g) =

∑ni=1 λ

(g)i ei . For eachi ∈ {1, . . . , n}, letyi : T(X#)→ K be given byyi(T(g)) = λ(g)

ifor all g ∈ X#. The uniqueness of the representation of each element ofT(X#) implies that eachyi is linear. Henceyi is continuous onT(X#) since the linear spaceT(X#) is finite-dimensional. Then eachTi = yi ◦ T belongs to(X#, τp)∗ andT(g) =

∑ni=1 Ti(g)ei for all g ∈ X#. Since (X#, τp)∗ = lin({δx : x ∈ X})(⊂ (X#)′) by Lemma 1.9, for

eachi ∈ {1, . . . , n} there arem(i) ∈ N, λ(i)1 , . . . , λ

(i)m(i) ∈ K andx(i)

1 , . . . , x(i)m(i) ∈ X such thatTi =

∑m(i)j=1 λ

(i)j δx(i)

j. Then,

for eachg ∈ X#, we may write

T(g) =n∑

i=1

Ti(g)ei =

m∑

j=1

g(x j)u j = J

m∑

j=1

δ(x j ,0) ⊠ u j

(g)

for certainm ∈ N, x1, . . . , xm ∈ X andu1, . . . , um ∈ E. This proves thatJ is onto. �

Corollary 1.11. Let X be a pointed metric space. If E is a finite-dimensional Banach space, then X⊠ E is linearlyisomorphic toL((X#, τp); E). In particular, X⊠ K is linearly isomorphic to(X#, τp)∗ = lin({δx : x ∈ X}).


2. Lipschitz tensor product functionals

We introduce the concept of Lipschitz tensor product functional of a Lipschitz functional and a bounded linearfunctional.

Definition 2.1. Let X be a pointed metric space and E a Banach space. Let g∈ X# and φ ∈ E∗. The mapg⊠ φ : X ⊠ E→ K, given by

(g⊠ φ)(u) =n∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉

for u =∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, is called the Lipschitz tensor product functional of g andφ.

By Lemma 1.4, note that

(g⊠ φ)(u) =n∑

i=1

〈(g · φ)(xi) − (g · φ)(yi), ei〉 = u(g · φ).

The following result which follows easily from this formulagathers some properties of these functionals.

Lemma 2.1. Let g ∈ X# andφ ∈ E∗. The functional g⊠ φ : X ⊠ E → K is a well-defined linear map satisfyingλ(g⊠ φ) = (λg) ⊠ φ = g⊠ (λφ) for anyλ ∈ K. Moreover,(g1 + g2) ⊠ φ = g1 ⊠ φ + g2 ⊠ φ for all g1, g2 ∈ X# andg⊠ (φ1 + φ2) = g⊠ φ1 + g⊠ φ2 for all φ1, φ2 ∈ E∗.

Definition 2.2. Let X be a pointed metric space and E a Banach space. The space X#i E∗ is defined as the linear

subspace of(X ⊠ E)′ spanned by the set{g⊠ φ : g ∈ X#, φ ∈ E∗

}. This space is called the associated Lipschitz

tensor product of X⊠ E.

From the aforementioned formula we also derive easily the following fact.

Lemma 2.2. For any∑m

j=1 g j ⊠ φ j ∈ X#i E∗ and

∑ni=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, we have

m∑

j=1

g j ⊠ φ j

n∑

i=1

δ(xi ,yi ) ⊠ ei

=

n∑

i=1

δ(xi ,yi) ⊠ ei

m∑

j=1

g j · φ j

.

Each elementu∗ in X#i E∗ has the formu∗ =

∑mj=1 λ j(g j ⊠ φ j), wherem ∈ N, λ j ∈ K, g j ∈ X# andφ j ∈ E∗, but

this representation is not unique. Sinceλ(g⊠ φ) = (λg) ⊠ φ = g⊠ (λφ), each element ofX#i E∗ can be expressed

as∑m

j=1 g j ⊠ φ j . This representation can be improved as follows.

Lemma 2.3. Every nonzero element u∗ in X#i E∗ has a representation

∑mj=1 g j ⊠ φ j such that the functions

g1, . . . , gm in X# are nonzero and the functionalsφ1, . . . , φm in E∗ are linearly independent.

Proof. Let u∗ ∈ X#i E∗, u∗ , 0. Since 0⊠ φ = 0, we can take a representation foru∗,

∑ni=1 hi ⊠ φi , where

h1, . . . , hn are nonzero. If the vectorsφ1, . . . , φn are linearly independent, we have finished. Otherwise, takeF = lin({φ1, . . . , φn}) and choose a subset of{φ1, . . . , φn}, which is a basis forF, φ1, . . . , φp (after reordering) forsomep < n. For eachi ∈ {p + 1, . . . , n} we can express the vectorφi as a unique linear combination in the form


φi =∑p

k=1 λ(i)k φk, whereλ(i)

1 , . . . , λ(i)p ∈ K. Using Lemma 2.1, we can write

u∗ =p∑

i=1

hi ⊠ φi +

n∑

i=p+1

hi ⊠ φi

=

p∑

i=1

hi ⊠ φi +

n∑

i=p+1

hi ⊠

p∑

k=1

λ(i)k φk

=

p∑

i=1

hi ⊠ φi +

n∑

i=p+1

p∑

k=1

λ(i)k (hi ⊠ φk)

=

p∑

i=1

hi ⊠ φi +

p∑

k=1

n∑

i=p+1

λ(i)k (hi ⊠ φk)

=

p∑

i=1

hi ⊠ φi +

p∑

k=1

n∑

i=p+1

λ(i)k hi

⊠ φk

=

p∑

j=1

h j +

n∑

i=p+1

λ(i)j hi

⊠ φ j .

Denoteg j = h j +∑n

i=p+1 λ(i)j hi for eachj ∈ {1, . . . , p}. Sinceu∗ , 0, after reordering, we can takem≤ p for which

g j , 0 for all j ≤ mandg j = 0 for all j > m+ 1. Then∑m

j=1 g j ⊠ φ j is a representation ofu∗ satisfying the requiredconditions. �

Our next aim is to show that the associated Lipschitz tensor productX#i E∗ is linearly isomorphic to the space

of Lipschitz finite-rank operators fromX to E∗. This class of Lipschitz operators appears in [17, 16].Let us recall that ifX is a set andE is a vector space, then a mapf : X → E is said to have finite-dimensional

rank if the subspace ofE generated byf (X), lin( f (X)), is finite-dimensional in whose case the rank off , denotedby rank(f ), is defined as the dimension of lin(f (X)).

For a pointed metric spaceX and a Banach spaceE, we denote by Lip0F (X,E∗) the set of all Lipschitz finite-rank operators fromX to E∗. Clearly, Lip0F (X,E∗) is a linear subspace of Lip0(X,E∗). For anyg ∈ X# andφ ∈ E∗,we consider in Lemma 1.5 the elementsg · φ of Lip0F(X,E∗) defined by (g · φ)(x) = g(x)φ for all x ∈ X. Note thatrank(g · φ) = 1 if g , 0 andφ , 0. Now we prove that these elements generate linearly the space Lip0F (X,E∗).

Lemma 2.4. Every element f∈ Lip0F (X,E∗) has a representation in the form f=∑m

j=1 g j ·φ j, where m= rank(f ),g1, . . . , gm ∈ X# andφ1, . . . , φm ∈ E∗.

Proof. Suppose that lin(f (X)) is m-dimensional and let{φ1, . . . , φm} be a basis of lin(f (X)). Then, for eachx ∈ X,the elementf (x) ∈ f (X) is expressible in a unique form asf (x) =

∑mj=1 λ

(x)j φ j with λ(x)

1 , . . . , λ(x)m ∈ K. For each

j ∈ {1, . . . ,m}, define the linear mapy j : lin( f (X)) → K by y j( f (x)) = λ(x)j for all x ∈ X. Let g j = y j ◦ f . Clearly,

g j ∈ X# and, givenx ∈ X, we havef (x) =∑m

j=1 λ(x)j φ j =

∑mj=1 g j(x)φ j. Hencef =

∑mj=1 g j · φ j . �

Theorem 2.5. The map K: X#i E∗ → Lip0F (X,E∗), defined by

K

m∑

j=1

g j ⊠ φ j

=m∑

j=1

g j · φ j ,

is a linear isomorphism.

Proof. The mapK is well defined by applying Lemma 2.2 and Theorem 1.7. Clearly, K is linear. Moreover, it issurjective by Lemma 2.4 and injective by Lemma 2.2. �


3. Lipschitz tensor product operators

We introduce the concept of Lipschitz tensor product operator of a Lipschitz operator and a bounded linearoperator.

Definition 3.1. Let X,Y be pointed metric spaces and E, F Banach spaces. Let h∈ Lip0(X,Y) and T ∈ L(E, F).The map h⊠ T : X ⊠ E→ Y⊠ F, given by

(h⊠ T)(u) =n∑

i=1

δ(h(xi ),h(yi )) ⊠ T(ei)

for u =∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, is called the Lipschitz tensor product operator of h and T.

Lemma 3.1. Let h∈ Lip0(X,Y) and T ∈ L(E, F). Then h⊠ T : X ⊠ E→ Y⊠ F is a well-defined linear operator.

Proof. Let u =∑n

i=1 δ(xi ,yi) ⊠ ei andv =∑m

i=1 δ(x′i ,y′i ) ⊠ e′i be in X ⊠ E. If u = v, then Proposition 1.6 says us

that∑n

i=1(g(xi) − g(yi))ei =∑m

i=1(g(x′i ) − g(y′i ))e′i for all g ∈ BX#. In particular, this holds for all function in

BX# of the form (f ◦ h)/(1 + Lip(h)) with f varying in BY#. It follows that∑n

i=1 ( f (h(xi)) − f (h(yi))) T(ei) =∑mi=1

(f (h(x′i )) − f (h(y′i ))

)T(e′i ) for all f ∈ BY#, and this implies that

∑ni=1 δ(h(xi ),h(yi ))⊠T(ei) =

∑mi=1 δ(h(x′i ),h(y′i ))⊠T(e′i )

again by Proposition 1.6. Hence the maph⊠ T is well defined.We see thath⊠ T is linear. Letλ ∈ K. Thenλu =

∑ni=1 δ(xi ,yi) ⊠ (λei), and Definition 3.1 and Lemma 1.1 give

(h⊠ T)(λu) =n∑

i=1

δ(h(xi ),h(yi )) ⊠ T(λei)

=

n∑

i=1

δ(h(xi ),h(yi )) ⊠ λT(ei)

= λ

n∑

i=1

δ(h(xi ),h(yi)) ⊠ T(ei)

= λ(h⊠ T)(u).

Takeu+ v =∑n+m

i=1 δ(x′′i ,y′′i ) ⊠ e′′i as in Lemma 1.3. Then we have

(h⊠ T)(u+ v) =n+m∑

i=1

δ(h(x′′i ),h(y′′i )) ⊠ T(e′′i )

=

n∑

i=1

δ(h(x′′i ),h(y′′i )) ⊠ T(e′′i ) +n+m∑

i=n+1

δ(h(x′′i ),h(y′′i )) ⊠ T(e′′i )

=

n∑

i=1

δ(h(xi ),h(yi )) ⊠ T(ei) +n+m∑

i=n+1

δ(h(x′i−n),h(y′i−n)) ⊠ T(e′i−n)

=

n∑

i=1

δ(h(xi ),h(yi )) ⊠ T(ei) +m∑

i=1

δ(h(x′i ),h(y′i )) ⊠ T(e′i )

= (h⊠ T)(u) + (h⊠ T)(v).

�

4. Lipschitz cross-norms

We denote the linear spaceX ⊠ E endowed with a normα by X ⊠α E, and its completion byX⊠αE. We arelooking for a norm on the linear spaceX⊠ E, and for our purposes it is convenient to work with norms thatsatisfythe following conditions.


Definition 4.1. Let X be a pointed metric space and E a Banach space. We say thata normα on X⊠ E is aLipschitz cross-norm if

α(δ(x,y) ⊠ e

)= d(x, y) ‖e‖

for all (x, y) ∈ X2 and e∈ E.A Lipschitz cross-normα on X⊠ E is said to be dualizable if given g∈ X# andφ ∈ E∗, we have

∣∣∣∣∣∣∣

n∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉

∣∣∣∣∣∣∣≤ Lip(g) ‖φ‖α

n∑

i=1

δ(xi ,yi) ⊠ ei

for all∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E.A Lipschitz cross-normα on X⊠ E is called uniform if given h∈ Lip0(X,X) and T ∈ L(E,E), we have

α

n∑

i=1


≤ Lip(h) ‖T‖α

n∑

i=1

δ(xi ,yi) ⊠ ei

for all∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E.

The dualizable Lipschitz cross-norms onX⊠E may be characterized by the boundedness of the Lipschitz tensorproduct functionals.

Proposition 4.1. A Lipschitz cross-normα on X⊠ E is dualizable if and only if, for each g∈ X# andφ ∈ E∗, thelinear functional g⊠ φ : X ⊠α E→ K is bounded and‖g⊠ φ‖ = Lip(g) ‖φ‖.

Proof. Let α be a Lipschitz cross-norm onX ⊠ E. Giveng ∈ X# andφ ∈ E∗, if α is dualizable, we have∣∣∣∣∣∣∣(g⊠ φ)

n∑

i=1

δ(xi ,yi) ⊠ ei

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣

n∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉

∣∣∣∣∣∣∣≤ Lip(g) ‖φ‖α

n∑

i=1

δ(xi ,yi) ⊠ ei

for all∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E. Hence the linear functionalg⊠ φ is bounded onX ⊠α E and‖g⊠ φ‖ ≤ Lip(g) ‖φ‖.The opposite inequality Lip(g) ‖φ‖ ≤ ‖g⊠ φ‖ is deduced from the fact that

|g(x) − g(y)| |〈φ, e〉| =∣∣∣(g⊠ φ)(δ(x,y) ⊠ e)

∣∣∣ ≤ ‖g⊠ φ‖α(δ(x,y) ⊠ e

)= ‖g⊠ φ‖d(x, y) ‖e‖

for all x, y ∈ X ande ∈ E.Conversely, if for anyg ∈ X# andφ ∈ E∗, the linear functionalg⊠ φ : X ⊠α E → K is bounded and‖g⊠ φ‖ =

Lip(g) ‖φ‖, then∣∣∣∣∣∣∣

n∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣(g⊠ φ)

n∑

i=1

δ(xi ,yi) ⊠ ei

∣∣∣∣∣∣∣

≤ ‖g⊠ φ‖α

n∑

i=1

δ(xi ,yi) ⊠ ei

= Lip(g) ‖φ‖α

n∑

i=1

δ(xi ,yi) ⊠ ei

for all∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, and soα is dualizable. �

Similarly, the boundedness of the Lipschitz tensor productoperators characterizes the uniform Lipschitz cross-norms onX ⊠ E.

Proposition 4.2. A Lipschitz cross-normα on X ⊠ E is uniform if and only if, for each h∈ Lip0(X,X) andT ∈ L(E,E), the linear operator h⊠ T : X ⊠α E→ X ⊠α E is bounded and‖h⊠ T‖ = Lip(h) ‖T‖.


Proof. Let α be a Lipschitz cross-norm onX ⊠ E. If α is uniform, givenh ∈ Lip0(X,X) andT ∈ L(E,E), we have

α

(h⊠ T)

n∑

i=1

δ(xi ,yi) ⊠ ei

= α

n∑

i=1

δ(h(xi),h(yi )) ⊠ T(ei)

≤ Lip(h) ‖T‖α

n∑

i=1

δ(xi ,yi) ⊠ ei

for all∑n

i=1 δ(xi ,yi)⊠ei ∈ X⊠E. It follows that the linear operatorh⊠T is bounded onX⊠αE and‖h⊠ T‖ ≤ Lip(h) ‖T‖.For the reverse inequality, notice that

d(h(x), h(y)) ‖T(e)‖ = α(δ(h(x),h(y)) ⊠ T(e)

)

= α((h⊠ T)

(δ(x,y) ⊠ e

))

≤ ‖h⊠ T‖α(δ(x,y) ⊠ e

)

= ‖h⊠ T‖d(x, y) ‖e‖

for all x, y ∈ X ande ∈ E, and therefore Lip(h) ‖T‖ ≤ ‖h⊠ T‖.Conversely, if for eachh ∈ Lip0(X,X) andT ∈ L(E,E), the linear maph⊠ T : X ⊠α E → X ⊠α E is bounded

and‖h⊠ T‖ = Lip(h) ‖T‖, then

α

n∑

i=1


= α(h⊠ T)

n∑

i=1

δ(xi ,yi) ⊠ ei

≤ ‖h⊠ T‖α

n∑

i=1

δ(xi ,yi) ⊠ ei

= Lip(h) ‖T‖α

n∑

i=1

δ(xi ,yi) ⊠ ei

for all∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, and soα is uniform. �

Remark 4.1. A reading of the proofs of the two preceding propositions shows that a Lipschitz cross-normα onX ⊠ E is dualizable (uniform) if for each g∈ X# andφ ∈ E∗, then g⊠ φ ∈ (X ⊠α E)∗ and ‖g⊠ φ‖ ≤ Lip(g) ‖φ‖(respectively, if for each h∈ Lip0(X,X) and T ∈ L(E,E), then h⊠T ∈ L(X⊠αE,X⊠αE) and‖h⊠ T‖ ≤ Lip(h) ‖T‖).

5. Associated Lipschitz cross-norms

We introduce the concept of Lipschitz cross-norm on the associated Lipschitz tensor product ofX ⊠ E.

Definition 5.1. Let X be a pointed metric space and E a Banach space. We say thata normβ on X#i E∗ is a

Lipschitz cross-norm ifβ(g⊠ φ) = Lip(g) ‖φ‖ for all g ∈ X# andφ ∈ E∗.

We now construct the following Lipschitz cross-norm onX#i E∗.

Proposition 5.1. Letα be a dualizable Lipschitz cross-norm on X⊠ E. The mapα′ : X#i E∗ → R, given by

α′(u∗) = sup

∣∣∣∣∣∣∣∣

m∑

j=1

g j ⊠ φ j

n∑

i=1

δ(xi ,yi) ⊠ ei

∣∣∣∣∣∣∣∣: α

n∑

i=1

δ(xi ,yi) ⊠ ei

≤ 1

for u∗ =∑m

j=1 g j ⊠ φ j ∈ X#i E∗, is a Lipschitz cross-norm on X#

i E∗.

Proof. Let u∗ =∑m

j=1 g j ⊠ φ j ∈ X#i E∗. If

∑ni=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, we have

m∑

j=1

g j ⊠ φ j

n∑

i=1

δ(xi ,yi) ⊠ ei

=m∑

j=1

(g j ⊠ φ j)

n∑

i=1

δ(xi ,yi) ⊠ ei

,


and sinceα is dualizable, by applying Proposition 4.1 gives∣∣∣∣∣∣∣∣

m∑

j=1

g j ⊠ φ j

n∑

i=1

δ(xi ,yi) ⊠ ei

∣∣∣∣∣∣∣∣≤

m∑

j=1

∣∣∣∣∣∣∣(g j ⊠ φ j)

n∑

i=1

δ(xi ,yi) ⊠ ei

∣∣∣∣∣∣∣

≤

m∑

j=1

∥∥∥g j ⊠ φ j

∥∥∥α

n∑

i=1

δ(xi ,yi ) ⊠ ei

=

m∑

j=1

Lip(g j)∥∥∥φ j

∥∥∥α

n∑

i=1

δ(xi ,yi) ⊠ ei

.

Then the supremum on the right side of the equality in the statement exists. It is immediate thatα′ is a well-definedmap. Givenλ ∈ K, it is plain thatλu∗ =

∑mj=1 g j ⊠ (λφ j) and therefore

α′(λu∗) = sup

∣∣∣∣∣∣∣∣

m∑

j=1

g j ⊠ (λφ j)

n∑

i=1

δ(xi ,yi) ⊠ ei

∣∣∣∣∣∣∣∣: α

n∑

i=1

δ(xi ,yi) ⊠ ei

≤ 1

.

Clearly, the supremum on the right side above is equal to|λ|α′(u∗) and thusα′(λu∗) = |λ|α′(u∗).Let v∗ =

∑rj=1 g′j ⊠ φ

′j ∈ X#

i E∗. Thenu∗ + v∗ =∑m

j=1 g j ⊠ φ j +∑r

j=1 g′j ⊠ φ′j . Since

∣∣∣∣∣∣∣∣

m∑

j=1

g j ⊠ φ j +

r∑

j=1

g′j ⊠ φ′j

n∑

i=1

δ(xi ,yi) ⊠ ei

∣∣∣∣∣∣∣∣

≤

∣∣∣∣∣∣∣∣

m∑

j=1

g j ⊠ φ j

n∑

i=1

δ(xi ,yi) ⊠ ei

∣∣∣∣∣∣∣∣+

∣∣∣∣∣∣∣∣

r∑

j=1

g′j ⊠ φ′j

n∑

i=1

δ(xi ,yi ) ⊠ ei

∣∣∣∣∣∣∣∣

≤(α′(u∗) + α′(v∗)

)α

n∑

i=1

δ(xi ,yi) ⊠ ei

for all∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, it follows thatα′(u∗ + v∗) ≤ α′(u∗) + α′(v∗).Assumeu∗ , 0. By Lemma 2.3, we can take a representation ofu∗,

∑mj=1 g j ⊠ φ j , such thatg1, . . . , gm in

X# are nonzero andφ1, . . . , φm in E∗ are linearly independent. Theng1(x) , 0 for somex ∈ X. The linearindependence of theφ j ’s yields that

∑mj=1 g j(x)φ j , 0. Now, takee ∈ E for which

∑mj=1 g j(x)〈φ j, e〉 , 0, that is,(∑m

j=1 g j ⊠ φ j

)(δ(x,0) ⊠ e) , 0. This implies thatδ(x,0) ⊠ e, 0, and so

α′(u∗) ≥

∣∣∣∣(∑m

j=1 g j ⊠ φ j

) (δ(x,0) ⊠ e

)∣∣∣∣α(δ(x,0) ⊠ e)

> 0.

In this way, we have proved thatα′ is a norm onX#i E∗.

In order to show thatα′ is a Lipschitz cross-norm onX#i E∗, let g ∈ X# andφ ∈ E∗. For all x, y ∈ X ande∈ E,

we have

|(g(x) − g(y)) 〈φ, e〉| =∣∣∣∣(g⊠ φ)

(δ(x,y) ⊠ e

)∣∣∣∣ ≤ α′(g⊠ φ)α(δ(x,y) ⊠ e) = α′(g⊠ φ)d(x, y) ‖e‖ ,

and from this we infer that Lip(g) ‖φ‖ ≤ α′(g⊠ φ). For the reverse, sinceα is dualizable, we have∣∣∣∣∣∣∣(g⊠ φ)

n∑

i=1

δ(xi ,yi) ⊠ ei

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣

n∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉

∣∣∣∣∣∣∣≤ Lip(g) ‖φ‖α

n∑

i=1

δ(xi ,yi) ⊠ ei

for any∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, and this implies thatα′(g⊠ φ) ≤ Lip(g) ‖φ‖.�


Definition 5.2. Let X be a pointed metric space and E a Banach space. Letα be a dualizable Lipschitz cross-normon X⊠ E. The normα′ on X#

i E∗ is called the associated Lipschitz norm ofα. The vector space X# i E∗ with thenormα′ will be denoted by X# iα′ E∗ and its completion by X#iα′E∗.

Remark 5.1. Note that ifα is a dualizable Lipschitz cross-norm on X⊠ E, then X#iα′E∗ is a linear subspace of

(X⊠αE)∗.

6. The induced Lipschitz dual norm

Definition 6.1. For each u=∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, define:

L(u) = sup

∣∣∣∣∣∣∣

n∑

i=1

〈 f (xi) − f (yi), ei〉

∣∣∣∣∣∣∣: f ∈ Lip0(X,E∗), Lip( f ) ≤ 1

.

Note that the supremum on the right side above exists andL(u) ≤∑n

i=1 d(xi , yi) ‖ei‖ because∣∣∣∣∣∣∣

n∑

i=1

〈 f (xi) − f (yi), ei〉

∣∣∣∣∣∣∣≤

n∑

i=1

|〈 f (xi) − f (yi), ei〉|

≤

n∑

i=1

‖ f (xi) − f (yi)‖ ‖ei‖

≤ Lip( f )n∑

i=1

d(xi, yi) ‖ei‖

for all f ∈ Lip0(X,E∗). Moreover,L defines a map fromX ⊠ E to R by Lemma 1.4.

Theorem 6.1. The linear space X⊠ E is contained inLip0(X,E∗)∗ and L is the dual norm of the normLip ofLip0(X,E∗) induced on X⊠ E. Moreover, L is a Lipschitz cross-norm on X⊠ E.

Proof. Let x, y ∈ X ande ∈ E. Sinceδ(x,y) ⊠ e is a linear map on Lip0(X,E∗) and∣∣∣(δ(x,y) ⊠ e)( f )

∣∣∣ = |〈 f (x) − f (y), e〉| ≤ ‖ f (x) − f (y)‖ ‖e‖ ≤ Lip( f )d(x, y) ‖e‖

for all f ∈ Lip0(X,E∗), thenδ(x,y)⊠e ∈ Lip0(X,E∗)∗ and thusX⊠E ⊂ Lip0(X,E∗)∗. For everyu =∑n

i=1 δ(xi ,yi)⊠ei ∈

X ⊠ E, we haveL(u) = sup

{|u( f )| : f ∈ Lip0(X,E∗), Lip( f ) ≤ 1

}

by Lemma 1.4, and thereforeL is the dual norm of the norm Lip of Lip0(X,E∗) induced onX ⊠ E. Finally, weprove thatL is a Lipschitz cross-norm. By above-proved,

∣∣∣(δ(x,y) ⊠ e)( f )∣∣∣ ≤ d(x, y) ‖e‖ for all f ∈ Lip0(X,E∗) with

Lip( f ) ≤ 1, and henceL(δ(x,y) ⊠ e) ≤ d(x, y) ‖e‖. For the reverse estimate, takeφ ∈ E∗ with ‖φ‖ = 1 satisfying|〈φ, e〉| = ‖e‖, and consider the mapf : X→ E∗ given by

f (z) = (d(0, x) − d(z, x))φ (z ∈ X).

An easy verification shows thatf is in Lip0(X,E∗) with Lip( f ) ≤ 1 and∣∣∣(δ(x,y) ⊠ e)( f )

∣∣∣ = |〈 f (x) − f (y), e〉| = |〈d(x, y)φ, e〉| = d(x, y) |〈φ, e〉| = d(x, y) ‖e‖ ,

and therefored(x, y) ‖e‖ =∣∣∣(δ(x,y) ⊠ e)( f )

∣∣∣ ≤ L(δ(x,y) ⊠ e). �

The following result is essentially known. For completeness we include it here with an alternate proof.

Theorem 6.2. [17, Theorem 4.1]Let X be a pointed metric space and let E be a Banach space. ThenLip0(X,E∗)is isometrically isomorphic to(X⊠LE)∗, via the mapΛ : Lip0(X,E∗)→ (X⊠LE)∗ given by

Λ( f )(u) =n∑

i=1

〈 f (xi) − f (yi), ei〉


for f ∈ Lip0(X,E∗) and u=∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E. Its inverseΛ−1 : (X⊠LE)∗ → Lip0(X,E∗) is defined by⟨Λ−1(ϕ)(x), e

⟩=

⟨ϕ, δ(x,0) ⊠ e

⟩

for ϕ ∈ (X⊠LE)∗, x ∈ X and e∈ E.

Proof. By Corollary 1.8, the mapf 7→ Λ( f ) is a linear monomorphism from Lip0(X,E∗) into (X ⊠ E)′. In fact,Λ( f ) ∈ (X⊠LE)∗ and‖Λ( f )‖ ≤ Lip( f ). Note that

|Λ( f )(u)| =

∣∣∣∣∣∣∣

n∑

i=1

〈 f (xi) − f (yi), ei〉

∣∣∣∣∣∣∣≤ Lip( f )L(u)

for all u =∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X⊠ E. ThereforeΛ( f ) is bounded onX⊠L E, hence onX⊠LE by the denseness of theformer set in the latter one. In order to see thatΛ is a surjective isometry, letϕ be an element of (X⊠LE)∗. Definef : X→ E∗ by

〈 f (x), e〉 = ϕ(δ(x,0) ⊠ e) (x ∈ X, e ∈ E) .

It is plain that f (x) is a well-defined bounded linear functional onE and that f is well defined. Observe that〈 f (x) − f (y), e〉 = ϕ(δ(x,y) ⊠ e) for all x, y ∈ X ande ∈ E. Fix x, y ∈ X. It follows that

|〈 f (x) − f (y), e〉| =∣∣∣ϕ(δ(x,y) ⊠ e)

∣∣∣ ≤ ‖ϕ‖ L(δ(x,y) ⊠ e) = ‖ϕ‖d(x, y) ‖e‖

for all e ∈ E, and so‖ f (x) − f (y)‖ ≤ ‖ϕ‖d(x, y). Hence f ∈ Lip0(X,E∗) and Lip(f ) ≤ ‖ϕ‖. For anyu =∑ni=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, we get

Λ( f )(u) =n∑

i=1

〈 f (xi) − f (yi), ei〉 =

n∑

i=1

ϕ(δ(xi ,yi) ⊠ ei) = ϕ

n∑

i=1

δ(xi ,yi) ⊠ ei

= ϕ(u).

HenceΛ( f ) = ϕ on a dense subspace ofX⊠LE and, consequently,Λ( f ) = ϕ. Moreover, Lip(f ) ≤ ‖ϕ‖ = ‖Λ( f )‖.This completes the proof of the theorem. �

7. The Lipschitz injective norm

We introduce the Lipschitz injective norm onX ⊠ E.

Definition 7.1. For each u=∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, define:

ε(u) = sup

∣∣∣∣∣∣∣

n∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉

∣∣∣∣∣∣∣: g ∈ BX#, φ ∈ BE∗

.

Notice that the supremum on the right side in the previous definition exists since∣∣∣∣∣∣∣

n∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉

∣∣∣∣∣∣∣≤

n∑

i=1

|(g(xi) − g(yi)) 〈φ, ei〉|

≤

n∑

i=1

Lip(g)d(xi, yi) ‖φ‖ ‖ei‖

≤

n∑

i=1

d(xi, yi) ‖ei‖

for all g ∈ BX# andφ ∈ BE∗ . Note thatn∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉 = (g⊠ φ)

n∑

i=1

δ(xi ,yi) ⊠ ei

,

and, consequently,ε defines a map fromX ⊠ E toR by Lemma 2.1.

Theorem 7.1. ε is a uniform and dualizable Lipschitz cross-norm on X⊠ E.


Proof. Let u =∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E. Suppose thatε(u) = 0. Then∑n

i=1(g(xi) − g(yi))〈φ, ei〉 = 0 for all g ∈ BX#

andφ ∈ BE∗ , and this happens if and only ifu = 0 by Proposition 1.6.For anyλ ∈ K, we haveλu =

∑ni=1 δ(xi ,yi) ⊠ (λei), and therefore

ε(λu) = sup

∣∣∣∣∣∣∣

n∑

i=1

(g(xi) − g(yi)) 〈φ, λei〉

∣∣∣∣∣∣∣: g ∈ BX#, φ ∈ BE∗

= |λ| ε(u).

Let v =∑m

i=1 δ(x′i ,y′i ) ⊠ e′i ∈ X ⊠ E. Then

∑n+mi=1 δ(x′′i ,y

′′i ) ⊠ e′′i , being as in Lemma 1.3, is a representation foru+ v.

For anyg ∈ BX# andφ ∈ BE∗ , it holds that∣∣∣∣∣∣∣

n+m∑

i=1

(g(x′′i ) − g(y′′i )

) ⟨φ, e′′i

⟩∣∣∣∣∣∣∣≤

∣∣∣∣∣∣∣

n∑

i=1

(g(x′′i ) − g(y′′i )

) ⟨φ, e′′i

⟩∣∣∣∣∣∣∣+

∣∣∣∣∣∣∣

n+m∑

i=n+1

(g(x′′i ) − g(y′′i )

) ⟨φ, e′′i

⟩∣∣∣∣∣∣∣

=

∣∣∣∣∣∣∣

n∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉

∣∣∣∣∣∣∣+

∣∣∣∣∣∣∣

n+m∑

i=n+1

(g(x′i−n) − g(y′i−n)

) ⟨φ, e′i−n

⟩∣∣∣∣∣∣∣

=

∣∣∣∣∣∣∣

n∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉

∣∣∣∣∣∣∣+

∣∣∣∣∣∣∣

m∑

i=1

(g(x′i ) − g(y′i )

) ⟨φ, e′i

⟩∣∣∣∣∣∣∣

≤ ε(u) + ε(v),

and thereforeε(u+ v) ≤ ε(u) + ε(v). Henceε is a norm onX ⊠ E.We claim thatε is a Lipschitz cross-norm. Takeδ(x,y) ⊠ e ∈ X ⊠ E. For anyg ∈ BX# andφ ∈ BE∗ , we have

|(g(x) − g(y)) 〈φ, e〉| ≤ Lip(g)d(x, y) ‖φ‖ ‖e‖ ≤ d(x, y) ‖e‖ ,

and soε(δ(x,y) ⊠ e) ≤ d(x, y) ‖e‖. For the converse inequality, we can findg0 ∈ BX# andφ0 ∈ BE∗ such that|g0(x) − g0(y)| = d(x, y) and〈φ0, e〉 = ‖e‖. For example,g0(z) = d(0, x) − d(z, x) for all z ∈ X. Then

ε(δ(x,y) ⊠ e

)≥ |(g0(x) − g0(y)) 〈φ0, e〉| = d(x, y) ‖e‖ ,

and this proves our claim.Now takeg ∈ X# andφ ∈ E∗. By Definition 7.1, we have

∣∣∣∣∣∣∣

n∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉

∣∣∣∣∣∣∣≤ Lip(g) ‖φ‖ ε

n∑

i=1

δ(xi ,yi) ⊠ ei

for all∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E. Hence the normε is dualizable. Then, by Proposition 5.1,ε′ is a Lipschitzcross-norm onX#

i E∗.Finally, we prove that the normε is uniform. Leth ∈ Lip0(X,X) andT ∈ L(E,E). Let T∗ denote the adjoint

operator ofT. We now recall that the Lipschitz adjoint maph# : X# → X#, given byh#(g) = g ◦ h for all g ∈ X#,is a continuous linear operator and

∥∥∥h#∥∥∥ = Lip(h). Indeed, it is clear thath# is linear. Letg ∈ X# andx, y ∈ X. We

have∣∣∣h#(g)(x) − h#(g)(y)

∣∣∣ = |g(h(x)) − g(h(y))| ≤ Lip(g)d(h(x), h(y)) ≤ Lip(g)Lip(h)d(x, y),

henceh#(g) ∈ X# and Lip(h#(g)) ≤ Lip(g)Lip(h). It follows thath# is bounded and∥∥∥h#

∥∥∥ ≤ Lip(h). Taking thefunction defined onX by g(z) = d(h(x), 0)− d(h(x), z) which is inBX#, we get

d(h(x), h(y)) = |g(h(x)) − g(h(y))| =∣∣∣h#(g)(x) − h#(g)(y)

∣∣∣ ≤ Lip(h#(g))d(x, y),

which gives Lip(h) ≤ Lip(h#(g)) ≤∥∥∥h#

∥∥∥ Lip(g) ≤∥∥∥h#

∥∥∥ and so∥∥∥h#

∥∥∥ = Lip(h).


Given∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, we have∣∣∣∣∣∣∣

n∑

i=1

(g(h(xi)) − g(h(yi))) 〈φ,T(ei)〉

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣(h#(g) ⊠ T∗(φ))

n∑

i=1

δ(xi ,yi) ⊠ ei

∣∣∣∣∣∣∣

≤ ε′(h#(g) ⊠ T∗(φ)

)ε

n∑

i=1

δ(xi ,yi) ⊠ ei

= Lip(h#(g)) ‖T∗(φ)‖ ε

n∑

i=1

δ(xi ,yi) ⊠ ei

≤ Lip(h)Lip(g) ‖T‖ ‖φ‖ ε

n∑

i=1

δ(xi ,yi) ⊠ ei

≤ Lip(h) ‖T‖ ε

n∑

i=1

δ(xi ,yi) ⊠ ei

for all g ∈ BX# andφ ∈ BE∗ , and hence

ε

n∑

i=1


≤ Lip(h) ‖T‖ ε

n∑

i=1

δ(xi ,yi) ⊠ ei

,

which proves that the normε is uniform. �

Theorem 7.2. ε is the least dualizable Lipschitz cross-norm on X⊠ E.

Proof. According to Theorem 7.1,ε is a dualizable Lipschitz cross-norm onX⊠E. Letα be a dualizable Lipschitzcross-norm onX ⊠ E and assume, for contradiction, that

α

n∑

i=1

δ(xi ,yi) ⊠ ei

< ε

n∑

i=1

δ(xi ,yi) ⊠ ei

for some∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E. By the definition ofε, there existg ∈ BX# andφ ∈ BE∗ such that

α

n∑

i=1

δ(xi ,yi) ⊠ ei

<∣∣∣∣∣∣∣

n∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉

∣∣∣∣∣∣∣.

By Proposition 5.1,α′ is a Lipschitz cross-norm onX#i E∗, and we have

∣∣∣∣∣∣∣

n∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣(g⊠ φ)

n∑

i=1

δ(xi ,yi) ⊠ ei

∣∣∣∣∣∣∣≤ α′(g⊠ φ)α

n∑

i=1

δ(xi ,yi) ⊠ ei

.

Henceα′(g⊠ φ) > 1 and thus Lip(g) ‖φ‖ < α′(g⊠ φ). This contradicts thatα′ is a Lipschitz cross-norm. Thereforeα ≥ ε and this proves the theorem. �

The completionX⊠εE of X⊠ε E is called the injective Lipschitz tensor product ofX andE. Next we justify thisterminology.

Theorem 7.3. Let X be a pointed metric space and let E be a Banach space. Let X0 ⊂ X be a subset of Xcontaining0, and let E0 be a closed linear subspace of E. Then X0⊠εE0 is a linear subspace of X⊠εE.


Proof. Let u =∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X0 ⊠ E0. It is sufficient to prove thatεX⊠E(u) = εX0⊠E0(u), where

εX0⊠E0(u) = sup

∣∣∣∣∣∣∣

n∑

i=1

(g0(xi) − g0(yi)) 〈φ0, ei〉

∣∣∣∣∣∣∣: g0 ∈ BX#

0, φ0 ∈ BE∗0

,

εX⊠E(u) = sup

∣∣∣∣∣∣∣

n∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉

∣∣∣∣∣∣∣: g ∈ BX#, φ ∈ BE∗

.

By applying the classical Hahn–Banach theorem we can extendeachφ0 ∈ BE∗0to aφ ∈ BE∗ , and by applying the

nonlinear Hahn–Banach theorem we can extend eachg0 ∈ BX#0

to ag ∈ BX#, hence we see thatεX0⊠E0(u) ≤ εX⊠E(u).Conversely, by restricting the functionalsφ ∈ BE∗ to E0 and the Lipschitz functionsg ∈ BX# to X0, we obtain thatεX⊠E(u) ≤ εX0⊠E0(u). �

We can identifyX⊠εE with the space of all approximable bounded linear operatorsof (X#, τp) to E.

Proposition 7.4. The map J: X ⊠ε E→ F ((X#, τp); E), defined by

J(u)(g) =n∑

i=1


for u =∑n

i=1 δ(xi ,yi) ⊠ei ∈ X⊠E and g∈ X#, is an isometric isomorphism. As a consequence, X⊠εE is isometricallyisomorphic to the closure in the operator norm topology ofF ((X#, τp); E).

Proof. By Theorem 1.10,J is a linear bijection. Ifu =∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, we have

‖J(u)‖ = sup{‖J(u)(g)‖ : g ∈ BX#}

= sup

∣∣∣∣∣∣∣φ

n∑

i=1


∣∣∣∣∣∣∣: g ∈ BX#, φ ∈ BE∗

= sup

∣∣∣∣∣∣∣

n∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉

∣∣∣∣∣∣∣: g ∈ BX#, φ ∈ BE∗

= ε(u).

The consequence is immediate. �

LetF (X) be the Lipschitz-free Banach space over a pointed metric spaceX. Let us recall thatF (X) is the closedlinear subspace of (X#)∗ spanned by the set{δx : x ∈ X}, where for eachx ∈ X, δx is the evaluation functional atthe pointx defined onX#. Combining Proposition 7.4 and Corollary 1.11, we can provethat if X is a pointedmetric space, thenX⊠εK is isometrically isomorphic toF (X); in fact, much more is true. We show below that thespaceX⊠εE can be identified with the injective Banach-space tensor productF (X)⊗εE. First, let us recall somefundamental properties of the spaceF (X).

Theorem 7.5. [1],[23, pp. 39-41]Let X, Y be pointed metric spaces, and E a Banach space.

(i) The dual ofF (X) is (canonically) isometrically isomorphic to X#, with the duality pairing given by〈g, δx〉 = g(x) for all g ∈ X# and x ∈ X. Moreover, on bounded subsets of X#, the weak* topologycoincides with the topology of pointwise convergence.

(ii) The mapιX : x 7→ δx is an isometric embedding of X intoF (X).(iii) For any Lipschitz map T: X → Y with T(0) = 0, there is a unique linear mapT : F (X) → F (Y) such

that T ◦ ιX = ιy ◦ T. Furthermore,∥∥∥T

∥∥∥ = Lip(T).(iv) For any Lipschitz map T: X → E with T(0) = 0, there is a unique linear mapT : F (X) → E such that

T ◦ ιX = T. Furthermore,∥∥∥T

∥∥∥ = Lip(T).


It is because of the universal properties above that the spaceF (X) is called the Lipschitz-free space overX, orsimply the free space overX. These spaces have been recently used as tools in nonlinear Banach space theory, see[13, 19] and the survey [14].

Proposition 7.6. The map I: X ⊠ε E→ F (X) ⊗ε E, defined by

I (u) =n∑

i=1

(δxi − δyi ) ⊗ ei

for u =∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, is a linear isometry. As a consequence, X⊠εE is isometrically isomorphic toF (X)⊗εE.

Proof. Let u =∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X⊠ E. SinceF (X)∗ ≡ X#, note that the norm of∑n

i=1(δxi − δyi ) ⊗ ei in F (X)⊗ε Eis given by

sup

∣∣∣∣∣∣∣

n∑

i=1

〈g, δxi − δyi 〉〈φ, ei〉

∣∣∣∣∣∣∣: g ∈ BX#, φ ∈ BE∗

.

Since〈g, δxi − δyi 〉 is preciselyg(xi) − g(yi), Proposition 1.6 shows thatI is well defined (and thus linear) andmoreover a quick glance at Definition 7.1 shows thatI is an isometry.

Recall that the linear span of{δx}x∈X is dense inF (X), hence the tensors of the form∑n

i=1(δxi − δyi ) ⊗ ei , withxi , yi ∈ X andei ∈ E, are dense inF (X)⊗εE. This shows that the mapI has dense range, and thusX⊠εE isisometrically isomorphic toF (X)⊗εE. �

8. The Lipschitz projective norm

We introduce the Lipschitz projective norm onX ⊠ E.

Definition 8.1. For each u∈ X ⊠ E, define:

π(u) = inf

n∑

i=1

d(xi , yi) ‖ei‖ : u =n∑

i=1

δ(xi ,yi) ⊠ ei

,

the infimum being taken over all representations of u.

Theorem 8.1. π is a uniform and dualizable Lipschitz cross-norm on X⊠ E such that L≤ π.


i=1 δ(xi ,yi) ⊠ ei be a representation ofu. Using Definition 6.1, we have seen thatL(u) ≤

∑ni=1 d(xi , yi) ‖ei‖. Since this holds for every representation ofu, it follows thatL(u) ≤ π(u). Suppose that

π(u) = 0. SinceL(u) ≤ π(u) andL is a norm onX ⊠ E, thenu = 0.We check thatπ(λu) = |λ| π(u). If λ ∈ K, thenλu =

∑ni=1 δ(xi ,yi) ⊠ (λei) and so

π(λu) ≤n∑

i=1

d(xi, yi) ‖λei‖ = |λ|

n∑

i=1

d(xi , yi) ‖ei‖ .

Since the representation ofu is arbitrary, this implies thatπ(λu) ≤ |λ| π(u). If λ = 0, we haveπ(λu) = 0 = |λ| π(u)sinceπ(u) ≥ 0 for all u ∈ X ⊠ E. Assume thatλ , 0. Similarly, we haveπ(u) = π(λ−1(λu)) ≤

∣∣∣λ−1∣∣∣ π(λu), thus

|λ| π(u) ≤ π(λu) and henceπ(λu) = |λ| π(u).We show thatπ(u + v) ≤ π(u) + π(v) for all u, v ∈ X ⊠ E. Let ε > 0. Then there are representationsu =∑n

i=1 δ(xi ,yi)⊠ei andv =∑m

i=1 δ(x′i ,y′i ) ⊠e′i such that

∑ni=1 d(xi , yi) ‖ei‖ < π(u)+ ε/2 and

∑mi=1 d(x′i , y

′i )

∥∥∥e′i∥∥∥ < π(v)+ ε/2.

We can concatenate these representations to get a representation∑n+m

i=1 δ(x′′i ,y′′i ) ⊠ e′′i for u+ v as in Lemma 1.3. By


Definition 8.1, it follows that

π(u+ v) ≤n+m∑

i=1

d(x′′i , y′′i )

∥∥∥e′′i∥∥∥

=

n∑

i=1

d(x′′i , y′′i )

∥∥∥e′′i∥∥∥ +

n+m∑

i=n+1

d(x′′i , y′′i )

∥∥∥e′′i∥∥∥

=

n∑

i=1

d(xi , yi) ‖ei‖ +

n+m∑

i=n+1

d(x′i−n, y′i−n)

∥∥∥e′i−n

∥∥∥

=

n∑

i=1

d(xi , yi) ‖ei‖ +

m∑

i=1

d(x′i , y′i )

∥∥∥e′i∥∥∥

< π(u) + π(v) + ε.

By the arbitrariness ofε, we deduce thatπ(u+ v) ≤ π(u) + π(v). Henceπ is a norm onX ⊠ E.We now prove thatπ is a Lipschitz cross-norm. Let (x, y) ∈ X2 ande ∈ E. It is immediate thatπ(δ(x,y) ⊠ e) ≤

d(x, y) ‖e‖. Conversely, sinceL is a Lipschitz cross-norm onX ⊠ E and L ≤ π, it follows that d(x, y) ‖e‖ =L(δ(x,y) ⊠ e) ≤ π(δ(x,y) ⊠ e).

Let g ∈ X# andφ ∈ E∗. For any∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, we have∣∣∣∣∣∣∣(g⊠ φ)

n∑

i=1

δ(xi ,yi) ⊠ ei

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣

n∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉

∣∣∣∣∣∣∣≤ Lip(g) ‖φ‖

n∑

i=1

d(xi, yi) ‖ei‖ .

Since the value of (g⊠φ)(∑n

i=1 δ(xi ,yi) ⊠ ei

)does not depend on the representation of

∑ni=1 δ(xi ,yi) ⊠ei by Lemma 2.1,

it follows that ∣∣∣∣∣∣∣(g⊠ φ)

n∑

i=1

δ(xi ,yi) ⊠ ei

∣∣∣∣∣∣∣≤ Lip(g) ‖φ‖ π

n∑

i=1

δ(xi ,yi) ⊠ ei

.

Therefore the Lipschitz cross-normπ is dualizable by Proposition 4.1 and Remark 4.1.Similarly, by applying Proposition 4.2 and Remark 4.1, we see that the Lipschitz cross-normπ is uniform. Let∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E. For everyh ∈ Lip0(X,X) andT ∈ L(E,E), we have

π

(h⊠ T)

n∑

i=1

δ(xi ,yi) ⊠ ei

= π

n∑

i=1

δ(h(xi ),h(yi )) ⊠ T(ei)

≤

n∑

i=1

d(h(xi), h(yi)) ‖T(ei)‖

≤

n∑

i=1

Lip(h)d(xi, yi) ‖T‖ ‖ei‖

= Lip(h) ‖T‖n∑

i=1


The value of (h⊠ T)(∑n

i=1 δ(xi ,yi) ⊠ ei

)is independent of the representation of

∑ni=1 δ(xi ,yi) ⊠ ei by Lemma 3.1, and

therefore we conclude that

π

(h⊠ T)

n∑

i=1

δ(xi ,yi) ⊠ ei

≤ Lip(h) ‖T‖ π

n∑

i=1

δ(xi ,yi) ⊠ ei

.

�


The Lipschitz projective norm onX ⊠ E and the dual norm of the norm Lip of Lip0(X,E∗) induced onX ⊠ Ecoincide as we see next.

Corollary 8.2. Let X be a pointed metric space and E a Banach space. Thenπ = L on X⊠ E.

Proof. By Theorem 8.1,L ≤ π. To prove thatL ≥ π, suppose by contradiction thatL(u0) < 1 < π(u0) for someu0 ∈ X⊠E. DenoteB = {u ∈ X⊠E : π(u) ≤ 1}. Clearly,B is a closed and convex set inX⊠πE. Applying the Hahn–Banach separation theorem toB and{u0} and taking into account Theorem 6.2, there exists somef ∈ Lip0(X,E∗)with Lip( f ) = 1 such that Re(Λ( f ))(u0) > 1 ≥ sup{Re(Λ( f ))(u) : u ∈ B}. SinceL(u0) ≥ |Λ( f )(u0)| ≥ Re(Λ( f ))(u0),thenL(u0) > 1 and this is a contradiction. �

Theorem 8.3. π is the greatest Lipschitz cross-norm on X⊠ E.

Proof. We have seen in Theorem 8.1 thatπ is a Lipschitz cross-norm onX⊠E. Now, letα be a Lipschitz cross-normon X ⊠ E and letu ∈ X ⊠ E. If

∑ni=1 δ(xi ,yi) ⊠ ei is a representation ofu, we have

α(u) = α

n∑

i=1

δ(xi ,yi) ⊠ ei

≤n∑

i=1

α(δ(xi ,yi) ⊠ ei

)=

n∑

i=1


Now the very definition ofπ givesα(u) ≤ π(u). �

Dualizable Lipschitz cross-norms onX ⊠ E are characterized by being between the Lipschitz injectiveandLipschitz projective norms.

Proposition 8.4. A normα on X⊠ E is a dualizable Lipschitz cross-norm if and only ifε ≤ α ≤ π.

Proof. If α is a dualizable Lipschitz cross-norm onX ⊠ E, thenε ≤ α ≤ π by Theorems 7.2 and 8.3. Conversely,if α is a norm onX ⊠ E that lies betweenε andπ, thenα(δ(x,y) ⊠ e) = d(x, y) ‖e‖ follows immediately from the factthatε andπ are Lipschitz cross-norms. Letg ∈ X# andφ ∈ E∗. Then∣∣∣∣∣∣∣

n∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉

∣∣∣∣∣∣∣≤ Lip(g) ‖φ‖ ε

n∑

i=1

δ(xi ,yi) ⊠ ei

≤ Lip(g) ‖φ‖α

n∑

i=1

δ(xi ,yi) ⊠ ei

for all∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, and so the Lipschitz cross-normα is dualizable. �

Let us recall that a completion of a normed spaceE is a Banach spaceE that includes a dense linear subspaceisometric toE. Every normed space has a completion and the completion is unique up to isometric isomorphism.By [9, Lemma 3.100], every elemente of the completionE of E can be written ase=

∑∞n=1 en, whereen ∈ E and∑∞

n=1 ‖en‖ < ∞. Moreover,‖e‖ = inf{∑∞

n=1 ‖en‖}, where the infimum is taken over all series inE summing up toe.

Combining this with Definition 8.1, we obtain the following result.

Theorem 8.5. Every element u∈ X⊠πE admits a representation

u =∞∑

i=1

δ(xi ,yi) ⊠ ei

such that∞∑

i=1

d(xi, yi) ‖ei‖ < ∞.

Moreover,

π(u) = inf

∞∑

i=1

d(xi , yi) ‖ei‖ : u =∞∑

i=1

δ(xi ,yi) ⊠ ei ,

∞∑

i=1

d(xi , yi) ‖ei‖ < ∞

.

The dual pairing satisfies the formula⟨ ∞∑

i=1

δ(xi ,yi) ⊠ ei , f

⟩=

∞∑

i=1

〈 f (xi) − f (yi), ei〉


for all f ∈ Lip0(X,E∗).

Next we consider the boundedness of the linear operatorh ⊠ T : X ⊠ E → Y ⊠ F for the Lipschitz projectivenorms.

Proposition 8.6. Let X,Y be pointed metric spaces and E, F Banach spaces. Let h∈ Lip0(X,Y) and T ∈ L(E, F).Then there exists a unique bounded linear operator h⊠π T : X⊠πE→ Y⊠πF such that(h⊠π T)(u) = (h⊠ T)(u) forall u ∈ X ⊠ E. Furthermore,‖h⊠π T‖ = Lip(h) ‖T‖.



π((h⊠ T)(u)) = π

n∑

i=1


≤

n∑

i=1

d(h(xi), h(yi)) ‖T(ei)‖

≤ Lip(h) ‖T‖n∑

i=1


An appeal to Definition 8.1 yieldsπ((h⊠ T)(u)) ≤ Lip(h) ‖T‖ π(u). Thereforeh ⊠ T is bounded fromX ⊠π E toY⊠π F and‖h⊠ T‖ ≤ Lip(h) ‖T‖. Moreover, the converse estimate follows easily from

d(h(x), h(y)) ‖T(e)‖ = π(δ(h(x),h(y)) ⊠ T(e)) = π((h⊠ T)(δ(x,y) ⊠ e)) ≤ ‖h⊠ T‖d(x, y) ‖e‖

for all x, y ∈ X ande ∈ E. Thus, we have‖h⊠ T‖ = Lip(h) ‖T‖. Finally, it is well known thath⊠ T has a uniquebounded linear extension to an operatorh⊠π T : X⊠πE→ Y⊠πF with ‖h⊠π T‖ = Lip(h) ‖T‖. �

It turns out that there is a very close relationship between the Lipschitz projective norm and the projective tensorproduct of Banach spaces. In fact, just as it was the case for the injective norm in Proposition 7.6, the Lipschitzprojective norm onX ⊠ E can be identified with the projective norm on the tensor product of F (X) andE. Theauthors wish to thank Richard Haydon for suggesting that this might be true.

Proposition 8.7. The map I: X ⊠π E→ F (X) ⊗π E, defined by

I (u) =n∑

i=1


for u =∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, is a linear isometry. Moreover, X⊠πE is isometrically isomorphic toF (X)⊗πE.

Proof. As in the proof of Proposition 7.6, Proposition 1.6 guarantees that the mapI is well defined (and it is clearlylinear). Lettingu =

∑ni=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, note that using the fact that the mapx 7→ δx is an isometry fromX

intoF (X), ∥∥∥∥∥∥∥

n∑

i=1


∥∥∥∥∥∥∥F (X)⊗πE

≤

n∑

i=1

∥∥∥δxi − δyi

∥∥∥F (X)‖ei‖ =

n∑

i=1


Taking the infimum over all representations ofu, we conclude that‖I (u)‖ ≤ ‖u‖. Let η > 0 be given. FromCorollary 8.2 there existsf ∈ Lip0(X,E∗) with Lip( f ) ≤ 1 such that〈u, f 〉 > ‖u‖ − η, where the pairing is theone given in Theorem 8.5. By Theorem 7.5, the linear extension f : F (X) → E∗ of f has norm at most one.From the properties of the projective tensor product of Banach spaces, the dual ofF (X)⊗πE can be identified withL(F (X),E∗), where the pairing is given by

⟨ n∑

i=1

γi ⊗ ei ,T⟩=

n∑

i=1

〈Tγi , ei〉 (γi ∈ F (X), ei ∈ E, T ∈ L(F (X),E∗)) .


In particular,

‖I (u)‖ ≥ 〈I (u), f 〉 =n∑

i=1

〈 f (δxi − δyi ), ei〉 =

n∑

i=1

〈 f (xi) − f (yi), ei〉 = 〈u, f 〉 > ‖u‖ − η.

Letting η go to 0, we obtain that‖I (u)‖ ≥ ‖u‖, and thusI is a linear isometry. Since the sums of the form∑ni=1 δ(xi ,yi) ⊠ ei and

∑ni=1(δxi − δyi ) ⊗ ei are dense respectively inX⊠πE andF (X)⊗πE, I extends to an isometric

isomorphism betweenX⊠πE andF (X)⊗πE. �

Proposition 8.7 implies in particular that given a pointed metric spaceX, there is a Banach spaceA such thatX⊠πE is isometric toA⊗πE for every Banach spaceE. The authors would like to thank Jesus Castillo for pointingout a result in categorical Banach space theory that shows this was to be expected. Without going into all thedetails, let us outline the argument. First, a theorem of Fuks [11, Section 6] (a nice presentation can be found in[3, Proposition 5.6], where the reader can also find the definitions of the categorical terms we use below) statesthe following: if F, G are two covariant Banach functors such that for any Banach spacesE andF, we have thatL(F(E), F) is linearly isometric toL(E,G(F)), then there exists a Banach spaceA such that for every Banach spaceE, F(E) is linearly isometric toA⊗πE andG(F) is linearly isometric toL(A, F). Now consider a fixed pointedmetric spaceX. Note that it induces two covariant Banach functorsX⊠π(·) and Lip0(X, ·). Arguments closelyrelated to those that led us to prove Corollary 8.2 show that,for any Banach spacesE andF, we haveL(X⊠πE, F)is linearly isometric toL(E, Lip0(X, F)), so Fuks’ result applies.

The spaceX⊠πE is called the projective Lipschitz tensor product ofX and E. This term derives from thefollowing result. Before stating it, recall that a Lipschitz map f : X→ Z is calledC-co-Lipschitz if for everyx ∈ Xandr > 0, f (B(x, r)) ⊃ B( f (x), r/C). Moreover, it is called a Lipschitz quotient if it is surjective, Lipschitz andco-Lipschitz [2].

Theorem 8.8. Let X,Y be pointed metric spaces and q: X→ Z a Lipschitz quotient that is 1-Lipschitz and C-co-Lipschitz for every C> 1. Let E be a Banach space, E0 a closed linear subspace of E and Q: E → E/E0 thenatural quotient map. Then q⊠ Q: X⊠πE→ Z⊠π(E/E0) is a quotient operator.

Proof. Thanks to Proposition 8.7 and the behavior of the projectivetensor norm with respect to quotients [20,Proposition 2.5], it suffices to prove that ifq: X→ Z is such a Lipschitz quotient then the induced map ˜q: F (X)→F (Z) is a linear quotient operator. Notice that from Theorem 7.5, ‖q‖ = Lip(q) = 1. Now letu ∈ F (Z), and letε > 0. From Proposition 8.7,F (X) ≡ F (X)⊗πK ≡ X⊠πK. Thus from Theorem 8.5, there exists a representationu =

∑∞j=1 δ(zj ,z′j) ⊠ a j such that (1+ ε)π(u) ≥

∑∞j=1 |a j |d(zj, z′j). For eachj, choosex j , x′j ∈ X such thatq(x j) = zj ,

q(x′j) = z′j andd(x j, x′j) ≤ (1+ ε)d(zj, z′j). Settingu′ =∑∞

j=1 δ(x j ,x′j) ⊠ a j, clearly q(u′) = u (so it follows thatq issurjective) and

π(u′) ≤∞∑

j=1

|a j |d(x j, x′j) ≤ (1+ ε)

∞∑

j=1

|a j |d(zj, z′j) ≤ (1+ ε)2π(u).

Since this holds for allε > 0, it follows thatπ(u) = inf {π(u′) : q(u′) = u}. �

In a similar manner, the projective norm respects complemented subspaces. We say that a subsetZ ⊂ X thatcontains the point 0 is a Lipschitz retract ofX, or that it is Lipschitz complemented inX, if there exists a Lipschitzmap (called a Lipschitz retraction)r : X→ Z such thatr(z) = z for all z ∈ Z.

Proposition 8.9. Let Z be a Lipschitz retract of X, and let F be a complemented subspace of E. Then Z⊠πF iscomplemented in X⊠πE and the norm on Z⊠πF induced by the Lipschitz projective norm of X⊠πE is equivalentto the Lipschitz projective norm on Z⊠πF. If Z is Lipschitz complemented with a Lipschitz retraction of Lipschitzconstant one and F is complemented by a linear projection of norm one, then Z⊠πF is a subspace of X⊠πE and isalso complemented by a projection of norm one.


Proof. This follows from the corresponding result for the projective tensor product (see [20, Proposition 2.4]),after noting that a Lipschitz retractionr : X → Z (that in particular sends 0 to 0) extends to a linear projectionr : F (X)→ F (Z) ⊂ F (X) with ‖r‖ = Lip(r). �

Calculating the projective norm of an element in a tensor product of Banach spaces is generally difficult, butthere is a particular case where the calculation is relatively easy: for any Banach spaceE, ℓ1⊗πE is isometricallyisomorphic toℓ1(E) (see [20, Example 2.6]). In the nonlinear setting, trees play a role analogous to that ofℓ1 inthe linear theory, so the following result is not surprising.

Proposition 8.10. Let T = (X,E) be a graph with finite vertex set X and edge setE which is a tree, that is,it contains no cycles. Consider T as a pointed metric space, with distance function given by the shortest-pathdistance and a distinguished fixed point0 ∈ X. Let G be a Banach space. Then T⊠πG is isometrically isomorphicto ℓ1(E; G).

Proof. We say that a vertexx ∈ X is positive (negative) if it is at an even (respectively, odd) distance from 0∈ X.Note that, sinceT is a tree, the endpoints of every edge inE have different parities. Therefore every edge{x, y} inE will be written as (x, y) with x negative andy positive.

Considerx, y ∈ X. Let n = d(x, y) and{x = z0, z1, . . . , zn = y} be the unique minimal-length path inT joining xandy. Since, for eachv ∈ G,

‖v‖d(x, y) =n∑

i=1

‖v‖d(zi , zi−1),

in order to calculateπ(u) for u ∈ T⊠G, it suffices to consider only representations involvingδ(xi ,yi) with (xi , yi) ∈ E.By the triangle inequality, in the representation we can consolidate all terms corresponding to the same edge(xi , yi) ∈ E, so we can consider only representations of the form

u =∑

(x,y)∈E

δ(x,y) ⊠ v(x,y).

But, for eachu ∈ T ⊠G, there is only one such representation, something easily seen by induction on the size ofthe tree. If we defineJ : T ⊠π G→ ℓ1(E; G) by u 7→ (v(x,y))(x,y)∈E, J is then clearly a linear isometry that extends toan isometric isomorphism betweenT⊠πG andℓ1(E; G). �

More generally, in the linear case we have thatL1(µ)⊗πE is isometrically isomorphic toL1(µ; E) for any measureµ (see [20, Example 2.19]). In our nonlinear setting, a possible analogue will be given by a generalization ofProposition 8.10 to a more general class of metric trees. This will depend heavily on the identification of theLipschitz-free space over such trees carried out in [12]. Before stating the result, let us recall a definition. AnR-tree is a metric spaceX satisfying the following two conditions: (1) For any pointsa andb in X, there exists aunique isometryφ of the closed interval [0, d(a, b)] into X such thatφ(0) = a andφ(d(a, b)) = b; (2) Any one-to-one continuous mappingϕ : [0, 1]→ X has the same range as the isometryφ associated to the pointsa = ϕ(0) andb = ϕ(1).

Corollary 8.11. Let X be anR-tree and E a Banach space. Then there exists a measureµ such that X⊠πE isisometric to L1(µ; E).

Proof. By [12, Corollary 3.3], there exists a measureµ such thatF (X) is isometrically isomorphic toL1(µ). FromProposition 8.7,X⊠πE is isometrically isomorphic toF (X)⊗πE. Finally, from [20, Example 2.19],L1(µ)⊗πE isisometric toL1(µ; E). �

9. The Lipschitz p-nuclear norms

We introduce now the Lipschitzp-nuclear normsdp onX⊠E for 1 ≤ p ≤ ∞. They are Lipschitz versions of theknown tensor norms of Chevet [8] and Saphar [21]. Similar versions were introduced in [4] on spaces ofE-valuedmolecules onX, where they were shown to be in duality with spaces of Lipschitz p′-summing maps.


Definition 9.1. Let 1 ≤ p ≤ ∞. Let E be a Banach space and let e1, . . . , en ∈ E. Define:

‖(e1, . . . , en)‖p =

n∑

i=1

‖ei‖p

1p

if 1 ≤ p < ∞,

max1≤i≤n ‖ei‖ if p = ∞.

Let X be a pointed metric space and let x1, . . . , xn, y1, . . . , yn ∈ X. Define:

∥∥∥(δ(x1,y1), . . . , δ(xn,yn))∥∥∥Lw

p=

supg∈BX#

n∑

i=1

|g(xi) − g(yi)|p

1p

if 1 ≤ p < ∞,

supg∈BX#

(max1≤i≤n|g(xi) − g(yi)|

)if p = ∞.

Let p′ be the conjugate index of p defined by p′ = p/(p− 1) if p , 1, p′ = ∞ if p = 1, and p′ = 1 if p = ∞.For each u∈ X ⊠ E, define:

dp(u) = inf

∥∥∥(δ(x1,y1), . . . , δ(xn,yn))

∥∥∥Lw

p′‖(e1, . . . , en)‖p : u =

n∑

i=1

δ(xi ,yi) ⊠ ei

,

the infimum being taken over all representations of u.

Theorem 9.1. For 1 ≤ p ≤ ∞, dp is a uniform and dualizable Lipschitz cross-norm on X⊠ E.


i=1 δ(xi ,yi) ⊠ ei be a representation ofu. Clearly,dp(u) ≥ 0. Let λ ∈ K. Since∑ni=1 δ(xi ,yi) ⊠ (λei) is a representation ofλu, Definition 9.1 yields

dp(λu) ≤∥∥∥(δ(x1,y1), . . . , δ(xn,yn))

∥∥∥Lw

p′‖(λe1, . . . , λen)‖p = |λ|

∥∥∥(δ(x1,y1), . . . , δ(xn,yn))∥∥∥Lw

p′‖(e1, . . . , en)‖p .

If λ = 0, it follows thatdp(λu) = 0 = |λ|dp(u). For λ , 0, since the preceding inequality holds for everyrepresentation ofu, we deduce thatdp(λu) ≤ |λ|dp(u). For the converse estimate, note thatdp(u) = dp(λ−1(λu)) ≤∣∣∣λ−1

∣∣∣ dp(λu) by using the proved inequality, thus|λ|dp(u) ≤ dp(λu) and hencedp(λu) = |λ|dp(u).We prove the triangular inequality fordp as follows. Letu, v ∈ X ⊠ E and letε > 0. If u = 0 or v = 0, there is

nothing to prove. Assumeu , 0 , v. We can choose representations foru andv, say

u =n∑

i=1

δ(xi ,yi) ⊠ ei , v =m∑

i=1

δ(x′i ,y′i ) ⊠ e′i ,

such that ∥∥∥(δ(x1,y1), . . . , δ(xn,yn))∥∥∥Lw

p′‖(e1, . . . , en)‖p ≤ dp(u) + ε

and ∥∥∥(δ(x′1,y′1), . . . , δ(x′m,y′m))∥∥∥Lw

p′

∥∥∥(e′1, . . . , e′m)∥∥∥

p≤ dp(v) + ε.

Fix r, s ∈ R+ arbitrarily and define

δ(x′′i ,y′′i ) =

{r−1δ(xi ,yi) if i = 1, . . . , n,s−1δ(x′i−n,y

′i−n) if i = n+ 1, . . . , n+m,

e′′i =

{rei if i = 1, . . . , n,se′i−n if i = n+ 1, . . . , n+m.

It is plain thatu+ v =∑n+m

i=1 δ(x′′i ,y′′i ) ⊠ e′′i and therefore we have

dp(u+ v) ≤∥∥∥(δ(x′′1 ,y′′1 ), . . . , δ(x′′n+m,y

′′n+m))

∥∥∥Lw

p′

∥∥∥(e′′1 , . . . , e′′n+m)

∥∥∥p.

We distinguish three cases.


Case 1.1 < p < ∞. An easy verification yields(∥∥∥(δ(x′′1 ,y′′1 ), . . . , δ(x′′n+m,y

′′n+m))

∥∥∥Lw

p′

)p′

≤

(r−1

∥∥∥(δ(x1,y1), . . . , δ(xn,yn))∥∥∥Lw

p′

)p′

+

(s−1

∥∥∥(δ(x′1,y′1), . . . , δ(x′m,y′m))∥∥∥Lw

p′

)p′

and (∥∥∥(e′′1 , . . . , e′′n+m)

∥∥∥p

)p=

(r ‖(e1, . . . , en)‖p

)p+

(s∥∥∥(e′1, . . . , e

′m)

∥∥∥p

)p.

Using Young’s inequality, it follows that

dp(u+ v) ≤∥∥∥(δ(x′′1 ,y′′1 ), . . . , δ(x′′n+m,y

′′n+m))

∥∥∥Lw

p′

∥∥∥(e′′1 , . . . , e′′n+m)∥∥∥

p

≤1p′

(∥∥∥(δ(x′′1 ,y′′1 ), . . . , δ(x′′n+m,y′′n+m))

∥∥∥Lw

p′

)p′

+1p

(∥∥∥(e′′1 , . . . , e′′n+m)∥∥∥

p

)p

≤r−p′

p′

(∥∥∥(δ(x1,y1), . . . , δ(xn,yn))∥∥∥Lw

p′

)p′

+r p

p

(‖(e1, . . . , en)‖p

)p

+s−p′

p′

(∥∥∥(δ(x′1,y′1), . . . , δ(x′m,y′m))∥∥∥Lw

p′

)p′

+sp

p

(∥∥∥(e′1, . . . , e′m)∥∥∥

p

)p.

Since r, s were arbitrary inR+, taking

r = (dp(u) + ε)−1/p′∥∥∥(δ(x1,y1), . . . , δ(xn,yn))

∥∥∥Lw

p′,

s= (dp(v) + ε)−1/p′∥∥∥(δ(x′1,y′1), . . . , δ(x′m,y′m))

∥∥∥Lw

p′,

we obtain that dp(u+ v) ≤ dp(u) + dp(v) + 2ε.

Case 2. p = 1. Now we have

d1(u+ v) ≤∥∥∥(δ(x′′1 ,y′′1 ), . . . , δ(x′′n+m,y

′′n+m))

∥∥∥Lw

∞

∥∥∥(e′′1 , . . . , e′′n+m)

∥∥∥1

=

(max

{r−1

∥∥∥(δ(x1,y1), . . . , δ(xn,yn))∥∥∥Lw

∞, s−1

∥∥∥(δ(x′1,y′1), . . . , δ(x′m,y′m))∥∥∥Lw

∞

})

·(r ‖(e1, . . . , en)‖1 + s

∥∥∥(e′1, . . . , e′m)

∥∥∥1

),

and taking, in particular, r=∥∥∥(δ(x1,y1), . . . , δ(xn,yn))

∥∥∥Lw

∞and s=

∥∥∥(δ(x′1,y′1), . . . , δ(x′m,y′m))∥∥∥Lw

∞, we infer that d1(u+ v) ≤

d1(u) + d1(v) + 2ε.

Case 3. p = ∞. We have

d∞(u+ v) ≤∥∥∥(δ(x′′1 ,y′′1 ), . . . , δ(x′′n+m,y

′′n+m))

∥∥∥Lw

1

∥∥∥(e′′1 , . . . , e′′n+m)∥∥∥∞

≤

(r−1

∥∥∥(δ(x1,y1), . . . , δ(xn,yn))∥∥∥Lw

1+ s−1

∥∥∥(δ(x′1,y′1), . . . , δ(x′n,y′n))∥∥∥Lw

1

)

·(max

{r ‖(e1, . . . , en)‖∞ , s

∥∥∥(e′1, . . . , e′m)∥∥∥∞

}),

and for r= ‖(e1, . . . , en)‖−1∞ and s=

∥∥∥(e′1, . . . , e′m)∥∥∥−1

∞, we deduce that d∞(u+ v) ≤ d∞(u) + d∞(v) + 2ε.

In any case,dp(u+ v) ≤ dp(u) + dp(v) + 2ε and sodp(u+ v) ≤ dp(u) + dp(v) by the arbitrariness ofε. Hencedp

is a seminorm for 1≤ p ≤ ∞. Now, we claim thatε(u) ≤ dp(u) ≤ π(u). Indeed, we have∣∣∣∣∣∣∣

n∑

i=1

(g(xi) − g(yi)) 〈φ, ei〉

∣∣∣∣∣∣∣≤

n∑

i=1

|g(xi) − g(yi)| ‖ei‖ ≤∥∥∥(δ(x1,y1), . . . , δ(xn,yn))

∥∥∥Lw

p′‖(e1, . . . , en)‖p

for everyg ∈ BX# andφ ∈ BE∗ , where we have used Holder’s inequality in the case 1< p < ∞. Therefore

ε(u) ≤∥∥∥(δ(x1,y1), . . . , δ(xn,yn))

∥∥∥Lw

p′‖(e1, . . . , en)‖p, and since it holds for each representation ofu, we deduce that

ε(u) ≤ dp(u). Sinceε is a Lipschitz cross-norm, this implies thatdp is a norm and that

‖e‖d(x, y) = ε(δ(x,y) ⊠ e) ≤ dp(δ(x,y) ⊠ e)


for all x, y ∈ X ande ∈ E. Moreover, Definition 9.1 gives

dp(δ(x,y) ⊠ e) ≤ ‖e‖ supg∈BX#

|g(x) − g(y)| = ‖e‖d(x, y).

Hencedp is a Lipschitz cross-norm. Thendp ≤ π by Theorem 8.3 as we wanted. Now our claim implies thatdp isdualizable by Proposition 8.4.

Finally, to prove thatdp is uniform, takeh ∈ Lip0(X,X) andT ∈ L(E,E). Letu ∈ X⊠E and pick a representation∑ni=1 δ(xi ,yi) ⊠ ei for u. We have

∥∥∥(δ(h(x1),h(y1)), . . . , δ(h(xn),h(yn)))∥∥∥Lw

p′‖(T(e1), . . . ,T(en))‖p ≤ Lip(h)

∥∥∥(δ(x1,y1), . . . , δ(xn,yn))∥∥∥Lw

p′‖T‖ ‖(e1, . . . , en)‖p .

Taking infimum over all the representations ofu, it follows thatdp((h⊠ T)(u)) ≤ Lip(h) ‖T‖dp(u). �

Next we show that the Lipschitz 1-nuclear normd1 is justly the Lipschitz projective normπ.

Proposition 9.2. For every u∈ X ⊠ E,

d1(u) = inf

n∑

i=1

d(xi , yi) ‖ei‖ : u =n∑

i=1

δ(xi ,yi) ⊠ ei

taking the infimum over all representations of u.



π(u) ≤n∑

i=1

d(xi , yi) ‖ei‖

=

n∑

i=1

supg∈BX#

|g(xi) − g(yi)|

‖ei‖

≤

n∑

i=1

max1≤i≤n

supg∈BX#

|g(xi) − g(yi)|

‖ei‖

=∥∥∥(δ(x1,y1), . . . , δ(xn,yn))

∥∥∥Lw

∞‖(e1, . . . , en)‖1

and thereforeπ(u) ≤ d1(u). The converse inequality follows from Theorems 9.1 and 8.3. �

10. Lipschitz approximable operators

The notions of Lipschitz compact operators and Lipschitz approximable operators fromX to E were introducedin [16]. Let us recall that a Lipschitz operatorf ∈ Lip0(X,E) is said to be Lipschitz compact if its Lipschitz image{( f (x) − f (y))/d(x, y) : x, y ∈ X, x , y} is relatively compact inE and f is said to be Lipschitz approximable if it isthe limit in the Lipschitz norm Lip of a sequence of Lipschitzfinite-rank operators fromX to E.

We show that the spaces of Lipschitz finite-rank operators and Lipschitz approximable operators can be identi-fied as spaces of continuous linear functionals.

Theorem 10.1. The map K: X#iπ′ E∗ → Lip0F (X,E∗), defined by

K

m∑

j=1

g j ⊠ φ j

=m∑

j=1

g j · φ j ,

is an isometric isomorphism. As a consequence, the space of all Lipschitz approximable operators from X to E∗ isisometrically isomorphic to X#iπ′E∗.


Proof. By Theorem 2.5,K is a linear bijection. For any∑m

j=1 g j ⊠ φ j ∈ X#i E∗, we have

π′

m∑

j=1

g j ⊠ φ j

= sup

∣∣∣∣∣∣∣∣

m∑

j=1

g j ⊠ φ j

n∑

i=1

δ(xi ,yi) ⊠ ei

∣∣∣∣∣∣∣∣: π

n∑

i=1

δ(xi ,yi) ⊠ ei

≤ 1

= sup

∣∣∣∣∣∣∣∣

n∑

i=1

⟨m∑

j=1

g j · φ j

(xi) −

m∑

j=1

g j · φ j

(yi), ei

⟩∣∣∣∣∣∣∣∣: π

n∑

i=1

δ(xi ,yi) ⊠ ei

≤ 1

= sup

∣∣∣∣∣∣∣∣Λ

m∑

j=1

g j · φ j

n∑

i=1

δ(xi ,yi) ⊠ ei

∣∣∣∣∣∣∣∣: L

n∑

i=1

δ(xi ,yi) ⊠ ei

≤ 1

=

∥∥∥∥∥∥∥∥Λ

m∑

j=1

g j · φ j

∥∥∥∥∥∥∥∥

= Lip

m∑

j=1

g j · φ j

.

by using Proposition 5.1, Lemmas 2.2 and 1.4, Corollary 8.2 and Theorem 6.2. HenceK is an isometry. Theconsequence follows from a known result of Functional Analysis. �

From Theorems 6.2 and 10.1 and Corollary 8.2, we infer the next consequence.

Corollary 10.2. The space X#iπ′E∗ is isometrically isomorphic to(X⊠πE)∗ if and only if each Lipschitz operatorfrom X to E∗ is Lipschitz approximable.

We recall that a Banach spaceE is said to have the approximation property if given a compactsetK ⊂ E andε > 0, there is a finite-rank bounded linear operatorT : E → E such that‖T x− x‖ < ε for every x ∈ K. Theapproximation property was thoroughly studied by Grothendieck in [15]. In [16, Corollary 2.5], it was shown thatX# has the approximation property if and only if the space of allLipschitz approximable operators fromX to Eis the space of all Lipschitz compact operators fromX to E. Using this fact and Theorem 10.1, we derive thefollowing result.

Corollary 10.3. Let X be a pointed metric space such that X# has the approximation property. Then, for anyBanach space E, the space of all Lipschitz compact operatorsfrom X to E∗ is isometrically isomorphic to X#iπ′E∗.

By Theorem 6.2 and Corollary 10.3, we have the following.

Corollary 10.4. Let X be a pointed metric space such that X# has the approximation property and let E be aBanach space. Then X#

iπ′E∗ is isometrically isomorphic to(X⊠πE)∗ if and only if each Lipschitz operator from Xto E∗ is Lipschitz compact.

We close this section with a new formula for the normπ′. By [16, Lemma 1.1], the closed unit ball of theLipschitz-free Banach spaceF (X) over a pointed metric spaceX coincides with the closure of the convex balancedhull of the set{(δx − δy)/d(x, y) : x, y ∈ X, x , y} in (X#)∗, whereδx is the evaluation functional atx defined onX#.It is well known that every element in the convex balanced hull of that set is of the form

n∑

i=1

λiδxi − δyi

d(xi , yi)

for somen ∈ N, λ1, . . . , λn ∈ K,∑n

i=1 |λi | ≤ 1 and (x1, y1), . . . , (xn, yn) ∈ X2 with xi , yi for i ∈ {1, . . . , n}.


Definition 10.1. For each∑m

j=1 g j ⊠ φ j ∈ X#i E∗, define:

ε

m∑

j=1

g j ⊠ φ j

= sup

∣∣∣∣∣∣∣∣

m∑

j=1

γ(g j)⟨φ j , e

⟩∣∣∣∣∣∣∣∣: γ ∈ BF (X), e ∈ BE

.

Note that this supremum exists since, for allγ ∈ BF (X) ande ∈ BE,∣∣∣∣∣∣∣∣

m∑

j=1

γ(g j)⟨φ j, e

⟩∣∣∣∣∣∣∣∣≤

m∑

j=1

∣∣∣∣γ(g j)⟨φ j , e

⟩∣∣∣∣ ≤m∑

j=1

‖γ‖Lip(g j)∥∥∥φ j

∥∥∥ ‖e‖ ≤m∑

j=1

Lip(g j)∥∥∥φ j

∥∥∥ .

Theorem 10.5. The associated Lipschitz normπ′ of π on X⊠ E isε on X#i E∗.

Proof. Let∑m

j=1 g j ⊠ φ j ∈ X#i E∗. We have

π′

m∑

j=1

g j ⊠ φ j

= sup

∣∣∣∣∣∣∣∣

m∑

j=1

g j ⊠ φ j

n∑

i=1

δ(xi ,yi) ⊠ ei

∣∣∣∣∣∣∣∣: π

n∑

i=1

δ(xi ,yi ) ⊠ ei

≤ 1

= sup

∣∣∣∣∣∣∣∣

m∑

j=1

g j ⊠ φ j

(δ(x,y) ⊠ e

)∣∣∣∣∣∣∣∣: π

(δ(x,y) ⊠ e

)≤ 1

= ε

m∑

j=1

g j ⊠ φ j

.

In order to justify these equalities, denote byα(∑m

j=1 g j ⊠φ j) andβ(∑m

j=1 g j ⊠φ j) the first and the second supremumwhich appear above. The first equality follows from Proposition 5.1. To see thatα(

∑mj=1 g j ⊠φ j) ≤ β(

∑mj=1 g j ⊠φ j),

we will use the easy fact that ifn ∈ N, a1, . . . , an ∈ R+0 andb1, . . . , bn ∈ R

+, then

a1 + · · · + an

b1 + · · · + bn≤ max

{a1

b1, · · · ,

an

bn

}.

Fix∑n

i=1 δ(xi ,yi) ⊠ ei ∈ X ⊠ E, nonzero. If∑p

i=1 δ(x′i ,y′i )⊠ e′i =

∑ni=1 δ(xi ,yi) ⊠ ei , we have

∣∣∣∣(∑m

j=1 g j ⊠ φ j

) (∑ni=1 δ(xi ,yi) ⊠ ei

)∣∣∣∣∑p

i=1 d(x′i , y′i )

∥∥∥e′i∥∥∥

=

∣∣∣∣(∑m

j=1 g j ⊠ φ j

) (∑pi=1 δ(x′i ,y

′i ) ⊠ e′i

)∣∣∣∣∑p

i=1 d(x′i , y′i )

∥∥∥e′i∥∥∥

≤

∑pi=1

∣∣∣∣(∑m

j=1 g j ⊠ φ j

)(δ(x′i ,y′i ) ⊠ e′i )

∣∣∣∣∑p

i=1 d(x′i , y′i )

∥∥∥e′i∥∥∥

≤ max

∣∣∣∣(∑m

j=1 g j ⊠ φ j

)(δ(x′i ,y′i ) ⊠ e′i )

∣∣∣∣d(x′i , y

′i )

∥∥∥e′i∥∥∥

: 1 ≤ i ≤ p

≤ β

m∑

j=1

g j ⊠ φ j

.

By the definition ofπ, it follows that∣∣∣∣∣∣∣∣

m∑

j=1

g j ⊠ φ j

n∑

i=1

δ(xi ,yi) ⊠ ei

∣∣∣∣∣∣∣∣≤ β

m∑

j=1

g j ⊠ φ j

π

n∑

i=1

δ(xi ,yi) ⊠ ei

.

This ensures thatα(∑m

j=1 g j ⊠ φ j) ≤ β(∑m

j=1 g j ⊠ φ j). The converse inequality is clearly certain.


Now, given anyδ(x,y) ⊠ e ∈ X ⊠ E with 0 < π(δ(x,y) ⊠ e) ≤ 1, we obtain∣∣∣∣∣∣∣∣

m∑

j=1

g j ⊠ φ j

(δ(x,y) ⊠ e

)∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣∣

m∑

j=1

(g j(x) − g j(y))⟨φ j , e

⟩∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣∣

m∑

j=1

(δx − δy

d(x, y)

)(g j)

⟨φ j , d(x, y)e

⟩∣∣∣∣∣∣∣∣≤ ε

m∑

j=1

g j ⊠ φ j

since (δx − δy)/d(x, y) ∈ SF (X) and d(x, y)e ∈ BE. Passing to the supremum we arrive atβ(∑m

j=1 g j ⊠ φ j) ≤ε(

∑mj=1 g j ⊠ φ j).

Finally, we show thatε(∑m

j=1 g j⊠φ j) ≤ π′(∑m

j=1 g j⊠φ j). For anyn ∈ N, λ1, . . . , λn ∈ K,∑n

i=1 |λi | ≤ 1, (xi , yi) ∈ X2

andxi , yi for eachi ∈ {1, . . . , n}, ande∈ BE, we have∣∣∣∣∣∣∣∣

m∑

j=1

n∑

i=1

λiδxi − δyi

d(xi , yi)

(g j)⟨φ j , e

⟩∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣∣

m∑

j=1

n∑

i=1

λig j(xi) − g j(yi)

d(xi , yi)

⟨φ j , e

⟩

∣∣∣∣∣∣∣∣

=

∣∣∣∣∣∣∣∣

m∑

j=1

n∑

i=1

λi

d(xi , yi)(g j ⊠ φ j)(δ(xi ,yi) ⊠ e)

∣∣∣∣∣∣∣∣

=

∣∣∣∣∣∣∣∣

m∑

j=1

g j ⊠ φ j

n∑

i=1

δ(xi ,yi) ⊠λie

d(xi , yi)

∣∣∣∣∣∣∣∣

≤ π′

m∑

j=1

g j ⊠ φ j

since

π

n∑

i=1

δ(xi ,yi) ⊠λie

d(xi , yi)

≤n∑

i=1

π

(δ(xi ,yi) ⊠

λied(xi , yi)

)=

n∑

i=1

d(xi , yi)|λi | ‖e‖d(xi , yi)

= ‖e‖n∑

i=1

|λi | ≤ 1.

By the density of the elements∑n

i=1 λi(δxi − δyi )/d(xi, yi) in BF (X), we infer that∣∣∣∣∣∣∣∣

m∑

j=1

γ(g j)⟨φ j, e

⟩∣∣∣∣∣∣∣∣≤ π′

m∑

j=1

g j ⊠ φ j

for all γ ∈ BF (X) ande ∈ BE. Taking supremum over all suchγ and e, we conclude thatε(∑m

j=1 g j ⊠ φ j) ≤π′(

∑mj=1 g j ⊠ φ j) and this completes the proof. �

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Departamento de Matematicas, Universidad de Almerıa, 04120 Almerıa, SpainE-mail address: m [email protected]

Department ofMathematics, University of Texas at Austin, Austin, TX 78712-1202.E-mail address: [email protected]

Departamento de Matematicas, Universidad de Almerıa, 04120 Almerıa, SpainE-mail address: [email protected]

Campus Universitario de Puerto Real, Facultad de Ciencias, Universidad de Cadiz, 11510 Puerto Real, Cadiz, SpainE-mail address: [email protected]

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