Extreme points in Lipschitz-free spaces over compact metric spaces
EEXTREME POINTS IN LIPSCHITZ-FREE SPACES OVERCOMPACT METRIC SPACES
RAM ´ON J. ALIAGA
Abstract.
We prove that all extreme points of the unit ball of a Lipschitz-free space over a compact metric space have finite support. Combined withprevious results, this completely characterizes extreme points and implies thatall of them are also extreme points in the bidual ball. For the proof, we developsome properties of an integral representation of functionals on Lipschitz spacesoriginally due to K. de Leeuw. Introduction
This short paper deals with the extremal structure of the unit ball of Lipschitz-free Banach spaces. Let us begin by establishing our work setting. Given a pointedmetric space (
M, d ), where we choose an arbitrary element 0 ∈ M as a base point,we consider the Lipschitz space Lip ( M ) of all real-valued Lipschitz functions f on M such that f (0) = 0. This space is a Banach space endowed with the norm (cid:107) f (cid:107) L given by the Lipschitz constant of f . More than that, Lip ( M ) is (isometrically) adual Banach space. Its canonical predual F ( M ) can be constructed by consideringthe evaluation operators δ ( x ) : f (cid:55)→ f ( x ) for x ∈ M , which are obviously elementsof Lip ( M ) ∗ , and then taking F ( M ) = span δ ( M ) . The space F ( M ) is usually called the Lipschitz-free space over M , and the weak ∗ topology induced by it agrees with the topology of pointwise convergence on boundedsubsets of Lip ( M ).Lipschitz and Lipschitz-free spaces are objects of intrinsic interest, as they canbe regarded as the metric versions of the usual C ( K ) spaces and their duals; seeWeaver’s monograph [12] for an exhaustive account of their basic properties (notethat Lipschitz-free spaces are referred to as Arens-Eells spaces in it). However, theircurrent main interest to Banach Space theory is the following linearization property:any Lipschitz mapping M → M between two metric spaces can be extended toa bounded linear operator between F ( M ) and F ( M ) (we identify each M i with δ ( M i ) ⊂ F ( M i )). See [10] for a survey of applications of this principle to nonlinearfunctional analysis.The roots of the study of the extremal structure of the unit ball B F ( M ) canbe traced back to the 1990’s. The first important general result was obtained byWeaver [11]: every preserved extreme point of B F ( M ) (i.e. such that it is also anextreme point of the bidual ball B F ( M ) ∗∗ ) is an elementary molecule , i.e. an elementof F ( M ) of the form m pq = δ ( p ) − δ ( q ) d ( p, q ) Mathematics Subject Classification.
Primary 46B20; Secondary 46E27.
Key words and phrases.
Lipschitz-free space, extreme point, de Leeuw map. a r X i v : . [ m a t h . F A ] F e b R. J. ALIAGA for some p (cid:54) = q ∈ M . As an easy consequence, all finitely supported extreme pointsmust be elementary molecules, too. Note that elementary molecules are the canoni-cal norming elements for Lip ( M ), which renders them natural suspects for extremepoints of the unit ball (compare to the extreme points of B C ( K ) ∗ being multiplesof the point masses). Moreover, their simplicity allows for elegant geometric char-acterizations of extremal elements in terms of an equivalent condition on the pair p, q that only involves the metric structure of M .After Weaver’s early result almost no progress was made in this topic for about20 years. The first new results to appear were due to Garc´ıa-Lirola, Petitjean,Proch´azka and Rueda Zoca, who proved that the sets of preserved extreme pointsand denting points of B F ( M ) coincide [8], and that an elementary molecule m pq isa strongly exposed point of B F ( M ) if and only if there is C > d ( p, x ) + d ( q, x ) − d ( p, q ) ≥ C · min { d ( p, x ) , d ( q, x ) } for all x ∈ M [9]. Shortly afterwards the author of this paper, in collaboration withE. Perneck´a [4] and A. J. Guirao [2], respectively, gave similar characterizations ofelementary molecules that are extreme or preserved extreme points. Specifically, m pq is an extreme, resp. preserved extreme point of B F ( M ) if and only if there isno point x (cid:54) = p, q in M , resp. βM (the Stone- ˇCech compactification of M ) suchthat d ( p, x ) + d ( q, x ) − d ( p, q ) > . Finally, Petitjean and Proch´azka proved that any extreme molecule is also an ex-posed point of B F ( M ) ; this result appeared in the joint paper [5].These results complete the characterization of extreme molecules in Lipschitz-free spaces, which covers all preserved extreme points and stronger notions. How-ever, the problem of determining all extreme points of B F ( M ) is not completelysolved yet because the following question still remains open: Question.
Must every extreme point of B F ( M ) be an elementary molecule? Only partial answers to this question have been obtained so far. For instance,this was recently shown to be true for all 0-hyperbolic metric spaces M in [6]. Ithas also been known since the 1990’s that the question has a positive answer forcompact spaces M such that the space lip ( M ) of locally flat Lipschitz functionson M is an isometric predual of F ( M ) (see [12, Corollary 4.41]). Very recently, ithas been announced that this condition is equivalent to F ( M ) simply being a dualspace [1]. This includes compact spaces with null 1-Hausdorff measure or with aH¨older metric.Here we provide a positive answer to the question for all compact spaces M , withno additional assumptions about duality. In fact, our answer covers the slightlymore general case where M is a proper space, i.e. such that every closed andbounded subset is compact. Theorem 1.
Let M be a proper metric space. Then the extreme points of B F ( M ) are exactly the elementary molecules m pq where p (cid:54) = q ∈ M are such that d ( p, x ) + d ( q, x ) > d ( p, q ) for all x ∈ M \ { p, q } . Moreover, all of them are also exposed points and preservedextreme points of B F ( M ) . For the rest of this paper, M will denote a complete metric space with metric d (which will not be proper unless specified explicitly) and we will assume withoutmention that an arbitrary base point 0 ∈ M has been chosen. The closed ball withcenter x and radius r will be denoted as B ( x, r ). By the linearization propertymentioned above, the Lipschitz-free space F ( N ) over a subset N ⊂ M can be XTREME POINTS IN LIPSCHITZ-FREE SPACES OVER COMPACT METRIC SPACES 3 identified with a closed subspace of F ( M ), namely F ( N ) = span δ ( N ). Let usrecall that, for m ∈ F ( M ), its support supp( m ) is defined as the smallest closedsubset of M such that m ∈ F (supp( m )); the existence of such a set is proved in[4, 5].For a Hausdorff space X , M ( X ) will stand for the space of Radon measures (i.e.finite regular signed Borel measures) on X . Recall that, given two Hausdorff spaces X, Y , a Borel mapping f : X → Y , and a Borel measure µ on X , the pushforwardmeasure f (cid:93) µ is defined by f (cid:93) µ ( E ) = µ ( f − ( E ))for every Borel E ⊂ Y . Pushforwards clearly preserve positivity and the totalvariation norm, and f (cid:93) µ ∈ M ( Y ) if µ ∈ M ( X ) and Y is normal. It also holds that (cid:82) Y g d ( f (cid:93) µ ) = (cid:82) X ( g ◦ f ) dµ for every Borel function g on Y .2. Integral representation
Our proof of Theorem 1 will be based on the integral representation of func-tionals in F ( M ) and its bidual. In [3], the author and E. Perneck´a studied thefunctionals that can be expressed as integration against a measure on M or somecompactification of M . In general this is not possible for all elements of F ( M ),only for those that can be written as the difference of two positive elements. Butthere is an alternative representation, originally due to de Leeuw [7], that coversall functionals in Lip ( M ) ∗ . The key idea is representing a functional not as an“integral of evaluation functionals”, but as an “integral of elementary molecules”instead.Let us introduce this representation. Following [12], denote (cid:102) M = { ( x, y ) ∈ M × M : x (cid:54) = y } . The de Leeuw map is the mapping Φ : Lip ( M ) → C ( (cid:102) M ) defined byΦ f ( x, y ) = f ( x ) − f ( y ) d ( x, y )for ( x, y ) ∈ (cid:102) M . Clearly (cid:107) Φ f (cid:107) ∞ = (cid:107) f (cid:107) L , so every Φ f is bounded and admits aunique continuous extension to β (cid:102) M , the Stone- ˇCech compactification of (cid:102) M , thatwe will still denote by Φ f for simplicity. Thus we may consider Φ as a linearisometric embedding of Lip ( M ) into C ( β (cid:102) M ). Its adjoint operator is therefore aquotient map from C ( β (cid:102) M ) ∗ = M ( β (cid:102) M ) onto Lip ( M ) ∗ = F ( M ) ∗∗ . That is, forevery φ ∈ F ( M ) ∗∗ there exists a representing measure µ ∈ M ( β (cid:102) M ) such thatΦ ∗ µ = φ , i.e. (cid:104) f, φ (cid:105) = (cid:90) β (cid:102) M (Φ f ) dµ for any f ∈ Lip ( M ). It should be obvious that the elementary molecule m xy isrepresented by the Dirac measure δ ( x,y ) , for any ( x, y ) ∈ (cid:102) M .The following mappings defined on β (cid:102) M will be used often: • Let p : (cid:102) M → M × M by the identity mapping and extend it to a continuousmapping p : β (cid:102) M → βM × βM . We will call this the projection mapping,and denote its first and second coordinates by p and p , respectively. Fora set E ⊂ β (cid:102) M we will denote p s ( E ) = p ( E ) ∪ p ( E ) ⊂ βM and call thisthe shadow of E . • Let r : (cid:102) M → (cid:102) M be the reflection mapping given by r ( x, y ) = ( y, x ), andextend it to a continuous mapping r : β (cid:102) M → β (cid:102) M . Notice that r ◦ r is theidentity on (cid:102) M and hence on all of β (cid:102) M , thus r is a bijection. R. J. ALIAGA
The representing measure µ ∈ M ( β (cid:102) M ) for φ ∈ F ( M ) ∗∗ is not unique. Since Φ ∗ is a quotient map, we may always choose it such that (cid:107) µ (cid:107) = (cid:107) φ (cid:107) . We will now seethat we may additionally choose µ to be positive. The underlying idea is that wemay replace a negative point mass − δ ( x,y ) by the positive point mass δ ( y,x ) as bothrepresent the same functional m yx . We may formalize this in terms of pushforwardmeasures as follows. Lemma 2.
For any µ ∈ M ( β (cid:102) M ) we have Φ ∗ ( µ + r (cid:93) µ ) = 0 .Proof. If µ = δ ( x,y ) for ( x, y ) ∈ (cid:102) M , the identity holds as Φ ∗ δ ( x,y ) = m xy andΦ ∗ ( r (cid:93) δ ( x,y ) ) = Φ ∗ δ ( y,x ) = m yx = − m xy . Now notice that M ( β (cid:102) M ) = span w ∗ ext B M ( β (cid:102) M ) = span w ∗ (cid:110) δ ζ : ζ ∈ β (cid:102) M (cid:111) = span w ∗ (cid:110) δ ζ : ζ ∈ (cid:102) M (cid:111) . Indeed, the first equality follows from the Krein-Milman theorem, the second onefrom the well-known fact that the extreme points of B M ( β (cid:102) M ) are the positive andnegative Dirac measures, and the last one from the defining properties of the Stone-ˇCech compactification β (cid:102) M . The identity now follows for all µ ∈ M ( β (cid:102) M ) by theweak ∗ continuity of Φ ∗ . (cid:3) Proposition 3.
For every φ ∈ F ( M ) ∗∗ there is a positive µ ∈ M ( β (cid:102) M ) such that Φ ∗ µ = φ and (cid:107) µ (cid:107) = (cid:107) φ (cid:107) .Proof. Since Φ ∗ is a quotient map, we may find µ that satisfies all requirementsexcept possibly being positive. In order to achieve that, let µ = µ + − µ − bethe Jordan decomposition of µ and let µ (cid:48) = µ + + r (cid:93) µ − . By Lemma 2 we have0 = Φ ∗ ( µ − + r (cid:93) µ − ), thereforeΦ ∗ ( µ (cid:48) ) = Φ ∗ µ + + Φ ∗ ( r (cid:93) µ − ) = Φ ∗ µ + − Φ ∗ µ − = Φ ∗ µ as we needed. It is clear that µ (cid:48) ≥ (cid:107) µ (cid:48) (cid:107) = (cid:13)(cid:13) µ + (cid:13)(cid:13) + (cid:13)(cid:13) r (cid:93) µ − (cid:13)(cid:13) = (cid:13)(cid:13) µ + (cid:13)(cid:13) + (cid:13)(cid:13) µ − (cid:13)(cid:13) = (cid:107) µ (cid:107) . (cid:3) In general, measures µ ∈ M ( β (cid:102) M ) represent functionals Φ ∗ µ that belong to F ( M ) ∗∗ . In our argument, we will need to determine whether a given measure µ represents a weak ∗ continuous functional on Lip ( M ). This is usually not aneasy task. A natural sufficient condition would be that µ is concentrated on (cid:102) M .In [12, Lemma 4.36], it is proved that this is indeed sufficient under an additionalhypothesis of local compactness. We provide here an argument that avoids thisextra hypothesis: Proposition 4. If µ ∈ M ( β (cid:102) M ) is concentrated on (cid:102) M then Φ ∗ µ ∈ F ( M ) .Proof. We will show that Φ ∗ µ is weak ∗ continuous. By the Banach-Dieudonn´etheorem, it is enough to check that its restriction to B Lip ( M ) is weak ∗ continuous.That is: fix a net ( f i ) in B Lip ( M ) that converges weak ∗ , i.e. pointwise, to f ∈ B Lip ( M ) , then we need to show that (cid:104) f i , Φ ∗ µ (cid:105) → (cid:104) f, Φ ∗ µ (cid:105) .Suppose first that M is separable. Then B Lip ( M ) is weak ∗ metrizable, so it isenough to prove the result for a sequence ( f n ) instead of a net. In that case wehave Φ f n → Φ f pointwise on (cid:102) M , and | Φ f n | ≤ µ -integrable, hencelim n →∞ (cid:104) f n , Φ ∗ µ (cid:105) = lim n →∞ (cid:90) (cid:102) M (Φ f n ) dµ = (cid:90) (cid:102) M (Φ f ) dµ = (cid:104) f, Φ ∗ µ (cid:105) by Lebesgue’s dominated convergence theorem.For the general case, fix ε >
0. By regularity there is a compact set K ⊂ (cid:102) M such that (cid:107) µ − µ | K (cid:107) = | µ − µ | K | ( (cid:102) M ) ≤ ε , and so (cid:107) Φ ∗ µ − Φ ∗ ( µ | K ) (cid:107) ≤ ε . Then XTREME POINTS IN LIPSCHITZ-FREE SPACES OVER COMPACT METRIC SPACES 5 S = p s ( K ) is a compact subset of M , hence separable. Since K ⊂ (cid:101) S we may identify µ | K with a Radon measure on (cid:101) S . Obviously f i | S → f | S pointwise, thereforelim i (cid:104) f i , Φ ∗ ( µ | K ) (cid:105) = lim i (cid:90) (cid:101) S (Φ f i ) dµ | K = (cid:90) (cid:101) S (Φ f ) dµ | K = (cid:104) f, Φ ∗ ( µ | K ) (cid:105) where the second equality holds by the previous paragraph. Thus Φ ∗ ( µ | K ) ∈ F ( M )and so dist(Φ ∗ µ, F ( M )) ≤ ε . Letting ε → ∗ µ ∈ F ( M ). (cid:3) The converse of Proposition 4 is false. For instance, take any ζ ∈ β (cid:102) M \ (cid:102) M ,then µ and µ + δ ζ + δ r ( ζ ) represent the same functional by Lemma 2. But evenif we avoid this trick by restricting our attention to measures with minimal norm,i.e. satisfying the conditions in Proposition 3, we can still find counterexamples toProposition 4: Example 5.
Consider M = [0 ,
1] and m = δ (1) ∈ F ( M ). For every n ∈ N , let µ n = 1 n n (cid:88) k =1 δ ( k − n , kn ) ∈ M ( (cid:102) M ) . It should be clear that (cid:107) µ n (cid:107) = 1 and Φ ∗ µ n = m for every n . Now let µ be a weak ∗ cluster point of the sequence ( µ n ). Then Φ ∗ µ = m , but supp( µ ) is clearly containedin the “diagonal” β (cid:102) M \ (cid:102) M .While it is in general not possible to guarantee that a measure µ representing anelement of F ( M ) will be concentrated on (cid:102) M , we can at least prove that it will beconcentrated “away from infinity”. More specifically, for proper M we can provethat µ is concentrated on p − ( M × M ). In the proof we will require the followingsimple fact. Lemma 6.
Suppose that M is proper and let h be a Lipschitz function on M withbounded support. Then Φ h vanishes outside of p − ( M × M ) .Proof. Let ζ ∈ β (cid:102) M \ p − ( M × M ), choose a net ( x i , y i ) in (cid:102) M that converges to ζ , and select a subnet such that d ( x i , y i ) either tends to infinity or converges to afinite value. In the former case | Φ h ( x i , y i ) | ≤ (cid:107) h (cid:107) ∞ /d ( x i , y i ) converges to 0 andthus Φ h ( ζ ) = 0. In the latter case, notice that either both p ( ζ ) and p ( ζ ) belongto M , or neither of them does. The first case implies ζ ∈ p − ( M × M ), whereasthe second case implies h ( x i ) = h ( y i ) = 0 eventually as h vanishes outside of abounded set, and so Φ h ( ζ ) = 0. (cid:3) Proposition 7.
Suppose that M is proper and µ ∈ M ( β (cid:102) M ) is such that Φ ∗ µ ∈F ( M ) , (cid:107) Φ ∗ µ (cid:107) = (cid:107) µ (cid:107) , and µ ≥ . Then µ is concentrated on p − ( M × M ) .Proof. For n ∈ N , let H n be the function defined by H n ( x ) = d ( x, ≤ n − − n d ( x,
0) , if 2 n ≤ d ( x, ≤ n +1 n +1 ≤ d ( x, x ∈ M . By [5, Lemma 2.3], for every f ∈ Lip ( M ) we have f H n ∈ Lip ( M )with (cid:107) f H n (cid:107) L ≤ (cid:107) f (cid:107) L and therefore f H n w ∗ −→ f . It follows that (cid:90) β (cid:102) M (Φ f ) dµ = lim n →∞ (cid:90) β (cid:102) M Φ( f H n ) dµ = lim n →∞ (cid:90) S Φ( f H n ) dµ = (cid:90) S (Φ f ) dµ where S = p − ( M × M ). Indeed, the first equality holds because Φ ∗ µ is weak ∗ continuous, the second one is a consequence of Lemma 6 as each f H n has boundedsupport, and the third one follows from Lebesgue’s dominated convergence theorem R. J. ALIAGA since Φ( f H n ) → Φ f pointwise on S . We conclude that µ and µ | S represent thesame functional in F ( M ). But then (cid:107) µ (cid:107) = (cid:107) Φ ∗ µ (cid:107) = (cid:107) Φ ∗ ( µ | S ) (cid:107) ≤ (cid:107) µ | S (cid:107) , and so µ must be concentrated on S . (cid:3) We will also need some understanding of the relationship between the support of µ and the support of the represented functional when the latter belongs to F ( M ). Lemma 8.
Let m ∈ F ( M ) and µ ∈ M ( β (cid:102) M ) be such that Φ ∗ µ = m . Then supp( m ) ⊂ p s (supp( µ )) .Proof. Recall that by [5, Proposition 2.7] a point x ∈ M belongs to supp( m ) if andonly if for every r > f ∈ Lip ( M ) that is supported on B ( x, r ) and such that (cid:104) m, f (cid:105) (cid:54) = 0. We will show that no point x ∈ M \ p s (supp( µ ))satisfies that condition, and this will prove the lemma.Indeed, fix one such x . Notice that, since supp( µ ) is a compact subset of β (cid:102) M , p s (supp( µ )) is also a compact subset of βM . Therefore there is r > K = B ( x, r ) βM does not intersect p s (supp( µ )). Suppose that f ∈ Lip ( M ) issupported in B ( x, r ) and let W = p − ( K ) ∪ p − ( K )which is a closed subset of β (cid:102) M that does not intersect supp( µ ). If ζ ∈ supp( µ ) thenthere is a neighborhood U of ζ such that U ∩ W = ∅ . Hence, if ( x, y ) ∈ (cid:102) M ∩ U then x, y / ∈ B ( x, r ) and f ( x ) = f ( y ) = 0, and it follows by density that Φ f ( ζ ) = 0.This shows that Φ f = 0 on supp( µ ) and therefore (cid:104) m, f (cid:105) = (cid:82) β (cid:102) M (Φ f ) dµ = 0. Asexplained above, this proves that x / ∈ supp( m ). (cid:3) Since the shadow of supp( µ ) is a compact subset of βM , by Lemma 8 it mustalways contain the closure supp( m ) βM . It is easy to see that, in fact, we can alwaysfind a representing measure whose support has minimal shadow. Proposition 9.
For every m ∈ F ( M ) there is a positive µ ∈ M ( β (cid:102) M ) such that Φ ∗ µ = m , (cid:107) µ (cid:107) = (cid:107) m (cid:107) , and p s (supp( µ )) = supp( m ) βM .Proof. Let S = supp( m ), then m ∈ F ( S ) and so Proposition 3 yields a positivemeasure µ ∈ M ( β (cid:101) S ) with (cid:107) µ (cid:107) = (cid:107) m (cid:107) and such that (Φ | Lip ( S ) ) ∗ µ = m . It is easilychecked that β (cid:101) S and the closure (cid:101) S β (cid:102) M of (cid:101) S in β (cid:102) M are equivalent compactifications of (cid:101) S , and that βS can be similarly identified with S βM . Therefore µ can be identifiedwith an element of M ( β (cid:102) M ) that is supported on (cid:101) S β (cid:102) M and such that Φ ∗ µ = m .Moreover Lemma 8 implies that S βM ⊂ p s (supp( µ )). But clearly p s (supp( µ )) ⊂ βS = S βM as well. Thus µ is the sought measure. (cid:3) The representation from Proposition 9 is not unique in general. For instance, let M = { , , } ⊂ R and m = δ (1) + δ (2), with (cid:107) m (cid:107) = 3. Then µ = δ (1 , + 2 δ (2 , and µ (cid:48) = 2 δ (1 , + δ (2 , are two positive representations of m with minimal normand minimal shadow.Let us remark here that many of the previous statements admit generalizations.For instance, when M is not proper, the same arguments show that Lemma 6 andProposition 7 hold with p − ( M R × M R ) in place of p − ( M × M ), where M R isthe subset of βM consisting of points that are limits of bounded nets. For proper M , that set is simply M . There are also versions of Lemma 8 and Proposition 9that hold for functionals in F ( M ) ∗∗ that are not weak ∗ continuous, as long as weconsider an appropriate definition of support for such functionals (see [3, Section3]). However, introducing this definition is a harder task that requires several new XTREME POINTS IN LIPSCHITZ-FREE SPACES OVER COMPACT METRIC SPACES 7 notions and replacing references to βM with the lesser-known Samuel (or uniform)compactification of M . We have thus decided not to include the details here asthey will not be needed for the proof of Theorem 1 in the proper case.3. Proof of the main result
Let us now proceed with the proof of Theorem 1. We start with the followingfact:
Lemma 10.
Let m ∈ ext B F ( M ) and µ ∈ M ( β (cid:102) M ) be such that Φ ∗ µ = m , (cid:107) µ (cid:107) = 1 ,and µ ≥ . Suppose that λ ∈ M ( β (cid:102) M ) is such that ≤ λ ≤ µ and Φ ∗ λ ∈ F ( M ) .Then Φ ∗ λ = (cid:107) λ (cid:107) · m .Proof. Let ν = µ − λ , so that ν ≥ ∗ ν ∈ F ( M ). If either λ = 0 or ν = 0 then the lemma holds trivially, so assume otherwise. By positivity we have (cid:107) λ (cid:107) + (cid:107) ν (cid:107) = (cid:107) µ (cid:107) = 1. Using the fact that (cid:107) Φ ∗ (cid:107) = 1 we get1 = (cid:107) Φ ∗ µ (cid:107) ≤ (cid:107) Φ ∗ λ (cid:107) + (cid:107) Φ ∗ ν (cid:107) ≤ (cid:107) λ (cid:107) + (cid:107) ν (cid:107) = 1 . All inequalities are therefore equalities. It follows that (cid:107) Φ ∗ λ (cid:48) (cid:107) = (cid:107) Φ ∗ ν (cid:48) (cid:107) = 1, where λ (cid:48) = λ/ (cid:107) λ (cid:107) and ν (cid:48) = ν/ (cid:107) ν (cid:107) . Now notice that m = Φ ∗ λ + Φ ∗ ν = (cid:107) λ (cid:107) · Φ ∗ λ (cid:48) + (cid:107) ν (cid:107) · Φ ∗ ν (cid:48) is a convex combination of Φ ∗ λ (cid:48) , Φ ∗ ν (cid:48) ∈ B F ( M ) . Hence Φ ∗ λ (cid:48) = m as claimed. (cid:3) This lemma tells us that, if we use Proposition 3 to choose a representation µ of m ∈ ext B F ( M ) that is positive and has minimal norm, then every measure λ defined by dλ = h dµ where 0 ≤ h ≤ m , as long as we know that it represents an element of F ( M ).It is easy to force such a measure λ to be concentrated on certain regions bychoosing h appropriately, and this will yield upper bounds for supp( m ) by Lemma8. The difficult part is to ensure that λ represents a w ∗ -continuous functional –the condition from Proposition 4 is too simple to be useful here. The next lemmashows how we can accomplish this when M is proper. Lemma 11.
Suppose that M is proper, and let h be a Lipschitz function on M with bounded support. Then, for every µ ∈ M ( β (cid:102) M ) such that Φ ∗ µ ∈ F ( M ) , wealso have Φ ∗ λ ∈ F ( M ) where dλ = ( h ◦ p ) dµ . By symmetry (or by analogous construction), the same is true if we replace p with p . Proof.
Let us start by noticing that the identityΦ( f h )( x, y ) = Φ f ( x, y ) · h ( x ) + Φ h ( x, y ) · f ( y )holds for any f ∈ Lip ( M ) and any ( x, y ) ∈ (cid:102) M . It follows that the continuousextensions Φ( f h ) = (Φ f ) · ( h ◦ p ) + (Φ h ) · ( f ◦ p )also agree in β (cid:102) M . Since f h ∈ Lip ( M ), we may integrate against µ to obtain (cid:104) f h, Φ ∗ µ (cid:105) = (cid:90) β (cid:102) M (Φ f )( h ◦ p ) dµ + (cid:90) β (cid:102) M ( f ◦ p )(Φ h ) dµ = (cid:104) f, Φ ∗ λ (cid:105) + (cid:90) β (cid:102) M ( f ◦ p ) dν = (cid:104) f, Φ ∗ λ (cid:105) + (cid:90) βM f d (( p ) (cid:93) ν ) R. J. ALIAGA where dν = (Φ h ) dµ . Notice that λ, ν ∈ M ( β (cid:102) M ), as h ◦ p and Φ h are continuousand bounded. By Lemma 6, ν is in fact concentrated on p − ( M × M ) and therefore( p ) (cid:93) ν is concentrated on M .Recall that the functional f (cid:55)→ (cid:104) f h, Φ ∗ µ (cid:105) is weak ∗ continuous (see e.g. [5, Lemma2.3]). Thus, in order to prove that Φ ∗ λ ∈ F ( M ) it will be enough to show that thefunctional f (cid:55)→ (cid:90) βM f d (( p ) (cid:93) ν ) = (cid:90) M f d (( p ) (cid:93) ν )also belongs to F ( M ). By [3, Proposition 4.4], it suffices to show that the function ρ ∈ Lip ( M ) given by ρ ( x ) = d ( x,
0) is integrable against ( p ) (cid:93) ν , i.e. that theintegral (cid:90) β (cid:102) M ( ρ ◦ p )(Φ h ) dµ is finite. Thus it will be enough to prove that g = ( ρ ◦ p ) | Φ h | is bounded. In thecase where M is compact, this is immediate as ρ is bounded. Let us now verify itin the unbounded case. Fix r > h ) ⊂ B (0 , r ). Let ( x, y ) ∈ (cid:102) M ,and consider three cases: • If y ∈ B (0 , r ), then g ( x, y ) = d ( y, | Φ h ( x, y ) | ≤ r (cid:107) h (cid:107) L . • If y / ∈ B (0 , r ) and x ∈ B (0 , r ), then h ( y ) = 0 and g ( x, y ) = d ( y, | h ( x ) | d ( x, y ) ≤ d ( y, d ( x, y ) · r (cid:107) h (cid:107) L ≤ (cid:18) d ( x, d ( x, y ) (cid:19) · r (cid:107) h (cid:107) L ≤ r (cid:107) h (cid:107) L . • If y / ∈ B (0 , r ) and x / ∈ B (0 , r ), then h ( y ) = h ( x ) = 0 and g ( x, y ) = 0.Thus (cid:107) g (cid:107) ∞ ≤ r (cid:107) h (cid:107) L on (cid:102) M and therefore also on β (cid:102) M . This finishes the proof. (cid:3) We can now finally prove Theorem 1, following the argument outlined beforeLemma 11.
Proof of Theorem 1.
Let m ∈ ext B F ( M ) . Apply Proposition 3 to find µ ∈ M ( β (cid:102) M )such that Φ ∗ µ = m , (cid:107) µ (cid:107) = 1 and µ ≥
0. By Proposition 7, µ is concentrated on theopen set p − ( M × M ), therefore supp( µ ) must intersect that set. Hence we maychoose points x ∈ p (supp( µ )) ∩ M and y ∈ p (supp( µ )) ∩ M (we allow the case x = y ). Fix z ∈ M \ { x, y } , and we will show that z / ∈ supp( m ).Construct a Lipschitz function h on M with bounded support and such that0 ≤ h ≤ h = 1 in a neighborhood of x , and h = 0 in a neighborhood of z . If λ ∈ M ( β (cid:102) M ) is given by dλ = ( h ◦ p ) dµ , then 0 ≤ λ ≤ µ and moreoverΦ ∗ λ ∈ F ( M ) by Lemma 11. It is clear that λ (cid:54) = 0 as x ∈ p (supp( λ )). Thus,writing λ (cid:48) = λ/ (cid:107) λ (cid:107) , we have Φ ∗ λ (cid:48) = m by Lemma 10. We may therefore replaceour representing measure µ with λ (cid:48) , and hence assume that z / ∈ p (supp( µ )).A similar construction with y and p in place of x and p , respectively, shows thatwe may replace µ again and assume that z / ∈ p (supp( µ )) either. This implies that z / ∈ p s (supp( µ )) and therefore z / ∈ supp( m ) by Lemma 8. Since z was arbitrary, wehave thus proved that supp( m ) ⊂ { x, y } is finite.The main result now follows from [4, Theorem 1.1], which characterizes extremepoints with finite support. The fact that m is an exposed point follows from [5,Theorem 3.2], and it is a preserved extreme point by either [2, Theorem 4.2] forthe compact case or [4, Proposition 2.3(a)] for the proper case. (cid:3) The following corollary is immediate:
Corollary 12.
Let M be a complete metric space and m ∈ ext B F ( M ) . If supp( m ) is proper, then m is an elementary molecule.Proof. Let S = supp( m ), then m ∈ F ( S ) by the definition of support and obviously m ∈ ext B F ( S ) , thus m is a molecule by Theorem 1. (cid:3) XTREME POINTS IN LIPSCHITZ-FREE SPACES OVER COMPACT METRIC SPACES 9
We finish this paper by remarking that most of the arguments developed hereare general and work for any choice of complete metric space M . The only placewhere compactness is used in an essential way is Lemma 11. So finding a differentargument or construction that yields measures λ ≤ µ representing weak ∗ continuousfunctionals might provide a way to extend Theorem 1 to the noncompact case. Acknowledgments
The author was partially supported by the Spanish Ministry of Economy, Indus-try and Competitiveness under Grant MTM2017-83262-C2-2-P.
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Instituto Universitario de Matem´atica Pura y Aplicada, UniversitatPolit`ecnica de Val`encia, Camino de Vera S/N, 46022 Valencia, Spain
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