doi:10.7151/dmdico.1170
SPACES OF LIPSCHITZ FUNCTIONS ON METRIC SPACES
Diethard Pallaschke Institute of Operations Research
University of Karlsruhe – KIT D–76128 Karlsruhe, Germany e-mail: diethard.pallaschke@kit.edu
and
Dieter Pumpl¨un
Faculty of Mathematics and Computer Science FernUniversitaet Hagen
D–58084 Hagen, Germany e-mail: dieter.pumpluen@fernuni-hagen.de
Dedicated to Zbigniew Semadeni, the founder of Categorical Functional Analysis
Abstract
In this paper the universal properties of spaces of Lipschitz functions, defined over metric spaces, are investigated.
Keywords: categories of Lipschitz spaces, Saks spaces, base normed spaces.
2010 Mathematics Subject Classification: 46M99, 26A16, 46B40.
1. Basic properties of Lipschitz functions
Let (X, d) be a semimetric space, i.e., a metric space for which the condition d(x, y) = 0 does not imply x = y. If there is no confusion, we will omit the semimetric and will only write X instead of (X, d). Moreover, to exclude the trivial case we will assume that d ≡ 0 implies card(X) = 1. If several semimetric spaces occur, we will write (X, dX), i.e., take the space as index.
Definition 1.1. If X, Y are (semi) metric spaces with (semi) metrics dX, dY, a mapping f : X −→ Y is called Lipschitz iff there exists an M ≥ 0 such that (L) dX(f (x), f (y)) ≤ M dY(x, y) for all x, y ∈ X.
One puts
(1) L(f ) := inf{M | M ≥ 0 and dY(f (x), f (y)) ≤ M dX(x, y) for all x, y ∈ X}.
L(f ) is called the Lipschitz constant of f. If L(f ) ≤ 1, then f is called a contrac- tion.
Special cases of Definition 1.1 are if Y is a normed or a seminormed linear space and dY the metric induced by the norm; in the following we will mostly study the case Y = R or Y = CI. If, for a semimetric space X, edX denotes the equiv- alence relation on X induced by the semimetric dX then X/∼
dX
, the set of equiv- alence classes, carries a canonical structure of a metric space. A Lipschitz map- ping f : X −→ Y between two semimetric spaces induces a Lipschitz mapping f : X/ˆ ∼
dX
−→ Y/∼
dY
with L(f ) = L( ˆf ). Hence, the theory of Lipschitz mappings between semimetric spaces cannot yield more information than the theory of Lip- schitz mappings between metric spaces. This is the reason why, in the following, only Lipschitz mappings on metric spaces are investigated.
Lemma 1.2. Let A ⊂ X be a non-empty subset of a metric space (X, d) and let distA: X −→ R with distA(v) = inf
x∈Ad(v, x) be the distance function of A. Then:
• the distance function is a Lipschitz function with Lipschitz constant 0 ≤ L ≤ 1.
• if A = X then L = 0 and if A 6= X and there exists a y ∈ X \ A which has a closed point to A, i.e., there exists a z ∈ A with d(y, z) = distA(y), then L = 1 and distA is an isometry.
Proof. Let x, y ∈ E and ε > 0 be given. By the definition of the infimum there exists a point z ∈ A with distA(y) ≥ d(y, z) − ε. We get
distA(x) ≤ d(x, z) ≤ d(x, y) + d(y, z)
≤ d(x, y) + distA(y) + ε.
By interchanging x and y
(2) |distA(x) − distA(y)| ≤ d(x, y)
follows. Hence, the distance function is a Lipschitz function with Lipschitz con- stant L ≤ 1.
If A = X then distA= 0. Now assume that there exists an y ∈ X \ A which has a closed point to A, i.e., there exists a z ∈ A with d(y, z) = distA(y), then
|distA(z) − distA(y)| = |0 − distA(y)| = d(z, y), which implies L = 1 by (2).
Remark 1.3. The Lipschitz functions on X separate points:
More precisely, let (X, d) be a metric space and let {x1, . . . , xn} ⊂ X be a finite subset of pairwise disjoint points, i.e., xi 6= xj for i 6= j. Then there exist Lipschitz functions ϕi : X −→ R such that for i ∈ {1, . . . , n}
ϕi(xj) =
1 : i = j
0 : i 6= j .
Proof. For i ∈ {1, . . . , n} put Ai := {x1, . . . , xn} \ {xi} and because of Lemma 1.2 define ϕi: X −→ R by ϕi(x) = distdistAi(x)
Ai(xi). Then ϕiis a Lipschitz function with these properties.
2. Spaces of Lipschitz functions
In the following let Lip denote the category of metric spaces and Lipschitz maps and define
LIip(X) := Lip(X, R), X ∈ Lip.
Furthermore, we introduce the following subcategories of Lip called Lip1 which is the subcategory defined by all contractions, Lip∞ the subcategory generated by all metric spaces (X, d) of finite diameter, i.e., diam(X) = sup{d(x, y) | x, y ∈ X} < ∞ and finally:
Lip∞1 = Lip1∩ Lip∞.
For a metric space X consider the space L∞(X) of all real valued bounded Borel- measurable functions endowed with the supremum norm k k∞. This is a Banach space for any metric space X. Put for a metric space X
Lip(X) := LIip(X) ∩ L∞(X) and take as norm
kf kL:= max{L(f ), kf k∞}.
Note that the same definition makes sense for f : X −→ E in Lip and E ∈ Vec, where Vec denotes the category of real normed linear spaces and continuous linear maps and Vec1 the subcategory defined by linear contractions (For more details see [5] and [9]).
Lip(X) carries two norms k k∞ and k kL and we have obviously
(∗) k k∞≤ k kL.
L(Lip(X)) denotes the closed unit ball with respect to k kL and ∞(Lip(X)) the closed unit ball with respect to k k∞. Obviously
L(Lip(X)) ⊂ ∞(Lip(X)).
holds. One defines
C(Lip(X)) := {f | f ∈ Lip(X), f (x) ≥ 0 for all x ∈ X} .
C(Lip(X)) is a proper, generating cone in Lip(X), i.e., Lip(X) = C(Lip(X)) − C(Lip(X)). That C(Lip(X)) is proper is trivial. For every f ∈ Lip(X) and every x ∈ X one has
(∗∗) −kf k∞≤ f (x) ≤ kf k∞,
which implies that C(Lip(X)) is generating. Furthermore, C(Lip(X)) is closed with respect to k k∞because C(Lip(X)) =T
x∈X{f ∈ Lip(X) | f (x) ≥ 0} and, moreover, because of (∗) also with respect to k kL.
In order to avoid mixing up both normed spaces, let Lip∞(X) := Lip(X), k k∞
and LipL(X) := Lip(X), k kL . Moreover, let us point out that the product, as well as the pointwise maximum and minimum of two bounded Lipschitz functions is again a Lipschitz function and let us denote by 1I ∈ Lip(X) the constant function 1I(x) = 1, for all x ∈ X.
1I ∈ Lip(X) is an order unit with respect to k k∞ (not k kL).
Furthermore, observe that Lip∞(X) is in general not a Banach space. To see this, take X := [0, 1] the unit interval with d(x, y) = |x − y|. Then the function f : [0, 1] −→ R with f (x) := √
x is not Lipschitz, but it is the uniform limit of the Lipschitz functions fn : [0, 1] −→ R with fn(x) := min{nx,√
x}, because supt∈[0,1]|fn(t) − f (t)| = 4n12.But LipL(X) is a Banach space ([12] 1.6.2 ).
Let us recall some notations: If C is an arbitrary cone of a vector space E, then a convex subset B ⊂ C is called a base of C if every z ∈ C \ {0} has a unique representation z = λb with λ > 0 and b ∈ B. Every cone C ⊂ E induces a
partial order ≤ by x ≤ y if and only if y − x ∈ C. A partial order ≤ is called archimedian if, for some y ≥ 0 and all λ > 0 x ≤ λy implies x ≤ 0. For a, b ∈ E let [a, b] := {z ∈ E | a ≤ z ≤ b}. Moreover, we use the notation a ∨ b = max{a, b}
if the maximum exists and correspondingly the minimum a ∧ b = min{a, b}, and write |a| for a ∨ (−a). A norm k k for E is called a Riesz norm if, for all a, b ∈ E, the inequality |a| ≤ |b| implies kak ≤ kbk. An element e ∈ C is called an order unit if for every z ∈ E there exists a λ > 0 such that λe ≤ z ≤ λe.
Remark 2.1. Let E be a vector space and C ⊂ E a generating cone.
• If C has an order unit e ∈ C then the function
z 7→ kzk = inf{λ > 0 | − λe ≤ z ≤ λe}
is a norm for E, which is called an order unit norm.
• If C has a base B ⊂ C such that the set S = conv (B ∪ −B) is order bounded, then the Minkowski functional
p(z) = inf{λ > 0 | z ∈ λS}
is a norm for E. We call this norm p a base norm and call E a base normed space. The base is denoted by B = Bs(E) and C = R+Bs(E) holds with R+ = [0, +∞).
For a real vector space E we denote by E∗ the algebraic dual, that is the vector space of all linear forms from E to R. If E is endowed with a locally convex Haus- dorff linear topology τ then the pair (E, τ ) is called a locally convex topological vector space and we denote by E0 its topological dual, that is the vector space of all continuous linear forms from E to R.
A Saks space is a triple (E, k k, τ ) where k k is a norm on the real linear topological space E and τ is a locally convex Hausdorff linear topology τ on E such that the unit ball k k(E) is τ -closed and τ -bounded. For any normed vector space (E, k k) the triple (E0, k k0, σ(E0, E)) is a Saks space, where k k0 is the dual norm and σ(E0, E) the weak-* topology on E0.
Proposition 2.2. For a metric space X the space Lip∞(X) := Lip(X), k k∞ endowed with the pointwise order of functions is a regular ordered order unit normed space with the closed and generating order cone C(Lip(X)) the order unit 1I ∈ Lip(X) and ∞(Lip(X)) = [−1I, 1I]. Lip∞(X) is in general not complete.
Proof. The equation ∞(Lip(X)) = [−1I, 1I] follows from (∗∗). The proof of the remaining assertions is straightforward.
Proposition 2.3. LipL(X) is a Banach space and the cone C(Lip(X)) is gener- ating and k kL-closed but k kL is not a Riesz-norm with respect to C(Lip(X)).
Proof. The completeness of LipL(X) is shown in [12], Proposition 1.6.2, and that C(Lip(X)) is a proper generating cone was shown above as was the closedness.
Example 2.4. Define the function
f : R −→ R given by f (x) :=
2x : x ∈ [−12,12] 1 : x ≥ 12
−1 : x ≤ −12 ,
then L(f ) = 2 and kf k∞ = 1 which implies kf kL = 2, i.e., L(Lip(R)) ⊂6=
∞(Lip(R)).
Note furthermore, that if A ⊂ Lip(X) is k k∞-closed, then A is also k kL-closed, i.e., for the topologies,
τk k∞ ⊂ τk kL, holds which follows from (∗).
Remark 2.5. For a metric space X, LipL(X), k kL, k k∞ is a Saks space with the topology of k k∞, i.e., a 2−normed linear space in the sense of Semadeni [10], who investigated these spaces which led to the introduction of Saks spaces.
C(Lip(X)) is k kL-closed, proper and generating but does not make LipL(X) a regular ordered Saks space because k kL is not a Riesz norm with respect to C(Lip(X)) (see [7]).
As we have seen, k k∞ ≤ k kL implies L(Lip(X)) ⊂ ∞(Lip(X)), i.e., L(Lip(X)) is k k∞-bounded and ∞(Lip(X)) is also k kL-closed.
Additionally it should be noted, for further use, that C(Lip(X)) is 1-normal (see [13], Prop. 9.2 (e), p. 86), i.e., g ≤ f ≤ h implies kf k∞≤ max {kgk∞, khk∞} . Moreover, Lip∞(X) is a Stonian vector lattice (see [2] p. 186).
As for any normed linear space (E, k k), (E, k k, σ(E, E0)) is a Saks space, we have the canonical Saks spaces Lip(X), k k∞, σ (Lip∞(X), Lip∞(X)0) and Lip(X), k kL, σ (LipL(X), LipL(X)0) . The first one is even a regularly ordered Saks space because Lip∞(X) ∈ Vec+1 (see [7], Example 3.2 iii) ).
3. The duals of Lipschitz functions spaces
Now the connection between Lip∞(X) and LipL(X) will be investigated as well as between their (topological) dual spaces Lip∞(X)0 and LipL(X)0 which are
both linear subspaces of Lip(X)∗. Inequality (∗) implies that in the commutative diagram
LipL(X ) -
Id
#
Lip∞(X )
λ ∈ Lip∞(X )0
R Q
Q Q
Q Q
Q Q
QQs λ ◦ Id
the identity map is a contraction but not a quasi-isometry( see [12], pp. 3–4).
The dual norm of Lip∞(X)0 will be denoted by k k0∞ and the dual norm of LipL(X)0 by k k0L. Now Id induces an injective contraction κX : Lip∞(X)0 −→
LipL(X)0 with κX(λ) := λ ◦ Id, which may be considered as an inclusion, hence we often write κX(λ) := λ.
For λ ∈ Lip∞(X)0 one has:
kκX(λ)k0L = sup {|λ(f )| | kf kL≤ 1}
≤ sup {|λ(f )| | kf k∞≤ 1} (because of (∗))
= kλk0∞, i.e., taking κX as inclusion, then
(∗0) k k0L≤ k k0∞.
Hence, we have
0∞(Lip(X) ⊂ 0L(Lip(X)), i.e., one may regard Lip∞(X)0 as a subspace of LipL(X)0.
In the following we will show that in general Lip∞(X)0 is a proper subspace of LipL(X)0, Lip∞(X)0⊂6=LipL(X)0. For this we use the construction of point derivations from D.R. Sherbert [11] (see also [12] Chapter 7) which we briefly outline.
Consider the real Banach space
l∞:= {x := (xn)n∈N | (xn)n∈N bounded sequence } endowed with the supremum norm
kxk∞:= sup
n∈N
|xn|.
Let c ⊂ l∞ denote the closed subspace of all convergent sequences and let lim : c −→ R be the continuous linear functional which assigns to every convergent sequence its limit. Consider a norm-preserving Hahn-Banach extension ”LIM” of the functional ”lim” to l∞:
c -
⊂
l∞
? LIM
R Q
Q Q
Q Q
Q Q
Q s lim
with the following additional properties:
(i) LIMn→∞xn= LIMn→∞xn+1, (ii) lim inf
n→∞ xn≤ LIMn→∞xn≤ lim sup
n→∞
xn.
We shall use the the notation LIMn→∞xn= LIM(x) for x := (xn)n∈N∈ l∞. These functionals ”LIM” are called translation invariant Banach limits (see [4], Chapter II.4, Exercise 22).
Now let (X, d) be a metric space and ∆ := {(x, x) ∈ X × X | x ∈ X} the diagonal of X × X. For a double sequence w ∈ ((X × X) \ ∆)N, w := (xn, yn)n∈N) one defines the sequence
Tw(f ) := f (yn) − f (xn) d(yn, xn)
| n ∈ N
, f ∈ Lip(X).
This yields a mapping
Tw : Lip(X) −→ l∞ with Tw(f ) := f (yn) − f (xn) d(yn, xn)
n∈N
,
which satisfies kTw(f )k∞ ≤ L(f ) ≤ kf kL, i.e., a continuous linear map Tw : LipL(X) −→ l∞. Hence for any Banach limit LIM the composition Dw= LIM◦Tw
is a continuous linear functional
Dw : LipL(X) −→ R with Dw(f ) = LIM(Tw(f )), i.e., Dw ∈ LipL(X)0.
As the definition of Dw resembles the classical definition of the derivation of functions, one is interested if and under which assumptions Dw is a derivation in the sense of Bourbaki [3] i.e., for f, g ∈ Lip(X) and x ∈ X,
(P R0) Dw(f g) = f Dw(g) + gDw(f )
holds, i.e., in our case, as the left side does depend (directly) on x ∈ X (P R) Dw(f g) = f (x)Dw(g) + g(x)Dw(f )
where the dependence of Dw(f g) on x ∈ X has to be specified. The answer to this is (see [11], Prop. 8.5)
Proposition 3.1. If (X, d) is a metric space and for w ∈ ((X × X) \ ∆)N, w := (xn, yn)n∈N) satisfies x0 = limn→∞xn = limn→∞yn, i.e., x0 ∈ X is non- isolated, then (P R) is satisfied and Dw is called a point derivation in x0.
Proof. If (an)n∈N, (bn)n∈N ∈ l∞, then, for convergent (an)n∈N, with a = limn→∞an one has for any Banach limit LIM
LIMn→∞(an· bn− αbn) = 0,
because |an· bn− αbn| = |an − α||bn| ≤ B|an− α| if k(bn)n∈Nk∞ = B, which implies limn→∞(an· bn− αbn) = LIMn→∞(an· bn− αbn) = 0.
Hence
LIMn→∞(an· bn) = αLIMn→∞bn follows. This implies, for f, g ∈ Lip(X)
Dw(f g) = LIM(Tw(f g))
= LIMn→∞
(f g)(yn) − (f g)(xn) d(yn, xn)
= LIMn→∞
f (yn)g(yn) − g(xn)
d(yn, xn) + g(xn)f (yn) − f (xn) d(yn, xn)
= f (x0)LIMn→∞
g(yn) − g(xn) d(yn, xn)
+ g(x0)LIMn→∞
f (yn) − f (xn) d(yn, xn)
= f (x0)Dw(g) + g(x0)Dw(f ) which completes the proof.
In [11], Proposition 8.5, it is shown that for x0 isolated any Dw ≡ 0 if w fulfills the condition of Proposition 3.1.
Proposition 3.2. In general, Lip∞(X)0 is a proper subspace of LipL(X)0, Lip∞(X)0 ⊂6=LipL(X)0.
Proof. Consider the metric space X := [0, 1] with the usual metric d(x, y) =
|x − y|. Let Dw ∈ LipL([0, 1])0 be any point derivation for x = 0 and define the sequence ϕk, k ∈ N, k > 1, in LipL(X) by
ϕk(x) := max (√
kx,
√k
1 − k(x − 1) )
.
Then
max ϕk(x) = ϕk(1 k) = 1
√k.
This implies that the sequence (ϕk)nk∈N converges to 0 with respect to k k∞. A trivial calculation shows Dw(ϕk) = √
k. Since the constant function 0 has Dw(0) = 0 and the sequence (Dw(ϕk))k∈N is unbounded, it follows that Dw 6∈
Lip∞(X)0 which completes the proof.
4. The space of point functionals
A central role in our investigations plays, for a metric space X, x ∈ X, the mapping
δX : X −→ Lip(X)∗ with δX(x)(f ) := f (x), f ∈ Lip(X).
First note that for every x ∈ X the linear functional δX(x) ∈ Lip(X)∗ is con- tinuous with respect to both norms k k∞ and k kL. The restriction of δX to Lip∞(X)0 is denoted by δ∞X and to LipL(X)0 by δLX. The upper indicees are omitted if misunderstandings are not possible.
Proposition 4.1. Let (X, d) be a metric space. Then (a) For all x ∈ X, kδX∞(x)k0∞= kδXL(x)k0L= 1 holds.
(b) δXL : X −→ LipL(X)0 is a contraction.
(c) δX is injective and δX(X) is a linearly independent set.
Proof. (a): For both norms k k∞and k kL one has kδLX(x)(f )k0L = sup {|δX(x)| | f ∈ L(Lip(X))}
= sup {|f (x)| | f ∈ L(Lip(X))} ≤ sup {|f (x)| | kf k∞≤ 1}
= kδ∞X(x)k0∞≤ max {kf k∞, 1} ≤ max {kf kL, 1} ≤ 1.
Hence, as k1I(x)k = 1, kδX(x)k0L= kδX(x)k0∞= 1 follows.
(b): One has
kδLX(x) − δXL(y)k0L = sup|δXL(x)(f ) − δXL(y)(f )| | kf kL≤ 1
= sup {|f (x) − f (y)| | kf kL≤ 1}
≤ sup {kf kLd(x, y) | kf kL≤ 1} = d(x, y).
To prove (c) let α1, α2, . . . , αn∈ R and x1, x2, . . . , xn∈ X be given with xi 6= xk for i 6= k, and assume thatPn
j=1αjδX(xj) = 0 holds. By Remark 1.3 there exist ϕ1, ϕ2, . . . , ϕn∈ Lip(X) with
ϕi(xj) = δij =
1 : i = j
0 : i 6= j NowPn
j=1αjδX(xj)(ϕi) = 0 yields αi= 0.
Next we consider the linear subspace generated by the point functionals
D(X) := hδXi :=
( λ =
n
X
i=1
αiδX(xi) | x1, . . . , xn∈ X, α1, . . . , αn∈ R, n ∈ N )
of Lip(X)∗.
Remark 4.2. By DL(X) we denote the space D(X) endowed with the dual norm k k0L and by D∞(X) we denote the space D(X) endowed with the dual norm k k0∞. The canonical injections are δXL and δX∞. At the beginning of Section 2 it was already pointed out, that a normed linear space E and a metric space X with metric d, the notions Lipschitz function for a mapping f : X −→ E and kf k∞as well kf kL are defined analogously to the case E = R. Hence, for δX∞ and δLX one may try to compute these norms. As δX∞ is not Lipschitz only δXL remains. For the sake of brevity we denote the ∞-norm of δXL : X −→ LipL(X)0 with kδXLk∞ and get from Proposition 4.1
kδLXk∞= sup
kδX(x)k0L | x ∈ X
= 1, which implies kδLXkL= 1 for the L-norm, as L δLX ≤ 1.
Theorem 4.3. Let X be a metric space, E ∈ Vec1 with norm k k and ϕ : X −→ E, ϕ ∈ Lip with kϕkL≤ 1. Then there exists a unique ϕ0 : DL(X) −→ E in Vec1 with ϕ = (ϕ0)δXL and kϕ0k = kϕkL such that
X -
δLX
(DL(X))
? (ϕ0)
(E) Q
Q Q
Q Q
Q Q
Q s ϕ
commutes.
Proof. We may assume that ϕ 6= 0 because otherwise the statement is trivially true. The assumption kϕkL ≤ 1 implies ϕ(X) ⊂ (E), hence we may restrict ϕ to (E) in its image domain. As misunderstandings are not possible we will denote the restriction also by ϕ. As {δX(x) | x ∈ X} is a basis of DL(X) the linear mapping ϕ0: D(X) −→ E is well defined by
ϕ0
n
X
i=1
αiδX(xi)
! :=
n
X
i=1
αiϕ(xi)
and satisfies ϕ = ϕ0◦δX. A routine calculation shows that ϕ0is a linear mapping.
In order to prove kϕ0k = kϕkL, let λ ∈ E0 with λ 6= 0, then kλ ◦ ϕkL ≤ kλk0kϕkL and hence λϕ
kλk0kϕkL ∈ (LipL(X)) . For ξ =Pn
i=1αiδLX(xi) ∈ DL(X) there exists a λξ ∈ E0 withkλξk0 = 1 and λξ(ϕ0(ξ)) = kϕ0(ξ)kE. Now:
kϕ0(ξ)kE = λξ ϕ0 n
X
i=1
αiδXL(xi)
!!
= λξ n
X
i=1
αiϕ(xi)
!
=
n
X
i=1
αiλξ◦ ϕ(xi)
≤ kϕkL
n
X
i=1
αi
λξ◦ ϕ kϕkL (xi)
= kϕkL
n
X
i=1
αiδXL(xi)
! λξ◦ ϕ kϕkL
= kϕkLsup
|ξ(f )| = |
n
X
i=1
αif (xi)|
f ∈ L(Lip(X))
≤ kϕkLkξkL
which implies kϕ0k ≤ kϕkL≤ 1.
On the other hand, kϕkL= kϕ0◦ δXLkL≤ kϕ0kkδXLkL= kϕ0k (because of Remark 4.2) and hence kϕkL= kϕ0k follows.
Remark 4.4. For a metric space X, Theorem 4.3 means that the mapping δXL −→ (DL(X)) is universal with respect to all Lipschitz mappings ϕ : X −→ E, kϕkL ≤ 1 where E is a norned real linear space. This is equivalent to the following statement:
The unit ball functor : Vec1 −→ Lip1 has DL : Lip1 −→ Vec1 as a left adjoint.
Proof. The proof is elementary because maps a contraction in Vec1 to a contraction in Lip1and Theorem 4.3 states that δXL −→ (DL(X)) is a canonical universal embedding for a metric space into a normed real linear space.
Proposition 4.5. Let (X, d) be a metric space. Then D∞(X) is a base normed linear space with base Bs (D∞(X)) = conv ({δX(x) | x ∈ X}) , the convex hull of δX(X), the order cone C (D∞(X)) = R+Bs (D∞(X)) and the base norm
kλkB =
k
X
i=1
αiδX(xi) B
=
k
X
i=1
|αi|
if λ =Pk
i=1αiδX(xi)is the representation of λ in the basis δX(X).
Proof. Lip∞(X) is an order unit normed linear space (see Proposition 2.2) with order unit 1I. The order in Lip∞(X) is pointwise. Hence, well-known results (cp, e.g. [13], Chapt. 9) imply that Lip∞(X)0 is a base ordered linear space with cone
C(Lip∞(X)0) :=λ | λ ∈ Lip∞(X)0, λ(f ) ≥ 0 for all f ∈ C(Lip∞(X)) and base
Bs Lip∞(X)0 = λ | f ∈ Lip∞(X)0, λ ≥ 0 and λ(1I) = 1
= C(Lip(X)0) ∩λ | λ ∈ Lip∞(X)0 and λ(1I) = 1 . We define
C (D∞(X)) := C Lip∞(X)0 ∩ D∞(X) and B := Bs Lip∞(X)0 ∩ D∞(X).
Of course, B is a base set in D∞(X) with B ⊂ C (D∞(X)) , hence, R+B ⊂ C (D∞(X)) .
If λ :=Pn
i=1αiδX(xi) ∈ C (D∞(X)) , then Remark 1.3 yields at once αi ≥ 0, 1 ≤ i ≤ n. The converse implication is obviously true. If λ > 0, then kαk :=
Pn
i=1αi> 0 follows and
(1) λ = kαk
n
X
i=1
αi
kαkδX(xi) holds. On the other hand, λ =Pn
i=1αiδX(xi) ∈ B is equivalent to λ(1I) = 1, i.e., λ(1I) =Pn
i=1αi = 1. This shows that B is the convex hull of {δX(x) | x ∈ X} , i.e., B = conv {δX(x) | x ∈ X} and because of (1), C (D∞(X)) = R+B, hence Bs (D∞(X)) := B is a base for C (D∞(X)) .
For a basis representation of λ =Pn
i=1αiδX(xi) ∈ D∞(X), λ 6= 0 put kα+k :=
n
X
i=1 αi≥0
αi , kα−k :=
n
X
i=1 αi<0
αi.
Then
(2) λ = kα+kλ+− kα−kλ−
with
λ+=
n
X
i=1 αi>0
αi
kα+kδX(xi) for kα+k 6= 0 and λ+ := 0 else, and analogously for λ−.
As a subset of Bs (Lip∞(X)0) the set Bs (D∞(X)) is linearly bounded. Hence the base seminorm induced by Bs (D∞(X)) is a norm (cp. [13]). We denote this norm by k k0 for the moment. One of the possible representations for a base norm is:
kλk0 = inf {β + γ | β, γ ≥ 0, λ = βξ − γη, ξ, η ∈ Bs (D∞(X)) } . For λ :=Pn
i=1αiδX(xi) (2) implies
kλk0 ≤ kα+k + kα−k =
n
X
i=1
|αi|.
Now, let λ = βφ − γψ be a second representation. We may take the union of δX(x), x ∈ X, which appear in the basis representation of λ, φ, and ψ and denote it by {δX(xi) | xi∈ X, 1 ≤ i ≤ n} . Let λ =Pn
i=1αiδX(xi), φ =Pn
i=1ϕiδX(xi), and ψ =Pn
i=1ψiδX(xi) where ϕi ≥ 0 and ψi ≥ 0 for 1 ≤ i ≤ n. Then
n
X
i=1
αiδX(xi) =
n
X
i=1
(βϕi− γψi) δX(xi)
hence αi = (βϕi− γψi) and |αi| = |βϕi− γψi| ≤ βϕi+ γψi. As φ, ψ ∈ Bs (D∞(X)) one hasPn
i=1ϕi =Pn
i=1ϕi= 1 andPn
i=1|αi| ≤ β + γ follows such that kλkB=Pn
i=1|αi| bas been proved.
In view of Theorem 4.3 it is natural to ask if a similar result can be proved for δX∞ : X −→ Bs (D∞(X)) . This is indeed the case as the following Proposition shows.
Let us denote the category of base-normed linear spaces and linear base pre- serving mappings (which are, by the way contractions) by BN-Vec1. If, for E ∈ BN-Vec1, Bs (E) denotes the base of E, then
Proposition 4.6. Let X be a metric space, E ∈ BN-Vec1 and ϕ : X −→ Bs(E) a Lipschitz mapping. Then there exists a unique ϕ0 : D∞(X) −→ E in BN-Vec1 such that
X -
δX∞
Bs(D∞(X))
? Bs(ϕ0)
Bs(E) Q
Q Q
Q Q
Q Q
Q s ϕ
commutes, and 1 = kϕ0k = kϕk∞, where kϕk∞= sup {kϕkE | x ∈ X} .
Proof. As δX∞(X) is a basis of D∞(X) the linear mapping ϕ0 : D∞(X) −→ E is well defined by
ϕ0 n
X
i=1
αiδX(xi)
! :=
n
X
i=1
αiϕ(xi) and makes the above diagram commutative.
The proof of the equality of norms is trivial. As ϕ0 is a contraction kϕ0 ≤ 1 holds. Also kϕk∞k = 1 as ϕ(X) ⊂ Bs(E). Moreover
1 = kϕ(x)kE ≤ kϕ0kkδ∞X(x)k0∞= kϕ0k ≤ 1, i.e., kϕ0k = 1.
Despite the fact that the Lipschitz constant or the norm k kL does not appear explicitly in Proposition 4.6 the result is nonetheless quite interesting for metric spaces.
Corollary 4.7. Let (X, dX) be a metric space. Then
(i) δX∞: X −→ Bs (D∞(X)) is a universal embedding into the base of the based normed linear space D∞(X), i.e. the canonical functor
Bs : BN-Vec1 −→ Lip1 has the functor D∞ : Lip1 −→ BN-Vec1 as a left adjoint with the adjunction morphism δX∞: X −→ Bs (D∞(X)) . (ii) δX∞: X −→ Bs (D∞(X)) is a universal contractive embedding into a metric
convex module, i.e., if C is a convex module (see [6]) and f : X −→ C is in Lip1, then there is a unique affine mapping f0 : Bs (D∞(X)) −→ C with f = f0◦ δX∞.
If Met-Conv denotes the category of metric convex modules, or, what is the same, of metric convex subsets of real linear spaces [6] and affine mappings, Bs (D∞(X)) may be regarded as a functor BsD∞ : Lip1 −→
Met-Conv which is left adjoint to the canonical forgetful functor U : Met-Conv −→ Lip1 assigning to every C ∈ Met-Conv its underlying metric space and δX∞ induces the adjunction morphism.
Proof. (i) is just a reformulation of Proposition 4.6 (ii) results by straightforward arguments from Proposition 4.6 and the results in §2 of [6].
Corollary 4.7 (ii) shows an interesting fact, namely the canonical and close con- nection between metric and convex structures.
5. The predual
There is another interesting topology, which was first investigated by R.F. Arens and J. Eells [1], and which will be discussed in this section.
The following proof uses a method completely different from the one used in [1] and is considerably shorter.
Define
ג : LipL(X) −→ D0L(X) and k : D0L(X) −→ LipL(X) by
ג(f )(λ) := λ(f ), f ∈ LipL(X), λ ∈ D0L(X) and
k(λ)(x) := λ(δXL(x)), λ ∈ D0L(X), x ∈ X.
Theorem 5.1. ג : LipL(X) −→ D0L(X) and k : DL0 (X) −→ LipL(X) are in Vec1 and
ג ◦ k = idDL0(X) and k ◦ ג = idLipL(X), x ∈ X holds. Hence, ג and k are isometries.
Proof. Let us denote the norm dual to k k0L of DL(X) on D0L(X) by k k#L. For x, y ∈ X one has if λ ∈ DL0 (X) :
|k(λ)(x) − k(λ)(y)| = |λ δLX(x) − λ δXL(y) |
≤ kλk#Lk(δXL(x) − δLX(y)k0L≤ kλk#Ld(x, y) (because of Proposition 4.1).
Hence L(k(λ)) ≤ kλk#L follows. Moreover
|k(λ)(x)| = |λ δLX(x) | ≤ kλk#LkδXL(x)kL= kλk#L and we have
kk(λ)k∞= sup {|k(λ)(x)| | x ∈ X} = sup {|λ δXL(x) | | x ∈ X}
≤ kλk#Lsup {|δLX(x)| | x ∈ X} ≤ kλk#L. This yields
(∗) kk(λ)kL= max {kk(λ)k∞, L (k(λ)) } ≤ kλk#L i.e., kkk ≤ 1, k is a contraction.
For f ∈ LipL(X) one gets kג(f)k#L = supn
|ג(f)(λ)|
λ ∈ DL(X) and kλk0L≤ 1o
= sup n
|λ(f )|
λ ∈ DL(X) and kλk0L≤ 1o
≤ kλk0Lsupn kf kL
kλk0L≤ 1o
≤ kf kL, which gives
(∗∗) kגk = supn
kג(f)k#L
kf kL≤ 1o
≤ 1.
i.e., ג is also a contraction, kגk ≤ 1.
For λ ∈ DL0 (X) and x ∈ X ג(k(λ)) δLX(x)
= δLX(x) ((k(λ)) = k(λ)(x) = λ δXL(x) , holds for all x ∈ X, which yields ג◦k(λ) = λ, because δLX(x) | x ∈ X is a basis of DL(X) and hence
ג ◦ k = idD0L(X). Also for f ∈ LipL(X) and any x ∈ X
k(ג(f ))(x) = ג(f ) δLX(x) = δXL(x)(f ) = f (x) which results in k(ג)(f ) = f and hence
k ◦ ג = idLipL(X),
i.e., k and ג are inverse to each other. This together with (∗) and (∗∗) yields the assertion
Remark 5.2. To show the dependence of X, an index X will be added: גX
and kX, because both are natural transformations between interesting functors.
The interesting topology mentioned at the beginning of this section is the dual topology σ (D0L(X), DL(X)) transferred by k (and ג) to LipL(X).
References
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Received 31 May 2015
