arXiv:1911.07680v2 [math.PR] 20 Apr 2020 h oooyon topology the ter ftelte nerleit ntewa es,ta s ftefollowing the if is, that sense, weak the in exists integral latter the if o l Λ all for esrsspotdo lsdcne e.Let set. convex closed a on supported measures 1. n[,Catr3]. Chapter [2, in n,b h anBnc eaainterm n has one theorem, separation Hahn–Banach the by and, ovxhl.Fo o n ewl s h a vrasti re oden to order in set a over bar the use will we on, now From hull. convex hoe 1. Theorem ttmnsaeequivalent: are statements (ii) (i) ednt h e fRdnpoaiiymaue on measures probability Radon of set the denote We oeta f()eit,then exists, (1) if that Note olwn stetermta ie hrceiaino barycente of characterization a gives that theorem the is Following a h anga ftepeetnt st hrceiebarycenters characterize to is note present the of goal main The ∈ where pcs ipecaatrzto sotie ntrso th of terms in obtained is probabi me characterization of simple Radon set particula a convex signed the of spaces, finite is case of support A corresponding space the the space. where given itself a is of space subset underlying convex closed a on n has One hr exist There X ∈ hrceiaini rsne fbrcneso h Radon the of barycenters of presented is characterization A fameasure a of X V ∗ a where , Let = X nbrcneso rbblt measures probability of barycenters On { µ M x sgnrtdb h rnlto-nain metric translation-invariant the by generated is ∈ ∈ ⊂ P X M ( X µ ∗ X ∃ | stetplgclda of dual topological the is ∈ egyBrznAa Miftakhov Azat Berezin Sergey ) eannepycmatcne e,adlet and set, convex compact non-empty a be > α with P ( X : 0 is ) supp − αx µ = 1+ (1 + Λ a a a = M = = supp Abstract n ihbarycenter with and Z X X Z Z α M µ 1 X ) Λ µ x µ x a µ x = X eaF´ce pc.Wtotls fgenerality, of loss Without Fr´echet space. a be ∈ ( ( dx oedtiso uhitgascnb found be can integrals such on details More . V ( dx M dx a ) , a ) } (2) ) , . X ∈ co(supp by eaieinterior. relative e iymaue.Frlclycompact locally For measures. lity srso erccmatand compact metric a on asures rbblt esrssupported measures probability P neeti tde,weethe where studied, is interest r ( a X ρ ; .B ento,tebarycen- the definition, By ). µ so h esrsfrom measures the of rs holds on ,weeco( where ), a t t oooia closure. topological its ote X fteRdnprobability Radon the of ∈ M frdtisse[2]). see details (for hn h following the Then, . · tnsfrthe for stands ) P ( X (4) (1) (3) ). Remark 1. We note that the condition (4) is non-local and concerns the whole set M. Remark 2. We require M to be compact in order to ensure the separability of M and existence of weak integrals (e.g., see [2, Theorem 3.27]). If X is finite-dimensional, the theorem holds without this requirement.

Proof of Theorem 1. (a) First, we prove that (i) ⇒ (ii). Let c ∈ M, and let Uδ(c) be the open ball of radius δ > 0 centered at c. Because M is the support of µ, one has µ(Uδ(c)) > 0. Also, since M is compact, so is Uδ(c) ∩ M, and 1 cδ = xµ(dx) ∈ M (5) µ(Uδ(c)) UδZ(c) is well-defined altogether. It is easy to show that

lim cδ = c (6) δ→+0

in the σ(X,X∗). Indeed, take any Λ ∈ X∗. Since Λ is continuous, for every ε>

0 there exist δ0 > 0 such that x ∈ Uδ0 (c) implies |Λ(x) − Λ(c)| = |Λ(x − c)| <ε. And it follows from the definition of the weak integral that 1 |Λ(cδ − c)| ≤ |Λ(x − c)| µ(dx) <ε (7) µ(Uδ(c)) UδZ(c)

whenever δ ∈ (0, δ0). This means that cδ → c in the weak topology as δ → +0. Further, for any δ > 0, either µ(Uδ(c))=1or 0 < µ(Uδ(c)) < 1. We show that in both cases cδ ∈ Va. Indeed, if µ(Uδ(c)) = 1, then cδ = a and thus cδ ∈ Va. If0 <µ(Uδ(c)) < 1, set 1 c˜δ = xµ(dx) ∈ M. (8) µ(X \ Uδ(c)) X\UZδ(c)

Clearly, αcδ + (1 − α)˜cδ ∈ M, α ∈ [0, 1], by convexity. Moreover,

a = µ(Uδ(c))cδ + 1 − µ(Uδ(c)) c˜δ. (9)

Therefore, by a simple geometric argument and by the definition of Va, it is clear that cδ ∈ Va. Since X is a locally convex space and since Va is convex, the closures of Va in the weak and original topologies coincide. Consequently, by passing to the limit δ → +0, one arrives at

c = lim cδ ∈ Va. (10) δ→+0 This concludes the proof of the claim. (b) Next, we prove that (ii) ⇒ (i) by constructing µ ∈ P(X) with support M and barycenter a. Being a metric compact, M is separable, hence there exist M0 such that M0 = M = Va. ∞ ∞ And without loss of generality, one can think that M0 = {xn}n=1 ⊂ Va and {xn}n=1 = M. By ∞ the definition of Va, there exist {αn}n=1 such that αn > 0 and −αnxn +(1+ αn)a ∈ M.

2 Let us define the discrete measure

∞ 1 αnδxn + δ−αnxn+(1+αn)a µ = n · , (11) 2 1+ αn Xk=1

where δx is the delta-measure at x. Clearly, this is a Radon probability measure, and a simple computation shows that its barycenter is a. Indeed, for every Λ ∈ X∗ one has

∞ 1 αnΛxn +(−αnΛxn +(1+ αn)Λa) Λxµ(dx)= n · =Λa. (12) 2 1+ αn XZ Xk=1 ∞ It remains to prove that supp µ = M. First, we note that {xn}n=1 ⊂ supp µ. Conse- ∞ quently, M = {xn}n=1 ⊂ supp µ, and therefore M ⊂ supp µ. By the definition (11), one also has supp µ ⊂ M, which concludes the proof.

Further, we will use the following standard notation from convex analysis. For a, b ∈ X we define the (line) segment [a, b] and the open (line) segment (a, b) by

[a, b]= {x ∈ X|x = (1 − λ)a + λb, λ ∈ [0, 1]}, (13) (a, b)= {x ∈ X|x = (1 − λ)a + λb, λ ∈ (0, 1)}.

Let us recall that the relative of a set M is

relint(M)= {x ∈ M| ∃U(x): U(x) ∩ aff(M) ⊂ M}, (14) where U(x) is an open neighborhood of x and aff(M) is the affine hull of M. Also, we recall that the relative algebraic interior of M is

core(M)= {x ∈ M| ∀y ∈ aff(M) ∃α> 0 : [x, −αy +(1+ α)x] ⊂ M}. (15)

It is well-known that any locally compact topological is finite-dimensional (e.g., see [2]), in which case the following corollary takes place.

Corollary 1.1. If X is a locally compact space and M ⊂ X is a non-empty closed , then the set of barycenters of the Borel probability measures with support Mcoincides with the relative interior of M.

Proof. We note that in finite-dimensional spaces any probability Borel measure is Radon. It is also well-known (see [3]) that in such spaces the relative interior and the relative algebraic interior of M coincide and are non-empty. Now, let a ∈ relint(M) = core(M) be any point. By the definition of the relative algebraic interior, for every y ∈ M ⊂ aff(M) the segment [y, a] can be prolonged beyond the point a within M. This means that y ∈ Va, and thus M ⊂ Va. Hence, by Theorem 1 (see also Remark 2) there exist a Radon probability measure on X with supp µ = M and barycenter a. It is left to prove that if for some a ∈ M one has Va = M, then a ∈ relint(M). Notice that Va is a non-empty convex set. Since we deal with the finite-dimensional space, Va has a non-empty

3 relative interior; and relint(Va) = relint(Va) = relint(M). Let x belong to relint(Va) ⊂ Va. It follows from the definition of Va that there exist a segment [x, y] ⊂ M such that a ∈ (x, y). Since x also belongs to relint(M), there exist an open neighborhood U(x) of x, satisfying U(x) ∩ aff(M) ⊂ M. By convexity of M, one obtains

(1 − λ)(U(x) ∩ aff(M)) + λy ⊂ M, λ ∈ [0, 1]. (16)

It is also easy to verify directly that

(1 − λ)(U(x) ∩ aff(M)) + λy = (1 − λ)U(x)+ λy ∩ aff(M), λ ∈ [0, 1]. (17)

Combining (16) and (17), and noticing that for λ ∈ [0, 1) the set (1 − λ)U(x)+ λy is an open neighborhood of the point (1 − λ)x + λy, one sees that any point of (x, y) belongs to relint(M) by the very definition (14). In particular, this means that a ∈ relint(M). 2. It is tempting to think that Corollary 1.1 takes place in infinite-dimensional spaces as well. Unfortunately, this is not the case even for Hilbert spaces, as the following counterexample shows. Let X be the of real sequences endowed with the l2-scalar product, and let M be the Hilbert cube, a compact convex set,

∞ 1 1 M = − , . (18) k k kY=1   ∞ 1 P R We take a = {ak}k=1 ∈ M, where ak = k+1 . It is easy to construct a measure µk ∈ ( ) with supp µk = [−1/k, 1/k] such that 1 xµ (dx)= . (19) k k +1 [−1/k,Z 1/k]

Having done that, consider the product of measures restricted to X

∞

µ = µk . (20)

k=1 X O

One usually defines the product of measures on the product of spaces, having in mind the product topology. Even though the corresponding induced topology on X is strictly coarser than the l2- topology, they both generate the same Borel sigma-algebra on X. Thus, it is clear that µ can be regarded as a Borel measure on the Hilbert space X; moreover, since X is a complete and separable metric space, µ is Radon. It is clear by construction that supp µ ⊂ M. We prove that the other inclusion by reductio ad absurdum. Let b ∈ M and suppose that µ(Uε(b)) = 0 for some ε > 0, where Uε(b) is the ball of radius ε centered at b. One can choose N such that 4 ε2 < . (21) n2 2 n>N X

4 Then ∞ N ε2 0= µ x ∈ X (x − b )2 <ε2 ≥ µ x ∈ M (x − b )2 < n n n n 2 ( n=1 ) ( n=1 ) X X (22) N N 2 2 ε = µk x ∈ M (xn − bn) < . ( 2 ) k=1 n=1 O X The latter is clearly positive, which gives a contradiction and yields supp µ = M. Now, we prove that a is the barycenter of µ. Thanks to the Riesz representation theorem, there ∞ ∞ exist {λk}k=1 ∈ X such that for every x = {xk}k=1 ∈ X one has ∞

Λx = λkxk. (23) Xk=1 By the definition of the barycenter we write ∞ ∞ ∞

Λxµ(dx)= λkxk µ(dx)= λk xk µ(dx)= λkak =Λa, (24) XZ MZ Xk=1 Xk=1 MZ Xk=1 where one can interchanged the sum and the integral by dominated convergence since M is a bounded set in X. This shows that a is indeed the barycenter of µ. Next, we recall that in infinite-dimensional spaces the relative interior and relative algebraic interior do not necessarily coincide (see [3]). However, from (14) and (15) one sees that the former is a subset of the latter. Thus, it is sufficient to show that a does not belong to the relative algebraic interior of M. We prove this claim, again, by contradiction. Suppose that a ∈ core(M). Then the segment [0, a] can be prolonged beyond the point a within M. In other words, there exist α> 0 such that (1 + α)a ∈ M. The latter is equivalent to 1 1 − ≤ (1 + α)a ≤ , k =1, 2,.... (25) k k k Multiplying by k + 1 and letting k →∞ yield − 1 ≤ (1 + α) ≤ 1, (26) which contradicts α> 0 and concludes the proof. 3. Now, we describe the set of barycenters of measures on the space of probability measures. Let K be a metric compact space and X = M(K) the space of signed finite Radon measures on K. By the Riesz–Markov theorem, X can be identified with the topological dual C∗(K) of the space C(K) of continuous functions on K. We endow C∗(K) with the weak-* topology σ(C∗(K),C(K)). Having in mind the canonical embedding C(K) ֒→ C∗∗(K), one can say that this topology is the weakest topology which makes continuous all the functionals from C∗∗(K) that correspond to elements of C(K). This topology is locally convex as is the corresponding topology τw on X. The restriction of τw to the convex set M = P(K) ⊂ X of probability measures on K produces the usual topology of weak convergence on M and thus makes this set compact. The barycenter µ ∈ X of a measure η ∈ P(X) is, by definition,

µ = ν η(dν) (27)

XZ

5 if the latter integral exists in the weak sense. That is, since (C∗(K))′ = C(K), where (·)′ is the topological dual in the weak-* topology, µ is the barycenter of η if and only if for every f ∈ C(K),

f(x)µ(dx)= f(x)ν(dx) η(dν). (28)   KZ XZ KZ   Also, note that µ = ν η(dν), (29)

suppZ η and, by the Hahn–Banach separation theorem, one has µ ∈ co(supp η). The following result characterizes measures from X with support M.

Theorem 2. The set of barycenters of the measures from P(X) with support M coincides with the set of the measures from M with support K.

Proof. (a) First, we prove that the barycenter of a measure from P(X) with support M is a measure from M with support K. Take any η ∈ P(X) such that supp η = M, and let µ ∈ M be its barycenter. We prove that supp µ = K by contradiction. Indeed, suppose this is not the case. Then, there exist a non-zero nonnegative continuous bounded function f ∈ Cb(K) such that

f(x)µ(dx)=0. (30)

KZ

Using (28) one gets f(x)ν(dx)η(dν)=0, (31)

MZ KZ and since the integrand is non-negative,

f(x)ν(dx)=0 (32)

KZ η-almost surely on M. The latter, in fact, holds for all ν ∈ M = P(K) due to continuity in ν of the left-hand side of (32) with respect to the topology of weak convergence. Consequently, by choosing ν to be the delta measure at an arbitrary point in K, one immediately obtains f(x)=0, x ∈ K. (33) This contradicts the fact that f is non-zero and concludes the proof of the claim.

6 (b) Now, assume that µ ∈ M and supp µ = K. Let

∞ ∞ A = (a1, a2,...) ∈ [0, 1] aj ≥ 0, aj =1 (34) ( ) j=1 X ∞ be a closed subset of [0, 1] endowed with the l1 -norm. Since A is separable, there exist a Radon probability measure λ on [0, 1]∞ with support A (e.g., see the proof of Theorem 1). ∞ ∞ ∞ Let us also introduce the Radon probability measure λ ⊗ µ = λ ⊗ j=1 µj on A × K = ∞ A × Kj, where the µj are copies of µ and the Kj are copies of K. It is easy to see that j=1 N Q supp λ ⊗ µ∞ = A × K∞. (35)

∞ Indeed, for any open neighborhood U(c) of c =(ca; c1,...) ∈ A × K , by the definition of the product topology, there exist an open set of the form

∞

Ua(ca) × Uj(cj), (36) j=1 Y

where Ua(ca) ⊂ A and Uj(cj) ⊂ Kj are open neighborhoods of ca and cj, respectively, such that Uj(cj) =6 Kj only for finitely many j ∈ N. Then, for large enough N one has

N ∞ (λ ⊗ µ )(U(c)) ≥ λ(Ua(ca)) µ(Uj(cj)) > 0, (37) j=1 Y which proves (35). The next step is to define the map F : A × K∞ → M by

∞

F (a, x)= ajδxj . (38) j=1 X (n) ∗ (n) ∗ It is easy to show that F is continuous. Indeed, let a → a ∈ A in l1-norm and x → x ∈ K∞ in the product topology. We will prove that F (a(n), x(n)) converges to F (a∗, x∗) weakly. For every f ∈ C(K),

∞ f(y) F (a(n), x(n))(dy) − f(y)F (a∗, x∗)(dy) ≤ |a(n)f(x(n)) − a∗f(x∗)| j j j j Z Z j=1 K K X (39) ∞ (n) ∗ ∗ (n) ∗ ≤ sup |f(x)|ka − a kl1 + aj |f(xj ) − f(xj )|→ 0, x∈K j=1 X where the latter term tends to zero thanks to the dominated convergence theorem. This proves the continuity of F . Now, let us define the measure η to be the pushforward of λ ⊗ µ∞ under F :

η =(λ ⊗ µ∞) ◦ F −1, (40)

7 which is readily verified to be a Radon probability measure. We prove that this measure is supported on M. Indeed, since it is known (e.g., see [1, Ex. 8.1.6]) that F (A × K∞)= M, (41) for every open neighborhood U(ν) of ν ∈ M there exist (a, x) ∈ A × K∞ such that F (a, x) ∈ U(ν). Consequently, due to F being continuous and due to (35), one has η(U(ν)) > 0, and thus, since ν is arbitrary, supp η = M. It is left to check that the barycenter of η is µ. One can write

f(y)ν(dy)η(dν)= f(y) F (a, x)(dy)(λ ⊗ µ∞)(da, dx)

MZ KZ A×ZK∞ KZ ∞ ∞ (42) ∞ = ajλ(da) f(xj)µ (dx)= ajλ(da) f(x)µ(dx)= f(x)µ(dx), j=1 ∞ j=1 X ZA KZ X ZA KZ KZ where we use the definition (40) of η, Fubini’s theorem, and the dominated convergence to interchange the sum and the integrals. According to (28), the formula (42) means exactly that the barycenter of η is µ. This concludes the proof of the theorem.

As the final remark we point out that our proof relies heavily on the fact that K is compact; even though, barycenters are well-defined for a wider class of Radon probability measures (with finite first moments). An open question of interest is to characterize such measures as well.

Acknowledgments. We would like to thank V. Bogachev for bringing the problem considered in this note to our attention and A. Borichev for helpful discussions and valuable comments. Also, we greatly appreciate the detailed responses of the reviewers. Their remarks and comments, without a doubt, helped us improve our paper. S.B. is supported by the European Research Council (ERC) under the European Unions Horizon 2020 research and innovation programme, grant 647133 (ICHAOS). A.M. is supported by the RFBR grants 14-01-90406, 14-01-00237 and the SFB 701 at Bielefeld University.

References

[1] V.I. Bogachev, Measure theory, Vol. 2, Springer, Berlin, 2007.

[2] W. Rudin, , McGraw-Hill, New York, 1991.

[3] C. Z˘alinescu, Convex analysis in general vector spaces, World Sci., River Edge, NJ, 2002.

8 Sergey Berezin Aix-Marseille Universit´e, Centrale Marseille, CNRS, Institut de Math´ematiques de Marseille, UMR7373, 39 Rue F. Joliot Curie 13453, Marseille, France; St. Petersburg Department of V.A. Steklov Mathematical Institute of RAS, 27 Fontanka 191023, St. Petersburg, Russia E-mail address: [email protected], [email protected] Azat Miftakhov Moscow State University, Faculty of Mechanics and Mathematics, 1 Leninskiye Gory 119991, Moscow, Russia E-mail address: [email protected]

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