Any law of group metric invariant is an inf-convolution.

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Any law of group metric invariant is an inf-convolution.

Mohammed Bachir

To cite this version:

Mohammed Ba hir

June 30, 2015

Laboratoire SAMM4543, UniversitéParis1Panthéon-Sorbonne, CentreP.M.F. 90rue Tolbia 75634Paris edex13

Email : Mohammed.Ba hiruniv-paris1.fr

Abstra t. Inthisarti le,webringanewlightonthe on eptoftheinf- onvolutionoperation

⊕

andprovidesadditionalinformationstotheworkstartedin[1℄and[2℄. Itisshownthat any internallawofgroupmetri invariant(evenquasigroup) anbe onsideredasaninf- onvolution. Consequently,theoperationoftheinf- onvolutionoffun tionsonagroupmetri invariantisin realityanextensionoftheinternallawof

X

tospa esoffun tionson

X

. Wegiveanexampleof monoid

(S(X), ⊕)

fortheinf- onvolutionstru ture,(whi hisdenseinthesetofall

1

-Lips hitz bounded from bellow fun tions) forwhi h, themap

arg min : (S(X), ⊕) → (X, .)

is a(single valued) monoid morphism. It is also proved that, given a group omplete metri invariant

(X, d)

, the ompletemetri spa e

(K(X), d

∞

)

ofallKatetovmapsfrom

X

to

R

equiped with theinf- onvolutionhasanaturalmonoidstru turewhi hprovidesthefollowingfa t: thegroup ofallisometri automorphisms

Aut

Iso

(K(X))

ofthemonoid

K(X)

,isisomorphi to thegroup of allisometri automorphisms

Aut

Iso

(X)

of thegroup

X

. Ontheother hand,weprovethat thesubset

K

C

(X)

of

K(X)

of onvexfun tions onaBana hspa e

X

, an beendowedwitha onvex onestru turein whi h

X

embedsisometri allyasBana hspa e.

Keyword,phrase: Inf- onvolution;groupandmonoidstru ture;Katetovfun tions. 2010Mathemati sSubje t: 46T99;26E99;20M32.

Contents

1 Introdu tion. 2

2 The inf- onvolutionon ompletemetri spa e. 4 3 The monoidsstru turefor the inf- onvolution. 8 4 Metri properties and the densityof

S(X)

in

Lip

1

_(X)

. 10

5 The map

arg min(.)

asmonoidmorphism. 11

6 Examplesof inf- onvolutionmonoidin thedis rete ase. 13 6.1 Theinf- onvolutionmonoid

(l

∞

dis

(Z), ⊕)

. . . 13 6.2 Theinf- onvolutionmonoid

(l

∞

dis

(Z/pZ), ⊕)

. . . 13

7 The set ofKatetov fun tions. 13 7.1 Themonoidstru tureof

K(X)

. . . 13 7.2 The onvex onestru tureof

K

C

(X)

. . . 16

This arti lebrings someadditionalinformations to thestudy of theinf- onvolutionstru ture developedin[1℄and[2℄. Givenaset

X

,amap

α : X × X → X

andtworealvaluedfun tions

f

and

g

dened on

X

. The inf- onvolutionof

f

and

g

with respe t themap

α

is dened as follows

f ⊕

|{z}

α

g(x) :=

inf

y,z∈X/α(y,z)=x

{f (y) + g(z)} ; ∀x ∈ X.

(1)

Histori ally, theinf- onvolutionappearedas atooloffun tional analysisand optimization andstartswiththeworksofMa Shane[10℄,Fen hel,MoreauandRo kafellar;see[9℄for refer-en es,seealsothebookofJ.-B.Hiriart-UrrutyandC.Lemare hal[5℄. Weprovedin[1℄and[2℄, thattheinf- onvolutionalsoenjoysaremarkablealgebrai properties. Forexample,weproved that the set

(Lip

1

+

(X), ⊕)

of all nonegativeand

1

-Lips hitzfun tions dened ona omplete metri invariantgroup

(X, d)

, isamonoidand itsgroupof unitis isometri allyisomorphi to

X

. This resultmeans that themonoid stru ture of

(Lip

1

+

(X), ⊕)

ompletely determines the groupstru tureof

X

whenever

X

isangroupmetri invariant.

Inthis paper, wegiveadditional lightingto the understanding of the inf- onvolution op-eration. Indeed, itseems that theinf- onvolutionis notan external operationto the spa e

X

a ting on it, but is in reality a anoni alextension of the internal lawof

X

to the spa e

Lip

1

+

(X)

, whenever

X

isagroup metri invariant. In oderwords,anyinternal law ofmetri invariantgroup(evenquasigroup)isaninf- onvolution. Thisapproa hismotivatedby Propo-sition1and Theorem1below.

Ametri spa e

(X, ., d)

equippedwith aninternallaw

. : (y, z) 7→ y.z

denedfrom

X × X

into

X

issaidtobemetri invariant,if

d(x.y, x.z) = d(y.x, z.x) = d(y, z) ∀x, y, z ∈ X.

Note that everygroupis metri invariantforthe dis reetmetri . For examplesof nottrivial group metri invariant, see [2℄ (For informationson group omplete metri invariantsee [7℄). Let us denote by

γ : x ∈ X 7→ δ

x

theKuratowskioperator, where

δ

x

: t ∈ X 7→ d(x, t).

We denoteby

X

ˆ

theimageof

X

undertheKuratowskioperator,

X := γ(X)

ˆ

. Theset

X

ˆ

isendowed withthesup-metri

d

∞

(γ(a), γ(b)) := sup

x∈X

|γ(a)(x) − γ(b)(x)|.

ItiswellknownandeasytoseethattheKuratowskioperator

γ

isanisometry: forall

a, b ∈ X

d

∞

(γ(a), γ(b)) = d(a, b).

Wedenetheinf- onvolutionon

X

ˆ

as intheformula(1). Fortwoelement

γ(a), γ(b) ∈ ˆ

X

,

(γ(a) ⊕ γ(b)) (x) :=

inf

y,z∈X/y.z=x

{γ(a)(y) + γ(b)(z)} .

We obtain the following result whi h say that

(X, .)

and

( ˆ

X, ⊕)

has in generalthe same algebrai stru ture. Re all that aquasigroupis anonempty magma

(X, .)

su h that forea h pair

(a, b)

theequation

a.x = b

hasauniquesolutionon

x

andtheequation

y.a = b

hasaunique solutionon

y

. Aloopis anquasigroupwithanidentityelementand agroupisanasso iative loop.

Proposition1 Let

(X, ., d)

be a metri invariant spa e. Then, the following assertions are equivalent.

(2) ( ˆ

X, ⊕)

is aquasigroup (respe tively,loop,group, ommutativegroup).

Inthis ase,theKuratowskioperator

γ : (X, ., d) → ( ˆ

X, ⊕, d

∞

)

isanisometri isomorphism of quasigroups (respe tively,loops,groups, ommutativegroups).

We then ask whether the operation

⊕

of

( ˆ

X, ⊕)

naturally extends to the whole spa e

(Lip

1

+

(X), ⊕)

. An answer is given by the following result. The part

(1) ⇒ (2)

was estab-lished in [1℄ forBana hspa esin onvexsetting andin [2℄in thegroupframework(as well as thedes riptionofthegroupofunitof

(Lip

1

+

(X), ⊕)

).

Theorem 1 Let

(X, ., d)

bea ompletemetri invariantquasigroup. Then thefollowing asser-tionsare equivalent.

(1) (X, .)

isa( ommutative) group.

(2) (Lip

1

+

(X), ⊕)

isa( ommutative) monoid.

Inthis ase,the identityelementof

(Lip

1

+

(X), ⊕)

is

γ(e)

where

e

isthe identityelement of

X

anditsgroupof unitis

X

ˆ

whi h isisometri ally isomorphi to

X

.

TheProposition1andTheorem 1are inouropinion theargumentsshowingthat themonoid stru ture of

(Lip

1

+

(X), ⊕)

is in realitya naturalextension of thegroup stru ture of

(X, .)

to theset

Lip

1

+

(X)

.

WeusethefollowingresultintheproofofTheorem1. Thisresultisthekeyofthisalgebrai theoryoftheinf- onvolution. Thepart

I) ⇒ II)

wasprovedin[2℄. Thepart

II) ⇒ I)

isnew. A moregeneralform inmetri spa eframeworknotne essarilygroupis givenin se tion2. Theorem 2 Let

(X, ., d)

beagroup ompletemetri invariant andlet

a ∈ X

. Let

f

and

g

be two lower semi ontinuousfun tionson

(X, d)

. Then,the following assertionsareequivalent

I)

themap

x 7→ f ⊕ g(x)

has astrongminimum at

a

II)

thereexists

(˜

y, ˜

z) ∈ X × X

su hthat

y ˜

˜

z = a

and:

f

has astrongminimumat

y

˜

and

g

has atstrongminimuma

z

˜

.

Theorem2alsogivesthefollowing orollary. Considerthefollowingsubmonoidof

Lip

1

_(X)

S(X) :=



f ∈ Lip

1

(X)/ f

hasastrongminimum

andthemetri

ρ

dened for

f, g ∈ Lip

1

_(X)

by

ρ(f, g) = sup

x∈X

|f (x) − g(x)|

1 + |f (x) − g(x)|

Forareal-valuedfun tion

f

withdomain

X

,

arg min(f )

isthesetofelementsin

X

thatrealize theglobalminimumin

X

,

arg min(f ) = {x ∈ X : f (x) = inf

y∈X

f (y)}.

For the lassof fun tions

f ∈ S(X)

,

arg min(f ) = {x

f

}

isasingleton, where

x

f

is thestrong minimumof

f

. Weidentifythesingleton

{x}

withtheelement

x

.

Corollary 1 Let

(X, ., d)

be a group omplete metri invariant having

e

as identity element. Then,

(S(X), ρ)

isadense subsetof

Lip

1

_(X)

andforall

f, g ∈ S(X)

wehave

Inother words, themap

arg min : (S(X), ⊕, ρ) → (X, ., d)

is ontinuousmonoid morphismand onto. Wehavethefollowing ommutativediagram,where

I

denotestheidentitymap on

X

and

γ

the Kuratowskioperator

(X, .)

γ

//

I

%%

❑

❑

❑

❑

❑

❑

❑

❑

❑

(S(X), ⊕)

arg min



(X, .)

Wearealsointerestedonthemonoidstru tureoftheset

K(X)

ofKatetovfun tions.There are lotofliteratureonthemetri andthetopologi al stru tureofthis spa e(See forinstan e [3℄,[6℄and[8℄). Wegiveinthisse tionsomeresultsaboutthemonoidstru tureof

K(X)

when

X

isagroup,andthe onvex onestru tureofthesubset

K

C

(X)

of

K(X)

(of onvexfun tions) when

X

isaBana hspa e. Let

(X, d)

beametri spa e; wesaythat

f : X → R

isaKatetov mapif

|f (x) − f (y)| ≤ d(x, y) ≤ f (x) + f (y); ∀x, y ∈ X.

(2)

These maps orrespondto one-pointmetri extensions of

X

. Wedenote by

K(X)

theset of allKatetovmapson

X

;weendowitwiththesup-metri

d

∞

(f, g) := sup

x∈X

|f (x) − g(x)| < +∞

whi hturns itinto a ompletemetri spa e. Re allthat

X

isometri allyembedsin

K(X)

via the Kuratowski embedding

γ : x → δ

x

, where

δ

x

(y) := d(x, y)

, and that one has, for any

f ∈ K(X)

, that

d

∞

(f, γ(x)) = f (x)

. It is shown in Se tion 7 that

(K(X), ⊕)

has a monoid stru ture and

K

C

(X)

hasa onvex one stru ture. Weobtainthe followinganalogous to the Bana h-Stonetheoremwhi hsaythatthemetri monoid

(K(X), ⊕, d

∞

)

, ompletelydetermine the omplete metri invariant group

(X, d)

. Note that in the following result, any monoid isometri isomorphism has the anoni al form. We do not know if this is the ase for other monoidsas theset ofall onvex

1

-Lips hitzboundedfrom bellowfun tionsdened onBana h spa e(See Problem2. in[1℄).

Theorem 3 Let

(X, d)

and

(Y, d

′

₎

be two omplete metri invariant groups. Then, a map

Φ : (K(X), ⊕, d

∞

) → (K(Y ), ⊕, d

∞

)

is a monoid isometri isomorphism if, and only if there exists a group isometri isomorphism

T : (X, d) → (Y, d

′

₎

su h that

Φ(f ) = f ◦ T

−1

for all

f ∈ K(X)

. Consequently,

Aut

Iso

(K(X))

(the group of all isometri automorphism of the monoid

K(X)

)isisomorphi asgroup to

Aut

Iso

(X)

(thegroup ofall isometri automorphism of the group

X

).

2 The inf- onvolution on omplete metri spa e.

The main theorem of this se tion (Theorem 4) extend[Theorem 3, [2℄℄ and [Corollary3, [2℄℄ to omplete metri invariant spa e. In [Theorem 3, [2℄℄ and [Corollary 3, [2℄℄, only the part

I) ⇒ II)

wasprovedinthegroup ontext. Here,wegiveane essarilyand su ient ondition in themoregeneralmetri ontext.

Weneed somenotationsand denitions. Let

X

beaset and

α : X × X → X

beamap. Given

x ∈ X

wedenoteby

∆

α

(x)

thefollowingsetdependingon

α

∆

α

(x) := {(y, z) ∈ X × X : α(y, z) = x} ⊂ X × X.

Note that

∆

α

(α(s, t)) 6= ∅

for all

s, t ∈ X

. Wealsodenote by

∆

1,α

(x)

(respe tively,

∆

2,α

(x)

) theproje tionof

∆

α

(x)

ontherst(respe tively,these ond) oordinate:

∆

2,α

(x) := {z ∈ X/∃y

z

∈ X : α(y

z

, z) = x} ⊂ X.

Denition1 Let

(X, d)

be metri spa e and

α : X × X → X

, be a map. We say that

α

is

d

-invariantat

x ∈ X

,if

∆

α

(x) 6= ∅

andthereexists

L

1

, L

2

, L

′

1

, L

′

2

> 0

su hthat

L

2

d(y

1

, y

2

) ≤ d(α(y

1

, z), α(y

2

, z)) ≤ L

1

d(y

1

, y

2

); ∀y

1

, y

2

∈ ∆

1,α

(x); z ∈ ∆

2,α

(x).

and

L

′

2

d(z

1

, z

2

) ≤ d(α(y, z

1

), α(y, z

2

)) ≤ L

1

′

d(z

1

, z

2

); ∀z

1

, z

2

∈ ∆

2,α

(x); y ∈ ∆

1,α

(x).

Theset

∆

α

(x)

is endowedwiththemetri indu edbytheprodu tmetri topologyof

X × X

i.e

d ((y, z), (y

˜

′

_{, z}

′

_{)) := d(y, y}

′

_{) + d(z, z}

′

₎

forall

(y, z), (y

′

_{, z}

′

_{) ∈ X × X}

.

Proposition2 Let

(X, d)

be metri spa e and

α : X × X → X

bea map. Suppose that

α

is

d

-invariantat

x ∈ X

. Then the restri tion of

α

tothe set

∆

α

(x)

is ontinuous.

Proof. Let

(y, z), (y

0

, z

0

) ∈ ∆

α

(x)

. Then,

d(α(y, z), α(y

0

, z

0

)) ≤

d(α(y, z), α(y, z

0

)) + d(α(y, z

0

), α(y

0

, z

0

))

≤ L

′

1

d(z, z

0

) + L

1

d(y, y

0

)

≤ max(L

′

1

, L

1

) (d(z, z

0

) + d(y, y

0

))

Thisinequalityshowsthattherestri tionof

α

totheset

∆

α

(x)

is ontinuous.

Fortwofun tions

f

and

g

on

X

,wedenethemap

η

f,g

dependingon

f

and

g

by

η

f,g

: X × X

→ R ∪ {+∞}

(y, z) 7→ f (y) + g(z)

Notethattheinf onvolutionof

f

and

g

at

x ∈ X

,withrespe ttothelaw

α

, oin idewith theinnimumof

η

f,g

on

∆

α

(x)

f ⊕

|{z}

α

g(x) :=

inf

y,z∈X/α(y,z)=x

{f (y) + g(z)} :=

(y,z)∈∆

inf

α

(x)

η

f,g

(y, z).

Exemples 1 TheDenition 1issatisedin the following ases.

1)

Let

(X, k.k)

be ave tornormedspa eand

α : X × X → X

bethe mapdenedby

α(y, z) :=

y+z

. Inthis ase,theinf- onvolution orrespondtothe lassi aldenitionoftheinf- onvolution on ve torspa e andwehave

∆

1,α

(x) = ∆

2,α

(x) = X

for all

x ∈ X

and

α

satises

kα(y, x) − α(z, x)k = kα(x, y) − α(x, z)k = ky − zk; ∀x, y, z ∈ X.

2)

Let

(C, k.k)

bea onvexsubsetof ave tor normedspa e

(X, k.k)

andlet

λ ∈]0, 1[

be axed real number. Let

α : C × C → C

be the map dened by

α(y, z) := λy + (1 − λ)z

. Then,

{(x, x)} ⊂ ∆

α

(x)

and

∆

α

(x) = {(x, x)}

if, andonlyif

x

isan extremepointof

C

andwehave

kα(y, x) − α(z, x)k = λky − zk; ∀x, y, z ∈ C.

and

kα(x, y) − α(x, z)k = (1 − λ)ky − zk; ∀x, y, z ∈ C.

3)

If

(X, ., d)

isametri group,(

.

isthelawofinternal ompositionof

X

)and

α : (y, z) 7→ y.z

, then

∆

1,α

(x) = ∆

2,α

(x) = X

forall

x ∈ X

. Moreover,

α

is

d

-invariantat

x

forea h

x ∈ X

if andonly if,

(X, ., d)

ismetri invariant. Were allthatametri groupis saidtobemetri in-variant,if

d(y.x, z.x) = d(x.y, x.z) = d(y, z)

forall

x, y, z ∈ X

. Everygroupismetri invariant forthedis reetmetri . We anndexamplesof groupmetri invariantin [2℄.

4)

However, there exists examplesof metri monoids

(M, d)

with alaw

.

whi h is notmetri invariantbutsu hthat

.

is

d

-invariantatea helementofthegroupofunitof

M

(SeeProposition 5andRemark1).

Denition2 Let

(X, d)

beametri spa e, wesaythat afun tion

f

has astrongminimumat

x

0

∈ X

, if

inf

X

f = f (x

0

)

andfor all

ǫ > 0

,there exists

δ > 0

su hthat

0 ≤ f (x) − f (x

0

) ≤ δ ⇒ d(x, x

0

) ≤ ǫ.

A strongminimum isin parti ularunique. By

dom(f )

wedenote the domain of

f

, dened by

dom(f ) := {x ∈ X : f (x) < +∞}

. Allfun tionsinthearti learesupposedsu hthat

dom(f ) 6=

∅

.

Inwhatfollows,anelement

α(y, z) ∈ X

willsimplybenotedby

yz

andtheinf- onvolution oftwofun tions

f

and

g

willsimplybedenotedby

f ⊕ g(x) := inf

yz=x

{f (y) + g(z)} .

For

a ∈ X

,wesaythat

f ⊕ g(a)

is stronglyattainedat

(y

0

, z

0

)

, iftherestri tionof

η

f,g

tothe set

∆

α

(a)

hasastrongminimumat

(y

0

, z

0

) ∈ ∆

α

(a)

.

Theorem 4 Let

(X, d)

bea ompletemetri spa es. Let

α : X × X → X

, beamap(

α(y, z) :=

yz

for all

y, z ∈ X × X

). Let

f

and

g

be two lower semi ontinuous fun tionson

(X, d)

. Let

a ∈ X

and suppose that the map

α

is

d

-invariant at

a

. Then, the following assertions are equivalent.

I)

the map

x 7→ f ⊕ g(x)

has astrong minimumat

a ∈ X

II)

thereexists

(˜

y, ˜

z) ∈ ∆

α

(a)

i.e

y ˜

˜

z = a

, su hthat :

f

hasastrongminimum at

y

˜

and

g

has atstrongminimum a

z

˜

.

Moreover, inthis ase,wehave

(1)

the restri tedmap

η

f,g

: ∆

α

(a) → R ∪ {+∞}

has astrong minimumat

(˜

y, ˜

z) ∈ ∆

α

(a)

i.e

f ⊕ g(a)

isstronglyattainedat

(˜

y, ˜

z)

.

(2) f (x) − f (˜

y) ≥ f ⊕ g(x˜

z) − f ⊕ g(a)

and

g(x) − g(˜

z) ≥ f ⊕ g(˜

yx) − f ⊕ g(a)

for all

x ∈ X

.

Proof. First,fromthedenitionoftheinf- onvolution,forall

y, y

′

_{, z, z}

′

_{∈ X}

,

f ⊕ g(yz

′

) ≤

f (y) + g(z

′

)

(3)

f ⊕ g(y

′

z) ≤

f (y

′

) + g(z).

(4)

Byaddingbothinequalilies(3)and(4)aboveweobtain

f ⊕ g(yz

′

) + f ⊕ g(y

′

z) ≤

(f (y) + g(z)) + (f (y

′

) + g(z

′

)) .

(5)

I) ⇒ II)

. Repla ing

f

by

f −

1

2

f ⊕ g(a)

and

g

by

g −

1

2

f ⊕ g(a)

, we an assumewithoutloss ofgeneralitythat

f ⊕ g

hasastrongminimumat

a

and

f ⊕ g(a) = 0

. Let

(y

n

)

n

; (z

n

)

n

⊂ X

be su hthatforall

n ∈ N

∗

,

y

n

z

n

= a

and

0 = f ⊕ g(a) ≤

f (y

n

) + g(z

n

) < f ⊕ g(a) +

1

n

.

in otherwords

0 ≤ f (y

n

) + g(z

n

)

<

1

n

.

(6) Byappllaying(5)with

y = y

n

;

z = z

n

;

y

′

_{= y}

p

and

z

′

_{= z}

p

wehave

f ⊕ g(y

n

z

p

) + f ⊕ g(y

p

z

n

)

≤ (f (y

n

) + g(z

n

)) + (f (y

p

) + g(z

p

))

0 = 2(f ⊕ g(a)) ≤ f ⊕ g(y

n

z

p

) + f ⊕ g(y

p

z

n

) ≤

1

n

+

1

p

.

(7)

Sin e

x 7→ f ⊕ g

hasastrongminimumat

a

,then

d(x, a) → 0

whenever

f ⊕ g(x) → 0

. Onthe otherhand,

f ⊕ g(x) ≥ f ⊕ g(a) = 0

forall

x ∈ X

. Thusfrom(7),wegetthat

f ⊕ g(y

n

z

p

) → 0

and

f ⊕ g(y

p

z

n

) → 0

when

n, p → +∞

. Wededu e that

d(y

n

z

p

, a) → 0

, when

n, p → +∞

. Sin e

y

p

z

p

= a

for all

p ∈ N

and

α

is

d

-invariant at

a

, then

d(y

n

, y

p

) ≤

1

L

2

d(y

n

z

p

, y

p

z

p

) =

1

L

₂

d(y

n

z

p

, a) → 0

, when

n, p → +∞

. Hen e

(y

n

)

n

is a Cau hy sequen e and so onverges to somepoint

y ∈ X

˜

sin e

(X, d)

is omplete metri spa e. Similarly, weprovethat

(z

n

)

n

on-vergesto somepoint

z ∈ X

˜

. Bythe ontinuityof themap

α : (y, z) 7→ yz

(SeeProposition2), wededu ethat

y ˜

˜

z = lim

n

(y

n

z

n

) = lim

n

(a) = a

.

Usingthelowersemi- ontinuityof

f

and

g

andtheformulas(6)weget

f (˜

y) + g(˜

z)

≤ lim inf

n→+∞

f (y

n

) + lim inf

n→+∞

g(z

n

)

≤ lim inf

n→+∞

(f (y

n

) + g(z

n

)) ≤ 0 = f ⊕ g(a).

Ontheotherhand,itisalwaystruethat

f ⊕ g(a) ≤ f (˜

y) + g(˜

z)

sin e

y ˜

˜

z = a

. Thus

f (˜

y) + g(˜

z) = f ⊕ g(a) = 0.

(8)

Using(8)weobtain

f ⊕ g(˜

yx)

≤ f (˜

y) + g(x) = g(x) − g(˜

z).

(9)

and

f ⊕ g(x˜

z) ≤

f (x) + g(˜

z) = f (x) − f (˜

y).

(10)

Using (10)and thefa t that

f ⊕ g

hasastrongminimumat

a

,wehavethat

f (x) − f (˜

y) ≥ 0

andif

f (y

n

) − f (˜

y) → 0

,then

f ⊕ g(y

n

z) → 0

˜

whi himpliesthat

d(y

n

z, a) → 0

˜

sin e

f ⊕ g

has astrongminimumat

a

. On theotherhandwehave

d(y

n

, ˜

y) ≤

1

L

2

d(y

n

˜

z, ˜

y ˜

z) =

1

L

2

d(y

n

z, a)

˜

by the

d

-invarian eof

α

at

a

. Thus

d(y

n

, ˜

y) → 0

and so

f

hasastrongminimumat

˜

y

. Thesame argument,byusing(9),showsthat

g

hasastrongminimumat

z

˜

II) ⇒ I)

. Werstprovethat

f ⊕ g

hasaminimumat

y ˜

˜

z = a

andthat

f (˜

y) + g(˜

z) = f ⊕ g(a)

. Indeed,sin e

f (y) ≥ f (˜

y)

and

g(z) ≥ g(˜

z)

forall

y, z ∈ X

,thenweget

f ⊕ g(a) := inf

yz=a

{f (y) + g(z)} ≥ f (˜

y) + g(˜

z)

Ontheotherhand,

f ⊕ g(a) ≤ f (˜

y) + g(˜

z)

sin e

y ˜

˜

z = a

. Thus,

f ⊕ g(a) = f (˜

y) + g(˜

z)

. Onthe other hand,usingagainthe fa tthat

f (y) ≥ f (˜

y)

and

g(z) ≥ g(˜

z)

for all

y, z ∈ X

, weobtain forall

x ∈ X

,

f ⊕ g(x) := inf

yz=x

{f (y) + g(z)} ≥ f (˜

y) + g(˜

z) = f ⊕ g(a).

Itfollowsthat

f ⊕g

hasaminimumat

a = ˜

y ˜

z

. Now,let

(x

n

)

n

⊂ X

beasequen ethatminimize

f ⊕ g

. Let

ǫ

n

→ 0

+

su hthat

f ⊕ g(a) ≤ f ⊕ g(x

n

) ≤ f ⊕ g(a) + ǫ

n

(11)

From the denition of

f ⊕ g(x

n

)

, for ea h

n ∈ N

∗

, there exists sequen es

(y

n

)

n

, (z

n

)

n

⊂ X

satisfying

y

n

z

n

= x

n

and

f ⊕ g(x

n

) −

1

n

≤ f (y

n

) + g(z

n

) ≤ f ⊕ g(x

n

) +

1

n

Sin e

f (˜

y) + g(˜

z) = f ⊕ g(a)

,itfollowsthat

f ⊕ g(x

n

) − f ⊕ g(a) −

1

n

≤ (f (y

n

) − f (˜

y)) + (g(z

n

) − g(˜

z)) ≤ f ⊕ g(x

n

) − f ⊕ g(a) +

1

n

Usingtheinequality(11)wegetforall

n ∈ N

∗

−

1

n

≤ (f (y

n

) − f (˜

y)) + (g(z

n

) − g(˜

z)) ≤ ǫ

n

+

1

n

Sin e

(f (y

n

) − f (˜

y)) ≥ 0

and

(g(z

n

) − g(˜

z)) ≥ 0

,wegetthat

0 ≤ f (y

n

) − f (˜

y) ≤ (f (y

n

) − f (˜

y)) + (g(z

n

) − g(˜

z)) ≤ ǫ

n

+

1

n

and

0 ≤ g(z

n

) − g(˜

z) ≤ (f (y

n

) − f (˜

y)) + (g(z

n

) − g(˜

z)) ≤ ǫ

n

+

1

n

Sending

n

to

+∞

, we have

lim

n→+∞

f (y

n

) = f (˜

y)

and

lim

n→+∞

g(z

n

) = g(˜

z)

. Sin e

f

and

g

has respe tively a strong minimum at

y

˜

and

z

˜

, we dedu e that

lim

n→+∞

y

n

= y

and

lim

n→+∞

z

n

= z

. By the ontinuity of the map

α : (y, z) 7→ yz

on

∆

α

(a)

, we obtain

lim

n→+∞

(y

n

z

n

) = ˜

y ˜

z = a

. Sin e

y

n

z

n

= x

n

forall

n ∈ N

∗

, wehave

lim

n→+∞

x

n

= a.

Thus

f ⊕ g

hasastrongminimumat

a

Moreover,wehavetheadditionalinformations:

(1) f ⊕ g(a)

is stronglyattained at

(˜

y, ˜

z)

(We assumeas in

I)

that

f ⊕ g(a) = 0

). We know from (8)that

f ⊕ g(a)

isattainedat

(˜

y, ˜

z)

. Tosee that

η

f,g

hasin fa t astrongminimum at

(˜

y, ˜

z)

,let

((y

n

, z

n

))

n

⊂ ∆

α

(a)

beanysequen esu hthat

f (y

n

) + g(z

n

) := η

f,g

(y

n

, z

n

) → inf

yz=a

{f (y) + g(z)} = η

f,g

(˜

y, ˜

z) = 0.

(12) Byapplying(5)with

y = ˜

y

,

y

′

_{= y}

n

,

z = ˜

z

and

z

′

_{= z}

n

andtheformulas(8)and(6),weobtain

0 ≤ f ⊕ g(˜

yz

n

) + f ⊕ g(y

n

z) ≤

˜

(f (˜

y) + g(˜

z)) + (f (y

n

) + g(z

n

))

=

(f (y

n

) + g(z

n

))

(13)

Thus

f ⊕ g(˜

yz

n

) → 0

(andalso

f ⊕ g(y

n

z) → 0

˜

)from(12)and(13)andthefa tthat

f ⊕ g(x) ≥

0 = f ⊕ g(a)

,forall

x ∈ X

. Itfollowsthat

d(˜

yz

n

, a) → 0

and

d(y

n

z, a) → 0

˜

, sin e

f ⊕ g

hasa strongminimumat

a

. Hen e,

d(y

n

, ˜

y) → 0

sin e

d(y

n

, ˜

y) ≤

1

L

2

d(y

n

z, ˜

˜

y ˜

z) =

1

L

2

d(y

n

z, a)

˜

bythe

d

-invarian eof

α

at

a

,andthefa t that

y ˜

˜

z = a

. Inasimilarwaywehave

d(z

n

, ˜

z) → 0

. Thus

(˜

y, ˜

z)

isastrongminimumof

η

f,g

.

(2)

Thispartfollowsfrom(9)and(10)

.

3 The monoids stru ture for the inf- onvolution.

The following orollary will permit to des ribe the group of unit of submonoids, for the inf- onvolutionstru ture,oftheset

Lip

1

_(X)

ofall

1

-Lips hitzandboundedfrombelowfun tions. Corollary 2 Let

(X, ., d)

be a group omplete metri invariant having

e

as identity element. Let

f

and

g

betwo

1

-Lips hitz fun tionson

X

. Then,the following assertionsareequivalent.

(1) f ⊕ g = d(e, .)

.

(2)

thereexists

y ∈ X

˜

and

c ∈ R

su hthat

f (.) = d(˜

y, .) + c := γ(˜

y) + c

and

g(.) = d(˜

y

−1

, .) − c = γ(˜

y

−1

) − c.

Proof.

(1) ⇒ (2)

. Sin ethelaw

.

isinparti ular

d

-invariantat

e

andthemap

d(e, .)

hasastrong minimumat

e

, by applyingTheorem 4,thereexists

y, ˜

˜

z ∈ X

su hthat

y ˜

˜

z = e

,

f (˜

y) + g(˜

z) =

f ⊕ g(e) = 0

and

f (x) − f (˜

y) ≥ d(e, x˜

z) = d(˜

y, x)

and

g(x) − g(˜

z) ≥ d(e, ˜

yx) = d(˜

y

−1

_{, x)}

, for all

x ∈ X.

On the other hand, sin e

f

and

g

are

1

-Lips hitz, wehave

f (x) − f (˜

y) ≤ d(˜

y, x)

and

g(x) − g(˜

z) ≤ d(˜

z, x) = d(˜

y

−1

_{, x)}

, for all

x ∈ X.

Thus,

f (x) − f (˜

y) = d(˜

y, x)

and

g(x) − g(˜

y

−1

_{) = g(x) − g(˜}

_{z) = d(˜}

_y

−1

_{, x)}

, for all

x ∈ X.

Inother words,

f (.) = d(˜

y, .) + c :=

γ(˜

y) + c

and

g(.) = d(˜

y

−1

_{, .) − c = γ(˜}

_y

−1

_{) − c}

,with

c = f (˜

y) = −g(˜

z)

.

(2) ⇒ (1)

. Suppose that

(2)

hold, then

f ⊕ g(x) = γ(˜

y) ⊕ γ(˜

y

−1

₎

_(x)

for all

x ∈ X.

Sin e

(X, ., d)

is group omplete metri invariant, by using Proposition 1. we get

f ⊕ g = γ(e) :=

d(e, .).

Lemma1 Let

(X, ., d)

beametri spa eand

. : (y, z) 7→ yz

bealaw of omposition of

X

.

1)

Supposethat

d(yx, zx) ≤ d(y, z)

and

d(xy, xz) ≤ d(y, z)

, for all

x, y, z ∈ X

. Then wehave, for all

a, b ∈ X

γ(ab) ≤ γ(a) ⊕ γ(b).

2)

Suppose

d(yx, zx) = d(xy, xz) = d(y, z)

, for all

x, y, z ∈ X

. Then, we have for all

x ∈ X

andall

a, b ∈ X

(γ(a) ⊕ γ(b)) (xb) = γ(ab)(xb)

and

(γ(a) ⊕ γ(b)) (ax) = γ(ab)(ax).

If moreover

X

isquasigroup,thenwe havefor all

a, b ∈ X

γ(a) ⊕ γ(b) = γ(ab).

Proof.

1)

Let

a, b, x ∈ X

,thenwehave

γ(a) ⊕ γ(b)(x) := inf

yz=x

{d(a, y) + d(b, z)} ≥

yz=x

inf

{d(az, yz) + d(ab, az)}

≥

inf

yz=x

d(ab, yz)

=

d(ab, x) := γ(ab)(x)

2)

Usingthemetri invarian e,wehaveforall

x ∈ X

andall

a, b ∈ X

(γ(a) ⊕ γ(b)) (xb)

:=

inf

yz=xb

{d(a, y) + d(b, z)}

≤

d(a, x)

=

d(ab, xb)

:= γ(ab)(xb)

Combiningthisinequalitywiththepart

1)

,weget

(γ(a) ⊕ γ(b)) (xb) = γ(ab)(xb)

. Inasimilar way, we prove

(γ(a) ⊕ γ(b)) (ax) = γ(ab)(ax)

. If moreover,

X

is a quasigroup,then for ea h

t, b ∈ X

,thereexists

x ∈ X

su hthat

t = xb

. Soweobtain

γ(a) ⊕ γ(b) = γ(ab).

Proof of Proposition 1.

(1) ⇒ (2)

. Suppose that

(X, α)

is quasigroup. Using Lemma 1, we havethat

γ(a) ⊕ γ(b) = γ(ab)

, for all

a, b ∈ X

. Using this formula and theinje tivity of

γ

, it is lear that

(γ(X), ⊕)

is quasigroup (respe tively, loop, group, ommutative group) whenever

(X, .)

isquasigroup(respe tively,loop,group, ommutativegroup).

Thelastpartofthetheorem,followsfromtheformula

γ(a) ⊕ γ(b) = γ(ab)

andthefa tthat

γ

isisometri .

(1) ⇒ (2)

. Suppose that

(γ(X), ⊕)

is a quasigroup. Let us provethat

(X, .)

is quasigroup. First, we showthat forall

a, b ∈ X

, we havethat

γ(a) ⊕ γ(b) = γ(ab)

. Indeed, let

a, b ∈ X

. Sin e

⊕

is aninternallawof

(γ(X), ⊕)

,thenthere exists

c ∈ X

su hthat

γ(a) ⊕ γ(b) = γ(c)

. Using Lemma 1, weobtain

γ(ab) ≤ γ(c)

. Hen e

0 ≤ d(ab, c) = γ(ab)(c) ≤ γ(c)(c) = 0

. This impliesthat

c = ab

. Finally wehave

γ(a) ⊕ γ(b) = γ(ab)

for

a, b ∈ X

. Fromthis formulaand theinje tivityof

γ

,itis learthat

(X, .)

isquasigroup(respe tively,loop,group, ommutative group)whenever

(γ(X), ⊕)

isquasigroup(respe tively,loop,group, ommutativegroup)

.

ThefollowingCorollaryis aparti ular aseoftheworkestablishedin[2℄.

Corollary 3 Let

(X, ., d)

be a group omplete metri invariant having

e

as identity element. Then,

(1)

the set

Lip

1

_(X)

of all

1

-Lips hitz and bounded from below fun tions, isa monoid having

γ(e) := d(e, .)

asidentity elementanditsgroup of unit

U(Lip

1

_(X))

oin ides with

X + R

ˆ

.

(2)

the set

Lip

1

+

(X)

ofall

1

-Lips hitzandpositive fun tions,isamonoidhaving

γ(e) := d(e, .)

asidentity elementanditsgroupof unit

U(Lip

1

+

(X))

oin ides with

X

ˆ

.

Proof.

(1)

The fa t that

Lip

1

_(X)

is a monoid having

γ(e) := d(e, .)

as identity element, followsfrom Proposition7. andLemma 3. in [2℄. UsingProposition1,wehavethat

X + R ⊂

ˆ

U(Lip

1

_(X))

. Thefa tthat

U(Lip

1

_{(X)) ⊂ ˆ}

_{X +R}

,followsfromCorollary2. Thus

U(Lip

1

_{(X)) =}

ˆ

X + R

.

(2)

Sin etheinf- onvolutionofpositivefun tionsisalsopositiveand

γ(e) := d(e, .) ∈ Lip

1

+

(X)

, then

Lip

1

+

(X)

isasubmonoidof

Lip

1

_(X)

. Ontheotherhand,

X ⊂ U(Lip

ˆ

1

+

(X)) ⊂ U(Lip

1

(X)) =

ˆ

X + R

. Sin etheelementof

U(Lip

1

+

(X))

arepositivefun tions weget

U(Lip

1

+

(X)) = ˆ

X.

WegivenowtheproofofTheorem1mentionedintheintrodu tion.

ProofofTheorem1. Thepart

(1) ⇒ (2)

anbededu edfrom[2℄(Seealso[1℄). Letusprove

(2) ⇒ (1)

. Sin e

(Lip

1

+

(X), ⊕)

is amonoid, thereexists and identityelement

f

0

∈ Lip

1

+

(X)

. Sin e

f

0

istheidentityelement,itsatisesin parti ular:

γ(a) ⊕ f

0

= γ(a)

forall

a ∈ X

. Sin e

γ(a) := d(a, .)

hasastrongminimumat

a

, applyingTheorem 4tothe fun tions

γ(a)

and

f

0

, thereexists

(˜

y, ˜

z) ∈ X × X

satisfying

y ˜

˜

z = a

su hthat

γ(a)

hasastrongminimumat

˜

y

and

f

0

hasastrongminimumat

z

˜

. Sin eastrongminimumisinparti ularunique,then

y = a

˜

. So,we have

a ˜

z = a

,forall

a ∈ X

. Thus

e := ˜

z

istheidentityelementof

X

. Fromtheasso iativityof

(Lip

1

+

(X), ⊕)

, weobtaininparti ular theasso iativityof

( ˆ

X, ⊕)

. Sin e

(X, .)

isaquasigroup (loop) then from Lemma 1, we have

γ(a) ⊕ γ(b) = γ(ab)

for all

a, b ∈ X

, so we dedu e by theinje tivityof

γ

,that

(X, .)

isalsoasso iative. Hen e,

(X, .)

is agroup. Thefa t that, the identityelementof

(Lip

1

+

(X), ⊕)

is

γ(e)

where

e

istheidentityelementof

X

anditsgroupof unit is

X

ˆ

,followsfromCorllary3

.

4 Metri properties and the density of

S(X)

in

Lip

1

_(X)

. Letus onsiderthefollowingsets

S(X) :=



f ∈ Lip

1

(X)/ f

hasastrongminimum

S

+

(X) :=



f ∈ Lip

1

+

(X)/ f

hasastrongminimum

Corollary 4 Let

(X, ., d)

be a group omplete metri invariant having

e

as identity element. Then,

S(X)

isasubmonoidof

(Lip

1

_{(X), ⊕)}

and

U(S(X)) = ˆ

X + R

. Ontheotherhand

S

+

(X)

isasubmonoid of

(Lip

1

+

(X), ⊕)

and

U(S

+

(X)) = ˆ

X

.

Proof. Sin e

(Lip

1

_{(X), ⊕)}

isamonoidhaving

γ(e) ∈ S(X)

asidentityelementandsin e

S(X)

is asubsetof

(Lip

1

+

(X), ⊕)

,itsu esto showthat

⊕

isaninternal lawof

S(X)

whi h isthe ase thanksto Theorem 4. On the other hand,

X + R ⊂ U(S(X)) ⊂ U(Lip

ˆ

1

_{(X)) = ˆ}

_{X + R}

. Hen e

U(S(X)) = ˆ

X + R

. Inasimilarwayweobtainthese ond partoftheCorollary

.

Considernowthemetri s

ρ

and

ρ

˜

on

Lip

1

_(X)

dened for

f, g ∈ Lip

1

_(X)

by

ρ(f, g) = sup

x∈X

|f (x) − g(x)|

1 + |f (x) − g(x)|

˜

ρ(f, g) = ρ(f − inf

X

f, g − inf

X

g) + | inf

X

f − inf

X

g|.

Proposition3 Let

(X, d)

bea ompletemetri spa e. Then

(1)

the sets

(Lip

1

_{(X), ρ)}

and

(Lip

1

_{(X), ˜}

_ρ)

(respe tively,

(Lip

1

+

(X), ρ)

and

(Lip

1

+

(X), ˜

ρ)

) are omplete metri spa es.

(2)

theset

(S(X), ρ)

isdense in

(Lip

1

_{(X), ρ)}

and

(S(X), ˜

ρ)

isdensein

(Lip

1

_{(X), ˜}

_ρ)

.

(3)

theset

(S

+

(X), ρ)

isdense in

(Lip

1

+

(X), ρ)

and

(S

+

(X), ˜

ρ)

isdensein

(Lip

1

+

(X), ˜

ρ)

.

Proof. The part

(1)

is similar to Proposition 5. in [1℄ (See also Lemma 1. in [1℄). Let us prove the part

(2)

. Indeed, let

f ∈ Lip

1

_(X)

and

0 < ǫ < 1

. Consider the fun tion

f

ǫ

:= (1 − ǫ)f

. Clearly,

f

ǫ

is

(1 − ǫ)

-Lips hitz and

ρ(f

ǫ

, f ) → 0

(respe tively

ρ(f

˜

ǫ

, f ) →

0

) when

ǫ → 0

. On the other hand, applying the variationalprin iple of Deville-Godefroy-Zizler[4℄tothe

(1 − ǫ)

-Lips hitzand bounded frombelowfun tion

f

ǫ

, thereexistsabounded Lips hitz fun tion

ϕ

ǫ

on

X

su h that

sup

x∈X

|ϕ

ǫ

(x)| ≤ ǫ

and

sup

x,y∈X/x6=y

|ϕ

ǫ

(x)−ϕ

ǫ

(y)|

|x−y|

≤ ǫ

and

f

ǫ

+ ϕ

ǫ

has a strong minimum at some point. We have that

f

ǫ

+ ϕ

ǫ

is

1

-Lips hitz and boundedfrombellowfun tionhavingastrongminimum,so

f

ǫ

+ϕ

ǫ

∈ S(X)

. Ontheotherhand,

ρ(f

ǫ

+ ϕ

ǫ

, f ) ≤ ρ(f

ǫ

+ ϕ

ǫ

, f

ǫ

) + ρ(f

ǫ

, f ) = ρ(ϕ

ǫ

, 0) + ρ(f

ǫ

, f )

. It followsthat

ρ(f

ǫ

+ ϕ

ǫ

, f ) → 0

when

ǫ → 0

(respe tively

ρ(f

˜

ǫ

+ ϕ

ǫ

, f ) → 0

). Thus

(S(X), ρ)

and

(S(X), ˜

ρ)

are respe tively densein

(Lip

1

_{(X), ρ)}

and

(Lip

1

_{(X), ˜}

_ρ)

.

(3)

Let

f ∈ Lip

1

+

(X)

, from

(2)

, there exists

f

ǫ

∈ S(X)

su h that

ρ(f

ǫ

, f ) → 0

when

ǫ → 0

. In parti ular,

inf

X

f

ǫ

→ inf

X

f

. If

inf

X

f > 0

, then for very small

ǫ

we have

f

ǫ

> 0

and so

f

ǫ

∈ S

+

(X)

. If

inf

X

f = 0

, sin e

inf

X

f

ǫ

→ inf

X

f = 0

then

ρ(f

ǫ

− inf

X

f

ǫ

, f ) ≤ ρ(f

ǫ

−

inf

X

f

ǫ

, f

ǫ

) + ρ(f

ǫ

, f ) → 0

when

ǫ → 0

and

(f

ǫ

− inf

X

f

ǫ

) ∈ S

+

(X)

. Thus,

(S

+

(X), ρ)

isdense in

(Lip

1

+

(X), ρ)

. Wededu ethenthat

(S

+

(X), ˜

ρ)

isalsodensein

(Lip

1

+

(X), ˜

ρ).

5 The map

arg min(.)

as monoid morphism.

Forareal-valuedfun tion

f

withdomain

X

,

arg min(f )

isthesetofelementsin

X

thatrealize theglobalminimumin

X

,

arg min(f ) = {x ∈ X : f (x) = inf

y∈X

f (y)}.

For the lassof fun tions

f ∈ S(X)

,

arg min(f ) = {x

f

}

isasingleton, where

x

f

is thestrong minimumof

f

. Inwhatfollows,weidentifythesingleton

{x}

withtheelement

x

. Wehavethe followingproposition.

Corollary 5 Let

(X, ., d)

be a group omplete metri invariant having

e

as identity element. Then, the map,

arg min : (S(X), ⊕, ρ) → (X, ., d)

is surje tive and ontinuous monoid morphism. We have the following ommutative diagram, where

I

denotesthe identity map on

X

and

γ

the Kuratowskioperator

(X, .)

γ

//

I

%%

❑

❑

❑

❑

❑

❑

❑

❑

❑

(S(X), ⊕)

arg min



(X, .)

Proof. Let

f, g ∈ S(X)

, then there exists

x

f

, x

g

∈ X

su h that

f

has a strong minimum at

x

f

= arg min(f )

and

g

has a strong minimum at

x

g

= arg min(g)

. Using Theorem 4,

f ⊕ g

has a strong minimum at

x

f

x

g

= (arg min(f )) (arg min(g))

. Thus

arg min(f ⊕ g) =

(arg min(f )) (arg min(g))

. Ontheotherhand,themap

arg min

sendtheidentityelement

d(e, .)

of

S(X)

totheidentityelement

e

of

X

,sin ethestrongminimumof

d(e, .)

is

e

. Hen e,

arg min

is amonoidmorphism. Forea h

x ∈ X

,

γ(x) ∈ ˆ

X ⊂ S(X)

and

arg min (γ(x)) = x

. Thus,

arg min

issurje tive. Letusprovenowthe ontinuityof

arg min

. First,notethatforall

f, g ∈ Lip

1

_(X)

andall

0 < α < 1

,

ρ (f, g) ≤ α ⇒ sup

x∈X

|f (x) − g(x)| ≤

α

1 − α

.

(14)

andin onsequen e,wealsohave

| inf

X

f − inf

X

g| ≤

α

1 − α

.

(15)

Let

(f

n

)

n

⊂ S(X)

and

f ∈ S(X)

. Let

x

n

:= arg min(f

n

)

and

x

f

= arg min(f )

. Sin e

f

hasa strongminimumat

x

f

, forall

ǫ > 0

,thereexists

δ > 0

su hthat forall

x ∈ X

,

|f (x) − f (x

f

)| ≤ δ ⇒ d(x, x

f

) ≤ ǫ.

Suppose that

ρ (f

n

, f ) ≤

δ

2+δ

. Using the triangular inequality and the inequations (14) and (15)with

α =

δ

2+δ

< 1

,wehave

|f (x

n

) − f (x

f

)| ≤

|f (x

n

) − f

n

(x

n

)| + |f

n

(x

n

) − f (x

f

)|

=

|f (x

n

) − f

n

(x

n

)| + | inf

X

f

n

− inf

X

f |

≤

2α

1 − α

=

δ

whi himplies that

d(arg min(f

n

), arg min(f )) := d(x

n

, x

f

) ≤ ǫ

. Thisimplies the ontinuityof

arg min

on

S(X).

Notethat,themap

ξ : (Lip

1

_{(X), ⊕, ρ) → R}

denedby

ξ : f 7→ inf

X

f

,is ontinuousmonoid morphism. Thefollowingproposition whi his a onsequen eoftheabove orollary,saysthat, in the aseof

(S(X), ⊕, ρ)

there areseveral ontinuousmonoidmorphismfrom

(S(X), ⊕)

into

K

with

K

= R

or

C

.

Proposition4 Let

(X, ., d)

be agroup omplete metri invariant and

χ : (X, ., d) → K

be a ontinuous group morphism. Then,

χ ◦ arg min : (S(X), ⊕, ρ) → K

is a ontinuous monoid morphism.

Let

X

beagroupwith the identity element

e

. We equip

X

with thedis rete metri

dis

. So

(X, dis)

is group metri invariantand

Lip

1

+

(X)

onsist in this ase on all positive fun tions su hthat

|f (x) − f (y)| ≤ 1

forall

x, y ∈ X

. TheKuratowskioperator

γ : x 7→ δ

x

isheredened by

δ

x

(y) = 1

if

y 6= x

and

δ

x

(x) = 0

, forall

x, y ∈ X

. Wetreatbelowthe asesof

X = Z

and

X = Z/pZ

.

6.1 The inf- onvolution monoid

(l

∞

dis

(Z), ⊕)

.

Let

X = Z

equipped with the dis rete metri

dis

whi h is invariant. Let

l

∞

dis

(Z)

the set of all sequen es

(x

n

)

n

of real positive numbers su h that

|x

n

− x

m

| ≤ 1

for all

n, m ∈ Z

. For

u = (u

n

)

n

and

v = (u

n

)

n

in

l

∞

dis

(Z)

,wedenethesequen e

(u ⊕ v)

n

:= inf

k∈Z

(u

n−k

+ v

k

);

∀n ∈ Z.

Theset

(l

∞

dis

(Z), ⊕)

isa ommutativemonoidhavingtheelement

δ

e

asidentityelementandits groupofunit

U(l

∞

dis

(Z))

is isomorphi to

Z

bytheisomorphism

I : Z → U(l

∞

dis

(Z))

,

k 7→ δ

k

.

6.2 The inf- onvolution monoid

(l

∞

dis

(Z/pZ), ⊕)

. Let

p ∈ N

∗

and

X = Z/pZ

equipped with the dis rete metri

dis

whi h is invariant. We denote by

l

∞

dis

(Z/pZ)

the set of all

p

-periodi sequen es

(x

n

)

n

of real positivenumbers su h that

|x

n

− x

m

| ≤ 1

forall

n, m ∈ {0, ..., p − 1}

. Weidentifyasequen e

(x

n

)

n

∈ l

∞

dis

(Z/pZ)

with

(x

0

, ..., x

p−1

)

. For

u = (u

n

)

n

and

v = (u

n

)

n

in

l

∞

dis

(Z/pZ)

,wedenethesequen e

(u ⊕ v)

n

:=

min

k∈{0,...,p−1}

(u

n−k

+ v

k

);

∀n ∈ {0, ..., p − 1} .

The set

(l

∞

dis

(Z/pZ), ⊕)

is a ommutative monoid having the element

δ

e

as identity element and its groupof unit

U(l

∞

dis

(Z/pZ))

is isomorphi to

Z

/pZ

by the isomorphism

I : Z/pZ →

U(l

∞

dis

(Z/pZ))

,

¯

k 7→ δ

k

.

7 The set of Katetov fun tions.

Wegivein thisse tionsomeresultsaboutthemonoidstru ture of

K(X)

when

X

isagroup, and the onvex one stru tureofthesubset

K

C

(X)

of

K(X)

(of onvexfun tions)when

X

is aBana hspa e. If

M

isamonoid, by

U(M )

wedenotethegroupofunit of

M

.

7.1 The monoid stru ture of

K(X)

.

Proposition5 Let

(X, d)

bea( ommutative)group metri invarianthaving

e

asidentity ele-ment. Then, themetri spa e

(K(X), ⊕, d

∞

)

isalso a( ommutative)monoidhaving

γ(e) = δ

e

asidentity elementandsatisfying:

(a) d

∞

(f ⊕ g, h ⊕ g) ≤ d

∞

(f, h)

and

d

∞

(g ⊕ f, g ⊕ h) ≤ d

∞

(f, h)

, for all

f, g, h ∈ K(X)

(b) d

∞

(δ

x

⊕ f, δ

x

⊕ h) = d

∞

(f ⊕ δ

x

, h ⊕ δ

x

) = d

∞

(f, h)

, for all

f, h ∈ K(X)

.

Proof. Sin e

K(X)

is a subset of the ( ommutative) monoid

Lip

1

_(X)

of

1

-Lips hitz and bounded frombelowfun tions,whi hhave

δ

e

as identityelement(See [2℄),itsu esto prove that, forall

f, g ∈ K(X)

andall

x

1

, x

2

∈ X

,wehave

invarian ethat,forall

n ∈ N

∗

,thereexists

y

n

, z

n

, y

′

n

, z

′

n

∈ X

su hthat

y

n

z

n

= x

1

and

y

′

n

z

′

n

= x

2

s.t

f ⊕ g(x

1

) + f ⊕ g(x

2

) ≥



f (y

n

) + g(z

n

) +

1

n



+



f (y

n

′

) + g(z

n

′

) +

1

n



=

(f (y

n

) + f (y

n

′

)) + (g(z

n

) + g(z

n

′

)) +

2

n

≥ d(y

n

, y

n

′

) + d(z

n

, z

n

′

) +

2

n

=

d(y

n

z

n

, y

′

n

z

n

) + d(y

′

n

z

n

, y

′

n

z

n

′

) +

2

n

=

d(x

1

, y

n

′

z

n

) + d(y

n

′

z

n

, x

2

) +

2

n

≥ d(x

1

, x

2

) +

2

n

Thus

f ⊕ g(x

1

) + f ⊕ g(x

2

) ≥ d(x

1

, x

2

)

by sending

n

to

+∞

. Hen e

(K(X), ⊕)

is a monoid having

δ

e

as identityelement.

We provenowthat

d

∞

(f ⊕ g, h ⊕ g) ≤ d

∞

(f, h)

. Let

f, g, h ∈ K(X)

and

x ∈ X

, thereexists

y

n

, z

n

su hthat

y

n

z

n

= x

and

h ⊕ g(x) > h(y

n

) + g(z

n

) −

1

n

. Hen e,forall

n ∈ N

∗

f ⊕ g(x) − h ⊕ g(x)

≤ (f (y

n

) + g(z

n

)) +



−h(y

n

) − g(z

n

) +

1

n



= f (y

n

) − h(y

n

) +

1

n

≤ d

∞

(f, h) +

1

n

Hen e,

d

∞

(f ⊕ g, h ⊕ g) ≤ d

∞

(f, h)

by sending

n

to

+∞

. In a similar way we provethat

d

∞

(g ⊕ f, g ⊕ h) ≤ d

∞

(f, h)

. For the part

(b)

, it su es to prove that

f ⊕ δ

a

(.) = f (.a

−1

₎

and

δ

a

⊕ f (.) = f (a

−1

_.)

for all

a ∈ X

,sin ethemap

x 7→ ax

and

x 7→ xa

are oneto one and onto from

X

to

X

whenever

a

is invertible. Indeed,

f ⊕ δ

a

(x) = inf

yz=x

{f (y) + d(z, a)} =

inf

(ya

−1

_)(az)=x



f (ya

−1

_{) + d(az, a)}

. Using the metri invarian e, we have for all

x ∈ X

,

f ⊕ δ

a

(x) = inf

yz=x



f (ya

−1

) + d(z, e)

:= f (.a

−1

) ⊕ δ

e

(x) = f (.a

−1

)(x)

, sin e

δ

e

is the i-dentity element. Similarlyweprovethat

δ

a

⊕ f (.) = f (a

−1

_.)

. This on ludethe proofof the proposition

.

Remark1 Ingeneral, one an not getequality in the part

(a)

of Proposition 5sin e the inf- onvolution doesnot havethe an ellationproperty ingeneral (See[11℄).

If

Y ⊂ X

and

f ∈ K(Y )

, dene

f : X → R

(the Katetov extension of f) by

f (x) =

inf

y∈Y

{f (y) + d(x, y)}

. Itiswellknownthat

f

isthegreatest1-Lips hitzmapon

X

whi his equalto

f

on

Y

;that

f ∈ K(X)

and

χ : f 7→ f

isanisometri embeddingof

K(Y )

into

K(X)

(seeforinstan e [6℄). Thankstothefollowinglemma (we an ndamoregeneralform in[2℄) we an assumewithoutlossofgeneralitythat

X

isa ompletemetri spa e.

Lemma2 Let

(X, d)

be a group whi h is metri invariant and

(X, d)

its group ompletion. Then,

(K(X), ⊕, d

∞

)

and

(K(X), ⊕, d

∞

)

areisometri ally isomorphi as monoids. More pre- isely, the map

χ : (K(X), ⊕, d

∞

) → (K(X), ⊕, d

∞

)

f

7→ f :=



x ∈ X 7→ inf

y∈X



f (y) + d(y, x)



(16)

Proof. Itsu estoshowsthat

χ

isasurje tivemorphismofmonoids. Thesurje tivityis lear, sin eif

F ∈ K(X)

,wetake

f = F

|X

,then

f = F

on

X

andso

f = F

on

X

by ontinuity. Let us show that

χ

is amonoid morphism. Indeed, let

f, g ∈ K(X)

. Usingthe ontinuity of

f

,

g

and

y 7→ y

−1

andthedensityof

X

in

X

,wehaveforall

x ∈ X

,

f ⊕ g(x)

=

inf

˜

y ˜

z=x



f (˜

y) + g(˜

z))

=

inf

˜

y∈X



f (˜

y) + g(˜

y

−1

x)

=

inf

y∈X



f (y) + g(y

−1

x)

.

= f ⊕ g(x)

Thus

f ⊕ g

oin idewith

f ⊕ g = f ⊕ g

on

X

. Hen e

f ⊕ g = f ⊕ g

.

The following theorem shows that, up to an isometri isomorphism of groups, the group of unit of

K(X)

andthegroupofunitof

X

arethesame.

Proposition6 Let

(X, d)

be agroup whi h is ompletemetri invariant. Then, the group of unit

U(K(X))

and

X

ˆ

(whi h isisometri allyisomorphi to

X

) oin ides.

Proof. Sin e

(X, d)

be a omplete metri invariant group, from Proposition 1 we get that

ˆ

X ⊂ U(K(X))

. Ontheotherhand,

U(K(X)) ⊂ U(Lip

1

+

(X)) = ˆ

X.

Wededu ethefollowinganalogoustotheBana h-Stonetheorem. Theorem 5 Let

(X, d)

and

(Y, d

′

₎

be two omplete metri invariant groups. Then, a map

Φ : (K(X), ⊕, d

∞

) → (K(Y ), ⊕, d

∞

)

is a monoid isometri isomorphism if, and only if there exists a group isometri isomorphism

T : (X, d) → (Y, d

′

₎

su h that

Φ(f ) = f ◦ T

−1

for all

f ∈ K(X)

. Consequently,

Aut

Iso

(K(X))

isisomorphi asgroup to

Aut

Iso

(X)

.

Proof. If

T : (X, d) → (Y, d

′

₎

isangroupisometri isomorphism, learly

Φ(f ) := f ◦ T

−1

gives an monoid isometri isomorphismfrom

(K(X), ⊕, d

∞

)

onto

(K(Y ), ⊕, d

∞

)

. For the onverse, let

Φ

bemonoid isometri isomorphismfrom

(K(X), d

∞

)

onto

(K(Y ), d

∞

)

, then

Φ

maps iso-metri ally thegroupofunit of

K(X)

onto thegroupofunit of

K(Y )

. UsingProposition 6,

Φ

mapsisometri allythegroup

X

ˆ

onto

Y

ˆ

. Then,themap

T := γ

−1

◦ Φ

_{| ˆ}

_X

◦ γ

givesanisometri groupisomorphismfrom

X

onto

Y

byProposition1,where

Φ

| ˆ

X

denotesthe restri tionof

Φ

to

X

ˆ

. Sin e

Φ

isisometri wehaveforall

f ∈ K(X)

andall

x ∈ X

f (x) = d

∞

f, δ

x

) = d

∞

(Φ(f ), Φ(δ

x

)) = d

∞

Φ(f ), δ

T (x)

= Φ(f ) (T (x))

whi h on ludetheproof

.

Lemma3 Let

(M, d)

bea metri monoid,

U(M )

itsgroup of unit. Suppose that

d(xu, yu) ≤

d(x, y)

and

d(ux, uy) ≤ d(x, y)

for all

x, y, u ∈ M

, and

d(xu, yu) = d(ux, uy) = d(x, y)

for all

x, y ∈ M

andall

u ∈ U(M )

. Then, for all

x ∈ X

and all

a, b ∈ U(X)

, wehave the following formula

d(x, ab) = inf

yz=x

{d(y, a) + d(z, b)} .

Proof. Let

a, b ∈ U(M )

and

x ∈ M

,wehave

inf

yz=x

{d(y, a) + d(z, b)} ≤ d(xb

−1

_{, a) (}

with

y = xb

−1

_{; z = b)}

=

d(x, ab)

inf

yz=x

{d(y, a) + d(z, b)} ≥

yz=x

inf

{d(yz, az) + d(az, ab)}

≥

inf

yz=x

d(yz, ab) (

byusingthetriangularinequality

)

=

d(x, ab).

Thus,

d(x, ab) = inf

yz=x

{d(y, a) + d(z, b)}.

Weobtainthefollowingformula.

Corollary 6 Let

(X, d)

be a group whi h is metri invariant. Let

f ∈ K(X)

and

a, b ∈ X

. Then

f (ab) =

inf

ϕ⊕ψ=f

{ϕ(a) + ψ(b)} .

Proof. Sin ethemonoid

(K(X), ⊕, d

∞

)

satisfytheProposition5,byapplyingLemma3tothe monoid

M = (K(X), ⊕, d

∞

)

and using thefa t that

X ⊂ U(K(X))

ˆ

(

= ˆ

X

) and

d

∞

(g, γ(x)) =

g(x)

forall

x ∈ X

andall

g ∈ K(X)

,weobtainforall

f ∈ K(X)

andall

a, b ∈ X

:

f (ab) = d

∞

(f, γ(ab)) = d

∞

(f, γ(a) ⊕ γ(b))

=

inf

ϕ⊕ψ=f

{d

∞

(ϕ, γ(a)) + d

∞

(ψ, γ(b))}

=

inf

ϕ⊕ψ=f

{ϕ(a) + ψ(b))} .

7.2 The onvex one stru ture of

K

C

(X)

.

Let

(X, k.k)

beaBana hspa e. Were all that

K

C

(X) := {f ∈ K(X) : f

onvex

} .

Sin e the inf- onvolutionof onvexfun tions is onvex,the set

(K

C

(X), ⊕)

is a omplete metri spa e and ommutativesubmonoid of

(K(X), ⊕)

. Weequip

K

C

(X)

withthe externallaw

⋆

dened as follows: forall

f ∈ K

C

(X)

andall

λ ∈ R

+

by

λ ⋆ f (x) := λf

 x

λ



; ∀x ∈ X if λ > 0

0 ⋆ f := γ(0) := k.k.

Were allbelowthedenitionofa onvex one.

Denition3 A ommutativemonoid

(C, ⊕)

equippedwith as alar multipli ation map

⋆ : R

+

_{× C}

_{→ C}

(λ, c)

7→ λ ⋆ c

asu hthat

1) 1 ⋆ c = c

and

0 ⋆ c = e

C

, for all

c ∈ C

where

e

C

denotesthe identity elementof

(C, ⊕)

.

2) (α + β) ⋆ c = (α ⋆ c) ⊕ (β ⋆ c)

for all

α, β ∈ R

+

and all

c ∈ C

.

3) λ ⋆ (c ⊕ c

′

_{) = (λ ⋆ c) ⊕ (λ ⋆ c}

′

₎

for all

λ ∈ R

+

andall

c, c

′

_{∈ C}

.

issaid tobea onvex one.

Thefollowingpropositioniseasilyveried.

Proposition7 Thespa e

(K

C

(X), ⊕, ⋆, d

∞

)

isa ompletemetri onvex onewiththeidentity element

γ(0)

.

The ompletemetri onvex onestru tureof

(K

C

(X), ⊕, ⋆, d

∞

)

indu e astru tureofBana h spa e on

X

ˆ

by setting

λ ⋆ γ(x) := (−λ) ⋆ γ(−x)

, if

λ < 0

and taking the norm

|||γ(x)||| :=

d

∞

(γ(x), γ(0))

for all

x ∈ X

. In fa t, we an also say that the Bana h spa e

X

extend its stru ture anoni allytosome onvex onestru tureon

K

C

(X)

.

Proposition8 The Kuratowski operator

γ : (X, +, ., k.k) →



ˆ

X, ⊕, ⋆, |||.|||



is an isometri isomorphism of Bana h spa es.

Proof. UsingProposition1,itjust remainsto provethat

γ(λx) = λ ⋆ γ(x)

forall

x ∈ X

and

λ ∈ R

. Indeeed, let

x ∈ X

and

λ ∈ R

∗+

, by denition

γ(λx) : y 7→ δ

λx

(y) = ky − λxk =

λk

_λ

y

− xk := λ ⋆ δ

x

(y)

. So,

γ(λx) = λ ⋆ γ(x)

. If

λ = 0

, then by denition

0 ⋆ γ(x) = γ(0)

. If

λ < 0

,

γ(λx) = γ((−λ)(−x)) = (−λ) ⋆ γ(−x) := λ ⋆ γ(x).

Theorem 6 Let

X

and

Y

twoBana h spa es. Then

(K

C

(X), ⊕, ⋆, d

∞

)

and

(K

C

(Y ), ⊕, ⋆, d

∞

)

areisometri allyisomorphi as onvex oneif, andonlyif,

X

and

Y

areisometri ally isomor-phi asBana h spa es.

Proof. SimilartotheproofofTheorem5

.

UsingthexedpointTheorem,weobtainthefollowingproposition.

Proposition9 Let

X

be a Bana h spa e,

g ∈ K

C

(X)

and

λ ∈ (0, 1)

. Then, there exists a uniquefun tion

f

0

∈ K

C

(X)

su hthat

(λ ⋆ f

0

) ⊕ g = f

0

.

Proof. Let us onsider the map

L : K

C

(X) → K

C

(X)

dened by

L(f ) = (λ ⋆ f ) ⊕ g

. Using Proposition5wehaveforall

f, f

′

_{∈ K}

C

(X)

,

d

∞

(L(f ), L(f

′

)) ≤ d

∞

(λ ⋆ f, λ ⋆ f

′

) = λd

∞

(f, f

′

).

Sin e

λ ∈ (0, 1)

, then

L

is ontra tive map. So by the xed point Theorem, there exists a uniquefun tion

f

0

∈ K

C

(X)

su hthat

L(f

0

) = f

0

.

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