The inf-convolution between algebra and optimization. Applications to the Banach-Stone theorem.

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The inf-convolution between algebra and optimization.

Applications to the Banach-Stone theorem.

Mohammed Bachir

To cite this version:

Mohammed Bachir. The inf-convolution between algebra and optimization. Applications to the

Banach-Stone theorem.. 2015. �hal-01164797�

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Mohammed Bachir. The inf-convolution between algebra and optimization. Applications to

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Appli ations to the Bana h-Stone theorem.

Mohammed Ba hir O tober 12, 2014

Laboratoire SAMM 4543, Université Paris 1 Panthéon-Sorbonne, Centre P.M.F.90 rue Tolbia 75634 Paris edex 13

Email : Mohammed.Ba hiruniv-paris1.fr

Abstra t. This work generalize and extend results obtained re ently in [2℄ from the Bana hspa es framework to the groupsframework. We studyabstra t lasses of fun -tions monoids for the inf- onvolution stru ture and give a omplete des ription of the group of unit of su h monoids. We then apply this results to obtain various versions of theBana h-Stonetheoremfor the inf- onvolution stru turein the group framework. Wealsogiveas onsequen eanalgebrai proofoftheBana h-Dieudonéetheorem. Our te hniques arebasedon anew optimization result.

Keyword, phrase: Inf- onvolution; group of unit; isomorphisms and isometries on metri groupsand monoids, the Bana h-Stone theorem.

2010 Mathemati s Subje t: 46B03-46B04-4620-20M32.

Contents

1 Introdu tion. 2

1.1 Motivationand example. . . 2

1.2 The mainalgebrai results forthe inf- onvolution. . . 3

1.3 The mainoptimization results for the inf- onvolution. . . 5

1.4 Organization ofthe paper. . . 6

1.5 A knowledgments. . . 6

2 Examples. 6 3 The proof of the main optimization results. 7 4 The inf- onvolution and algebra. 9 4.1 Properties anduseful lemmas. . . 9

4.1.1 The semigroup

F

0

(X)

. . . 9

4.1.2 The semigroup

F(X)

. . . 13

Weprovedre ently in[2 ℄aversionoftheBana h-Stone theoremfor theinf- onvolution stru ture. Morepre isely,if

(X, k.k)

isaBana hspa eand

CL

1

(X)

denotesthesetofall

1

-Lips hitz onvex and bounded from belowfun tions, then

(CL

1

(X), ⊕)

is a ommu-tativemonoidhaving

e = k.k

asidentityelement fortheoperation

⊕

ofinf- onvolution. Weproved thatthis monoidequipped with anatural metri , ompletelydetermine the Bana h stru ture of

X

. In [2 ℄ we used the Bana h-Dieudonné theorem(See Theorem 9) whi h appliesonly in this onvex framework.

In this arti le we establish general results in the the group framework instead of the Bana h spa es and we handle more general monoids. The tool usedin this paper is a new result of optimization. Our rst motivation is to prove a new versions of the Bana h-Stone theorem for the inf- onvolution stru ture in the group framework. For this purpose, wearegoing to studyand givea omplete andexpli itdes ription ofthe group ofunitofgeneral andabstra t lassofmonoidsfor theinf- onvolution stru ture. Histori ally, the inf- onvolution appeared as tool of fun tional analysis and optimiza-tion (See for instan e the work of [7℄, [10℄, [5℄, [12 ℄) but it turns out as we are going to reveal it in this arti le, that the inf- onvolution also enjoys a remarkable algebrai properties. Re allthatthe Bana h-Stone theoremassertsthat theBana hstru ture of the spa e

(C(K), k.k

∞

)

of ontinuous fun tionson a ompa t spa e

K

ompletely de-terminethetopologi alstru tureof

K

. Morepre isely,theBana hspa es

(C(K), k.k

∞

)

and

(C(L), k.k

∞

)

areisometri allyisomorphi ifandonlyifthe ompa t spa es

K

and

L

arehomeomorphi . TheBana h-Stone theoremhasbeen extendedon various dire -tions and other stru ture are onsidered by authors like the Bana h algebra stru ture or unital ve tor latti e stru ture. The literature being very ri h on this questions, we send ba kto the referen e [3℄ for a more omplete history and examples of extensions (See also [1℄ for the Bana h-Stone theorem for the Bana h stru ture on abstra t lass of fun tion spa es).

In all this paper, we assume that

(X, ., e

X

)

is a group (not ne essarily abelian) denoted multipli atively and having the identity element

e

X

. By

F(X)

we denote the set of all maps dened from

X

into

R

and bounded fom below. The inf- onvolution operation on

F(X)

(Seealso Moreau[7 ℄,[8℄) is dened by

(f ⊕ g)(x) =

inf

yz=x

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

=

inf

z∈X



f (xz

−1

) + g(z)

.

In general

f ⊕ g 6= g ⊕ f

if

X

is not assumedto be an abelian group.

Clearly,

(F(X), ⊕)

isa semigroup. If moreover

X

is anabelian group then

(F(X), ⊕)

is a ommutative semigroup. The semigroup

F(X)

is equiped with the useful metri :

d(f, g) := sup

x∈X

|(f (x) − inf

X

f ) − (g(x) − inf

X

g)|

1 + |(f (x) − inf

X

) − (g(x) − inf

X

g)|

+ | inf

X

f − inf

X

g|, ∀f, g ∈ F(X).

The following formulasis always true

inf

Thisguaranteesinparti ularthat

F

0

(X) := {f ∈ F(X) : inf

X

f = 0}

isasubsemigroup of

(F(X), ⊕)

.

1.1 Motivation and example.

The propositionbellowisthe kind ofresults thatwewishto show in thisarti le. Denition 1 Let

(X, m)

be a metri group. We say that

(X, m)

is a metri invariant group if the metri

m

isinvariant i.e

m(x, y) = m(ax, ay) = m(xa, ya) ∀x, y, a ∈ X.

If moreover

(X, m)

is omplete we saythat

(X, m)

is omplete metri invariantgroup. Remark 1 Every Fré het spa e is a omplete metri invariant group. For example of a non abelian omplete metri invariantgroup see Example 2.

We denote by

(Lip

0

(X), ⊕)

the semigroup of all Lips hitz and bounded from below fun tions

f

dened on

(X, m)

su hthat

inf

X

f = 0

and

Lip

1

0

(X)

the monoid in luded in

Lip

0

(X)

ofall

1

-Lips hitzmap. themonoid

Lip

1

0

(X)

hasthemap

ϕ

m

: x 7→ m(x, e

X

)

asidentityelement. Thesymbol

∼

=

denotes isometri ally isomorphi .

WeobtainthefollowingversionoftheBana h-Stonetheoremfortheinf- onvolution stru ture whi h say that the monoid

(Lip

1

0

(X), ⊕)

ompletely determine the omplete metri invariant group

(X, m)

.

Proposition 1 Let

(X, m)

and

(Y, m

′

₎

be omplete metri invariant group. Then the followingassertion are equivalent.

(1) (X, m) ∼

= (Y, m

′

)

as groups.

(2) (Lip

1

0

(X), d) ∼

= (Lip

1

0

(Y ), d)

as monoids.

(3)

There exits a semigroup isomorphism isometri

Φ : (Lip

0

(X), d) → (Lip

0

(Y ), d)

su h that

Φ(ϕ

m

) = ϕ

m

′

.

Theproof ofthis resultisbasedonthe followingtwo arguments(and followsfromthe more general Theorem ??):

(1)

An isomorphism of monoidssend the group of unit

U(Lip

1

0

(X))

of

Lip

1

0

(X)

on the group of unit

U(Lip

1

0

(Y ))

of

Lip

1

0

(Y )

.

(2)

Thegroupofunit

U(Lip

1

0

(X))

equippedwiththemetri

d

isisometri allyisomorphi to

(X,

m

1+m

)

. Thisisalsoequivalenttothefa tthat

U(Lip

1

0

(X))

equipedwiththemetri

d

∞

is isometri ally isomorphi to

(X, m)

, where

d

∞

(f, g) := sup

x∈X

{|f (x) − g(x)|} <

+∞

for all

f, g ∈ U (Lip

1

0

(X))

.

1.2 The main algebrai results for the inf- onvolution.

Thestudy ofabstra tsubsemigroups orsubmonoids of

F(X)

followsfromthe study of subsemigroups or submonoids of

F

0

(X)

.

Proposition 2

(F(X), ⊕, d) ∼

= (F

0

(X)×R, ⊕, d

1

)

assemigroups. Where

(f, t)⊕(g, s) :=

Our result in this paper also applies for general monoids in luded in

F

0

(X)

. Let

M

0,ϕ

(X)

beanabstra tmonoidof

F

0

(X)

andhaving

ϕ

asidentityelement,then

ϕ

isin parti ularanidempotentelementi.e

ϕ⊕ϕ = ϕ

. Wewondertheniftheresultobtainedin Proposition1 holdfor theabstra t lassofmonoid

M

0,ϕ

(X)

. Theanswer isarmative for idempotent element

ϕ

satisfying the ondition :

ϕ(x) = ϕ(x

−1

_{) = 0 ⇔ x = e}

X

. This motivates the following denition. Notethat every monoid

M

0,ϕ

(X)

having

ϕ

as identityelement isasubmonoid ofthe following formalmonoid

F

0,ϕ

(X) := {f ∈ F

0

(X) : f ⊕ ϕ = ϕ ⊕ f = f } .

Let us remark that sin e

inf

X

(f ⊕ g) = inf

X

f + inf

X

g; ∀f, g ∈ F(X)

, then every idempotent element of

F(X)

belongs ne essarily to

F

0

(X)

i.e

ϕ ⊕ ϕ = ϕ ⇒ ϕ ≥ 0 =

inf

X

ϕ

.

Denition 2 Let

ϕ ∈ F(X)

. We say that

ϕ

is a remarkable idempotent if

ϕ

is an idempotentelement and satisfy the followingtwo properties:

(1) ϕ(xy) = ϕ(yx)

pour tout

x, y ∈ X

(Always true if

X

is ommutative).

(2) ϕ(x) = ϕ(x

−1

) = 0 ⇔ x = e

X

.

We have the following more expli it hara terization of remarkable idempotent (see se tion 3.).

Proposition 3 Let

ϕ ∈ F(X)

. Then,

ϕ

is remarkable idempotent if and only if

ϕ

satisfay :

(1) ϕ(xy) = ϕ(yx)

for all

x, y ∈ X

.

(2) ϕ(x) = ϕ(x

−1

_{) = 0 ⇔ x = e}

X

.

(3) ϕ(xy) ≤ ϕ(x) + ϕ(y)

pour tout

x, y ∈ X

(i.e

ϕ

issubadditive).

Forea hremarkableidempotent

ϕ

we anasso iateina anoni alwaythemetri

∆

∞,ϕ

on

X

dened by

∆

∞,ϕ

(x, y) := max(ϕ(xy

−1

_{), ϕ(yx}

−1

₎₎

. We denote by

(X, ∆

∞,ϕ

)

the group ompletion of

(X, ∆

∞,ϕ

)

. We denote by

ϕ

the unique extension of

ϕ

to

(X, ∆

∞,ϕ

)

sin e

ϕ

is

1

-Lip hitz for the metri

∆

∞,ϕ

by subadditivity. Note that

(X, ∆

∞,ϕ

) =

(

X, ∆

∞,ϕ

)

and that

ϕ

is also a remarkable idempotent. We need the following setwhi h isa generalization ofthe set of

1

-Lips hitzfun tions :

Lip

1

_0,ϕ

(X) :=



f ∈ F

0

(X) : f (x) − f (y) ≤ ϕ(xy

−1

); ∀x, y ∈ X

.

Sin e

ϕ(xy

−1

_{) ≤ ∆}

∞,ϕ

(x, y)

and

∆

∞,ϕ

isametri invariant then

Lip

1

0,ϕ

(X)

isasubset of

Lip

1

0

(X)

ofall

1

-Lips hitz map

f

on

(X, ∆

∞,ϕ

)

su h that

inf

X

f = 0

.

Wehavethe following usefulidenti ation between the formalmonoid

F

0,ϕ

(X)

and the more expli itset

Lip

1

0,ϕ

(X)

.

Proposition 4 Let

ϕ ∈ F(X)

be aremarkableidempotent. Then

F

0,ϕ

(X) = Lip

1

0,ϕ

(X)

and so

Lip

1

0,ϕ

(X)

isa monoid having

ϕ

as identity element. Remark 2 Were over fromthe above propositionthe monoid

(Lip

1

0

(X), ⊕)

mentioned in the previous se tion from

F

0,ϕ

m

(X)

where

ϕ

m

: x 7→ m(x, e

X

)

whi h isa remarkable idempotent.

Proposition 5 Foreveryremarkableidempotent

ϕ ∈ F(X)

andeverymonoid

M

0,ϕ

(X)

we have that

M

0,ϕ

(X)

is a submonoidof themonoid

Lip

1

0,ϕ

(X)

.

The previousproposition explains thatthe study ofabstra t monoidsof

F

0

(X)

ensues from the study ofthe monoid

Lip

1

0,ϕ

(X)

. A. The group of unit.

Let us announ e now our main algebrai result. We denote by

U(Lip

1

0,ϕ

(X))

the group of unitof the monoid

(Lip

1

0,ϕ

(X), ⊕)

.

Theorem 1 Let

ϕ ∈ F(X)

a remarkable idempotent. Then,

(U (Lip

1

_0,ϕ

(X)), d

∞

) ∼

= (X, ∆

∞,ϕ

)

as groups. Thisis also equivalent to

(U (Lip

1

_0,ϕ

(X)), d) ∼

= (X,

∆

∞,ϕ

1 + ∆

∞,ϕ

).

Sin e the ompletion ofmetri spa es is unique up to isometry, the following orollary givesan alternative wayto onsiderer the ompletion of group metri invariant. Corollary 1 Let

(X, m)

a group metri invariant. Then

(X, m) ∼

= (U (Lip

1

0

(X)), d

∞

).

Letus hara terize nowthe group of unitof abstra tmonoid

M

0,ϕ

(X)

of

F

0

(X)

. Corollary 2 Let

ϕ ∈ F(X)

be a remarkable idempotent. Let

M

0,ϕ

(X)

be an abstra t monoid of

F

0

(X)

having

ϕ

as identityelement. Thenthe group of unit

U(M

0,ϕ

(X)), d)

of

M

0,ϕ

(X)

is isometri allyisomorphi toa subgroup of

(X,

∆

∞

,ϕ

1+∆

∞

,ϕ

)

. B. The Bana h-Stone theorem.

We obtain now the following general version of the Bana h-Stone theorem for the inf- onvolution stru ture.

Theorem 2 Let

X

and

Y

be tow groups and let

ϕ ∈ F(X)

and

ψ ∈ F(Y )

be two remarkable idempotents. Then

(1) ⇒ (2) ⇒ (3) ⇔ (4) ⇒ (5) ⇒ (6)

. If moreover we assume that

ϕ

and

ψ

are symetri (i.e

ϕ(x) = ϕ(x

−1

₎

and

ψ(y) = ψ(y

−1

₎

for all

x ∈ X

and all

y ∈ Y

),then

(1) − (6)

are equivalent.

(1)

There exista group isomorphism

T : X → Y

su h that

ψ ◦ T = ϕ

.

(2)

There existasemigroupisomorphismisometri

Φ : F(X) → F(Y )

sushthat

Φ(0) =

0

and

Φ(ϕ) = ψ

.

(3)

There exist a semigroup isomorphism isometri

Φ : F

0

(X) → F

0

(Y )

sush that

Φ(ϕ) = ψ

.

(4) (Lip

1

0,ϕ

(X), d) ∼

= (Lip

1

0,ψ

(Y ), d)

as monoids.

(5) (Lip

1

_0,ϕ

(X), d) ∼

= (Lip

1

_0,ψ

(Y ), d)

as monoids.

Thealgebrai main resultsofthe previousse tionsfollowsfromthe following optimiza-tion resultwhi h applieson a general group metri invariant (not ne essarilyabelian). This result have many of other appli ations of optimization in parti ular for the reso-lution of the inf- onvolution equations.

Denition 3 Let

(X, m)

be a metri spa e, we say that a fun tion

f

has a strong minimum at

x

0

∈ X

, if

inf

X

f = f (x

0

)

and

m(x

n

, x

0

) → 0

whenever

f (x

n

) → f (x

0

)

. A strong minimumis in parti ular unique.

Theorem 3 Let

(X, m)

be a omplete metri invariantgroup withthe identityelement

e

X

. Let

f

and

g

be two lower semi ontinuous fun tions on

(X, m)

. Suppose that the map

x 7→ f ⊕ g(x) + f ⊕ g(x

−1

₎

has a strongminimumat

e

X

and

f ⊕ g(e

X

) = 0

. Then there exists

z

0

∈ X

su hthat :

(1)

the map

η : z → f (z

−1

_{) + g(z)}

has a strong minimum at

z

0

∈ X

(we say that

f ⊕ g(e

X

)

isattained strongly at

z

0

).

(2) f (x) ≥ f ⊕ g(xz

0

) + f (z

0

−1

)

and

g(x) ≥ f ⊕ g(z

−1

0

x) + g(z

0

)

for all

z ∈ X

.

The following result shows that a strong linear perturbation of the onvolution

f ⊕ g

at some point

x

0

of two lower semi ontinuous fun tions

f

and

g

leads to a strong perturbationof

f

and

g

withthe sameperturbation onrespe tivelysomepoints

x

1

and

x

2

su h that

x

1

x

2

= x

0

.

Corollary 3 Let

(X, m)

bea omplete metri invariantgroupwiththeidentityelement

e

X

. Let

p : X → R

be a group morphism and

f

and

g

be two lower semi ontinuous fun tions on

(X, m)

. Supposethat the map

x 7→ f ⊕ g(x) − p(x)

has a strongminimum at

x

0

, then there exists

z

0

∈ X

su h that

(1)

the map

η : z → f (x

0

z

−1

_{) + g(z)}

has a strong minimum at

z

0

∈ X

i.e

f ⊕ g(x

0

)

is attained strongly at

z

0

.

(2) f − p

has a strong minimumat

x

0

z

−1

0

and

g − p

has a strong minimumat

z

0

.

1.4 Organization of the paper.

This arti le is organied as follow. In se tion 2. we give some examples of omplete metri invariant group, remarkable idempotent and monoids for the inf- onvolution stru ture. In se tion 3. we give the proof of our main optimization result Theorem 5 (Theorem 3 in the introdu tion). In se tion 4. we give several algebrai properties of the inf- onvolution stru ture and the proof of our main algebrai resultTheorem 6 (Theorem 1in the introdu tion). Inse tion5. Wegive variousversions ofthe Bana h-Stone theoremand the proof of the main resultof this se tion Theorem 7(Theorem 2 in the introdu tion). Finallyin se tion6. We give an algebrai proof ofthe well know Bana h-Dieudonnétheorem (SeeTheorem 9).

1.5 A knowledgments.

The author thanks Professor Gilles Godefroy for the diverse dis ussions as well asfor his invaluable advi e.

A. Complete metri invariant groups. Exemples 1 (Abelian group ase).

(1)

Every Fré het spa e is ompletemetri invariantgroup. In parti ularevery Bana h spa e equiped with the metri asso iated the the norm is a omplete metri invariant group.

(2)

Let

E

be a set of nite ardinal and

P(E)

the set of all subset of

E

. The set

(P(E), ∆)

is an abeliean group, where

∆

is the symmetri dieren e between two sets :

A∆B = (A ∪ B) \ (A ∩ B)

for all

A, B ∈ P(E)

. We denote by

|A|

the ardinal of

A

. Then

(P(E), m)

is a omplete metri group where

m

is the metri dened by

m(A, B) =

|A∆B|

_|E|

on

P(E)

.

(3)

Every group

X

is omplete metri invariantgroup for the dis rete metri .

Exemples 2 (Non abelian group ase). Let

H

be a real separable Hilbert spa e,

O(H)

be the orthogonal group on

H

and

I

be the identityoperator. We denote by

O

c

(H) := {T ∈ O(H) : I − T

isa oma t operator

}

and

O

h

(H) := {T ∈ O(H) : I − T

isaHilbert-Shmidt operator

} .

The metri s

d

c

and

d

h

as dened as follow :

d

c

(T, S) = kT − Sk

op

and

d

h

(T, S) =

kT − Sk

HS

where

k.k

op

isthe normoperator and

k.k

HS

isthe Hilbert-S hmidt norm

kAk

2

HS

= Tr|(A

∗

A)| :=

X

i∈I

kAe

i

k

2

where

k.k

isthenormof

H

and

{e

i

: i ∈ N}

anorthonormal basisof

H

. Thisdenition is independent of the hoi e of the basis.

Theorem 4 (See [[11℄, Theorem 1.1℄)

(O

C

(H), d

c

)

and

(O

h

(H), d

h

)

are omplete separable metri invariant(non abeliean) groups.

B. Remarkable idempotent. Exemples 3

(1)

Let

(X, m)

be a metri invariantgroupwiththe identityelement

e

X

and

0 < α ≤ 1

. Then the map

ϕ

in the following ases isremarkable idempotent in

F(X)

. In this as

ϕ

issymetri

ϕ(x) = ϕ(x

−1

₎

for all

x ∈ X

.

(a)

the map

ϕ(x) := m(e

X

, x)

α

for all

x ∈ X

.

(b)

Let

χ : R

+

_{→ R}

+

be an in reasingandsub-additivefun tionhavinga strong minimum at

0

and su h that

χ(0) = 0

. We dene

ϕ

as follow.

ϕ(x) :=

χ(m(e

X

, x)

α

)

.

(2)

Let

(X, k.k

X

)

be a real normed ve tor spa e. Let

C

be a onvex bounded sub-set of

X

ontaining the origin. Then the Minkowski fun tional

ϕ

C

(x) :=

(3)

Let

(X, k.k

X

)

be a ve tor normed spa e , and

K ⊂ X

∗

be a onvexweak-star losed and bounded su h that

int(K) 6= ∅

(

int(K)

denotes the interior of

K

for the normtopology). Thenthe supportfun tion dened by

σ

K

(x) := sup

p∈K

p(x)

is remarkable idempotent.

(4)

Let

X

be any group with the identity element

e

X

. Then the map

ϕ

e

X

dened by

ϕ

e

X

(e

X

) = 0

and

ϕ

e

X

(x) = 1

if

x 6= e

X

isremarkable idempotent. C. Examples of monoids for the inf- onvolution.

Exemples 4

(1)

Let

(X, k.k

X

)

be a Bana hspa e and and

K ⊂ X

∗

be a onvexweak-star losed and bounded su h that

int(K) 6= emptyset

. Let

LC(X)

the set of all bounded from below onvexandLips hitzfun tionson

X

. Then

(M

0,σ

K

(X), ⊕) := (LC(X)∩Lip

0,σ

K

(X), ⊕)

isa monoidhaving

σ

K

asidentityelement. If

K = B

X

∗

then

σ

K

= k.k

andinthis ase we re over the monoidstudied in [2℄.

(2)

Let

(X, m)

be a omplete metri invariant group with identity element

e

X

and let

0 < α ≤ 1

and

ϕ

m

(x) = m

α

_(e

x

; x)

forall

x ∈ X

. Let

Lip

α

_(X)

be the set of all bounded from below,

α

-H

o

¨

lder fun tions on

X

and

Lip

α

_(X)

is the set of all

α

-H

o

¨

lder fun tions on

X

su h that

inf

X

f = 0

. Then

Lip

α

_(X)

and

Lip

α

_(X)

are monoid having

ϕ

m

as identity element.

(3)

Let

IC(R

n

₎

bethesetofallinf- ompa tfun tionsfrom

R

n

into

R

andlet

0 < α ≤ 1

. Then

(IC(R

n

_{), ⊕)}

isa semigroup (See [7℄)and

IC(R

n

_{) ∩ Lip}

α

_(R

n

₎

isamonoid having

ϕ = k.k

α

as identityelement.

(4)

Let

X

be any algebrai group withthe identityelement

e

X

andlet

m

be the dis rete distan e on

X

and

δ

e

X

(x) = 0

if

x = e

X

andtake the value

1

otherwise. Let

F

1

_{(X) =}



f : X → R : sup

_x,y∈X

|f (x) − f (y)| ≤ 1

and

F

1

0

(X) := {f : X → [0, 1] : inf

X

f = 0}

. Then

F

1

_(X)

and

F

1

0

(X)

are monoids having

δ

e

X

as identityelement.

3 The proof of the main optimization results.

Theorem 5 Let

(X, m)

be a omplete metri invariantgroup withthe identityelement

e

X

. Let

f

and

g

be two lower semi ontinuous fun tions on

(X, m)

. Suppose that the map

x 7→ f ⊕ g(x) + f ⊕ g(x

−1

₎

has a strongminimumat

e

X

and

f ⊕ g(e

X

) = 0

. Then there exists

z

0

∈ X

su hthat :

(1)

the map

η : z → f (z

−1

_{) + g(z)}

has a strong minimumat

z

0

∈ X

.

(2) f (x) ≥ f ⊕ g(xz

0

) + f (z

₀

−1

)

and

g(x) ≥ f ⊕ g(z

−1

0

x) + g(z

0

)

for all

z ∈ X

. Proof.

(1)

Let

(z

n

)

n

⊂ X

be su hthat for all

n ∈ N

∗

,

f ⊕ g(e

X

) ≤ f (z

n

−1

) + g(z

n

) < f ⊕ g(e

X

) +

1

n

.

Sin e

f ⊕ g(e

X

) = 0

then

0 ≤ f (z

−1

n

) + g(z

n

) <

1

Onthe other handforall

x, y ∈ X

,

f ⊕ g(xy

−1

) ≤ f (x) + g(y

−1

)

(2)

f ⊕ g(yx

−1

) ≤ f (y) + g(x

−1

).

(3) By addingboth inequalilies(2)and (3)above weobtain for all

x, y ∈ X

f ⊕ g(xy

−1

) + f ⊕ g(yx

−1

) ≤

f (x) + g(x

−1

)

+ f (y) + g(y

−1

)

.

(4) By appllaying the above inequalitywith

x = z

−1

n

and

y = z

−1

m

, we have

f ⊕ g(z

_n

−1

z

m

) + f ⊕ g(z

m

−1

z

n

) ≤

f (z

n

−1

) + g(z

n

)

+ f (z

_m

−1

) + g(z

m

)

From our hypothesis we have that the map

z → f ⊕ g(z) + f ⊕ g(z

−1

₎

has a stong minimum at

e

X

with

0 = f ⊕ g(e

X

) + f ⊕ g(e

−1

X

)

. Sofromthe above inequalityand(1) we obtain

0 ≤ f ⊕ g(z

_n

−1

z

m

) + f ⊕ g(z

m

−1

z

n

) ≤

1

n

+

1

m

.

Thus

f ⊕ g(z

−1

n

z

m

) + f ⊕ g((z

n

−1

z

m

)

−1

) → 0

when

n, m → +∞

whi h implies that

m(e

X

, z

n

−1

z

m

) → 0

or equivalently

m(z

n

, z

m

) → 0

sin e

m

is invariant. Thus the sequen e

(z

n

)

n

is Cau hy in

(X, m)

and so onverges to some

z

0

sin e

(X, m)

is a omplete metri spa e. By the lower semi- ontinuity of

f

and

g

, the ontinuity of

z → z

−1

andbyusing the formulas (1)we get

f (z

₀

−1

) + g(z

0

) ≤ 0 = f ⊕ g(e

X

).

Onthe other hand,bydenitionwe have

f ⊕ g(e

X

) ≤ f (z

−1

0

) + g(z

0

)

. Thus

f (z

₀

−1

) + g(z

0

) = f ⊕ g(e

X

) = 0.

(5) Itfollowsthat

η

hasaminimum at

z

0

(bydenitionwehave

inf

z∈X

η(z) = f ⊕ g(e

X

)

). To see that

η

has a strong minimum at

z

0

, let

(x

n

)

n

be any sequen e su h that

f (x

−1

n

) + g(x

n

) → inf

z∈X



f (z

−1

) + g(z)

= 0

. By applying (4) with

x = z

−1

0

and

y = x

−1

_n

andthe formulas (5)and (1)with the fa t that

f ⊕ g(z) + f ⊕ g(z

−1

_{) ≥ 0}

for all

z ∈ X

, weobtainthat

f ⊕ g(z

−1

0

x

n

) + f ⊕ g(x

−1

n

z

0

) → 0

whi himpliesbyhypothesis that

m(x

n

, z

0

) → 0

. Thus

z

0

is astrong minimum of

η

.

(2)

Usingthe part

(a)

we have that

0 = f ⊕ g(e

X

) = f (z

−1

0

) + g(z

0

)

. We have

f ⊕ g(z

−1

0

x) =

inf

y∈X



f (z

0

−1

xy

−1

) + g(y)

≤ f (z

−1

0

) + g(x)

= −g(z

0

) + g(x).

(6) and

f ⊕ g(xz

0

) =

inf

y∈X



f (xz

0

y

−1

) + g(y)

≤ f (x) + g(z

0

)

= −f (z

0

−1

) + f (x).

(7)

Lemma 1 Let

(X, m)

beametri groupwiththeidentityelement

e

X

andlet

h : X → R

. Suppose that

h

has a strongminimum at

e

X

and

h(e

X

) = 0

, then the map

x 7→ h(x) +

h(x

−1

)

has a strongminimum at

e

X

and

h(e

X

) = 0

.

Remark 3 The onverse of the above proposition is not true in general (Take

h(x) =

x + |x|

on

X = (R, +)

).

Proof : Sin e

h

hasa strong minimum at

e

X

and

h(e

X

) = 0

then

h(x) ≥ h(e

X

) = 0

for all

x ∈ X

. So

h(x) + h(x

−1

_{) ≥ 0}

. On the other hand, we have

h(e

X

) + h(e

−1

X

) =

2h(e

X

) = 0

, and so

x → h(x) + h(x

−1

₎

has aminimum at

e

X

. Letus show that

e

X

is a strong minimum for

x → h(x) + h(x

−1

₎

. Indeed,sin e

h ≥ 0

, then wehave,

0 ≤ h(x) ≤ h(x) + h(x

−1

)

forall

x ∈ X

. If

(z

n

)

n

isasequen esu hthat

h(z

n

)+h(z

−1

n

) → inf

x∈X

h(x) + h(x

−1

)

=

0

then by the above inequalities we have that

h(z

n

) → 0

whi h implies that

z

n

→ e

X

sin e

h

has astrong minimum at

e

X

. Thus

x → h(x) + h(x

−1

₎

has astrong minimum at

e

X

and

h(e

X

) = 0

.

Corollary 4 Let

(X, m)

bea omplete metri invariantgroupwiththeidentityelement

e

X

. Let

p : X → R

be a group morphism and

f

and

g

be two lower semi ontinuous fun tions on

(X, m)

. Supposethat the map

x 7→ f ⊕ g(x) − p(x)

has a strongminimum at

x

0

, then there exists

z

0

∈ X

su h that

(1)

the map

η : z → f (x

0

z

−1

_{) + g(z)}

has a strong minimumat

z

0

∈ X

and

f (x

0

z

−1

0

) +

g(z

0

) = f ⊕ g(x

0

)

. In parti ular

f ⊕ g(x

0

)

is exa t at

x

0

.

(2) f − p

has a strong minimumat

x

0

z

−1

0

and

g − p

has a strong minimumat

z

0

. Proof. First,notethat

f ⊕ g(x) − p(x) = (f − p) ⊕ (g − p)(x)

for all

x ∈ X

sin e

p

isa group morphism. Letus set

c := f ⊕ g(x

0

) − p(x

0

)

and

h : t 7→ f ⊕ g(x

0

t) − p(x

0

t) − c

. Then

h

hasastrongminimumat

e

X

sin ebyhypothesis

x 7→ f ⊕g(x)−p(x)

hasastrong minimum at

x

0

. Let us denote by

f : t 7→ f (x

˜

0

t) − p(x

0

t) − c

and

˜

g : t 7→ g(t) − p(t)

. Then we have that

f

˜

and

˜

g

arelower semi- ontinuousand

f ⊕ ˜

˜

g = h

. We dedu ethen that the map

x 7→ ˜

f ⊕ ˜

g(x)

has a strong minimus at

e

X

. Thus by Lemma 1 we have that

x 7→ ˜

f ⊕ ˜

g(x) + ˜

f ⊕ ˜

g(x

−1

₎

hasastrongminimumat

e

X

andwe anapplyTheorem 5 to obtainthe existen e ofsome

z

0

∈ X

su h that:

(1)

the map

η : z → ˜

f (z

−1

_{) + ˜}

_g(z)

hasastrong minimum at

z

0

∈ X

.

(2) ˜

f (x) ≥ ˜

f ⊕ ˜

g(xz

0

) + ˜

f (z

0

−1

)

and

g(x) ≥ ˜

˜

f ⊕ ˜

g(z

−1

0

x) + ˜

g(z

0

)

for all

z ∈ X

.

Usingthefa tthat

p

isagroup morphismandbyrepla ing

f

˜

and

˜

g

bytheirexpression, we translate

(1)

and

(2)

respe tively asfollow

(1

′

)

the map

η : z → f (x

0

z

−1

_{) + g(z)}

hasa strong minimum at

z

0

∈ X

.

(2

′

_{) f (x) − p(x) ≥ (f ⊕ g(x) − p(x)) − (f ⊕ g(x}

0

) − p(x

0

)) + f (x

0

z

₀

−1

) − p(x

0

z

−1

₀

)

and

g(x) − p(x) ≥ f ⊕ g(x

0

z

0

−1

x) − p(x

0

z

−1

0

x)

− (f ⊕ g(x

0

) − p(x

0

)) + (g(z

0

) − p(z

0

)

, for all

x ∈ X

.

Using the fa t that

x 7→ f ⊕ g(x) − p(x)

has a strong minimum at

x

0

, this implies respe tively

(1

′′

)

the map

η : z → f (x

0

z

−1

_{) + g(z)}

hasastrong minimum at

z

0

∈ X

.

(2

′′

_{) f (x) − p}

hasastrong minimum at

x

0

z

−1

4.1 Properties and useful lemmas. 4.1.1 The semigroup

F

0

(X)

.

We have the following more expli it hara terization of remarkable idempotent. The proof followsimmediately fromLemma 2.

Proposition 6 Let

ϕ ∈ F(X)

. Then,

ϕ

is remarkable idempotent if and only if

ϕ

satisfay :

(1) ϕ(xy) = ϕ(yx)

for all

x, y ∈ X

.

(2) ϕ(x) = ϕ(x

−1

) = 0 ⇔ x = e

X

.

(3) ϕ(xy) ≤ ϕ(x) + ϕ(y)

pour tout

x, y ∈ X

(i.e

ϕ

issubadditive).

Lemma 2 Let

X

beagroupand

e

X

itsidentityelement. Supposethat

ϕ(e

X

) = 0

. Then

ϕ ⊕ ϕ = ϕ

if andonly if

ϕ

issub-additive i.e

ϕ(xy) ≤ ϕ(x) + ϕ(y)

for all

x, y ∈ X

.

Proof.

(⇒)

Suppose that

ϕ ⊕ ϕ = ϕ

. Then

ϕ(xy) =

inf

z∈X



ϕ(xyz

−1

) + ϕ(z)

≤ ϕ(y) + ϕ(x); ∀x, y ∈ X.

(⇐)

Forthe onverse supposethat

ϕ(xy) ≤ ϕ(x) + ϕ(y)

forall

x, y ∈ X

. Thenwe have

ϕ(x) = ϕ((xz

−1

)z) ≤ ϕ(xz

−1

) + ϕ(z); ∀x, z ∈ X

. Taking the innitum over

z ∈ X

we get

ϕ(x) ≤ ϕ ⊕ ϕ(x)

for all

x ∈ X

. Now

ϕ ⊕ ϕ(x) =

inf

z∈X



ϕ(xz

−1

) + ϕ(z)

≤ ϕ(x) + ϕ(e

X

)

= ϕ(x).

Thus

ϕ ⊕ ϕ = ϕ.

Lemma 3 Let

ϕ ∈ F(X)

be a remarkable idempotent. Then

(1)

Then for all

f ∈ F(X)

we have

f ⊕ ϕ = ϕ ⊕ f

(all elements

f ∈ F(X)

ommutes with

ϕ

).

(2) Lip

1

0,ϕ

(X) = F

0,ϕ

(X)

( in parti ular

(Lip

1

0,ϕ

(X), ⊕, ϕ)

isa monoid).

(3)

Every elements

f

of

F

0,ϕ

(X)

is

1

-Lips hitzfor the metri

∆

∞,ϕ

.

Proof.

(1)

Letus rstproves thatforall

f ∈ F(X)

we have

f ⊕ ϕ = ϕ ⊕ f

. Indeed, by using the fa t that

ϕ(xy) = ϕ(yx)

for all

x, y ∈ X

with the following variable hange

t = xy

−1

wehave for all

x ∈ X

,

f ⊕ ϕ(x) =

inf

y∈X



f (xy

−1

) + ϕ(y)

=

inf

y∈X



f (t) + ϕ(t

−1

x)

=

inf

t∈X



ϕ(xt

−1

) + f (t)

= ϕ ⊕ f (x)

(2)

We prove that

F

0,ϕ

(X) ⊂ Lip

1

0,ϕ

(X)

: Let

f ∈ F

0,ϕ

(X)

. Then by the denition of

F

0,ϕ

(X)

we have

ϕ ⊕ f = f ⊕ ϕ = f

. We are going to prove that

f ∈ Lip

1

0,ϕ

(X)

. Indeed, let

x, y ∈ X

andlet

(z

n

)

n

⊂ X

su h thatfor all

n ∈ N

∗

ϕ ⊕ f (y) > ϕ(yz

_n

−1

) + f (z

n

) −

1

n

(8)

Onthe other hand ,

ϕ ⊕ f (x) ≤ ϕ(xz

_n

−1

) + f (z

n

)

(9) By ombining (8)and (9)we have

ϕ ⊕ f (x) ≤ ϕ ⊕ f (y) + ϕ(xz

_n

−1

) − ϕ(yz

−1

_n

) +

1

n

(10)

Nowusing Lemma 2 wehave

ϕ ⊕ ϕ = ϕ

and sowe have

ϕ(xz

n

−1

) = ϕ ⊕ ϕ(xz

−1

n

)

=

inf

t∈X



ϕ(xz

_n

−1

t

−1

) + ϕ(t)

≤ ϕ(xy

−1

) + ϕ(yz

−1

_n

).

(11)

Combining (10)and (11)and sending

n

to

+∞

weobtain that

ϕ ⊕ f (x) ≤ ϕ ⊕ f (y) + ϕ(xy

−1

).

Thisshows that

ϕ ⊕ f ∈ Lip

1

0,ϕ

(X)

. But

ϕ ⊕ f = f

,thus

f ∈ Lip

1

0,ϕ

(X)

.

Weprove now that

Lip

1

0,ϕ

(X) ⊂ F

0,ϕ

(X)

: Let

f ∈ Lip

1

0,ϕ

(X)

. From part

(1)

we have

f ⊕ ϕ = ϕ ⊕ f

. We aregoingto prove that

f ⊕ ϕ = f

. By the denition of

Lip

1

0,ϕ

(X)

we have

f (x) ≤ f (y) + ϕ(xy

−1

); ∀x, y ∈ X.

Taking the innimum over

y ∈ X

, we get

f (x) ≤ f ⊕ ϕ(x)

for all

x ∈ X

. For the onverse inequality wehave

f ⊕ ϕ(x) =

inf

y∈X



f (y) + ϕ(xy

−1

)

≤ f (x) + ϕ(e

X

)

= f (x).

Thus

f ⊕ ϕ = ϕ ⊕ f = f

and so

f ∈ F

0,ϕ

(X)

.

(3)

Thispartfollowseasily from the part

(2)

and the denitionof

Lip

1

0,ϕ

(X).

.

Lemma 4 Let

ϕ ∈ F(X)

be a remarkable idempotent and

ϕ

be the unique extension of

ϕ

to the ompletion

(X, ∆

∞,ϕ

)

. Then

(Lip

1

0,ϕ

(X), d) ∼

= (Lip

1

0,ϕ

(X), d)

as monoids. More pre isely, the map

χ : (Lip

1

_0,ϕ

(X), ⊕, d) → (Lip

1

_0,ϕ

(X), ⊕, d)

f

7→ f =



¯

x 7→ inf

z∈X



ϕ(¯

xz

−1

) + f (z)



is an isometri isomorphism.

Proof. It iseasyto seethat

f ∈ Lip

1

0,ϕ

(X)

forall

f ∈ Lip

1

0,ϕ

(X)

and thatthemap

χ

is well dened sin e if

f = g

on

X

then learly

f = g

on

X

. Observe thatthe restri tion

f

_|X

of

f

to

X

oin ide with

ϕ ⊕ f

, by denition of

f

. The map

χ

isinje tive sin e, if

f = g

on

X

then bythe restri tion to

X

we obtain

ϕ ⊕ f = ϕ ⊕ g

. Thus

f = g

sin e

Lip

1

0,ϕ

(X) = F

0,ϕ

(X) := {f ∈ F

0

(X) : f ⊕ ϕ = ϕ ⊕ f = f }

by Lemma 3. The map

χ

is surje tive. Indeed, let

F ∈ Lip

1

0,ϕ

(X)

and set

f = F

|X

the restri tion of

F

to

X

. Then by denition

f (¯

x) := inf

z∈X



ϕ(¯

xz

−1

) + F (z)

. By the density of

X

in

X

and the ontinuityof

ϕ

and

F

on

X

wehave

f (¯

x) :=

inf

z∈X



ϕ(¯

xz

−1

) + F (z)

=

inf

z∈X



ϕ(¯

xz

−1

) + F (z)

=

ϕ ⊕ F

=

F

The last equality follows from the fa t that

F ∈ Lip

1

0,ϕ

(X) = F

0,ϕ

(X)

by Lemma 3. Letusshownowthat

χ

isamonoidmorphism. Indeed,let

f, g ∈ Lip

1

0,ϕ

(X)

. Usingthe ontinuity of

f

and

g

and the density of

X

in

X

, we easily see that

f ⊕ g

and

f ⊕ g

oin ide on

X

with

f ⊕ g

, sobythe inje tivityof

χ

−1

we have

f ⊕ g = f ⊕ g

. Thus

χ

isamonoid isomorphism. Thefa tthat

χ

isisometri followfromthe thedensityof

X

on

X

and a ontinuityargument

.

For the proofof the following lemmasee[[2 ℄, Lemma 1℄. Lemma 5 Let

f, g ∈ F

0

(X)

. Suppose that

d

∞

(f, g) := sup

x∈X

|f (x) − g(x)| < +∞

then

d(f, g) := sup

x∈X

|f (x) − g(x)|

1 + |f (x) − g(x)|

=

d

∞

(f, g)

1 + d

∞

(f, g)

.

Forea h xedpoint

x ∈ X

, the map

δ

ϕ

x

isdened on

X

by

δ

ϕ

_x

: X → R

z 7→ ϕ(zx

−1

).

Wedene the subset

G

ϕ

0

(X)

of

F

ϕ

(X)

by

G

ϕ

0

(X) := {δ

ϕ

x

: x ∈ X} .

Thefollowing Lemma isan adaptation to our framework of[[2℄, Lemma 3℄

.

Lemma 6 Let

ϕ ∈ F(X)

be a remarkable idempotent. Then,the map

γ

_X

ϕ

: (X, ∆

∞,ϕ

) → (G

ϕ

0

(X), d

∞

)

x 7→ δ

ϕ

_x

is a group isometri isomorphism. Or equivalently, themap

γ

_X

ϕ

: (X,

∆

∞,ϕ

1 + ∆

∞,ϕ

) → (G

0

ϕ

(X), d)

x 7→ δ

_x

ϕ

needtoprove justtherstpart. Indeed,let

x

1

, x

2

∈ X

, weprovethat

δ

ϕ

x

1

⊕ δ

ϕ

x

2

= δ

ϕ

x

1

x

2

. Indeed, let

x ∈ X

, we have :

δ

ϕ

_x

1

⊕ δ

ϕ

x

2

(x) =

inf

z∈X



δ

_x

ϕ

1

(xz

−1

_{) + δ}

ϕ

x

2

(z)

≤ δ

ϕ

x

1

(xx

−1

2

) + δ

x

ϕ

2

(x

2

)

= ϕ x(x

1

x

2

)

−1

+ ϕ(e

X

)

= δ

ϕ

x

1

x

2

(x).

For the onverse inequality, we use fa t that

ϕ(xy) = ϕ(yx)

for all

x, y ∈ X

and the sub-additivityof

ϕ

, to obtain for all

z ∈ X

δ

ϕ

x

1

x

2

(x) = δ

ϕ

x

1

x

2

(x)

= ϕ x(x

1

x

2

)

−1

= ϕ(xx

−1

₂

x

−1

₁

)

= ϕ(x

−1

₁

(xx

−1

₂

))

= ϕ((x

−1

1

xz

−1

)(zx

−1

2

))

≤ ϕ(x

−1

1

(xz

−1

)) + ϕ(zx

−1

2

)

=

ϕ((xz

−1

)x

−1

1

)

+ ϕ(zx

−1

2

)

= δ

_x

ϕ

1

(xz

−1

_{) + δ}

ϕ

x

2

(z)

Bytakingthe innimum over

z

in the last inequality, we obtain

δ

ϕ

x

1

x

2

(x) ≤ δ

ϕ

x

1

⊕ δ

ϕ

x

2

(x).

Thus,

δ

ϕ

x

1

⊕ δ

ϕ

x

2

= δ

ϕ

x

1

x

2

. Inother words,

γ

_X

ϕ

(x

1

x

2

) = γ

_X

ϕ

(x

1

) ⊕ γ

_X

ϕ

(x

2

), ∀x

1

, x

2

∈ X.

(14) Now by the denition of

G

ϕ

_(X)

,

γ

ϕ

X

is a surje tive map. Let us prove that

γ

ϕ

X

is one to one. Indeed, let

x

1

, x

2

∈ X

be su h that

δ

ϕ

x

1

= δ

ϕ

x

2

i.e

ϕ(xx

−1

1

) = ϕ(xx

−1

2

)

for all

x ∈ X

. Sin e

ϕ

satisfythe ondition:

ϕ(x) = ϕ(x

−1

_{) = 0 ⇔ x = e}

X

then byrepla ing

x

by

x

1

in arsttime and

x

by

x

2

inase ondtimeweobtain

0 = ϕ(e

X

) = ϕ(x

1

x

−1

2

) =

ϕ(x

2

x

−1

₁

) = ϕ( x

1

x

−1

₂

)

−1

whi h implies that

x

1

x

−1

2

= e

X

i.e

x

1

= x

2

. Now, sin e

X

is a group and

γ

ϕ

X

is a bije tive map satisfying the formula (14) then

(G

ϕ

0

(X), ⊕)

is a group asimage of group byan isomorphism. The identity element of

(G

ϕ

0

(X), ⊕)

is of ourse

γ

ϕ

X

(e

X

) = δ

ϕ

e

X

= ϕ

. Thus

γ

ϕ

X

isa group isomorphism. Letus show nowthat

γ

ϕ

X

is an isometry. By using the sub-additivity of

ϕ

and the fa tthat

ϕ(e

X

) = 0

we have :

d

∞

(δ

ϕ

x

1

, δ

ϕ

x

2

) = sup

x∈X

|δ

x

ϕ

1

(x) − δ

ϕ

x

2

(x)|

= sup

x∈X

|ϕ(xx

−1

1

) − ϕ(xx

−1

2

)|

= max



sup

x∈X

(ϕ(xx

−1

₁

) − ϕ(xx

−1

₂

)), sup

x∈X

(ϕ(xx

−1

₂

) − ϕ(xx

−1

₁

))



≤ max ϕ(x

2

x

−1

1

), ϕ(x

1

x

−1

2

)

= ∆

∞,ϕ

(x, y)

d

∞

(δ

x

1

(ϕ), δ

x

2

(ϕ)) = sup

x∈X

|δ

ϕ

x

1

(x) − δ

ϕ

x

2

(x)|

= sup

x∈X

|ϕ(xx

−1

1

) − ϕ(xx

−1

2

)|

= max



sup

x∈X

(ϕ(xx

−1

1

) − ϕ(xx

−1

2

)), sup

x∈X

(ϕ(x − x

2

) − ϕ(x − x

1

))



≥ max ϕ(x

2

x

−1

1

); ϕ(x

1

x

−1

2

)

= ∆

∞,ϕ

(x, y)

So

d

∞

(δ

ϕ

x

1

, δ

ϕ

x

2

) = ∆

∞,ϕ

(x, y)

. Thisshows that

γ

ϕ

X

isan isometry. These ond part of the Lemma follows fromLemma 5

.

4.1.2 The semigroup

F(X)

.

In this se tion we prove that using the following proposition, we an dedu t results in the semigroup

F(X)

anoni ally fromthe results of the semigroup

F

0

(X)

.

For all

(f, t), (g, s) ∈ F

0

(X) × R

, we denote by

(f, t)⊕(g, s) := (f ⊕ g, t + s)

and

d

1

((f, t); (g, s)) := d(f, g) + |t − s|

. If

ϕ

isidempotent element,wedenote by

F

ϕ

(X) := {f ∈ F(X) : f ⊕ ϕ = ϕ ⊕ f = f }

the monoid having the identity element

ϕ

and by

Lip

1

ϕ

(X)

the following set

Lip

1

ϕ

(X) :=



f ∈ F(X) : f (x) − f (y) ≤ ϕ(xy

−1

); ∀x, y ∈ X

.

If

M

is a monoid having

ϕ

as identity element, we denote by

U(M )

the group of unit of

M

i.e

U(M ) := {f ∈ M/∃g ∈ M : f ⊕ g = g ⊕ f = ϕ}

.

Proposition 7 Let

X

be a group. Thenthe following assertions hold,

(1)

The following map isan isometri isomorphismof semigroups

π : (F(X), ⊕, d) → (F

0

(X) × R, ⊕, d

1

)

f

7→ (f − inf

X

f, inf

X

f ).

(2)

Let

ϕ ∈ F(X)

be a remarkable idempotent. Then

(i) Lip

1

ϕ

(X) = F

ϕ

(X)

and

(Lip

1

ϕ

(X), ⊕, d) ∼

= (Lip

1

0,ϕ

(X)×R, ⊕, d

1

)

asmonoids.

(ii) (U (Lip

1

ϕ

(X)), ⊕, d) ∼

= (U (Lip

1

0,ϕ

(X)) × R, ⊕, d

1

)

as groups.

Proof. The part

(1)

is easy to verify. The part

(2)

follows from the fa t that the isomorphism

π

send the monoid

F

ϕ

(X)

on the monoid

F

0,ϕ

(X) × R

and the fa tthat

F

0,ϕ

(X) = Lip

1

0,ϕ

(X)

byLemma3. Notealsothat

f ∈ Lip

1

ϕ

(X)

ifandonlyif

f inf

X

f ∈

Lip

1

0,ϕ

(X).

Letusproofnowourrstmainalgebrai resultannoun edintheintrodu tion. A. The monoid

Lip

1

0,ϕ

(X)

.

Theorem 6 Let

ϕ ∈ F(X)

a remarkable idempotent. Then the following assertions hold.

(1) U (Lip

1

0,ϕ

(X)) = χ

−1

◦ γ

ϕ

X

(X).

Where

χ

is the isometri isomorphism of Lemma 4 and

γ

ϕ

X

is the isometri isomorphismof Lemma 6 applied to

(X, ∆

∞,ϕ

) = (X, ∆

∞,ϕ

)

.

(2) (U (Lip

1

0,ϕ

(X)), d) ∼

= (X,

∆

∞

,ϕ

1+∆

∞

,ϕ

)

as groups.

(3) (U (Lip

1

0,ϕ

(X)), d

∞

) ∼

= (X, ∆

∞,ϕ

)

as groups. Proof.

(1)

By usingLemma 4 we have that

χ(U (Lip

1

0,ϕ

(X))) = U (Lip

1

0,ϕ

(X))

and by using Lemma 6 we have that

γ

ϕ

X

(X) = G

ϕ

0

(X)

so we need to prove that the group of unit

U(Lip

1

0,ϕ

(X))

of

Lip

1

0,ϕ

(X)

oin idewith

G

ϕ

0

(X)

.

(∗) G

0

ϕ

(X) ⊂ U (Lip

1

0,ϕ

(X))

: this in lusion is lear sin e we now from Lemma 6 that

G

0

ϕ

(X)

isa group having

ϕ

asidentity element.

(∗∗) U (Lip

1

0,ϕ

(X)) ⊂ G

ϕ

0

(X)

: let

f ∈ U (Lip

1

0,ϕ

(X))

, there exists

g ∈ U (Lip

1

0,ϕ

(X))

su h that

f ⊕ g = ϕ

. Let us prove that the map

x 7→ ϕ(x) + ϕ(x

−1

₎

has a stong minimum at

e

X

on

(X, ∆

∞,ϕ

) = (X, ∆

∞,ϕ

)

. Indeed,sin e

ϕ

is remarkableidempotent then

ϕ ≥ 0 = ϕ(e

X

) = ϕ(e

−1

X

)

andso

x 7→ ϕ(x) + ϕ(x

−1

₎

hasaminimum at

e

X

. Onthe otherhand

∆

∞,ϕ

(x, e

X

) = max(ϕ(x), ϕ(x

−1

_{)) ≤ ϕ(x)+ϕ(x}

−1

₎

. Now,

ϕ(x

n

)+ϕ(x

−1

n

) →

0 ⇒ ∆

∞,ϕ

(x

n

, e

X

)

. Thus,the map

x 7→ ϕ(x) + ϕ(x

−1

₎

hasa strongminimum at

e

X

on the omplet metri invariant group

(X, ∆

∞,ϕ

)

. Sin e

ϕ = f ⊕ g

and sin e

f

and

g

are lower semi ontinuous (in fa t

1

-Lips hitz on

(X, ∆

∞,ϕ

)

), then we an apply Theorem 5to obtainsome

z

0

∈ X

su hthat

f (x) ≥ ϕ(xz

0

) + f (z

−1

0

)

forall

x ∈ X

. Onthe other hand sin e

f ∈ Lip

1

0,ϕ

(X)

, then we have

f (x) ≤ ϕ(xz

0

) + f (z

−1

0

)

for all

x ∈ X

. Thus

f (x) = ϕ(xz

0

) + f (z

0

−1

)

forall

x ∈ X

. Nowsin e

inf

X

f = 0 = inf

X

ϕ

then

f (z

−1

0

) = 0

. Finally, we have

f (x) = ϕ(xz

0

) = δ

ϕ

z

0

−1

(x)

for all

x ∈ X

i.e

f ∈ G

ϕ

0

(X)

.

Thepart

(2)

and

(3)

arejustinterpretationsofthepart

(1)

withthefa tthat

d =

d

∞

1+d

∞

on

G

ϕ

0

(X)

byLemma 5 sin e

d

∞

isnite on thisgroup byLemma 6

.

B. Abstra t monoid

M

ϕ

(X)

.

Denition 4 Let

S

be a subset of , we say that

S

satisfy the translation property

(T )

if the following propertyhold :

(T )

The maps

x 7→ f (zx)

and

x 7→ f (xz)

belongs to

M

ϕ

(X)

for all

f ∈ M

ϕ

(X)

andall

z ∈ X

.

Proposition 8 Let

ϕ ∈ F(X)

be a remarkable idempotent. Let

M

ϕ

(X)

be an abstra t monoid of

F

0

(X)

having

ϕ

asidentity element. Then

(1)

thegroupofunit

(U (M

ϕ

(X)), d)

of

M

ϕ

(X)

isisometri allyisomorphi toasubgroup

G

of

(X,

∆

∞

,ϕ

1+∆

∞

,ϕ

)

(2)

If

M

ϕ

(X)

satisfy the property

(T )

, then

(U (M

ϕ

(X)), d)

is isometri allyisomorphi to a subgroup of

G

su h that

X ⊂ G ⊂ X.

(3)

If thegroup

X

is omplete forthe metri

∆

∞,ϕ

and

M

ϕ

(X)

satisfytheproperty

(T )

then

U(M

ϕ

(X)), d) = (G

ϕ

0

(X), d)

isisometri ally isomorphi to

(X,

∆

∞

,ϕ

1+∆

∞

,ϕ

)

.

Proof.

(1)

Wehave

M

ϕ

(X) ⊂ F

0,ϕ

= Lip

0,ϕ

(X)

. So

U(M

ϕ

(X)) ⊂ U (Lip

0,ϕ

(X))

whi h is isometri allyisomorphi to

X

byTheorem 6. Sothe on lusion.

(2)

If

M

ϕ

(X)

satisfytheproperty

(T )

then

G

ϕ

0

(X) ⊂ U (M

ϕ

(X))

sin e

G

ϕ

0

(X)

isagroup in luded in

M

ϕ

(X)

. On the other hand

G

ϕ

0

(X)

is isometri ally isomorphi to

X

by Lemma 6. Thisgivesthe on lusion with the part

(1)

.

(3)

The on lusion follow fromthe part

(2)

sin e

X = X

in this ase

.

Corollary 5 Let

(X, m)

be omplete metri invariant group. Let

M

be an abstra t submonoidofthemonoid

Lip

1

0

(X)

satisfyingthetranslationproperty

(T )

Thenthegroup ofunit

(U (M ), d)

isisometri allyisomorphi to

(X,

m

1+m

)

. Thisshowthatallsubmonoid

M

of

Lip

1

0

(X)

satisng the property

(T )

, have the samegroup of unit.

Proof. Theproof follow fromthe part

(3)

of Proposition8 sin ein this ase

ϕ = ϕ

m

:

x 7→ m(x, e

X

)

and

(X, m) = (X, ∆

∞,ϕ

m

)

is omplete

.

5 Appli ations to the Bana h-Stone theorem.

Letusprove nowourversionofthe Bana h-Stone theoremwhi h statesthatthe stru -ture of the monoid

(Lip

1

0,ϕ

(X), ⊕, d)

ompletely determine the stru ture of the metri invariantgroup ompletion

(X, ∆

∞,ϕ

)

when

ϕ

isremarkableidempotentandsymmetri . Theorem 7 Let

X

and

Y

be tow groups and let

ϕ ∈ F(X)

and

ψ ∈ F(Y )

be two remarkable idempotents. Then

(1) ⇒ (2) ⇒ (3) ⇔ (4) ⇒ (5) ⇒ (6)

. If moreover we assume that

ϕ

and

ψ

are symetri (i.e

ϕ(x) = ϕ(x

−1

₎

and

ψ(y) = ψ(y

−1

₎

for all

x ∈ X

and all

y ∈ Y

),then

(1) − (6)

are equivalent.

(1)

There exista group isomorphism

T : X → Y

su h that

ψ ◦ T = ϕ

.

(2)

There existasemigroupisomorphismisometri

Φ : F(X) → F(Y )

sushthat

Φ(0) =

0

and

Φ(ϕ) = ψ

.

(3)

There exist a semigroup isomorphism isometri

Φ : F

0

(X) → F

0

(Y )

sush that

Φ(ϕ) = ψ

.

(4) (Lip

1

0,ϕ

(X), d) ∼

= (Lip

1

0,ψ

(Y ), d)

as monoids.

(5) (Lip

1

0,ϕ

(X), d) ∼

= (Lip

1

0,ψ

(Y ), d)

as monoids.

(6) (X, ∆

∞,ϕ

) ∼

= (Y , ∆

∞,ψ

)

as groups.

Proof. Notethatwe have

(X, ∆

∞,ϕ

) =

(

X, ∆

∞,ϕ

)

.

(1) ⇒ (2)

If

T : X → Y

is an isomorphism su h that

ψ ◦ T = ϕ

then the map

Φ : F(X) → F(Y )

dened by

Φ(f ) = f ◦ T

−1

is a semigroup isomorphism, isometri for the metri

d

andsatisfy

Φ(ϕ) = ψ

and

Φ(0) = 0

.

(2) ⇒

(3)

SinceΦ

is a semigroup isomorphism that

Φ(0 ⊕ f ) = Φ(0) ⊕ Φ(f )

for all

f ∈

F(X)

. Sin e

Φ(0) = 0

and

0 ⊕ f = inf

X

f

, then we obtainthat

Φ(inf

X

f ) = inf

Y

Φ(f )

for all

∈ F(X)

. In parti ular,

0 = Φ(0) = inf

Y

Φ(f )

for all

f ∈ F

0

(X)

, this show that

Φ

send

F

0

(X)

on

F

0

(Y ).

(3) ⇒ (4)

Sin e

Φ(ϕ) = ψ

and

Φ

isa semigroup isomorphism then learly

Φ

maps the monoid

F

0,ϕ

(X)

ontothemonoid

F

0,ψ

(Y )

. Sousingthe fa tthat

F

0,ϕ

(X) = Lip

1

0,ϕ

(X)

and

F

0,ψ

(Y ) = Lip

1

0,ψ

(Y )

byLemma 3,we obtain that

(Lip

1

0,ϕ

(X), d) ∼

= (Lip

1

0,ψ

(Y ), d)

asmonoids by

Φ

.

(4) ⇔ (5)

Follows fromLemma 4.

(5) ⇒ (6)

Sin e

(Lip

1

0,ϕ

(X), d) ∼

= (Lip

1

0,ψ

(Y ), d)

andsin eisomorphism ofmonoidssend the group of unit on the group of unit, we have

U(Lip

1

0,ϕ

(X)) ∼

= U (Lip

1

0,ϕ

(Y ))

. Using Theorem 6 we obtainthat

(X, ∆

∞,ϕ

) ∼

= (Y , ∆

∞,ψ

)

.

Supposenowthat

ϕ

and

ψ

aresymmetri ,then

∆

∞,ϕ

(x, e

X

) = ∆

∞,ϕ

(x, e

X

) = max(ϕ(x), ϕ(x

−1

) = ϕ(x)

for all

x ∈ X

and

∆

∞,ψ

(y, e

Y

) = ∆

∞,ψ

(y, e

Y

) = ψ(y)

for all

y ∈ Y

. We need to prove that

(6) ⇒ (1)

. Indeed, Let

T : (X, ∆

∞,ϕ

) → (Y , ∆

∞,ψ

)

be an isomorphism isometri . In parti ular we have

ψ(T (x)) = ∆

_∞,ψ

(T (x), e

Y

)

= ∆

_∞,ψ

(T (x), T (e

X

))

= ∆

∞,ϕ

(x, e

X

)

= ϕ(x).

This on lude the proof

.

Thefollowing orollaryshowsthatthemonoid

(Lip

1

0

(X), ⊕, d)

ompletelydetermine the stru ture ofthe ompletemetri invariant group

(X, m)

.

Corollary 6 Let

(X, m)

and

(Y, m

′

₎

be omplete metri invariant groups. Let

ϕ

m

:

x 7→ m(x, eX)

for all

x ∈ X

and

ψ

m

′

: y 7→ m

′

(y, e

_Y

)

forall

y ∈ Y

. Then,thefollowing assertions are equivalent.

(1) (X, m) ∼

= (Y, m

′

)

as groups.

(2)

There exist a semigroup isomorphism isometri

Φ : F(X) → F(Y )

sush that

Φ(ϕ

m

) = ψ

m

′

and

Φ(0) = 0

.

(3)

There exist a semigroup isomorphism isometri

Φ : F

0

(X) → F

0

(Y )

sush that

Φ(ϕ

m

) = ψ

m

′

.

(4) (Lip

1

0

(X), d) ∼

= (Lip

1

0

(Y ), d)

as monoids.

(5)

There exist a semigroup isomorphism isometri

Φ : (Lip

0

(X), d) → (Lip

0

(Y ), d)

sush that

Φ(ϕ

m

) = ψ

m

′

.

Proof. Thepart

(1) ⇔ (2) ⇔ (3) ⇔ (4)

followfromTheorem7andthefa tthat

(X, m)

and

(Y, m

′

₎

are omplete, and that

Lip

1