On isometries of Product of normed linear spaces

(1)

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On isometries of Product of normed linear spaces

Mohammed Bachir

To cite this version:

(2)

Mohammed Ba hir

November23, 2014

Laboratoire SAMM4543, UniversitéParis1Panthéon-Sorbonne, CentreP.M.F. 90rue

Tolbia 75634Paris edex13

Email: Mohammed.Ba hiruniv-paris1.fr,

Abstra t. We give a ondition on norms under whi h twove tornormed spa es

X

and

Y

areisometri allyisomorphi ifandonlyif

X

× R

and

Y

× R

areisometri allyisomorphi . We also provethat this resultfail forarbitrarynorms even if

X

= Y = R

2

bybuilding ageneri

ounterexamples.

Keyword,phrase: Normedve torspa eandisometries.

1 Introdu tion

We areinterestedin this paperin the followingquestion. Let

X

and

Y

beve torspa es and let

N

X

and

N

Y

betwonormson

(X × R, N

X

)

and

(Y × R, N

Y

)

respe tively. Thenorm

N

X

on

X

(andinasimilar way

N

Y

)denotes

N

X

(x, 0)

forall

x

∈ X

.

Problem. It is true that

(X × R, N

X

)

and

(Y × R, N

Y

)

are isometri ally isomorphi if and onlyif

(X, N

X

)

and

(Y, N

Y

)

areisometri allyisomorphi ?

Webeginbyshowingthat inthegeneral asetheanswertothisquestionisnoforarbitrary

norms

N

X

and

N

Y

,evenwhen

X

and

Y

aretwodimensionalve torspa es,by onstru tinga generi ounterexamples(SeeTheorem1). WeprovetheninTheorem2thattheresultistrue

forallnorms

(N

X

, N

Y

)

satisfyingthefollowingproperty

(P )

.

Denition1 Let

X

and

Y

betwove torspa es. Let

N

X

and

N

Y

betwo normson

X

× R

and

X

× R

respe tively. Wesaythethe pair

(N

X

, N

Y

)

satisfy theproperty

(P )

iffor all

x

∈ X

and all

y

∈ Y

:

N

X

(x, 0) = N

Y

(y, 0) ⇒ N

X

(x, λ) = N

Y

(y, λ), ∀λ ∈ R.

In allthe arti leweidentify

X

with

X

× {0}

andthe norm

N

X

on

X

denotes

N

X

(x, 0)

for all

x

∈ X

.

Exemples 1 Let

(X, k.k

X

)

and

(Y, k.k

Y

)

betwo normedve torspa es. Let

p

∈ [1, +∞[

and

N

X,p

(x, t) := (kxk

p

X

+ |t|

p

)

1 p

,

N

X,∞

(x, t) := max(kxk

X

,

|t|),

for all

(x, t) ∈ X × R.

Inasimilarwaywedene

N

Y,p

and

N

Y,∞

. Then thepairs

(N

X,p

, N

Y,p

)

and

(N

X,∞

, N

Y,∞

)

satisesthe property

(P )

.

(3)

Proposition1 Let

N

R

2

beanynormon

R

2

su hthat

N

X

(x, t) := N

R

2 (kxk

X

,

|t|)

forall

(x, t) ∈

X

× R

denedanormon

X

× R

(Similarly we dene

N

Y

on

Y

× R

). Then

(N

X

, N

Y

)

satisfy the property

(P )

.

Proof. Let

x

∈ X

and

y

∈ Y

be su h that

N

X

(x, 0) = N

Y

(y, 0)

. Then

N

R

2 (kxk

X

,

0) =

N

_R

2 (kyk

Y

,

0)

andso

kxk

X

N

R

2 (1, 0) = kyk

Y

N

_R

2 (1, 0)

,whi himpliesthat

kxk

X

= kyk

Y

. It fol-lowsthat

N

R

2 (kxk

X

,

|λ|) = N

_R

2 (kyk

Y

,

|λ|)

for all

λ

∈ R

. Inotherwords

N

X

(x, λ) = N

Y

(y, λ)

forall

λ

∈ R

.

Theproblemmentionedabovewasmotivatedatthersttimein[1℄byquestions onne ted

to the Bana h-Stone theorem, and solvedpositively only for the parti ular norms

N

X,p

and

N

Y,p

when

p

∈ [1, +∞[\ {2}

. The te hnique used in [1℄ did not in lude the ase p=2. The property

(P )

hear is moregeneraland allowed to in lude variednorms. We givein se tion4. othersimpleexamplesofappli ationsofTheorem2.

2 A generi ounterexample.

Theorem 1 Let

X

= Y = R

2

. For ea h norm

k.k

X

on

X

there existsa norm

k.k

Y

on

Y

, a norm

N

X

on

X

× R

andanorm

N

Y

on

Y

× R

su hthat:

(1) (X, k.k

X

)

is notisometri ally isomorphi to

(Y, k.k

Y

)

.

(2) (X × R, N

X

)

isisometri allyisomorphi to

(Y × R, N

Y

)

.

(3)

the restri tionof

N

X

to

X

oin idewith

k.k

X

andthe restri tion of

N

Y

to

Y

oin ide with

k.k

Y

.

Proof. Let

p

∈ [1, +∞[

. Letusdene

N

X

and

N

Y

as follow:

N

X

(x

1 , x

2 , t) := (k(x

1 , x

2 )k

p

X

+ |t|

p

₎

1 p

_,

∀(x

1 , x

2 , t) ∈ X × R

and

N

Y

(y

1 , y

2 , s) := (|y

2 |

p

+

k(y

1 , s)k

p

X

a

p

)

1 p

,

∀(y

₁

, y

₂

, s) ∈ Y × R.

Where

a

= k(1, 0)k

X

. Letus dene thenorm

k.k

Y,p

on

Y

as follows

k(y

1 , y

2 )k

Y,p

:= (|y

1 |

p

₊

|y

2 |

p

)

1 p

forall

(y

1 , y

2 ) ∈ Y

. Clearly,

N

X

(x

1 , x

2 ,

0) = k(x

1 , x

2 )k

X

,

∀(x

1 , x

2 ) ∈ X

and

N

Y

(y

1 , y

2 ,

0) = (|y

1 |

p

+ |y

2 |

p

)

1 p

:= k(y

₁

, y

₂

)k

_Y,p

,

∀(y

₁

, y

₂

) ∈ Y.

(Sin e

k(y

1 ,0)k

p

X

a

p

= |y

1 |

k(1,0)k

p

X

a

p

= |y

1 |

). Onthe otherhand,the followingmap isan isometri isomorphism:

Θ : (X × R, N

X

)

→ (Y × R, N

Y

)

(x

1 , x

2 , t) 7→ (ax

1 , t, ax

2 ).

Now,there existto ases:

Case 1: Ifeverypointof thesphere

S

X

of

X

isanextremepoint,we hoose

p

= 1

andso

S

Y

hasanon extremepointsin einthis ase

k(y

1 , y

2 )k

Y,1

= |y

1 | + |y

2 |

(Forexample

(

1

2 ,

1

2 )

isnot extremefor

k.k

Y,1

). Consequently

X

and

Y

annotbeisometri allyisomorphi .

Case 2: Ifthereexistssomepointofthesphere

S

X

whi hisnotextremepointthenwe hoose

p

= 2

andsoeverypointsof

S

Y

isanextremepointsin e

k(y

1 , y

2 )k

Y,2

= (|y

1 |

2 _{+ |y}

2 |

2 )

1

2

isthe eu lideannorm. Also

X

and

Y

annotbeisometri allyisomorphi .

(4)

Theorem 2 Let

X

and

Y

beave torspa es. Supposethat

(N

X

, N

Y

)

satisfythe property

(P )

. Then

(X × R, N

X

)

and

(Y × R, N

Y

)

areisometri ally isomorphi if and only if

(X, N

X

)

and

(Y, N

Y

)

areisometri ally isomorphi .

Theproofoftheabovetheoremisgivenin se tion3.2aftersomelemmas.

3.1 Notations and lemmas.

Weneedsomenotationsandlemmas. Let

Θ : (X × R, N

X

) → (Y × R, N

Y

)

beanisomorphism isometri . Weset

(a, u) = Θ

−1

_{(0, 1)}

and

(b, v) = Θ(0, 1)

. Let us denethe linear ontinuous map

χ

X

as follow:

χ

X

: X × R

−→ R

(x, t)

7→

t

Wedeneanalogouslythemap

χ

Y

by

χ

Y

: Y × R

−→ R

(y, t)

7→

t

Weobtainthefollowinglinearmapon

Y

× {0}

:

χ

X

◦ Θ

−1

: Y × {0} −→ R

Analogouslywehavealsothelinearmapon

X

× {0}

:

χ

Y

◦ Θ : X × {0} −→ R

Letusset

X

0 := Ker (χ

Y

◦ Θ)

and

Y

0 := Ker χ

X

◦ Θ

−1

.

Remark1 The linear spa es

X

0

and

Y

0

are not ne essarily losed sin e

χ

X

and

χ

Y

are not ne essarily ontinuous.

Lemma1

X

0

and

Y

0

are isometri ally isomorphi . Morepre isely,the map

Θ : (X

0 , N

X

)

→ (Y

0 , N

Y

)

(z, 0) 7→ Θ(z, 0)

(1)

isan isomorphism isometri .

Proof. Sin e

Θ

is anisomorphismisometri ,itsu esto showthattherestri tionof

Θ

to

X

0

is onto. Indeed, let

(y, 0) ∈ Y

0

. Clearly,

(z, 0) := Θ

−1

_{(y, 0) ∈ X}

0

sin e

χ

Y

◦ Θ(Θ

−1

_{(y, 0)) =}

χ

Y

(y, 0) = 0

andwehave

(y, 0) = Θ(z, 0)

.

Lemma2 Wehave onlytwo ases.

Case1:

u

6= 0

. In this ase,wehave

X

× {0} = X

0

. Case2:

u

= 0

. In this asewe have

Θ

−1

_{(0, 1) = (a, 0)}

and

X

× {0} = X

0 ⊕ R(a, 0)

. Similarly wehave,

Case1:

v

6= 0

. In this ase,wehave

Y

× {0} = Y

0

.

(5)

Proof. Forall

x

∈ X

thereexists

(y

x

, λ

x

) ∈ Y × R

su hthat

(x, 0) = Θ

−1

(y

x

, λ

x

)

= Θ

−1

(y

x

,

0) + λΘ

−1

(0, 1)

= Θ

−1

(y

x

,

0) + λ

x

(a, u)

(2)

= Θ

−1

_(y

x

,

0) + (λ

x

a, λ

x

u)

Sin e

Θ

−1

_(y

x

,

0) ∈ X

0 ⊂ X × {0}

andalso

(x, 0) ∈ X × {0}

,thenfromtheaboveequationwe obtainthat

(λ

x

a, λ

x

u) ∈ X × {0}

whi h impliesthat

λ

x

u

= 0

. Sowehave:

Case1:

u

6= 0

. In this ase,

X

× {0} = X

0

. Indeed, if

u

6= 0

then

λ

x

= 0

and so

(x, 0) =

Θ

−1

_(y

x

,

0) ∈ X

0

,forall

x

∈ X

i.e

X

×{0} ⊂ X

0

. Ontheotherhandweknowthat

X

0 ⊂ X ×{0}

. Case2:

u

= 0

. Inthis asewehave

Θ

−1

_{(0, 1) = (a, 0)}

andso

X

= X

0 ⊕ R(a, 0)

. Indeed,We have

X

0 ∩ R(a, 0) = (0, 0)

, sin eif

α

is areal numbersu h that

α(a, 0) ∈ X

0

then

0 = χ

Y

◦

Θ (α(a, 0)) = αχ

Y

(0, 1) = α

. Inotherwordsfrom(2),forall

x

∈ X

,thereexist

(y

x

, λ

x

) ∈ Y ×R

su hthat

(x, 0)

= Θ

−1

(y

x

,

0) + λ

x

(a, 0).

whith

Θ

−1

_(y

x

,

0) ∈ X

0

. Thus

X

× {0} ⊂ X

0 ⊕ R(a, 0) ⊂ X × {0}

andso

X

× {0} = X

0 ⊕ R(a, 0)

.

Inasimilarwayweobtainthese ondpartofthelemma.

Lemma3 Wehave,

u

= 0

ifandonly if

v

= 0

.

Proof. Suppose that

v

= 0

. Thenforall

(x, t) ∈ X × R

,wehave

Θ(x, t) = Θ(x, 0) + Θ(0, t) =

Θ(x, 0) + tΘ(0, 1) = Θ(x, 0) + t(b, v) = Θ(x, 0) + (tb, 0)

. Now,wearegoingtoprovethat

u

= 0

. Supposethatthe ontraryhold,that is

u

6= 0

. Then

X

× {0} = X

0

(Seethe ase1. inLemma 2). So

Θ(x, 0) ∈ Θ(X × {0}) = Θ(X

0 ) = Y

0

, sin e

Θ

is an isomorphism isometri from

X

0

onto

Y

0

(See the formula(1)). Nowsin e

Y

0 ⊂ Y × {0}

, then

Θ(x, 0) + t(b, 0) ∈ Y × {0}

. In other words,

Θ(x, t) ∈ Y × {0}

for all

(x, t) ∈ X × R

. So

Θ(X × R) ⊂ Y × {0}

. But

Θ

isan isomorphismbetween

X

× R

and

Y

× R

. Thisimpliesthat

Y

× {0} = Y × R

whi hisimpossible. Thus

u

= 0

. Inasimilarwayweobtainthe onverse.

3.2 Proof of Theorem 2 and some orollaries.

Wegivenowtheproofofthemain result.

Proof of Theorem 2. For the if part, let

T

: (X, N

X

) → (Y, N

Y

)

be an isomorphism iso-metri . Let usdene

Θ : (X × R, N

X

) → (Y × R, N

Y

)

by

Θ(x, λ) = (T (x), λ)

. Then, learly

Θ

is an isomorphismandby theproperty

(P )

itis alsoisometri . We provenowthe onlyif part. By ombiningLemma2and Lemma3wehavethat:

Case1. If

u

6= 0

and

v

6= 0

,then

X

× {0} = X

0

and

Y

× {0} = Y

0

. SobyLemma1we on lude that

X

× {0}

and

Y

× {0}

areisometri allyisomorphi forthenorms

N

X

and

N

Y

. So

(X, N

X

)

and

(Y, N

Y

)

areisometri allyisomorphi .

Case2. If

u

= 0

and

v

= 0

, using Lemma 2we havethat

Θ

−1

_{(0, 1) = (a, 0)}

and

X

× {0} =

X

0 ⊕ R(a, 0)

and

Θ(0, 1) = (b, 0)

and

Y

× {0} = Y

0 ⊕ R(b, 0)

. Nowweprovethat themap

ψ

: X × {0} = X

0 ⊕ R(a, 0) → Y × {0} = Y

0 ⊕ R(b, 0)

(z, 0) + λ(a, 0)

7→ Θ(z, 0) + λ(b, 0)

is anisomorphismisometri . Indeed, thefa t that

ψ

islinearand onto map is learby using Lemma 1. Letus provethat

ψ

isisometri forthenorms

N

X

and

N

Y

. Sin e

(z, 0) ∈ X

0

, by (1)there exist

(y, 0) ∈ Y

0

su hthat

Θ(z, 0) = (y, 0)

. Sin e

(z, 0) + λ(a, 0)

= Θ

−1

(Θ(z, 0)) + λΘ

−1

(0, 1)

= Θ

−1

(Θ(z, 0) + (0, λ))

= Θ

−1

(y, λ)

(6)

then,using thefa t that

Θ

−1

isisometri wehave

N

X

((z, 0) + λ(a, 0)) =

N

X

(Θ

−1

(y, λ))

=

N

Y

(y, λ).

(3)

On theother hand weknown that

(b, 0) = Θ(0, 1)

so

Θ(z, 0) + λ(b, 0) = Θ(z, 0) + λΘ(0, 1) =

Θ(z, λ)

. Thus,usingthefa tthat

Θ

isisometri wehave,

N

Y

(ψ ((z, 0) + λ(a, 0))) =

N

Y

(Θ(z, 0) + λ(b, 0))

=

N

Y

(Θ(z, λ))

=

N

X

(z, λ).

(4)

But

N

X

(z, 0) = N

Y

(y, 0)

sin e

Θ(z, 0) = (y, 0)

and

Θ

is isometri . Sin e

(N

X

, N

Y

)

satisfythe property

(P )

then

N

X

(z, λ) = N

Y

(y, λ)

. Thus,using theformulas(3)and (4)weobtainthat

ψ

is isometri .

Remark2 Byindu tion, we an easily extend the above theorem to

X

× R

n

(

n

∈ N

∗

) if we

assume that

(N

X

, N

Y

)

isapair ofnorms satisfyingthe following property

(P

n

₎

: for all

x

∈ X

all

y

∈ Y

, all

i

∈ {1, 2, ...n}

andall

(s

1 , s

2 , ..., s

i

); (s

′

1 , s

′

2 , ..., s

′

i

) ∈ R

i

: if

N

X

(x, s

1 , s

2 , ..., s

i

,

0, 0, ..., 0) = N

Y

(y, s

′

1 , s

′

2 , ..., s

′

i

,

0, 0, ..., 0)

then

N

X

(x, s

1 , s

2 , ..., s

i

, λ,

0, ..., 0) = N

Y

(y, s

′

1 , s

′

2 , ..., s

′

i

, λ,

0, ..., 0), ∀λ ∈ R

.

Exemples 2 Let

p

∈ [1, +∞[

, and

N

X,p

(x, s

1 , ..., s

n

) = (kxk

p

_X

+

n

X

k=1

|s

k

|

p

)

1 p

_,

N

X,∞

(x, s

1 , ..., s

n

) = max(kxk

X

,

|s

1 |, ..., |s

n

|)

for all

(x, s

1 , ..., s

n

) ∈ X × R

n

. In a similar way we dene

N

Y,p

and

N

Y,∞

. Then the pairs

(N

X,p

, N

Y,p

)

and

(N

X,∞

, N

Y,∞

)

satisesthe property

(P

n

₎

.

Corollary 1 Let

X

and

Y

be ave torspa es. Let

n

∈ N

∗

andsupposethat

(N

X

, N

Y

)

satisfy

(P

n

₎

. Then

(X ×R

n

_{, N}

X

)

and

(Y ×R

n

_{, N}

Y

)

(X, N

X

)

and

(Y, N

Y

)

As a remark we have the following orollary for inner produ t spa es. Note that a non

omplete inner produ t spa e has no orthonormal basis in general (See [5℄). The symbol

∼

=

meansisometri allyisomorphi .

Corollary 2 Let

(H, k.k

H

)

and

(L, k.k

L

)

betwo innerprodu tspa e(notne essary omplete). Then

(H, k.k

H

) ∼

= (L, k.k

L

)

ifandonlyifforallnitedimensionalsubspa es

E

⊂ H

and

F

⊂ L

su hthat

dim(E) = dim(F )

wehavethat

(E

⊥

_,

_k.k

H

) ∼

= (F

⊥

,

k.k

L

)

. Where

E

⊥

and

F

⊥

denotes

the orthogonal of

E

and

F

respe tively.

Proof. Let

E

⊂ H

and

F

⊂ L

su h that

dim(E) = dim(F ) = n

for

n

∈ N

. Bythe lassi al proje tion theorem on a omplete ve tor subspa e of an inner produ t spa e, we have

H

=

E

⊥

_{⊕ E}

and

L

= F

⊥

_{⊕ F}

. On theother hand itis learthat

(H, k.k

H

) ∼

= (E

⊥

_{× R}

n

_{, N}

E

⊥

,2

)

and

(L, k.k

L

) ∼

= (F

⊥

_{× R}

n

_{, N}

F

⊥

,2

)

, where

N

E

⊥

,2

and

N

F

⊥

,2

are dened as in theExample 2

with

p

= 2

. Sin e

(N

E

⊥

,2

, N

F

⊥

_,2

)

satisfy

(P

n

₎

thenfrom Corollary1weobtain

(E

⊥

_,

_k.k

H

) ∼

=

(F

⊥

_,

_k.k

(7)

We give in this se tion two appli ations of Theorem 2. We denote by

(C

1 _{[0, 1], N}

C

1 _[0,1]

)

the spa e of ontinuously dierentiable fun tions on

[0, 1]

endowed with thenorm

N

C

1 [0,1]

(f ) :=

N

_R

2 (kf

′

k

∞

,

|f (0)|)

,where

N

R

2

denotesanynormsatisfyingProposition 1. Let

(X, k.k

X

)

bea Bana hspa e. Wedenoteby

N

X

thenormdenedon

X

× R

by

N

X

(x, t) := N

R

2 (kxk

X

,

|t|)

for all

(x, t) ∈ X × R

. Finally, wedenote by

(C[0, 1], k.k

∞

)

thespa e of ontinuous fun tions on

[0, 1]

endowedwiththesupremumnorm.

Proposition2 Wehave

(X×R, N

X

) ∼

= (C

1 _{[0, 1], N}

C

1 [0,1]

)

,ifandonlyif

(X, k.k

X

) ∼

= (C[0, 1], k.k

∞

)

.

Proof. Let us dene thenorm

N

C[0,1]

on

C[0, 1] × R

by

N

C[0,1]

(g, t) := N

R

2 (kgk

∞

,

|t|)

for all

(g, t) ∈ C[0, 1] × R

. Letus onsider themap

χ

: (C

1 [0, 1], N

C

1 [0,1]

) → (C[0, 1] × R, N

C[0,1]

)

f

7→ (f

′

_{, f(0))}

Clearly,

χ

is an isomorphism isometri . So we have

(X × R, N

X

) ∼

= (C[0, 1] × R, N

C[0,1]

)

. Sin e

(N

X

, N

C[0,1]

)

satisfy the property

(P )

by Proposition 1 then using Theorem 2 we ob-tain that

(X, k.k

X

) ∼

= (C[0, 1], k.k

∞

)

, sin e

N

X

(kxk

X

,

0) = kxk

X

N

R

2 (1, 0)

and

N

C[0,1]

(g, 0) =

kgk

∞

N

R

2 (1, 0)

.

Let us re all some notions. Let

K

and

C

be onvex subsets of ve tor spa es. A fun tion

T

: K → C

is said to be ane if for all

x, y

∈ K

and

0 ≤ λ ≤ 1

,

T

(λx + (1 − λ)y) =

λT

(x) + (1 − λ)T (y)

. Thesetofall ontinuousreal-valuedanefun tions ona onvexsubset

K

ofatopologi alve torspa ewillbedenotedby

Af f

(K)

. Clearly,alltranslatesof ontinuous linearfun tionalsareelementsofA

(K)

,but the onversein nottruein general(see[4℄page 22.). However,wedohavethefollowingrelationship.

Proposition3 ([4 ℄,Proposition4.5)Assumethat

K

isa ompa t onvexsubsetofaseparated lo ally onvexspa e

X

then

n

a

∈ Af f (K) : a = r + x

∗

_|K

f or some x

∗

∈ X

∗

and some r

∈ R

o

isdense in

(Af f (K), k.k

∞

)

, where

k.k

∞

denotesthe normof uniform onvergen e. Butintheparti ular asewhen

X

isaBana hspa eand

K

= (B

X

∗

, w

∗

₎

istheunitballof

thedualspa e

X

∗

endowedwiththeweakstartopology,thewellknownresultduetoBana h

andDieudonné statesthat:

Theorem 3 (Bana h-Dieudonné). Thespa e

(Af f

0 (B

X

∗

), k.k

∞

)

isisometri ally identied to

(X, k.k)

. In other words,

Af f

0 (B

X

∗

) =

ˆ

z

|B

X∗

: z ∈ X .

Where

Af f

0 (B

X

∗

)

denotes the spa e of all ane weak star ontinuous fun tions that vanish at

0

and

z

ˆ

: p 7→ p(z)

for all

p

∈ X

∗

and

z

ˆ

|B

X∗

denotes the restri tionof

z

ˆ

to

B

X

∗

.

Now,let

X

and

Y

betwoBana hspa esandletusendowedthespa e

Af f

(B

X

∗

)

(andin a similar way the spa e

Af f

(B

Y

∗

)

) with the norm

N

(f ) := N

R

2 (kf − f (0)k

∞

,

|f (0)|)

for all

f

∈ Af f (B

X

∗

)

, where

N

R

2

denotes anynorm on

R

2

satisfying Proposition 1. We obtainthe

followingversionoftheBana h-Stonetheoremforanefun tions(Formoreinformationabout

theBana h-Stonetheoremsee[2℄and[3℄).

Proposition4 Let

X

and

Y

betwo Bana h spa es. Then thefollowingassertionsare equiva-lent.

(1) (Af f (B

X

∗

), N )

and

(Af f (B

Y

∗

), N )

(8)

Proof. Let

N

˜

be the norm on

Af f

0 (B

X

∗

) × R

dened by

N

˜

(f

0 , t) := N

R

2 (kf

0 k

∞

,

|t|)

for all

(f

0 , t) ∈ Af f

0 (B

X

∗

) × R

. Letus onsiderthemap,

χ

: (Af f (B

X

∗

), N ) → (Af f

0 (B

X

∗

) × R, ˜

N

)

f

7→ (f − f (0), f (0))

Clearly,

χ

isanisometri isomorphism. ThususingTheorem1wehavethat

(Af f (B

X

∗

), N )

and

(Af f (B

X

∗

), N )

(Af f

0 (B

X

∗

), k.k

∞

)

and

(Af f

0 (B

Y

∗

), k.k

∞

)

areisometri allyisomorphi ,whi hisequivalentbyTheorem3tothefa t that

(X, k.k

X

)

and

(Y, k.k

Y

)

areisometri allyisomorphi .

A knowledgments. Theauthor thanksProfessorJean-BernardBaillon andProfessorGilles

Godefroyforthediverseenri hingdis ussions.

Referen es

[1℄ M.Ba hir,Theinf- onvolutionasalawofmonoid.AnanaloguetoBana h-Stonetheorem

J.Math. Anal.Appl420,(2014)No.1,145-166.

[2℄ M. Ba hir, Surla dierentiabilité génériqueetle théorèmede Bana h-Stone C.R.A ad.

S i. Paris,330(2000),687-690.

[3℄ M.I. Garrido,J.A.Jaramillo, Variationsonthe Bana h-Stone Theorem,Extra taMath.

Vol.17,N. 3,(2002)351-383.

[4℄ R.Phelps,Le turesonChoquet'sTheorem,Se ondEdition,Le tureNotesinMath.Berlin,

(1997).

[5℄ P.Halmos,AHilbert Spa eProblem Book,Springer-Verlag,NewYork-Heidelberg-Berlin,