Dérivabilité d`un groupe additif continu à un paramètre d

−→
En⃗x 7→ ||⃗x||
||⃗x|| = 0 ⇔⃗x = 0 ,∀λ||λ⃗x|| =|λ|||⃗x||,||x1+⃗x2|| ≤ ||⃗x1|| +||⃗x2||
Σ
∀||||,∀||||1,∃a > 0,∃b > 0 : ∀⃗x , a||⃗x|| ≤ ||⃗x||1≤b||⃗x||
−→
En
||f|| = Max{||f(⃗x)||, ⃗x ∈Σ}
λ
||f|| = 0 ⇔f= 0;,||λ f|| =|λ|||⃗x|| ,||f+g|| ≤ ||f|| +||g||
||f◦g|| ≤ ||f||||g||
fa b
c d 
||f(cos θ⃗ i+ sin θ⃗ j)||2= L cos2θ+ 2M cos θsin θ+ N sin2θ
=a2+c2=ab +cd =c2+d2
2||f(cos θ⃗ i+ sin θ⃗ j)||2= (L −N) cos 2θ+ 2M sin 2θ+ (L + N)
2||f||2= (L + N) + (L −N)2+ 4M2mf||f(⃗x)||
2m2
f= (L + N) −(L −N)2+ 4M2:||f||2m2 f= det A2f mf= 0
f
a0
0d:||f|| = Max(|a|,|d|),a b
b a :||f|| =|a|+|b|,a−b
b a :||f|| =√a2+b2
s:z7→ (a+ib)z
s a +ib u |s(z)|=|(a+ib)u|=|a+ib|

e f ||f−e|| =k < 1
g=f−e Sn(g) = e−g+g2− ··· + (−1)igi+···(−1)ngn
fSn=e+ (−1)ngn+1 Sn
q < p Sp−Sq= (−1)q+1gq+1(e−g+···(−1)p−q−1gp−q−1)
Sp−Sq < kq+1/(1 −k)
n Snh gn
fh =e
−→
En
{f(s)}−→ E2
s
s= 0 s
f(s)−f(s0) = f(s0)f(s−so)−ee
||f(s)−f(s0)|| ≤ ||f(s0)||||f(s−so)−e||
s1
s0f(s)ds f(s)
s1
s0aij(s)ds f
s0< s1
s1
s0f(s)ds
≤s1
s0||f(s)||ds
s+t
sf(u)du
t
0f(s+u)du =f(s)t 0f(u)du ou f(s)h(t) avec h(t) = t 0f(u)du
t h(t)f t = 0
ε||f(t)−e|| <1/2t ε t ∈]0, ε[
]0, t[⊂]0, ε[
t
0f(u)−edu
≤t
0||f(s)−e)ds|| ≤ t/2
1
tt
0f(s)ds −e
≤1/21
tt
0f(s)ds h(t) = t 0f(s)ds t ̸= 0
f h
f(s) = h−1(t)s+t
sf(u)du
s+t
sf(u)du t f(t)−e
f′(s) = h−1(t)f(t)−ef(s)f′(s) = b f(s)
b f(0) = e b =f′(0)
b
f(t)eBtb h(t) (t)
•B = λ0
µ0B2=λB, eBt= I + eλt −1
λB,H(t) = tI + eλt −1−λt λ2B
•B = 0 0
µ0B2= 0 , eBt= I + tB,H(t) = tI + t2
2B
bh(t) = f(t)−e t b2=λb b2= 0
BH(t) = tB + eλt −1−λt
λB = eλt −1 λB = eBt−I
BH(t) = tB = eBt−I

R(
⃗
i, O)
gt:x7→ x′=ktx+ht
x′
1=ktht 0 1 x 1g′(0) : B = a α 0 0
•a̸= 0 2=a y′−ay = 0
eBt=I+eat −1
aB = 
eat eat −1
aα
0 1 
x′=eatx+eat −1
aα x′+α a=eatx+α ax0=−α a
eat
•a2= 0
eBt=I+tB = 1α t
0 1 :x′=x+α t
α g(t) = e
R(
⃗
i,⃗ j, O)
gt:x7→ x′=ktx+hty+lt, y′=mtx+nty+pt
B(
⃗
i,⃗ j, O)
x′
y′
1
=A1U
0 1 
x′
y′
1
1g′(0) = Σ V
0 0 
10 Σ = α
β
Σ
2=Σ2Σ V
0 0 3=Σ3Σ2V 0 0 
Σ
Σ (φ0, φ1)
(ψ0= 1, ψ1, ψ2)
•
Σ = 0−1
1 0 : Σ2+ I = 0 , φ0= cos t , φ1= sin t
3+ = 0 ψ0= 1, ψ1= sin t, ψ2= 1 −cos t

eBt=
cos t−sin t α sin t−β(1 −cos t)
sin tcos t α(1 −cos t) + βsin t
0 0 1
x′+β= cos t(x+β)−sin t(y−α)
y′−α= sin t(x+β) + cos t(y−α)
(−β, α)t
•Σ = 0 1
1 0: Σ2−I = 0 , φ0=t , φ1=t
3−= 0 ψ0= 1, ψ1=t, ψ2=t−1
eBt=
t t α t +β(t−1)
t t α(t−1) + β t
0 0 1
x′+β=t(x+β) + t(y+α)
y′+α=t(x+β) + t(y+α)
(−β, −α)t
•Σ2−aΣ = 0, , a ̸= 0
Σ = a0
0 0:φ0= 1 , φ1=t
3−a2= 0 ψ0= 1, ψ0= 1, ψ1=t, ψ2= ( ea t −1)/a
eBt=
1 0 ea t −1−at
a
0 1 βt
0 0 1
x′=xea t +ea t −1
aα
y′=y+βt
β̸= 0
•Σ2= 0 ,Σ̸= 0
Σ = 0 1
0 0:φ0= 1 , φ1=t
3= 0 ψ0= 1, ψ0= 1, ψ1=t, ψ2=t2/2
eBt=
1t α t +βt2
2
0 1 β t
0 0 1
x′=x+ty +α t +βt2
2
y′=y+β t
1/2β
Σ
•Σ = aI, a ̸= 0 : B2=aB
eBt=
ea t 0αea t −1
a
0ea t βea t −1
a
0 0 1
x′+α
a=x+α aea t
y′+β
a=y+β aea,t
−α/a, −β/a
•Ω = 0, B2= 0
eBt=
1 0 α t
0 1 β t
0 0 1
x′=x+α t
y′=y+β t
− − · · − · − · − · ··