Espaces normés : Exercices de mathématiques

E=C0([0 ; 1],R)
N(f) = Z1
0f(t)dt
a, b ≥0u, v > 0
√a+√b= 1 =⇒1
u+v≤a u+b v
f, g ∈E f, g > 0
N((f+g)−1)≤N(f)2N(f−1) + N(g)2N(g−1)
(N(f) + N(g))2
N(f+g)N((f+g)−1)≤max(N(f)N(f−1), N(g)N(g−1))
(E, k·k)K K =R C
x, y ∈E
kxk+kyk ≤ 2 maxkx+yk,kx−yk
x6= 0 y6= 0
x, y ∈E
kxk+kyk ≤ √2 maxkx+yk,kx−yk
√2
n∈Nn≥2k · k Mn(C)
∀(A, P )∈ Mn(C)×GLn(C),kAk=
P−1AP 
n∈N∗k·k [−1 ; 1]
Tnn
∀θ∈R, Tn(cos θ) = cos(nθ)
P n
kPk ≥ 1
2n−1
P Tncos(kπ/n)k∈Z
kPk=1
2n−1⇐⇒ P=1 2n−1Tn
A= (ai,j )∈ Mn,p(K)
kAk1=
n
X
i=1
p
X
j=1|ai,j |,kAk2=v
u
u
t
n
X
i=1
p
X
j=1|ai,j |2kAk∞= max
1≤i≤n,1≤j≤p|ai,j |
k·k1k·k2k·k∞Mn,p(K)
A= (ai,j )∈ Mn(R)
kAk= n
X
i,j=1
a2
i,j !1/2
k · k Mn(R)
∀A, B ∈ Mn(R),kABk≤kAkkBk

A= (ai,j )∈ Mn(C)
kAk= sup
1≤i≤n
n
X
j=1|ai,j |
k · k Mn(C)
∀A, B ∈ Mn(C),kABk≤kAkkBk
A= (ai,j )∈ Mn(C)
kAk= sup
1≤i≤n
n
X
j=1|ai,j |
k · k Mn(C)
λ A |λ| ≤ kAk
x= (x1, . . . , xn)∈Knp≥1
kxkp= n
X
i=1|xi|p!1/p
kxk∞= lim
p→+∞kxkp
f1, . . . , fn: [0 ; 1] →R
N: (x1, . . . , xn)7→ kx1f1+··· +xnfnk∞
Rn
N:R2→R
N(x1, x2) = sup
t∈[0;1]|x1+tx2|
R2
k·k∞
`1(N,K)u= (un)n∈N∈KN
`1(N,K) = nu∈KNX|un|<+∞o
`1(N,K)K
kuk1=
+∞
X
n=0|un|
IRL1(I, K)
f:I→K
L1(I, K) = f∈ C(I, K)ZI|f|<+∞
L1(I, K)K
kfk1=ZIf(t)dt
IRL2(I, K)
f:I→K
L2(I, K) = f∈ C(I, K)ZI|f|2<+∞

L2(I, K)K
kfk2=ZIf(t)2dt1/2
k · k Mn,1(R)
kXk= max
1≤i≤n|xi|
SMn,1(R)
A∈ Mn(R)
sup
X∈SkAXk
N(A) = sup
X∈SkAXk
X∈ Mn,1(R)kAXk ≤ N(A)kXk
NMn(R)
N(A) = sup
1≤i≤n
n
X
j=1|ai,j |
B(N,R)k·k∞
eC0
B(N,R)k·k∞
u= ((−1)n)n∈NC
B(N,R)k·k∞
x∈ B(N,R) ∆x
∆x(n) = x(n+ 1) −x(n)
F=∆xx∈ B(N,R)
e F
E[−1 ; 1] R
kfk∞= sup
x∈[−1;1]f(x)
f:x7→ 
1x∈]0 ; 1]
0x= 0
−1x∈[−1 ; 0[
F E [−1 ; 1] R
E=C0([0 ; 1],R)k·k1k·k2k·k∞
kfk1=Z1
0f(t)dt, kfk2=Z1 0
f(t)2dt1/2 kfk∞= sup
[0;1]|f|
k·k∞k·k1k·k2
k·k1k·k2
E=C1([−1 ; 1],R)N1, N2N3
N1(f) = sup
[−1;1]|f|, N2(f) = f(0)+ sup [−1;1]|f0|N3(f) = Z1
−1|f|

N1, N2N3E
N1N2N1N3
E=C1([0 ; 1],R)N:E→R+
N(f) = sf2(0) + Z1
0
f02(t) dt
N E
Nk·k∞
R[X]N1N2
N1(P) =
+∞
X
k=0P(k)(0)N2(P) = sup
t∈[−1,1]P(t)
N1N2R[X]
Pn=1
nXn
N1N2
R(N)
k·k1k·k2k·k∞R(N)
kuk1=
+∞
X
n=0|un|,kuk2= +∞
X
n=0
u2
n!1/2
kuk∞= sup
n∈N|un|
k·k1k·k∞
k·k1k·k2
`1(N,R)
kuk1=
+∞
X
n=0|un|
u∈`1(N,R)u
k·k∞
kuk∞= sup
n∈N|un|
k·k1k·k∞
u∈`1(N,R)u
k·k2
kuk2= +∞
X
n=0
u2
n!1/2
k·k1k·k2
B(N,R)k·k∞
a= (an)
a
Na:x7→
+∞
X
n=0
an|xn|
B(N,R)
Nak·k∞
E u = (un)n∈Nu0= 0
N∞(u) = sup
n∈N|un|N(u) = sup n∈N|un+1 −un|
E
N(u)≤2N∞(u)u∈E

E=f∈ C1([0 ; 1],R)f(0) = 0N1, N2
E
N1(f) = kf0k∞N2(f) = kf+f0k∞
N1N2E
N2N1
f(x)=e−xZx
0f(t) + f0(t)etdt
N1N2
ERf: [0 ; 1] →RC1
f(0) = 0 f∈E
N1(f) = sup
x∈[0;1]f(x)+ sup x∈[0;1]f0(x)N2(f) = sup x∈[0;1]f(x) + f0(x)
N1N2E
E=f∈ C2([0 ; π],R)f(0) = f0(0) = 0
N:f7→ kf+f00k∞
E
N
ν:f7→ kfk∞+kf00k∞
E=C([0 ; 1],R)E+E
ϕ∈E+
f∈E
kfkϕ= sup
t∈[0;1]nf(t)ϕ(t)o
k·kϕE
ϕ1ϕ2E+
k·kxk·kx2
NMn(R)c > 0
N(AB)≤cN(A)N(B)
d∈NE=Rd[X]X
d
ξ= (ξ0, . . . , ξd)d+ 1 P∈E
Nξ(P) =
d
X
k=0P(ξk)
NξE
(Pn)E n ∈N
Pn=
d
X
k=0
ak,nXk
(Pn)R
(Pn)R
k∈ {0, . . . , d}(ak,n)