TD M1 Analyse 2019 2020

Lip0[0,1] := {f: [0,1] →R/f(0) = 0 kfkL:= sup
x,y∈[0,1]:x6=y
|f(x)−f(y)|
|x−y|<+∞}.
(1) (Lip0[0,1],k.k∞) (B([0,1],R),k.k∞)
f: [0,1] −→ R
(2) (Lip0[0,1],k.kL)
(3) c > 0f∈Lip0[0,1]
kfk∞≤ckfkL.
(4) k.kLk.k∞
(X, k.k)K X
n∈N∗xn
1, ..., xn
in∈K K ⊂ ∪in
k=1Bf(xn
ik,1
n)
(X, k.k)X
(X, k.k)K X (xn)K
l∈K(xn) (xn)l l
(xn)
(X, k.k)Bf(0,1) (xn)
Bf(0,1) a > 0∀p, q ∈N
kxp−xqk ≥ a.
Bf(0,1)
(X, k.k) (Kn)
X Kn+1 ⊂Knn∈NK=∩n∈NKn
l∞(N)x= (xk)k∈l∞(N)
kxk∞= sup
k∈N
|xk|<+∞.
(1) (l∞(N),k·k∞)
(2) (xn)nl∞(N)M≥0
∀n∈N:kxnk∞≤M.
(xn)n
x l∞(N)
(X, k.k)A X
∀x∈X:d(x, A) := inf
a∈Akx−ak.
(1) x7→ d(x, A) 1
(2) A d(x, A)=0 x∈A
(3) A x ∈X a ∈A
d(x, A) = kx−ak

(X, k.kX) (Y, k.kY)A⊂X X
Ff:A−→ YF
a∈AFA
(X, k.kX) (Y, k.kY)F X
k≥0 0 < α ≤1
Lipk,α(F, Y ) := {f:X−→ Y:kf(x)−f(x0)kY≤kkx−x0kα
X;∀x, x0∈F}.
Lipk,α(F, Y )F
f: [a, b]−→ R
fsupx∈[a,b]|f(x)|<+∞Lip(f)
supx∈[a,b]|f(x)|
a f : [a+∞[−→ R
+∞f[a+∞[
Lc(X, Y )
X Y p ∈Lc(X, Y )
kpk:= sup
x∈X\{0}
kp(x)kY
kxkX
<+∞.
∀p∈Lc(X, Y )
kpk= sup
x∈X:kxkX=1
kp(x)kY
= sup
x∈Bf(0,1)
kp(x)kY.
E=R[X]k.k∞k.k∞:= sup{|pk(0)
k!|;k∈
N}
(1) k.k∞E
(2) f:E→E f(p) = Xp
p∈E f (E, k.k∞)kfk
E=C([0,1])
k.k1f∈Ekfk1:= R1
0|f(t)|dt
T:E−→ E
f7→ T(f)
T(f) : [0,1] →R
x7→ Zx
0
f(t)dt
T(E, k.k1)kTk

E=Mn(R)n
N A ∈E N(A) = sup1≤i≤nPn
j=1 |ai,j |N
E f E R ∀A∈E f(A) = T r(A)
f(E;N)kfk
BE(0,1) E= (C[0,1],k.k∞)
(fn)n⊂BE(0,1) x∈[0,1] n∈Nfn(x) = −nx + 1
x∈[0,1
n]fn(x)=0 x∈[1 n,1]
(X, k.kX)p:X→R
e∈X p(e)=1 X=Ker(p)⊕Re
(1) p
(2) Ker(p) := {x∈X:p(x)=0}=p−1({0})
p ker(p)X
n∈NE
n λ > 0P∈E
Z1
0
|P(t)|dt ≥λsup
t∈[0,1]
|P(t)|.
n∈NEnn
inf
P∈EnZ1
0
|P(t)|dt > 0.
E=C[0,1] kfk1:= R1
0|f(t)|dt kfk∞:= supx∈[0,1] |f(x)|
f∈Ek.k1k.k∞
E=R[X]kPk1=
Pn
i=0 |ai|P(X) = a0+a1X+... +anXnf:E→Rf(P) = P(2)
f
(l∞(N),k · k∞)
p∈[1,+∞[
lp(N) := {x= (xn) : kxkp:= (X
n≥0
|xn|p)
1
p<+∞}.
(lp(N),k·kp)

I: (lp(N),k·kp)−→ (l∞(N),k·k∞)
(xn)7→ (xn)
kIkop
(X, k·kX) (Y, k·kY)K
X k ≥0 0 < α ≤1
Lipk,α
0(K, Y ) := {f:K−→ Y:f(0) = 0 kf(x)−f(x0)kY≤kkx−x0kα
X;∀x, x0∈X}.
Lipk,α
0(K, Y ) (C(K, Y ),k·k∞)
K:C([a, b]) →C([a, b]) K(f)(s) = Rb
ak(s, t)f(t)dt
k: [a, b]×[a, b]→R(fn) (C([a, b]),k.k∞)
(1) k
(2) (K(fn))
(3) (K(fn)) X
fn(t) = pt+ 4(nπ)2t∈[0,+∞[
(1)
f= 0
(2) (fn) (C([0,+∞[),k.k∞)
K E f :K→K
∀x, y ∈K
kf(x)−f(y)k≤kx−yk.
a∈K
(1) (fn)
fn(x) = f(1
na+ (1 −1 n)x).
n∈N∗fntn
(2) f
f: [0,1] →RC1f(0) = 1
f0(x) = f(x−x2)
X= (C1([0,1]), N)N(f) = kfk∞+kf0k∞
f T
T(f)(x) = 1 + Zx
0
f(t−t2)dt.
T◦T