aborder en fin de première annéeUn problème corrigé d`analyse et

x
y
M(L)
M(0)
L
A
A ≤ L2
4π
ψ2πR
a
Za+2π
a
ψ(t)dt
ψ ψ
ψ=1
2πZa+2π
a
ψ(t)dt
a
Z2π
0
(f(t)−f)2dt ≤Z2π
0
f02(t)dt
f∈ C1(R) 2π
fC1(R)a b
a < b ≤a+π f(a) = f(b)=0
ϕ]a, b[
∀t∈]a, b[: ϕ(t) = f(t) cotan(t−a)
ϕ
[a, b]ϕ(a)ϕ(b)
ϕ[a, b]
ϕ
a b
u v a < u < v < b
fϕ u v
Zv
u
f02(t)dt −Zv
u
f2(t)dt −Zv
u
(ϕ(t)−f0(t))2dt
0 = Zb
a
f02(t)dt −Zb
a
f2(t)dt −Zb
a
(ϕ(t)−f0(t))2dt
Zb
a
f2(t)dt ≤Zb
a
f02(t)dt
f0(t) = ϕ(t)t∈[a, b]
λ f(t) = λsin(t−a)t[a, b]

fC1(R)
f π
Zb
a
f2(t)dt ≤Zb
a
f02(t)dt
a b f a < b
fC1(R)λ > 0fλ
∀t∈R:fλ(t) = f(t
λ)
Rλb
λa f2
λ(t)dt Rλb
λa fλ
02(t)dt Rb
af2(t)dt Rb af02(t)dt
fC1(R) 0 a b
Zb
a
f2(t)dt ≤Zb
a
f02(t)dt
nRc0, c1,· · · , cn
s1,· · · , sn
c0(t)=1, c1(t) = cos(t),· · · cn(t) = cos(nt)
s1(t) = sin(t),· · · sn(t) = sin(nt)
i∈ {0,· · · , n}j∈ {1,· · · , n}R2π
0ci(t)cj(t)dt R2π 0ci(t)sj(t)dt
R2π
0si(t)sj(t)dt
T= Vect(c0,· · · , cn, s1,· · · , sn)f∈ T f
Z2π
0
(f(t)−f)2dt ≤Z2π
0
f02(t)dt
x
y
M(0)
M(L)
L
A
A ≤ L2
2π
u w
uw ≤1
2(u2+w2)
MC1([0, L])
L
x y U =x◦M V =y◦M

C1([0, L])
U(0) = U(L)=0
Ay
A ≤ L2
2π
MR
L L U V
A
t→U(L
2πt)
T
A ≤ L2
4π

cotan ]0, π[a b
t∈]a, b[⇒t−a∈]0, b −a[⊂]0, π[
cotan 0 ϕ a
b−a<π cotan b−a
ϕ b
ϕ(b)=0 f(b) = 0
f(a) = f(b) = 0
cotan(y−π) = cotan(y)
a:cotan(x−a) = 1
x−a+o(1)
f(x) =f0(a)x+o(x−a)
⇒ϕ(x) = f0(a) + o(1)
b=a+π:cotan(x−a) = cotan(x−b) = 1
x−b+o(1)
f(x) =f0(b)x+o(x−b)
⇒ϕ(x) = f0(b) + o(1)
ϕ
ϕ(a) = f0(a)ϕ(b) = f0(b)f(b)=0
cotan0=−(1 + cotan2)
(fϕ)0(t)=2f0(t)f(t) cotan(t−a)−f2(t)(1 + cotan2(t−a))
=−(f(t) cotan(t−a)−f0(t))2−f2(t) + f02(t)
ϕ=fcotan u v
[ϕf]v
u=Zv
u
f02(t)dt −Zv
u
f2(t)dt −Zv
u
(ϕ(t)−f0(t))2dt (1)
f0f ϕ [a, b]
u0
v b (1) fϕ 0f
ϕ
0 = Zb
a
f02(t)dt −Zb
a
f2(t)dt −Zb
a
(ϕ(t)−f0(t))2dt (2)
Zv
u
(ϕ(t)−f0(t))2dt ≥0⇒Zv
u
f2(t)dt ≤Zb
a
f02(t)dt
(2)
Zv
u
(ϕ(t)−f0(t))2dt = 0 ⇒ ∀t∈[a, b] : ϕ(t)−f0(t)=0
(ϕ−f0)2
f f
y0(t)−cotan(t−a)y(t) = 0
y]a, b[
cotan(t−a) ln(sin(t−a))
f
λeln(sin(t−a)) =λsin(t−a)
a b
a b f [a, b]
π

u=λt
Zb
a
f2(t)dt =1
λZλb
λa
fλ2(u)du =1
λZλb
λa
fλ2(t)dt
u=λt fλ
0(x) = 1
λf0(x λ)
Zb
a
f02(t)dt =1
λZλb
λa
f02(u
λ)du =λZλb
λa
fλ2(u)du =λZλb
λa
fλ2(t)dt
f a b
Rb
af2(t)dt Rb af02(t)dt λ > 0
fλ
Rb
af2(t)dt
Rb
af02(t)dt =1
λ2Rλb
λa f2
λ(t)dt
Rλb
λa fλ
02(t)dt ≤1
λ2
fλλa λb
λ
∀i∈ {1,· · · , n}:Z2π
0
c0(t)ci(t)dt =Z2π
0
c0(t)si(t)dt = 0
Z2π
0
c2
0(t)dt =Z2π
0
c0(t)dt = 2π
∀(i, j)∈ {1,· · · , n}2, i 6=j:Z2π
0
ci(t)cj(t)dt =Z2π
0
si(t)sj(t)dt = 0
∀(i, j)∈ {1,· · · , n}2:Z2π
0
ci(t)sj(t)dt = 0
∀i∈ {1,· · · , n}:Z2π
0
c2
i(t)dt =Z2π
0
s2
i(t)dt =π
fT
∃(λ0,· · · , λn, µ1,· · · , µn)∈R2n+1 f=λ0c0+· · · +λncn+µ1s1+· · · +µnsn
f=λ0
f−f=λ1c1+· · · +λncn+µ1s1+· · · +µnsn
f0=µ1c1+· · · +nµncn−λ1s1− · · · − nλnsn
Z2π
0
(f−f)2=πλ2
1+· · · +λ2 n+µ2 1+· · · +µ2 n
Z2π
0
f02=πλ2
1+· · · +n2λ2 n+µ2 1+· · · +n2µ2 n
Z2π
0
f02−Z2π
0
(f−f)2=π(22−1)(λ2
2+µ2 2) + · · · + (n2−1)(λ2 n+µ2 n)≥0
(u−w)2≥0⇒uw ≤1
2(u2+w2)