Examen corrigé

R2π
0
dθ
1−2αcos(θ)+α2α∈C|α| 6= 1 z=eiθ dθ 1−2αcos(θ)+α2
f(z)dz f
[0,2π] 2P0 = c0< c1. . . c2P−1< c2P= 2π
2π f(x) = PP−1
p=0 1[c2p,c2p+1](x)−1[c2p+1,c2p+2](x)
x∈[0,2π]
cpf[−2π, 4π]P= 1 P= 2
f δper
c=Pk∈Zδc+2πk
d
∂xfnT∈ D0(R/(2πZ))
ˆ
Tn=heinx, T iR/(2πZ)
hϕ , ψiR/(2πZ)=Rr+2π
rϕ(x)ψ(x)dx
2π
ˆ
fnn∈Z
f∈Ws,2(Z/(2πZ)) s < 1/2
B={z∈C,|z|< π/2}
C\R−
f(z) = logR−[(1−z)/(1+z)] D(0,1) B
D(0,1) z7→ 1−z
1+z
D(0,1) g(z) = h1−z
1+zii= exp[ilogR−(1+z 1−z)]
g D(0,1)
C=z∈C, e−π/2<|z|< eπ/2
g
D(0,1) C
ΩR2∼C
z0∈Ω (rn)n∈Nrn>0{z∈C,|z−z0| ≤ rn} ⊂ Ω
∀n∈N, D(z0, rn)⊂Ω lim
n→∞ rn= 0 ,
∀n∈N, f(z0) = 1
2πZ2π
0
f(z0+rneiθ)dθ .
Ω

f
D(a, R) = {z∈C,|z−a| ≤ R} ⊂ Ω
h D(a, r)h|z−a|=R=f|z−a|=R
D(a, R)
v=f−h m D(a, R)
m > 0v−1(m)D(a, R)
m > 0zmv−1(m)|zm−a|
v m > 0
f=h D(a, R)fΩ
RC∞

z=eiθ dz =ieiθdθ
dθ
1−2αcos(θ) + α2=ieiθdθ
ieiθ(1 −α(eiθ +e−iθ) + α2)=dz
i(z−α(z2+ 1) + α2z)
=dz
−iα(z2−(α+1
α)z+ 1) =idz
α(z−1
α)(z−α).
I=Z2π
0
dθ
1−2αcos(θ) + α2= 2iπ 1 2iπ ZS1
idz
α(z−1
α)(z−α).
|α|<1f(z) = i
(z−1
α)(z−α)D(0,1)
α i(α−1/α)−1I=2π
1−α2|α|>1 1/α i(1/α −α)−1
I=2π
α2−1
∂xf=P2P−1
p=0 2(−1)pδper
cp
δper
ce−inc
2π
\
(∂xf)n=
2P−1
X
p=0
(−1)pe−incp
π.
f=Pn∈Zˆ fneinx D0(R/(2πZ)) ∂xD0(R/(2πZ))
X
n∈Z
in ˆ fneinx =∂xf=X
n∈Z
2P−1
X
p=0
(−1)pe−incp
π
einx .
∀n∈Z\ {0},ˆ fn=1
in
2P−1
X
p=0
(−1)pe−incp
π.
n= 0 R2π
0f(x)dx
2π
ˆ
f0=
P−1
X
p=0
(c2p+1 −c2p)−(c2p+2 −c2p+1) =
P−1
X
p=0
(−c2p+ 2c2p+1 −c2p+2).
Cf
∀n∈Z,|ˆ fn| ≤ Cf
1
hni.
s < 1/2Pn∈Zhni2s|ˆ fn|2<+∞f∈Ws,2(Z)

B
B
ϕ:z→1−z
1+zD(0,1) Z=ϕ(z) = 2 1+z−1
z=2
(Z+1) −1 = 1−Z
1+Z|z|<1|Z−1|<|Z+ 1|Z
[−1,1] RZ > 0ϕ D(0,1)
{RZ > 0}
ϕ0(z) = −2
(1+z)2D(0,1)
logR{RZ > 0}B
iB =z∈C,−π
2<R<π 2
Cg(z) = exp(if(z))
D(0,1) C
f0D(0,1)
iB g0
g iB C
Rω(∆u)v−u(∆v)dx =R∂ω(∂nu)v−u(∂nv)dσ n
ω=B(a, r)
f(a) = 1
|Sd−1|ZSd−1
f(a+rη)dη .
2f(a) = 1
2πR2π
0f(a+reiθ)dθ a ∈Ωr > 0
D(a, r)⊂Ωa∈Ωrn>0
limn→∞ rn= 0
F(z) = f(a+Rz)D(0,1)
H(reiθ) = 1
2πR2π
0P(r, θ −t)F(eit)dt P (r, θ) = Rh1+reiθ
1−reiθ i
D(0,1) D(0,1) H|z|=1 =F h(z) = H((z−
a)/R)C∞D(a, R) ∆
v=f−h z0∈Ω
f D(a, R)m
v|z−a|=R≡0m6= 0 m > 0v−1(m)
a R D(a, R)
m > 0z7→ |z−a|v−1(m)
zm∈v−1(m)|zm−a|= maxz∈v−1(m)|z−a|E=
nz∈C,R[(zm−a)z]>|zm−a|ov−1(m)v(z)<
m z ∈E∩D(a, R)v(z)≤m=v(zm)z∈D(a, R)
zm
max(v) = 0 v=f−h
w=−(f−h)f=h D(a, R)h f
D(a, R)D(a, R)⊂Ω ∆f(z) = 0 z∈Ωf
C∞Ω
f(x) = x1Q(x)R
C∞