Analyse - Université Lyon 1

f∈L1(0,2π)f
cn(f) = 1
2πZ2π
0
f(x)e−inx dx, n ∈Z.
f
∞
X
−∞
cn(f)einx.
Sn(f)f
Sn(f) =
n
X
k=−n
ck(f)eikx
Sn(f, x)x
Lp(0,2π)Lp
1≤p≤ ∞
kfkLp=1
2πZ2π
0|f|p1
p
p < ∞kfkL∞= sup ess |f|
p=∞LpR2π
2π
2π Cper
2π
L∞
kfgkLr≤ kfkLpkgkLq1
r=1 p+1 q
Lq⊂Lpk·kLp≤ k · kLqp≤q
L2
hf, gi=1
2πZ2π
0
fg
einx n∈Z
L1
|n|→∞
f∈L1
f C1cn(f0) = incn(f)
cn(f(−x)) = c−n(f)
cn(f) = c−n(f)
N
X
−N
αneinx →f N → ∞ f
f αn

f, g ∈L1f∗g
f∗g(x) = 1
2πZ2π
0
f(x−t)g(t)dt.
f∗g(x)x f ∗g∈L1
f, g, h ∈L1
f∗g=g∗f
(f∗g)∗h=f∗(g∗h)
f∗einx =cn(f)einx
1≤p, q, r ≤ ∞ 1 + 1
r=1 p+1 qf∈Lpg∈Lq
f∗g∈Lrkf∗gkLr≤ kfkLpkgkLq
Dn(x) =
n
X
k=−n
eikx
Dn
1
2πZ2π
0
Dn= 1
|x| ≤ π Dn(x) = 2 sin nx
x+rn(x)rnx n
kDnkL1=4
π2ln n+O(1) n→ ∞
Sn(f) = f∗Dn
Kn=D0+D1+··· +Dn−1
n
Kn=
n
X
k=−n
(1 −|k|
n)eikx =sin2(nx/2)
nsin2(x/2)
kKnkL1= 1
0< δ ≤πZδ<|x|<π Kn→0n→ ∞
f∗Kn=S0+S1+··· +Sn−1
n=
n
X
k=−n
(1 −|k|
n)ck(f)eikx
σn(f)
σn(f) = S0+S1+··· +Sn−1
n

Gn(x) =
n
X
k=1
sin kx
k
kGnkL∞≤π
2+ 1
0<x<2π Gn(x)→G(x) = π−x
2n→ ∞
lim inf
n→∞ kGnkL∞≥Zπ
0
sin t
t= 1,85 ··· >π 2=kGkL∞
Jn=K2
n
kK2
nkL1
Jn≥0kJnkL1= 1 Jn
2n
k∈ {0,1,2}R2π
0xkJn(x) = O(n−k)
Pz(x) =
∞
X
0
zne−inx +
∞
X
0
zneinx −1
z∈C|z|<1
Pz(x) = 1− |z|2
|eix −z|2
kPzkL1= 1
0≤r < 1Pr(x) = 1−r2
1−2rcos x+r20< δ < π
Zδ<|t|<π
Pr(t)→0r%1
Pzu
|z|<1|z| ≤ 1
u(z) = 1
2πZ2π
0
u(eit)Pz(t)dt.
z∈C|z|<1
x0f Sn(f, x0)

fkσn(f)kL∞≤ kfkL∞σn(f)→
f
f∈Lp1≤p < ∞ kσn(f)kLp≤ kfkLpσn(f)→f Lp
f∈L1x0∈Rf−, f+∈C
Z1
0
|f(x0+t)−f+|
tdt < ∞Z1
0
|f(x0−t)−f−|
tdt < ∞.
Sn(f, x0)→f++f−
2n→ ∞
f+f−f x0
G(x) = π−x
20<x<2π
L∞G L∞G
G x = 0
Lpp < ∞L∞
Λ⊂Z
λX
n∈Λ
aneinx
n6∈ Λ
f Sn(f, x0)→l l =f(x0)
f∈L1`1f
einx L2