Problème: Une équation fonctionnelle

◦
f(x+ 1) −f(x−1) = λf0(x).
k Ck(R)Ck
∆C0(R)
(∆f)(x) = f(x+ 1) −f(x−1) ;
λ
Eλ={f∈C1(R)|∆f=λf0}
f∈C0(R)f fa:R→Rfa(x) = f(x+a)a
f Eλ
fR
f
f Eλ
f(x+ 1) f(x−1)
f x
f∈C0(R)J(f)R
J(f)(x) = 1
2Z1
−1
f(x+t)dt .
k≥0
J(Ck(R)) ⊂Ck+1(R)
∆f J(f0)f∈C1(R)
J(f0) (J(f)))0
µ
|µ|>1f J(f) = µf
kfk= supx∈R|f(x)|f
|µ|>1f E2µf0
µ∈]1,+∞[E2µx7→ eKx K

h C0(R)J(h) = h
h un
u0(x) = 1
2Z1
−1
|h(x+t)−h(x)|2dt
un=J(un−1)∀n≥1.
u, v ∈R
|h(u)−h(v)|2≤1
2Z1
−1
|h(u+t)−h(v+t)|2dt
u1≥u0
(un)
u0J(|h|2)− |h|2
Tn=J◦J◦...◦J(|h|2) (Tn)x(Tn(x))
un(x)x
fR R f(P)
∀(x, y)∈R2, f(x+y) + f(x−y) = 2f(x)f(y)
(P)
f(P)f(0) = 0 f
f(P)
f(0) f
a > 0f(a)f(na)n∈Z
f(na)f(a)
a > 0∀x∈[0; a], f(x)>0f(a)
fpa
2qp∈Zq∈N
a > 0x>0∀q∈N, Iq={p∈N/ p. a
2q6x}
pq
q uq=pq.a
2q(uq)q∈Nx
fR
a
∀x∈[0; a], f(x)>0
f(a)f(x)x
f(a)=1 f
f(a)<1α∀x∈R, f(x) = cos(αx)