Continuité et équation fonctionnelle

f:R→R
∀x∈R, f(x) = f(x2)
f
f:R→R∀x∈R
fx+ 1
2=f(x)
f
f:R→R
∀x∈R, f(2x) = f(x) cos x
f
f:R→R
∀x, y ∈R, f(x+y) = f(x) + f(y)
f(0) x∈Rf(−x) = −f(x)
n∈Zx∈Rf(nx) = nf(x)
r∈Qf(r) = ar a =f(1)
x∈Rf(x) = ax
f:R→Rx, y ∈R
f(x+y) = f(x) + f(y)
f x0∈R
f
f:R→R
∀x, y ∈R, f x+y
2=1 2(f(x) + f(y))
f f(0) = f(1) = 0
f
∀x∈R,2f(x) = f(2x)
f
f
f:R→R
∀x, y ∈R, f x+y
2=1 2(f(x) + f(y))
f(0) = 0
∀x, y ∈R, f (x+y) = f(x) + f(y)
f

∀x∈R, f(−x) = f((−x)2) = f(x2) = f(x)
f
x > 0x1/2n−→
n→∞ 1f(x1/2n)−→ n→∞ f(1) f
f(x1/2n) = f(x1/2n−1) = · · · =f(x)
f(x) = f(1) x > 0x∈R∗
f(0) = limx→0+f(x) = f(1)
∀x∈R, f(x) = f(1)
x∈R(un)u0=x n ∈N
un+1 =un+ 1
2
x≥1 (un)
x≤1 (un)
(un)
n∈Nf(x) = f(un)f(x) = f(1)
f
f(x) = fx
2cos x 2=fx 4cos x 4cos x 2=. . . =fx 2ncos x 2n. . . cos x 2
sin x
2ncos x 2n. . . cos x 2=1 2nsin x
sin x
2nf(x) = sin x 2nfx 2n
x6= 0 n→+∞sin x
2n6= 0
f(x) = sin x
2nsin x
2nfx
2n→sin x xf(0)
∀x∈R, f(x) = sin x
x
x=y= 0 f(0) = 2f(0) f(0) = 0
y=−x f(0) = f(x) + f(−x)f(−x) = −f(x)
n∈Nf(nx) = nf(x)
n∈Z−n=−p p ∈N
f(nx) = −f(px) = −pf(x) = nf(x)
r∈Qr=p/q p ∈Zq∈N∗
f(r) = pf(1/q) = p
qqf(1/q) = p qf(1) = ar
x∈R(un)un→x un∈Q
f(un)→f(x)un∈Qf(un) = aun→ax
f(x) = ax
f(x+y) = f(x) + f(y)
∀r∈Q, f(r) = rf(1)
∀a∈R,∀n∈Z, f(na) = nf(a)
n= 0
f(0) = f(0) + f(0)
f(0) = 0
n∈N
f((n+ 1)a) = f(na) + f(a)
n∈Z
f(−x) = −f(x)
f(x) + f(−x) = f(0) = 0

r=p/q ∈Qp∈Zq∈N∗
f(r) = fp×1
q=pf 1 qf(1) = fq×1 q=qf 1 q
f(r) = p
qf(1) = rf(1)
x∈R
x∈R(xn)∈QNxn→x
xn+x0−x→x0f x0
f(xn+x0−x)→f(x0)
f(xn+x0−x) = f(x0)+(f(xn)−f(x))
f(xn)−f(x)→0
f(x) = lim
n→+∞f(xn) = xf(1)
f
f(2 −x) + f(x)=0 f(−x) + f(x)=0 f(x) = f(x+ 2) f
f(x/2) = f(x)/2f(2x) = 2f(x)
f f
f(2x) = 2f(x)f
f
a=f(1) −f(0) b=f(0) g(x) = f(x)−(ax +b)
g f
∀x∈R, f x
2=fx+ 0 2=1 2(f(x) + f(0)) = 1 2f(x)
∀x, y ∈R, f x+y
2=1 2f(x+y)
∀x, y ∈R, f (x+y) = f(x) + f(y)
f
f x 7→ ax
x7→ f(x)−f(0)
f x 7→ ax +b