Inégalités, fonctions continues, limites

p, q ∈]1,+∞[1
p+1 q= 1
X, Y ∈R+
XY 6Xp
p+Yq q.
x7→ −ln x
n∈N∗a1, . . . , an, b1, . . . , bn∈R+
n
X
i=1
ap
i=
n
X
i=1
bq
i= 1.
Pn
i=1 aibi61
n∈N∗x1, . . . , xn, y1, . . . , yn∈C
(∗)
n
X
i=1 |xiyi|6 n
X
i=1 |xi|p!1/p n
X
i=1 |yi|q!1/q
.
p, q ∈]1,+∞[1
p+1 q= 1
n∈N∗x1, . . . , xn, y1, . . . , yn∈C
X
i=1 |xi|·|xi+yi|p−16 n
X
i=1 |xi|p!1/p n
X
i=1 |xi+yi|p!1/q
X
i=1 |yi|·|xi+yi|p−16 n
X
i=1 |yi|p!1/p n
X
i=1 |xi+yi|p!1/q
.
n∈N∗x1, . . . , xn, y1, . . . , yn∈C
(∗∗) n
X
i=1 |xi+yi|p!1/p
6 n
X
i=1 |xi|p!1/p
+ n
X
i=1 |yi|p!1/p
.

n∈Nx∈R
sin x=x−x3
3! +··· + (−1)nx2n+1 (2n+ 1)! +Rn(x),|Rn(x)|6|x|2n+3 (2n+ 3)!
n∈Nx∈R
ex= 1 + x+x2
2! +··· +xn n!+En(x),|En(x)|6|x|n+1 (n+ 1)!e|x|.
x > 0θ=θ(x)∈]0,1[
sin x=x−x3
6cos xθ.
0< x 6π/2θ(x)
θ(x)x→0+
x∈[0, π/2] 2x
π6sin x6x
(E, k · k)RU E f :U→R
E=R2
(0,0)
f1(x, y) := xy
x2+y2, f2(x, y) := sin(xy) x2+y2, f3(x, y) := x5y3 x6+y4,
f4(x, y) := 1 + x+y
x2−y2, f5(x, y) := xy2 x2+y2, f6(x, y) := xsin y−ysin x x2+y2
E=R3
g1(x, y, z) := xyz
x2+y2+z2m, g2(x, y, z) := sin(|x|+|y|+|z|)
px2+y2+z2
f(x, y) := ((x2+y2) sin 1
xy xy 6= 0
0xy = 0.
f
f(x, y) = sin(xy)
yy6= 0, f(x, 0) = x.

Rnx7→ kxk∞:= max16i6n|xi|E=Mn(R)
nk·k :E→R+
kAk:= sup
kxk∞61kAxk∞.
kAkA∈Ek·k E
A, B ∈EkABk6kAkkBk
k∈N+A7→ AkE E
A, H ∈E m ∈N
k(A+H)m−Amk6(kAk+kHk)m−(kAk)m.
16i, j 6n A 7→ ai,j E ai,j
A i j
det : A7→ det A E
U E
U E
E=C1([0,1]; R)
f∈E7→ N1(f) := |f(0)|+ sup
t∈[0,1] |f0(t)|.
E
k·k:E→R+E
kxk= 0 x= 0
kλxk=|λ|kxk(λ, x)∈R×E
{x∈E, kxk61}E

E=C0([0,1]; R)F=C1([0,1]; R)f∈E
kfk∞:= sup
t∈[0,1] |f(t)|.
k·k∞E F
(E, k·k∞) (F, k·k∞)
E=C[X]k · k∞k · k1C R+
P=Pn
i=0 aiXi∈C[X]7−→ kPk∞:= max06i6n|ai|,
P=Pn
i=0 aiXi∈C[X]7−→ kPk1:= Pn i=0 |ai|.
k·k∞k·k1E
(E, k·k∞) (E, k·k1)
k·k∞k·k1
E=C0([0,1]; R)g∈E
Φ : E→R
f7→ R1
0f(t)g(t)dt.
E f 7→ kfk∞:= supt∈[0,1] |f(t)|Φ
(E, k·k∞)
(E, k · kE) (F, k · kF)u:E→F
u
(xn)n∈NE0u(xn)n∈N
F
E=Rnx= (x1, . . . , xn)7→ kxk1:= Pn
i=1 |xi|
x= (x1, . . . , xn)7→ kxk2:= (Pn
i=1 |xi|2)1/2x= (x1, . . . , xn)7→ kxk∞:= max16i6n|xi|E
x∈Rn\ {0}kxk∞6kxk16nkxk∞
kxk∞6kxk26(√n)kxk∞
1
√nkxk16kxk26kxk1.
E=C0([0,1]; R)f∈E
kfk∞:= sup
t∈[0,1] |f(t)| kfk1:= Z1
0|f(t)|dt.
k·k∞k·k1E

f:R→R
f(t) := exp(−1/t2)t6= 0 f(0) = 0.
f C∞R
f:R→R
f(x) := e−1/x2sin e1/x2x6= 0 f(0) = 0.
n∈Nf n x = 0
fR
f0x= 0
f
(E, k·kE)F:= Lc(E, E)
k·k k·kE
u∈F u0:= IdEn∈N∗un+1 := u◦un=un◦u
k·k F
u∈F n ∈Nkunk6kukn
u∈F
+∞
X
n=0
un
n!
Fexp u E
u v F u◦v=v◦uexp(u+v) =
exp u◦exp vu∈F f :R→F
f(t) := exp(tu),(t∈R).
fR R f0(t)t∈R
ϕ:R→FRt∈R
ϕ0(t) = u◦(ϕ(t)).
t∈Rξ(t) := exp(−tu)◦ϕ(t)
ξ0(t)t∈R