Géométrie [Lecture notes] by Antoine Chambert-Loir

By Antoine Chambert-Loir

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Supposons par l’absurde que m < n. L’application x → ( f 1 (x), . . , f m (x), 0, . . , 0) de U dans Rn est continue, injective, mais son image n’est pas ouverte, contradiction. 8. — Si Rn est homéomorphe à Rm , alors n = m. 9. — Soit A, B des parties de Rn et soit f : A → B un homéomor˚ = B˚ . phisme. Alors f ( A) ˚ est ouvert et contenu dans B , Démonstration. — Comme A˚ est ouvert dans Rn , f ( A) donc contenu dans B˚ . En considérant l’homéomorphisme réciproque de f , on prouve ˚ Comme f (A) = B , on en déduit que B˚ ⊂ f ( A), ˚ d’où finalede même que f −1 (B˚ ) ⊂ A.

X p−1 , a p , . . , a n ) xp = ap D p f (x 1 , . . , x p−1 , t , a p , . . , a n ) dt xp = (x p − a p )D p f (a) + ap D p f (x 1 , . . , x p−1 , t , a p , . . , a n ) − D p f (a) dt , si bien que n (x p − a p )D p f (a) f (x) − f (a) − p=1 ε n xp − ap p=1 pour tout x ∈ V . Comme ε est arbitraire, cela démontre que f est différentiable en a, de différentielle l’application linéaire (u 1 , . . , u n ) → np=1 u p D p f (a). Comme les applications a → D p f (a) sont continues, l’application a → D f (a) est continue.

Soit ε > 0. Comme f est uniformément continue sur la boule de centre 0 et de rayon R + 1, il existe un nombre réel δ strictement positif tel que f (x) − f (y) ε si x − y δ et x R. Par suite, si 0 < t δ/T et x R, on a ρ t ∗ f (x) − f (x) tn y tnε ε Rn tT y tT ρ(y/t ) ( f (x − y) − f (x)) dy ρ (y/t ) ρ . Cela prouve que quand t → 0, ρ t ∗ f converge uniformément vers f sur la boule de centre 0 et de rayon R. 3. 1. — Soit ϕ : Rn → Rn une application de classe C 1 ; on suppose qu’il existe R > 0 tel que ϕ(x) = x si x > R.

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