Dieudonne

Theorem (Ostrowski) Every nontrivial norm in Q is equivalent to ||-||_p for some prime p, or equivalent to |-|.

Proof:

Case (A):  Assume that ||n|| leq 1 for all natural numbers n， then we will show that for any x in Q, ||x||=||x||^lambda for some positive real number lambda, hence it is equivalent to some ||-||_p.

Case (B): Assume that there exists some natural number n such that ||n||>1, then we will show that for any natural number n, ||n||=n^s for some positive real s, hence it is equivalent to the absolute value (norm) |-|.

Proposition 2.2  Prod_{p leq  +infty} ||x||_p = 1.

Theorem (Ostrowski) Every nontrivial norm in Q is equivalent to ||-||_p for some prime p, or equivalent to |-|.

Proof:

Case (A):  Assume that ||n|| leq 1 for all natural numbers n， then we will show that for any x in Q, ||x||=||x||^lambda for some positive real number lambda, hence it is equivalent to some ||-||_p.

Case (B): Assume that there exists some natural number n such that ||n||>1, then we will show that for any natural number n, ||n||=n^s for some positive real s, hence it is equivalent to the absolute value (norm) |-|.

Proposition 2.2  Prod_{p leq  +infty} ||x||_p = 1.

The proof of p-adic norm is non-Archimedeam: write x and y in the general norm.

The proof of Proposition 1.1:  Step (a) ||x+y|| leq 2 max {||x||, ||y||};  Step (b) ||sum x_i|| leq 4n max {||x_i||};  Step (c) Use Newton-expansion and use (a) as an approximation of triangular inequality.

The proof of Proposition 1.2: Use Newton-expansion and triangular inequality.

The proof of Proposition 1.4:  For (2) => (3): there exists ||x_0||_1 >1, then for any x, there exists alpha such that ||x||_1 = ||x_0||_1^alpha . The condition (2) can be used as a bridge between ||-||_1  and ||-||_2 and one can show that ||x||_2 = ||x_0||_1^alpha. Now define lambda to be the real number such that ||x_0||_2=||x_0||_1^alpha and conclude easily ||x||_1^lambda=||x||_2.

The proof of p-adic norm is non-Archimedeam: write x and y in the general norm.

The proof of Proposition 1.1:  Step (a) ||x+y|| leq 2 max {||x||, ||y||};  Step (b) ||sum x_i|| leq 4n max {||x_i||};  Step (c) Use Newton-expansion and use (a) as an approximation of triangular inequality. 

The proof of Proposition 1.2: Use Newton-expansion and triangular inequality.

The proof of Proposition 1.4:  For (2) => (3): there exists ||x_0||_1 >1, then for any x, there exists alpha such that ||x||_1 = ||x_0||_1^alpha . The condition (2) can be used as a bridge between ||-||_1  and ||-||_2 and one can show that ||x||_2 = ||x_0||_1^alpha. Now define lambda to be the real number such that ||x_0||_2=||x_0||_1^alpha and conclude easily ||x||_1^lambda=||x||_2. 

Definition A norm over a field K. 

In particular, non-Archimedean norm.

Proposition 1.1 (criterion of a function being a norm) ||-|| is a norm if and only if ||1+x|| leq 2 for any ||x|| leq 1.

Proposition 1.2 (criterion of a norm being non-Archimedean) A norm is non-Archimedean if and only if ||n|| leq 1 for any n natural number.

Definition The distance associated to a normed field K.

Proposition 1.3 Certain maps are continuous.

Definition Two norms being equivalent.

Proposition 1.3 (criterion of two norms being equivalent)

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