# Re: two norms are equivalent ...

*From*: rusty <mr.rusty@xxxxxxx>*Date*: Sat, 18 Aug 2007 08:11:48 +0200

Proginoskes wrote:

On Aug 17, 8:37 am, Kiuhnm <kiuhnm03.4t.yahoo.it> wrote:

Proposition. Two norms |.| and ||.|| on E are equivalent iff there is a

constant M s.t., for all e in E, 1/M||e|| <= |e| <= M||e||.

Proof. Let

B_r^1(x) = {y in E | |y-x| <= r},

B_r^2(x) = {y in E | ||y-x|| <= r}.

Since neighborhoods of an arbitrary point are translates of

neighborhoods of the origin, the two topologies are the same iff for

every R > 0, there are constants M_1, M_2 > 0 s.t.

B_{M_1}^2(0) < B_R^1(0) < B_{M_2}^2(0). (1)

Then if ||x|| < M_1, then |x| <= R, i.e., if ||x|| < 1, then |x| <=

R/M_1. Thus, if e<>0, then

| e/||e|| | = |e|/||e|| <= R/M_1. (2)

[...]

First of all, I don't understand why (1) implies that B_R^1(0) is open

in (E,|.|) <=> it is open in (E,||.||).

I also can't figure out how (2) is derived.

Kiuhnm

Why the complicated approach? The definition of equivalent norms is

that there are constants A and B, with A not zero, such that

A |x| <= || x || <= B |x|.

Now just take M = max {B,1/A}, and you're done.

I think that the OP is quoting from a book and that the definition of

equivalence there is probably in terms of the topologies defined by the

norms being the same.

--

rusty

.

**Follow-Ups**:**Re: two norms are equivalent ...***From:*Kiuhnm

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