Re: two norms are equivalent ...

Proginoskes wrote:

On Aug 17, 8:37 am, Kiuhnm <> 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.


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.