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/Me <= e <= Me.
Proof. Let
B_r^1(x) = {y in E  yx <= r},
B_r^2(x) = {y in E  yx <= 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
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