0
$\begingroup$

Someone suggested today that $2$-normed spaces are actually equivalent to normed spaces. Can anyone who's familiar with the topic provide a counterexample? (I can't access Gähler's original paper introducing the notion, and hence have no way of telling whether the answer has already been provided there.)

[Recall: (Gähler, 1963) If $X$ is a vector space over $F$ (either the real or the complex field), then a real-valued, non-negative function $N$ on $X^2$ is said to be a $2$-norm on $X$ iff the following conditions are satisfied:

  1. $N(x,\ y)=0$ iff $x,\ y$ are linearly dependent vectors in $X$;

  2. $N(x,\ y)=N(y,\ x)$ for every $x,\ y \in X$;

  3. $N(\lambda x,\ y)=|\lambda|N(x,\ y)$ for every $\lambda \in F$ and for every $x,\ y \in X$;

  4. $N(x+y,\ z) \le N(x,\ z)+N(y,\ z)$ for all $x,\ y,\ z \in X$.]

Clarification: Gähler shows that (http://books.google.co.in/books?id=T7FMg6nT9KIC&lpg=PA219&ots=fSwZP9PFdY&dq=Geometry%20of%20Linear%202-Normed%20Spaces&pg=PA1#v=onepage&q=Geometry%20of%20Linear%202-Normed%20Spaces&f=false) linear $2$-normed spaces are normable and uniformizable provided the dimension of the space is greater than one. He also proves that if the space is a linear normed space, then it's possible to define a 2-norm on it. However, the converse is not true. This is the part I want the evidence of, preferably from the link to his original paper given below.

Edit: I finally got hold of the relevant paper by Gähler (http://www.mediafire.com/view/5s3x76daiodtkd7/Mathematische_Nachrichten_Volume_28_issue_1-2_1964_[doi_10.1002%2Fmana.19640280102]_Siegfried_Gähler_--_Lineare_2-normierte_Räume.pdf), but it's in German. Since my understanding of that particular language is limited to mere recognition of a few words, I would be grateful if someone helped me out by reading it and translating the answer therein.

P. S.: Please excuse my second link not rendering properly--I am too inexperienced, evidently. Somehow, the angular brackets don't seem to be working.

$\endgroup$
7
  • 1
    $\begingroup$ Would you explain what exactly does "equivalent" mean here? $\endgroup$ Feb 27, 2014 at 13:16
  • $\begingroup$ @PietroMajer: I'm not too sure on that either, if I may be honest. My best guess is that the person suggesting it meant that defining 2-norms is a redundant exercise , since this structure might ultimately be nothing but a special case/reworking of normed spaces. $\endgroup$
    – 12455421
    Feb 27, 2014 at 15:00
  • $\begingroup$ @PietroMajer: What I'm trying to see right now is whether a metric can be generated by a 2-norm, and, if so, then whether that same metric can be derived from a norm on that space as well. $\endgroup$
    – 12455421
    Feb 27, 2014 at 15:09
  • $\begingroup$ my first guess would have been that $N(x,y)$ is just another way of writing $\|x-y\|$ for a suitable norm $\|\cdot\|$, but then i saw (3) and (4)... $\endgroup$ Feb 28, 2014 at 9:15
  • $\begingroup$ @PietroMajer: I believe our doubt as to what "equivalence" means here is, sort of, cleared up now? $\endgroup$
    – 12455421
    Mar 13, 2014 at 11:30

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.