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.

  • 1
    $\begingroup$ Would you explain what exactly does "equivalent" mean here? $\endgroup$ – Pietro Majer Feb 27 '14 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 '14 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 '14 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)... $\endgroup$ – Delio Mugnolo Feb 28 '14 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 '14 at 11:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.