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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 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, 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.

Seirios
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12455421
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  • Let $x\in X$ be an arbitrary nonzero vector, and let $Y$ be an arbitrary complementary subspace of $\ {\rm span}(x)$. Then, for $v\in Y$, we can define $$|v|\ :=\ N(v, x),,$$ which is a norm on $Y$ by conditions 1., 3., 4. I'm not sure if we can bunch these together for a norm on the whole space $X$. – Berci Feb 27 '14 at 12:24
  • @Berci: With your choice of $x$, $|x |=0$ by 1. while $x \neq 0$. Taking $x=0$ seems to be better, no? – Seirios Feb 27 '14 at 12:31
  • If "someone" suggested it, maybe you should ask that person. – GEdgar Feb 27 '14 at 14:42
  • @GEdgar: Looking back, I sure wish I could've asked him, but he's a professor at a different university than where I'm a student at. A friend of mine attended a seminar where he gave a lecture, and in their conversations later, this topic turned up. – 12455421 Feb 27 '14 at 14:51
  • @12455421 It seems to me that Gähler shows that every 2-normed linear space of dimension $\neq 1$ is a locally convex topological vector space and as such a space uniformizable but he gives an example for a $2$-normed linear space that is not normable, I tried to translate the relevant part, the proof seems quite technical – Peter Melech Mar 10 '18 at 14:31
  • just realized You asked this 4 years ago :-) – Peter Melech Mar 10 '18 at 14:34

1 Answers1

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Gähler defines for $i=(i_1,i_2)\in I=(0,\infty)\times (0,\infty)$ the function $e_i=e_{(i_1,i_2)}\in (-\infty,\infty)^{I}$ to be one for $j=i$ and zero otherwise. He defines the space $L$ to be the linear hull of these functions and defines

$$||e_i,e_j||=0\text{ if } i=j,\sup\{i_2,j_2\}\text{ if }i_1=j_1\text{ and }i_2\neq j_2,1\text{ if }i_1\neq j_1.$$ Two elements of $L$ say $a,b$ can be written $a=\sum_{j\in I'}\alpha_je_j$ and $b=\sum_{j\in I'}\beta_je_j$ by using the same finite subset $I'\subset I$. He defines $$||a,b||=\frac{1}{2}\sum_{i,j\in I'}|\alpha_i\beta_j-\alpha_j\beta_i|||e_i,e_j||$$ and states that this is a $2$-norm on $L$ and that $L$ is equipped with a topology generated by $\sigma(a,b,c)=||b-a,c-a||$. He then proves that $L$ is not metrizable and thus not normable.

Peter Melech
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