Counting norms on an infinite dimensional vector space

Question

It is known that whenever E is a finite dimensional real vector space, there is only one norm on E up to equivalence (actually one non discrete vector space topology).

Is it known what happens when E is infinite dimensional? for sure, one can create two (infinitely many) non equivalent norms by using Hamel bases(*), but what about the precise cardinality (up to equivalence, or without taking into account equivalence at all) ?

(*) E.g. Let (e_i) a Hamel basis, write x = sum x_i e_i and put ||x|| = sum of |x_i| N(x) = sum of i|x_i|

Answer 1

Take a look at the papers "Über Normtopologien in linearen Räumen" (Link to the article) as well as "Über vollständige Normtopologien in linearen Räumen" (Link to the article) by D. Laugwitz.

He proves the following. Let $E$ be a vector space, denote by $a(E)$ the cardinality of a Hamel basis of $E$ and denote by $n(E)$ the number of mutually non-equivalent norms on $E$. Then

$$n(E) = 2^{a(E)}, \quad\text{if}~\mathfrak{c}\leq a(E),$$ $$ \mathfrak{c}\leq 2^{a(E)} \leq n(E) \leq 2^{\mathfrak{c}}, \quad\text{if}~\aleph_0\leq a(E)<\mathfrak{c},$$

and (of course)

$$ n(E) = 1 \quad\text{if}~1\leq a(E) < \aleph_0.$$

Here $\aleph_0$ denotes the cardinality of countable sets and $\mathfrak{c}$ the cardinality of the continuum. In the second article you can find the answer for complete norms.