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Let $(V,+,\cdot,\|.\|)$ be a normed vector space. Can we reconstruct addition $+$ of vectors and scalar multiplication $\cdot$ if we are given only the underlying set $V$ and the norm $\|\cdot\|\colon V\to\Bbb R$?

Clearly, we can find $0\in V$ as it is the only element of norm $0$, and we know $1\cdot v=v$ and $0\cdot v=0$. And we have the topology. But is that enough to reconstruct the missing operations?

lesderid
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Hagen von Eitzen
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    How do you know the topology? A norm induces a topology by defining a metric $d(x,y) = \lVert x - y \rVert$. But if you've forgotten the vector space structure, then you don't know how to subtract. – Ethan Dlugie May 03 '20 at 19:38
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    As a start on scalar multiplication, my intuition says that $2v$ is the element of largest norm in the $1$-ball around $v$. But I'm not even sure that's valid in your average function space. – JonathanZ May 03 '20 at 19:40
  • @EthanDlugie. So let's say that one has a metric space. Two questions arise: 1) How can we see if the metric space is given by a normed vector space? 2) And when it is, is the vector space unique? – md2perpe May 03 '20 at 19:49
  • My above comment was wrong. As @Julian Rosen's answer highlights, the "$1$-ball around $v$" is not available to us in this situation; we can only define balls around $0$. – JonathanZ May 04 '20 at 00:40

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No. If this were true, it would mean that every norm-preserving self bijection of a normed vector space preserved addition and scalar multiplication. A counterexample is $V=\mathbb{R}$ with $\|\cdot\|$ given by the usual absolute value. Then the self bijection $$ \varphi(x)=\begin{cases}x&:|x|\leq 1,\\ -x&:|x|>1\end{cases} $$ preserves the norm, but not the vector space structure.

Julian Rosen
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  • $d(1/2, 3/2) = 1, d(\varphi(1/2), \varphi(3/2)) = 2$, no? – JonathanZ May 04 '20 at 00:32
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    Yes, $\varphi$ does not preserve distance. But it does preserve the norm, and that is all the question requires. – Julian Rosen May 04 '20 at 00:34
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    D'oh, you're right! It's very hard to have a norm in the picture and not use the associated metric. – JonathanZ May 04 '20 at 00:36
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    As all spheres in $\Bbb R^n$, $n>1$ have continuum cardinality, it follows that knowing only the underlying set and the norm function, we cannot even guess $\dim V$; dimensions $0$ and $1$ can be inferred, of course, but all other finite dimensions are indistinguishable. – Hagen von Eitzen May 04 '20 at 20:36