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It is well known that for a bilinear form over an n-dimensional vector space, $n^2$ values (on all pairs of basis-vectors) determine it uniquely.

How many values do we need to specify in order to uniquely determine a norm? The first trivial observation is that homogeneity implies it's enough to specify all the values on the unit sphere.

Then, if we indeed assume $V$ is a finite-dimensional space, every two norms on $V$ are equivalent, and in particular are continuous w.r.t each other. Hence their set of mutual agreement is closed, so it is enough to specify a dense subset of the unit sphere. since a the unit sphere of a finite dimensional normed space is separable, we conclude it is enough to specify a countable number of values.

Is it possible to use only a finite number points? (the constant may depend on the dimension ofcourse).

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    $\begingroup$ There is a detailed answer below, but the rough answer is very simple. There is a bijective correspondence between norms and bounded symmetric convex sets containing a neighbourhood of the origin. (the set is the unit ball of the norm; the norm is the Minkowski functional of the set). $\endgroup$ – Anthony Quas Apr 12 '15 at 16:14
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    $\begingroup$ @AnthonyQuas: Minor nitpick: There should be an additional condition such as closedness or openness to make this a bijection. The open and the closed unit ball (and every set in between those two) have the same Minkowski functional. $\endgroup$ – Johannes Hahn Apr 14 '15 at 23:38
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Well, as you have certainly already remarked (reading your post, I assume this), bilinearity makes a big difference. For the "only norm" case, what you are looking for, if I understand correctly your question, is a set of uniqueness for the admissible norms on a given vector space $V$. Your demonstration establishes that values on a dense set $U$ (on the unit sphere) is sufficient (and then, we can reduce to countable). You can go further by choosing a set $U$ such that $U\cup (-U)$ is dense, then you cannot go further as, on $V$, a norm $p$ is the jauge function of the balanced convex $$ C=\{x\in V|\, p(x)\leq 1\} $$ then, if your set $U\cup (-U)$ is not dense, there is a point $M\in S_V$ (the unit sphere) and a neighbourhood $W$ of $\{M,-M\}$ such that $U\cap W=\emptyset$. Now, deforming the sphere around $\{M,-M\}$ in a convex way (you can find a one parameter deformation $C_t$ such), one obtains an infinite family of norms which coincide with $p$ on $U$ and differ from it. If the space $V$ is complex, just replace the "real balanced saturation" $U\mapsto U\cup (-U)$ by its analogue (the orbit of $U$ under the group of complex numbers of modulus one $\mathbb{U}=\{z\in\mathbb{C} | |z|=1\}$).

Proposition In order $U\subset S_V$ be a uniqueness set for the collection of all norms, it is necessary and sufficient that the orbit $G.U$ be dense in $S_V$ ($G=\{−1,1\}$ for the real case and $G=\mathbb{U}$ for the complex case).

[$G.U$ dense$\Longrightarrow$ $U$ is a uniqueness set] is clear.

Now, if $G.U$ is not dense, it exists a point $w\in S_V$ which does not belong to $\overline{G.U}=G.\overline{U}$ and then $$ max_{v\in U}\,\{|\langle w|v\rangle|\}=M<1 $$ Now, the convex sets $$ C_t=\{v\in V|\, ||v||\,\leq 1\ \mathit{and}\ |\langle w|v\rangle|\leq (1-t)M+t\} $$ for $0<t\leq 1$ give the desired deformation (for each $C_t$ contains $U$).

A bit more (Local implies global ?) For $p$ a norm, we will say that set $U_p\subset S_V$ is a relative uniqueness set w.r.t. $p$ if, for any norm $q$, $p|_{U_p}=q|_{U_p}\Longrightarrow p=q$. We have just seen that, if $p$ is the euclidean (resp. hermitian) norm then each $U_p$ is a uniqueness set for all norms. This is not the case for every norm (especially if it has many linear zones) for example take the two dimensional euclidean space $\mathbb{C}$ with the norm $|z|$ (real euclidean space). It can be seen easily that, with $p=l_1$ (see below), the set $$ U_{l_1}=\{e^{i\theta}\,|\, \theta\in T=[-\frac{\pi}{6},\frac{\pi}{6}]\cup [\frac{\pi}{3},\frac{2\pi}{3}]\} $$
is a relative uniqueness set w.r.t. the $l^1$ norm $||z||=|\Re(z)|+|\Im(z)|$ ; but $U\cup\, (-U)$ is not dense in the unit sphere(for example $e^{i\frac{\pi}{4}}$ is not in the closure of it).

Remark If you want $U_{l_1}$ to be denumerable replace $T$ by $T\cap \mathbb{Q}$, with the same closure.

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    $\begingroup$ an explicit deformation that works should be the set of $v$ such that $|v \cdot w| \leq 1$ for $w \in U$. $\endgroup$ – Will Sawin Apr 12 '15 at 15:40
  • $\begingroup$ can you please elaborate on your explicit example for deformation? I understand the intuitive idea that we can change the norm in a neigbourhood of $M$. I am still not sure about why the resulting function (the norm after the change) will satisfy the triangle inequality. $\endgroup$ – Asaf Shachar Apr 14 '15 at 13:12
  • $\begingroup$ I think I understood the general idea, I can just deform the original unit ball, to some other bounded, convex, symmetric shape, changin it only in a small enough neigbourhood of $M$ where I do not have any constraints about the norm's values. $\endgroup$ – Asaf Shachar Apr 14 '15 at 13:21
  • $\begingroup$ .@asaf shachar Exactly ! You are right. [can you please elaborate on your explicit example for deformation?]--> OK, I did it in the answer above. $\endgroup$ – Duchamp Gérard H. E. Apr 14 '15 at 16:35

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