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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 $p(z)=|z|$ (real space). It can be seen easily that 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.

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.

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 $p(z)=|z|$ (real space). It can be seen easily that the set $$ U=\{e^{i\theta}\,|\, \theta\in[-\frac{\pi}{6},\frac{\pi}{6}]\cup [\frac{\pi}{3},\frac{2\pi}{3}]\} $$
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).

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 $p(z)=|z|$ (real space). It can be seen easily that 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.

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 $q$, $p|_U=q|_U\Longrightarrow p=q$. 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 $p(z)=|z|$ (real space). It can be seen easily that the set $$ U=\{e^{i\theta}\,|\, \theta\in[-\frac{\pi}{6},\frac{\pi}{6}]\cup [\frac{\pi}{3},\frac{2\pi}{3}]\} $$
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).

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 $p(z)=|z|$ (real space). It can be seen easily that the set $$ U=\{e^{i\theta}\,|\, \theta\in[-\frac{\pi}{6},\frac{\pi}{6}]\cup [\frac{\pi}{3},\frac{2\pi}{3}]\} $$
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).