## New answers tagged norms

1

No, it isn't true.
Let $E$ be a non-Borel set in $X$ and consider the set of Dirac measures
$E' = \{ \delta_x : x \in E\} \subset S$. Note that for each Dirac measure $\mu$, the total-variation ball of radius 1/2 about $\mu$ contains no other Dirac measure. Hence $E'$ is closed in $T_N$, hence Borel in $T_N$.
On the other hand, it is well known that ...

5

For sake of completeness, I am writing a full answer based on the suggestion of
Pietro Majer.
The following are equivalent:
1) $A$ is an isometry w.r.to some norm.
2) $A$ is diagonalizable (over $\mathbb{C}$) , with all eigenvalues of modulus 1.
3) All orbits of $A$ are bounded ( $\sup_{k\in\mathbb{Z}}\|A^k x\|<+\infty$ for any $x\in ...

11

Operator norms are the same thing as injective tensor norms or, equivalently, smallest dual cross norms on $\operatorname{Hom}(V, W) \simeq V^\ast \otimes W$. This means that the dual norm $\Vert \cdot \Vert^\ast$ on $V \otimes W^\ast$ has to satisfy the following:
$$\Vert x \Vert^\ast = \inf_{x = \sum_i y_i \otimes z_i} \sum_i \Vert y_i \Vert \Vert z_i ...

6

Here is a non trivial constraint : ${\rm Hom}(V,W)$ contains (is spanned by) rank one morphisms
$$v\mapsto\ell(v)w,\qquad\ell\in V',w\in W.$$
If a given norm over ${\rm Hom}(V,W)$ is induced, then
$$\|w\otimes\ell\|=\|w\|_W\|\ell\|_*.$$
This yields the necessary condition
...

6

To get dimension 4, I think the norm
\begin{equation} \|a\| = \max_{\{i,j\}} |a_i-a_j| \vee \|a\|_\infty,
\end{equation}
where $a=(a_1,a_2,a_3,a_4)$,
works. Again $X$ samples the unit vector basis uniformly.
EDIT: It looks like a variation takes care of dimension 3. Use again
\begin{equation} \|a\| = \max_{\{i,j\}} |a_i-a_j| \vee \|a\|_\infty,
...

2

fedja: This is a very nice and elegant answer, thank you! (Have you seen/used such a norm ever before?) Before accepting your answer, I'd like to wait a bit, so that other participants feel more encouraged to present other answers.
In fact, let me give here a modification of your answer. I noticed that for $N=3$ your norm of $v=(v_1,\dots,v_N)$ is ...

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