## New answers tagged norms

1

If $f$ is homogeneous, you can just try to minimize it on the unit ball (which is a Lagrange multiplier problem, which does not mean it's easy), and see if any of your critical values are smaller than $1\dots$

19

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 ...

5

The best such inequality that depends only on $m$ and $n$ is:
$$
\frac{1}{\sqrt{mn}}\|A\| \leq \|A\|_* \leq \|A\|
$$
The right inequality is tight when $A$ is a matrix with a $1$ in the top-left corner and zeroes elsewhere. The left inequality is tight when $A$ is the matrix all of whose entries are $1$. These examples also show that you cannot get any ...

1

The first paper on characterization of inner product spaces was:
P. Jordan and J. Von Neumann, On inner products in linear, metric spaces, Ann. of Math. (2) 36 (1935), no. 3, 719–723.
There is a book on the subject:
Vasile Ion Istrăţescu. Inner Product Structures: Theory and Applications. Springer, 1987. Especially chapter 4.

12

Metric space $(X,\rho)$ satisfying Ptolemy inequality $\rho(a,b)\rho(c,d)+\rho(b,c)\rho(a,d)\geq \rho(a,c)\rho(b,d)$ is called ptolemaic space. A normed ptolemaic space must be inner product space.
Reference: I.J. Schoenberg, A remark on M. M. Day’s characterization of innerproduct spaces and a conjecture of L. M. Blumenthal. Proc. Am. Math. Soc. 3, 961–964 ...

Top 50 recent answers are included