## Hot answers tagged norms

21

So it turns out my earlier intuition was incorrect, and one can leverage the order properties of ${\bf R}$ to show:
Theorem. Let $X, Y$ be real Hilbert spaces, with the dimension of $X$ at least two, and let $f: X \to Y$ be a function such that $\|f(x)-f(y)\|=\|x-y\|$ whenever $\|x-y\|$ is a natural number. (We do not assume $f$ to be continuous.) Then ...

21

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 ...

18

If $E$ is to be a Hilbert space, a proof must establish more or less directly that the inequality implies the parallelogram law $\lVert x + y\rVert^2 + \lVert x-y\rVert^2 = 2\lVert x\rVert^2 + 2\lVert y\rVert^2$ for all $x,y \in E$. Since both, your hypothesis and the parallelogram law, are conditions on all $2$-dimensional subspaces of $E$, one can assume ...

16

I'm not in my best shape at the moment, so, please, check thoroughly what is written below.
The answer is "No".
The distribution does not matter much, but the norm does. So, we want to take $N\ge 3$ standard basis vectors $e_n$ and see if there is a chance to create a permutation invariant norm in $\mathbb R^N$ such that
$$(*)\qquad ...

13

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 ...

13

Let $X=Y=\mathbb R$ with the absolute value norm and define $n(a,b)=\sqrt{2a^2+2b^2-3ab}$. This is a norm on $\mathbb R^2$ because it is the quadratic form of the positive definite matrix $A=\left( \begin{smallmatrix} 2 & -3/2 \\ -3/2 & 2 \end{smallmatrix}\right)$.
Then $N(v) = n(|v_1|,|v_2|)$ is not a norm because the triangle inequality fails for ...

13

This is only an illustration to Christian Remling's beautiful answer; here are the concentric "balls" around the origin for this "norm":

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 ...

10

In many books$^*$ you can find the result that there is a projection of norm at most $\sqrt{n}$ onto any $n$ dimensional subspace of a Banach space. For reflexive spaces, this gives immediately that every $n$ codimensional subspace is the range of a projection that has norm at most $\sqrt{n} +1$. For non reflexive spaces, by using the principle of local ...

10

If you assume $f$ to be surjective then $f$ has to be linear without any assumptions on $V$ by the Mazur-Ulam theorem. Wikipedia doesn't offer much more information than a link to the beautiful recent proof by J. Väisälä.

8

The closest thing I know for induced norms is the Riesz–Thorin theorem. There are other Hölder-like inequalities for matrices, for example involving Schatten norms.

8

You might be interested in the whole (mostly Russian) literature on "Banach-Kantorovich" or "lattice-normed" spaces, which are:
"a triplet $(\mathcal U,E,\lambda)$ consisting of a vector space $\mathcal U$, a Dedekind complete vector lattice $E$ and a map $\lambda:\mathcal U\to E_+$ satisfying some natural conditions that allow one to consider $\lambda$ ...

8

Terry Tao's argument generalizes to show that $f$ must be an isometry whenever $X$ has dimension greater than 1 and $Y$ is strictly convex. We start by proving a series of lemmas:
Lemma 1: Let $x,y\in X$, and let $a,b\geq0$ be such that $a+b\geq\|x-y\|$ and $|a-b|\leq\|x-y\|$. Then there exists $z\in X$ such that $\|x-z\|=a$ and $\|y-z\|=b$.
Proof: Let ...

7

This is a comment rather than an answer but I am not entitled. The missing requirement on the two dimensional norm for the statement to be valid is that it be increasing in the natural sense that if $|x|\leq|x_1|, |y|\leq|y_1|$, then $n(x,y)\leq n(x_1,y_1)$. Norms without this property can easily be obtained by rotating a non-circular ellipse in standard ...

7

$\def\sign{\mathop{\rm sign}}$First of all, it is enough to prove the statement when $p=u/v$ is rational, $u$ is even and $v$ is odd (such numbers are dense on the real line). We need this to simplify the last argument.
Let the equation of the ellipse be $f(x,y)=ax^2+2bxy+cy^2=1$; one may assume that $b\neq 0$ (either by a small variation argument, or by ...

7

I got
$$r([A,B])\leq 4\sqrt{2} r(A) r(B).$$
It is lower than $8$ but still higher than the conjuncture $C_{nr}=4$.
I used the following facts:
For normal (i.e. $X^\*X=XX^\*$) matrices we have $r(X)=\sigma_1(X)$ (the largest singular value of X).
$\sigma_1(XY-YX)\leq 2 \sigma_1(X)\sigma_1(Y)$
Also, note that for $X$ and $Y$ hermitian (i.e. $X^\*= X$ ...

6

this is called the numerical radius of a matrix

6

$|||A|||$ is the largest s-number (modulus of gen. eigenvalue).
$f(A)$ is the smallest s-number. It is 0 if $A$ is not injective.

6

By the Courant-Fischer min-max theorem, if $A$ is Hermitian, then $f(A) = \lambda_n(A)$, the smallest eigenvalue of $A$.

6

(This answer expands the ideas in the comments by Mikael de la Salle and myself).
There is a family of matrix norms that can be defined as follows: given $p \geq 1,k \leq n$,
$$
\|A\|_{p,k} = \left(\sum_{i=1}^k \sigma_i(A)^p\right)^{1/p},
$$
where we denote by $\sigma_i(A)$ the $i$th singular value of $A$ (taken in nonincreasing order, $\sigma_1(A)\geq ...

6

Take a look at the papers "Über Normtopologien in linearen Räumen" (Link to the article) as well as "Über vollständige Normtopologien in linearen Räumen" (Link to the article) by D. Laugwitz.
He proves the following. Let $E$ be a vector space, denote by $a(E)$ the cardinality of a Hamel basis of $E$ and denote by $n(E)$ the number of mutually non-equivalent ...

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,
...

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

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 ...

6

Applying rescaling you can assume that $L=1$.
We look for a 1-Lipschitz piecewise linear maps $f\colon\mathbb{R}^m\to\mathbb{R}^n$.
If $m=n$ we get $f$ from Brehm's theorem;
it says that there is a piecewise distance preserving map of that type (in particular 1-Lipschitz and piecewise linear).
See Brehm, U., Extensions of distance reducing mappings to ...

6

Anton's answer is for the case where $X$ is a subset of a Hilbert space. For $X$ a general metric space (compactness does not help, BTW), $L$ must be increased by a factor of order $\log^{1/2}N$; see
$$$$
Johnson, William B.; Lindenstrauss, Joram Extensions of Lipschitz mappings into a Hilbert space. Conference in modern analysis and probability (New ...

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 ...

5

Here is a partial result.
Claim. $4 \le C_{nr} \le 8$.
Proof. The upper bound has already been shown by the OP. The lower-bound follows by
\begin{equation*}
A =
\begin{bmatrix}
0 & 1\\\\
0 & 0
\end{bmatrix},\qquad
B =
\begin{bmatrix}
0 & 0\\\\
-1 & 0
\end{bmatrix}
\end{equation*}
for which
...

5

You can use the surprising identity $(A^{-1}+B^{-1})^{-1}=A(A+B)^{-1}B$, and take the norms of the three factors separately.

5

The answer by Piotr Migdal can be modified to give the accurate inequality
$$r([A,B])\le4r(A)r(B),\qquad\forall A,B\in{\bf M}_n(\mathbb C).$$
The only new argument is that for every matrix $M$, there exists an angle $\theta$ such that $r(M)=\|{\rm Re}(e^{-i\theta}M)\|_2.$
Actually, we do have
$$r(M)=\sup_\alpha\|{\rm Re}(e^{-i\alpha}M)\|_2.$$
Hereabove, ...

Only top voted, non community-wiki answers of a minimum length are eligible