## Hot answers tagged norms

23

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 ...

19

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 N(N-1)\|e_1+e_2\|+2N\|...

14

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 $...

14

For those who don't have the book (or have the wrong version), here is the proof that the topological vector space of holomorphic functions on the unit disk is not normable (i.e. whose topology is not defined by a norm).
Definition: A topological vector space (over $\mathbb R$ or $\mathbb C$) is locally bounded iff there is a neighborhood $U$ of 0 such that ...

13

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

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 ...

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.

9

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 $...

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 $\lambda$ ...

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. $X^\*= X$ ...

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

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 ...

6

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

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

this is called the numerical radius of a matrix

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

(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

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

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,
\end{...

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
$$\|w_1\otimes\ell_1\|\cdot\|w_2\otimes\ell_2\|=\|w_1\otimes\ell_2\|\cdot\|w_2\otimes\...

6

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 \...

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 Haven,...

6

By Montel's theorem, every bounded set (w.r.t. the topology of uniform convergence on compact sets) in the space of holomorphic functions is relatively compact. If the space were normed its closed unit ball would be compact which implies that the space is finite dimensional.

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, ...

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