## Hot answers tagged norms

17

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 ...

10

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 ...

10

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

9

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 ...

9

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

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

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.

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

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

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 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 ...

5

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

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

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, ...

4

Here is a simple proof that the property holds only for Euclidean norms, at least if the norm in question is $C^1$ smooth and strictly convex. Surely it was known way before Gromov was born.
Let $S$ denote the unit sphere of the norm. First observe that, if $v\in S$ and $w$ are such that $\|w+tv\|\ge \|tv\|$ for all $t\in\mathbb R$, than $w$ is parallel to ...

4

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.

4

Yes, a classical result says that every 2-dimensional (real) normed space embeds into L^1.
Alternatively, if we assume that the unit ball is a $2n$-gon (the general case then follows by approximation), the corresponding space embeds into $\ell_1^n$. The dual picture is maybe even more transparent : any symmetric $2n$-gon is the Minkoswki sum of $n$ ...

4

The question that you are asking was asked in "On Generalizations of Minkowski's Inequality in the Form of a Triangle Inequality" by F. Mulholland (1949). In that paper, Mulholland established a sufficient condition on $f$, namely that it should satisfy $f(0)=0$, be increasing on $x \ge 0$ and be g-convex, i.e., $\log f(e^x)$ is convex on the reals.
...

4

(This isn't an answer, but I'm not allowed to comment yet and I think this is a worthwhile question, even if it means violating the rules; i'll delete the post when the time comes)
You claim that usual matrix norms do not work because of the $|f(x_1)-f(x_2)|$ criterion, but if $f$ is differentiable, then given any $x_1$ it is Lipschitz in some neighborhood ...

4

Maybe I am misinterpreting something, because according to my experiments, this function is neither convex nor concave.
The following is a counterexample (EDIT: I changed the example to use symmetric matrices):
\begin{equation*}
A=\begin{pmatrix}8 &4\\ 4 & 6\end{pmatrix},\quad B=\begin{pmatrix} 4 & 4\\ 4 & 6\end{pmatrix},\quad ...

4

After working a little bit, I solved the $2\times 2$ case. Surprisingly, ${\rm ext}(B_{\rm nr})$ is not closed. It consists of (recall that if $n=2$, the numerical range is an ellipse $E_A$ together with its interior; the ellipse may degenerate to a segment or a point)
homotheties $zI_2$ with $|z|=1$,
matrices $A$ whose ellipse $E_A$ is tangent interiorly ...

4

The simple example of strongly equivalent metrics in a comment of Willie Wong actually holds the key to what is going on:
Definition. Two generalized functions $\mu_1$ and $\mu_2$ on the unit sphere $S^n \subset \mathbb{R}^{n+1}$ will be called comparable if there exists a constant $\lambda \geq 1$ such that both
$$
\lambda \mu_1 - \mu_2 \; \; \hbox{and} ...

4

Sorry, my answer below is only partial, but I thought that it may still be somewhat interesting.
As far as I know, this inequality does not have a distinguished name. It is ultimately a consequence of the duality between $\|\cdot\|_\infty$ and $\|\cdot\|_1$, and the submultiplicativity of $\|\cdot\|_1$.
This is certainly known to you, but I wanted to ...

3

Thank you all for your interest. On ResearchGate Prof. Leonid Gurvits just pointed out an excellent 2 x 2 counterexample to the inequality I intended to prove.
I was tricked by the fact that the inequality was numerically satisfied for a huge number of 5 x 5, 7 x 7, 10 x 10 randomly generated matrices. I went straight to big dimensions, overlooking the test ...

3

Here is a somewhat silly way to do it. You need to prove that:
$$\frac{\left|\sum_{j=1}^l z_j\right|^p-\sum_{j=1}^l\left|z_j\right|^p}{\sum_{i\neq j}\left|z_i\right|\left|z_j\right|^{P-1}}$$
is bounded function on the unit sphere $S^{l-1}$. $S^{l-1}$ is compact, and this function is continuous away from the places where $z_i=\pm 1$, where the denominator ...

3

$f$ is convex separately for $A$ or $B$ but is not (in general) for the couple $(A,B)$.
Proof: The Hessian of $f$ in $X,Y$ is the following QUADRATIC form:
$Q(H,K)=2(||XHB+XAK||^2+2trace((XAB)^TXHK))-4trace(Y^TXHK)$.
Then $Q(H,0)=2||XHB||^2$ and $f$ is convex for $A$ (and similarly for $B$) - One must say a little more if there is $H\not=0$ s.t. $XHB=0$ ...

3

Computing such induced norms is a hard problem. For the case of $p=2$ and $q \ge 2$, have a look at this paper by Barak et al. to see how tricky the problem is.
Typically, for other than the nice cases of $1,2, \infty$ style, these norms are NP-hard to compute, with well-known results for the case $p \ge q$ (see also e.g.: "Matrix norms are NP-Hard to ...

3

The proof can be found in Golub, van Loan's "Matrix Computations", Sec 2.

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