# Tag Info

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 ...

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

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

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

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 ...

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 ...

6

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

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

this is called the numerical radius of a matrix

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

(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

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

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 ... 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 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. 3 The answer is no, for any n\ge 3 and any C^1-smooth strictly convex norm on \mathbb R^n. (Here "strictly convex" means that the triangle inequality is strict for any two non-collinear vectors. This is equivalent to the following: the norm restricted to any affine subspace not containing 0 is a strictly convex function on that subspace.) The claim you ... 3 Some experimentation led me to the following results: \alpha=1 yields the minimum value even for the operator-2 norm (certainly provable directly too) \|M-ee^T/n\|_2 can be larger than \sigma_2 + \cdots + \sigma_n \|M-2ee^T/n\|_2 = \sigma_1(M) (obviously, since M is rs) So the only reasonable bound is: \|M-ee^T/n\|_2 \le c_n\sigma_1(M), where ... 3 A useful and easy to compute bound is given by the reasonably well-known relation (see e.g., this Wikipedia section) \begin{equation*} \|A\| \le \sqrt{\|A\|_\infty \|A\|_1} \end{equation*} between the spectral norm, and the induced 1 and \infty norms of an arbitrary matrix A. Corollary: If A is elementwise nonnegative and row-stochastic, then \|A\| ... 3 For (i) the usual thing is to take the Lipschitz analogue of the Banach-Mazur distance; namely, the infimum over injective and surjective maps T from V to W of the Lipschitz constant of T times the Lipschitz constant of T^{-1}. Whether this is equivalent to the Banach-Mazur distance for separable Banach spaces is a well known open problem. See ... 3 The proof can be found in Golub, van Loan's "Matrix Computations", Sec 2. 3 First, the fact that UR implies URED is obvious (the definition of UR requires less from the sequences (x_n), (y_n) in order to conclude the convergence \|x_n-y_n\|\to 0.) Second, the notions LUR and URED are not related in general. The answers can be extracted from Chapter II of the monograph by Deville, Godefroy and Zizler Smoothness and renormings ... 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 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 ...

2

Your first integral is $|A|$ (volume), and similarly the other integrals. Cauchy-Schwarz inequality implies $|A\cap B|^2\leq |A||B|$ and nothing else can be said. Your inequality with 3 sets is not true. Take $A,B,C$ independent. That is the volume of each product set is equal to the product of the volumes. Such sets can be easily constructed, and can have ...

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