25
votes

Accepted

### Hölder's inequality for matrices

There are (at least two) "generalizations" of Hölder inequality to the non-commutative case. One is the so called tracial matrix Hölder inequality:
$$
|\langle A, B \rangle_{HS} |= |\mathrm{...

18
votes

### Matrix trace & norm

For hermitian $B$, the inequality is very easy to prove: use that $\|B\| - B$ is positive semidefinite together with the fact that the product of two positive semidefinite matrices has nonnegative ...

17
votes

### Matrix trace & norm

Expanding my comment into an answer, which offers a more general result.
Theorem (von Neumann). Let $A$ and $B$ be arbitrary $n\times n$ complex matrices. Then, $$|\text{trace}(AB)| \le \sum_{i=1}^...

12
votes

Accepted

### What is the $L^p$-norm of the (uncentered) Hardy-Littlewood maximal function?

Those are basic yet difficult questions. I don't know much about the uncentered case, but here is some information on the centered case.
A nonempty set $B \subseteq \mathbb{R}^d$ is centrally ...

12
votes

Accepted

### Norm of $n$-linear symmetric forms

After a bit of searching, I found that $\gamma_n=1$ for all $n$.
This is known as "van der Corput-Schaake inequality" (1935), discovered before by Szegö (1928), and mentioned to have been known to ...

11
votes

### which norms can be realized as operator norms?

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 \...

11
votes

### Are Hilbert-Schmidt operators on separable Hilbert spaces "Hilbert Schmidt" on the space of Hilbert Schmidt Operators?

No it's not.
If $s:H\to H$ is a rank one projection, then
$s\otimes\mathrm{id}:H\otimes \overline H\to H\otimes \overline H$
is a projection of rank $\dim(H)$.
In particular, if $H$ is infinite ...

11
votes

### Matrix trace & norm

A proof of
$$\lambda_n(B)\,{\rm tr}\,A\leq {\rm tr}\,(AB)\leq\lambda_1(B)\,{\rm tr}\,A$$
where $\lambda_n$ is the $n$-th largest eigenvalue of $B$ so $||B||=\lambda_1(B)$ and $A$, $B$ are positive ...

10
votes

Accepted

### Continuity of the product map

I think the answer is: It's always continuous if $A$ is subhomogeneous and never continuous otherwise(min or max).
First notice that if the product map is continuous for min, then it's continuous for ...

9
votes

Accepted

### Operator norm versus Hlawka inequality

It does not hold, at least for $n \geq 3$. The reason is that $M_n$ contains then a subspace isometric to $\ell_{\infty}^3$ (diagonal matrices with three non-zero entries) and for this space the ...

9
votes

Accepted

### Are Hilbert-Schmidt operators on separable Hilbert spaces "Hilbert Schmidt" on the space of Hilbert Schmidt Operators?

The answer is No, assuming of course that for you $s$ acts on $\mathcal{S}_\mathcal{H}$ by left-multiplication. There is no need to assume that $\mathcal{S}_\mathcal{H}$ is separable, since it ...

9
votes

Accepted

### Norm estimation of identity plus two non-commuting self-adjoint operators

The claim is false. Consider the following matrix argument.
\begin{eqnarray*}
\|(I+A+B)^{-1}A\| \le 1\quad\Leftrightarrow\quad \begin{bmatrix}I & (I+A+B)^{-1}A \\ A(I+A+B)^{-1} & I \end{...

9
votes

### Matrix trace & norm

We can work in an orthonormal basis in which $A$ is diagonal. In that case the diagonal entries of $AB$ are $a_{ii}b_{ii}$, so
$$
tr(AB) \leq|tr(AB)|=\left|\sum a_{ii}b_{ii}\right| \leq \sum a_{ii}|...

9
votes

Accepted

### Trace norm of operators obtained by restricting the matrix of a trace class operator

Here's an algorithm for testing an ad-hoc conjecture $C$ about Hilbert space operators. :-)
Set up the runtime environment correctly by loading the information "Most conjectures are false" ...

8
votes

### Operator norms of circulant matrices

The answer to Question 2 is "No". Note that the $2$-norm is just the maximum of $\left|\sum_{k=0}^{n-1}a_{k+1}z^k\right|$ over the $n$-th power roots of unity and the $1$-norm is just $\sum_k |a_k|$. ...

8
votes

Accepted

### Bounding the matrix norm of a commutator $[A,B]$ in terms of the norms of $A$ and $B$

A somewhat more general setting, namely, finding the best constant $C_{p,q,r}$ in
\begin{equation*}
\|AB-BA\|_p \le C_{p,q,r}\|A\|_q\|B\|_r,
\end{equation*}
for Schatten $p$,$q$,$r$-norms, is studied ...

8
votes

### Norm estimation of identity plus two non-commuting self-adjoint operators

(Unless I messed up the arithmetic, here's a counterexample.)
Let $B=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ and $A=\begin{bmatrix} 1 & 0 \\ 0 & R \end{bmatrix}$ where $R$ is a ...

8
votes

### Determinant of Jacobian and directional derivatives

Any matrix $A$ can be written as $B.U$ where $B=\sqrt{AA^\top}$ is posititive semidefinite and $U$ is orthogonal (polar decomposition). Thus $|\det(A)| =|\det(B)|.|\det(U)|$ is the product of the ...

8
votes

Accepted

### $\sup \left\| A x + B y\right\|_2$ subject to $\left\|x\right\|_2 = \left\|y\right\|_2 = 1$

The sets $\{Ax : \|x\|=1\}$ and $\{By : \|y\|=1\}$ are ellipsoids. Hence the set $\{Ax+By : \|x\|=\|y\|=1\}$ is the Minkowski sum of two ellipsoids. Googling for these terms returned this paper which ...

7
votes

### When is the norm of all positive operators on an ordered Banach space determined by their values on the positive cone?

Here are examples to show that neither of the two conditions 1. and 2. that you give as jointly sufficient can just be dropped.
Take $X$ to be two-dimensional real space with the pointwise order. ...

7
votes

### Operator norm versus Hlawka inequality

The inequality is true for $2 \times 2$ self-adjoint matrices, because of the following formula, which is reminiscent of $\max(|a|,|b|)=\frac{1}{2}|a+b|+\frac{1}{2}|a-b|$
$$ \|A\| = \frac{1}{2} | \...

7
votes

Accepted

### Completely bounded norm for unital maps with completely positive sections

Unfortunately, the answer is no.
Suppose $\Phi : M_4 \rightarrow M_2$ and $\Psi : M_2 \rightarrow M_4$ are given by
$$\Phi\left(\left[\begin{array}{cc} A & B\\ C& D\end{array}\right]\right) =...

7
votes

### The norm of tensor product operator on Lp spaces

For a Banach space $E$, let $L^p(X,E)$ be the completion of the simple functions $X\rightarrow E$ with the norm $\|f\| = \big(\int_X \|f(x)\|^p\big)^{1/p}$ (which reduces to a sum, as $f$ is simple). ...

6
votes

Accepted

### Equivalent Norms for the Dual of Sobolev / Bessel Spaces

Q1 has been addressed by Mark in the comments (absolute values are missing in your formula), but let me quickly look at this again: Just take Fourier transforms to see that
$$
\|g\|_{-s}^*=\sup_{\|h\|...

6
votes

Accepted

### Finding a norm on $ \mathbb{R}^X $ such that the "natural" embedding of a metric space $ X $ in $ \mathbb{R}^X $ becomes an isometry

Note that your embedding map $T$ actually takes values in the subspace
$\newcommand{\R}{{\mathbb R}}$
$c_{00}(X;\R)$ of finitely supported functions $X\to\R$. If you merely want a norm on this ...

6
votes

### Hölder's inequality for matrices

Let $A$ be a square matrix of dimension $n\times n$ and consider the following norm for $1< p<\infty$:
$$\|A\|_{p} = \max_{x \neq 0} \frac{\|Ax\|_p}{\|x\|_p}.$$
Let $\psi_p(x):=\big(|x_1|^{p-1}\...

6
votes

### which norms can be realized as operator norms?

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
$$\|...

6
votes

### Characterizing when matrices are 'dissipative'

The discrete-time analogue -- there exists a norm in which $|A_1^nx|\leq |x|$, $|A_2^nx|\leq |x|$ for all $n\geq 1$ and $x \in\mathbb{R}^d$ -- is equivalent to the property that the semigroup ...

6
votes

### Bounding the matrix norm of a commutator $[A,B]$ in terms of the norms of $A$ and $B$

Concerning your first question, as I noted in my comment above for any operator norm we have $\Vert [A,B]\Vert \leq 2\Vert A\Vert\Vert B\Vert$.
Conversely, let $A = \pmatrix{1 & 0 \\ 0 & -1}$,...

6
votes

### Norm estimation of identity plus two non-commuting self-adjoint operators

Though two answers are already posted, let me explain how to understand that it is false from general reasoning. The inequality $\|(I+A+B)^{-1}A\|\leqslant 1$ is equivalent to $\|(I+A+B)^{-1}A x\|\...

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