# Tagged Questions

**6**

votes

**2**answers

306 views

### $\|T\|_2 \le \sqrt{\|T\|_1\|T\|_\infty}$

Let $T$ be a linear operator acting on a finite-dimensional real or complex
vector space. As a direct consequence (or rather a particular case) of the
Riesz-Thorin theorem, we have
$$ \|T\|_2 \le ...

**10**

votes

**1**answer

321 views

### Operator norms of circulant matrices

The definition and basic properties of circulant matrices can be found here: http://en.wikipedia.org/wiki/Circulant_matrix.
For complex numbers $a_1,\ldots,a_n$, I will use the notation
$$
...

**5**

votes

**1**answer

149 views

### An extreme point of the ball of the space of compact operators

It is very easy to see that the unit ball of $c_0$ has no extreme points. I was trying to spot any extreme points in the unit ball of the space of compact operators on a Hilbert space (a ...

**1**

vote

**0**answers

179 views

### Tensors with low spectral norm

Consider a tensor $T$ with six indices, $T_{(ii')(jj')(kk')}$, where each index goes from $1$ to $n$. We can think of $T$ as a linear map from $\mathbb{R}^n \otimes \mathbb{R}^n \otimes \mathbb{R}^n$ ...

**4**

votes

**1**answer

434 views

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

I'm trying to investigate the interplay between the norm and cone of positive elements in ordered Banach spaces. In particular, I would like a nice characterization of when the norm of a positive ...

**3**

votes

**1**answer

335 views

### injective tensor norms for real tensors

If $A$ is an element of $\mathbb{R}^n \otimes\mathbb{R}^n \otimes\mathbb{R}^n$, then define its injective tensor norm to be
$$\|A\|_{\rm inj} := \max_{x,y,z\in \mathbb{R}^n, \|x\|=\|y\|=\|z\|=1}
...

**3**

votes

**2**answers

701 views

### How to prove the Hahn-Banach constructively

I am just wondering, how to prove the Hahn-Banach theorem constructively for a finite dimensional normed vector space.
Thanks in advance for any helpful answers.