## New answers tagged norms

2

I guess that putting an archimedean norm on a vector space over a nonarchimedean field gives just an uncorrelated product of something archimedean with something nonarchimedean. Number theorists sometimes look at all places of $\mathbb{Q}$ at once: all p-adic valuations and the archimedean valuation. I don't see a reason why one should want to look at just ...

5

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.

3

There is an elementary answer. Let $D$ be any domain of $\mathbb{C}$. The usual derivation operator $\partial : \mathcal{O}(D)\to \mathcal{O}(D)$ is continuous for the topology of uniform convergence (use Cauchy integral formula), but not for any norm. Consider indeed the sequence of functions $f_n:=\frac{e_n}{||e_n||}$ where $e_n~:~x\mapsto \exp(nx)$.

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 ...

0

In this paper http://arxiv.org/abs/1007.3418 one can find definitions for both inhomogenous and homogenous besov norms. The second definitions are the ones for the inhomogenous besov norms if you take away the ⟨⟩ brackets.
Furthermore it is obvious (since all norms on $\mathbb{R}^n$ are equivalent) that:
...

4

I will show $||B||<1$ (in the finite-dimensional case). Suppose $|Bv|=|v|$ for some nonzero $v$. Then as $|Bv|$ is the projection of $A^{\otimes 4} ( \sum_i v_i e_i^{\otimes 4})$ onto the subspace generated by $e_i^{\otimes 4}$, and $|A^{\otimes 4} ( \sum_i v_i e_i^{\otimes 4})|=| \sum_i v_i e_i^{\otimes 4}|=|v|$, it follows that $A^{\otimes 4} ( \sum_i ...

0

Here is the proof for $ ||B||\le 1$(This still doesn't answer the question):-
Consider the vector space $ \mathbb{C}^n\otimes \mathbb{C}^n\otimes \mathbb{C}^n\otimes \mathbb{C}^n$.
There consider the subspace spanned by $ e_i^{\otimes 4} $. Then $ e_i^{\otimes 4} $ forms a orthonormal basis for this subspace. And the matrix for $ A\otimes A\otimes A\otimes ...

0

No. Take the unitary
$
A=\begin{pmatrix}
0&1\\
-1&0
\end{pmatrix}
$
which satisfies your assumption. The matrix $B=\begin{pmatrix}
0&1\\
1&0
\end{pmatrix}$
has norm 1.

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