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bio website math.uni-trier.de/~wengenroth
location Universität Trier
age 47
visits member for 3 years, 5 months
seen yesterday
I am Professor for Mathematics at the Unversität Trier (Germany)

Jun
23
comment When does analytic in the operator norm imply analytic in the trace class norm?
Extending a bit Christian Remling's comment: Grothendieck proved that for a complete locally convex space $X$ a function $f:U\to X$ is holomorphic if $\varphi\circ f:U\to \mathbb C$ is holomorphic for all continuous linear functionals $\varphi$ on $X$. As far as I remember, one does not even need all $\varphi$.
Jun
15
comment Exponential rule for Whitney-$\mathcal{C}^{\infty}$-topology
Time to create an "Ask-Michor-tag". More seriously, Peter Michor, Andreas Kriegl, and collaborators did a lot on such questions for a big variety of function spaces. Look for "convenient calculus".
Jun
15
accepted The list of problems for Grothendieck's thesis
Jun
15
awarded  Nice Question
Jun
14
asked The list of problems for Grothendieck's thesis
Jun
11
reviewed Approve Lie Algebra, counterexample
Jun
9
comment Normed space that is sigma-totally-bounded but is not sigma-compact
Here is an idea (which might not yet really work) to find $c_0$ as a subspace of $C^1(I)$: Choose $\phi_n \in C^1(I)$ with disjoint supports (contained in $[1/n - 1/n^2, 1/n+1/n^2]$), $\phi_n(1/n)=1$ and $|\phi_n(x)|\le 1$ and define $T$ on $c_0$ by $T(\alpha)=\sum\limits_{n=2}^\infty \alpha_n \phi_n$. This will be isometric from $c_0$ to $C(I)$ and the only problem is, that $T(\alpha)$ is not differentiable at $0$ (only continuity follows from $\alpha_n\to 0$).
Jun
9
comment Normalized tight frame that is not orthonormal
My comment intended to show that the condition on the norms implies orthogonality.
Jun
8
comment Normalized tight frame that is not orthonormal
You mean $\|f\|_2^2$, don't you? For $f=\psi_{m,n}$ you then get $\|\psi_{m,n}\|_2^2= \sum_{j,k} |\langle \psi_{m,n},\psi_{j,k}\rangle|^2 = \|\psi_{m,n}\|_2^2 + \sum_{(j,k)\neq (n,m)} |\langle \psi_{m,n},\psi_{j,k}\rangle|^2 $ so that $\langle \psi_{m,n},\psi_{j,k}\rangle=0$ for all $(j,k)\neq (n,m)$.
Jun
2
awarded  Popular Question
May
30
comment topology of setwise convergence of measures
If you ask a question here it is not only a matter of politeness to show some reaction if your question is answered.
May
22
revised topology of setwise convergence of measures
corrected spelling
May
20
answered topology of setwise convergence of measures
May
11
revised Relation between dual of nuclear space $(\substack{\text{lim} \\ \leftarrow i} H_i)'$ and $\substack{\text{colim} \\ i \rightarrow } H_i$
Explanantion added.
May
7
answered Relation between dual of nuclear space $(\substack{\text{lim} \\ \leftarrow i} H_i)'$ and $\substack{\text{colim} \\ i \rightarrow } H_i$
May
6
comment Orthogonal complements of intersections of closed subspaces
On the other hand, the theorem of bipolars implies that $(H_1\cap \cdots \cap H_n)^\perp$ is always the closure of the sums of the orthogonal complements.
May
6
comment constant rank theorem for banach spaces
@TomekKania You are of course right. Replying to Benjamin's comment I though of Hilbert spaces.
May
5
comment constant rank theorem for banach spaces
I meant no countable Hamel basis, of course you can have a countable Schauder basis. I believe, that "x-dimensional" usually refers to Hamel bases. Having a countable Schauder basis is equivalent to separability.
May
4
comment constant rank theorem for banach spaces
There are no countable dimensional Banach spaces.
Apr
6
comment $\mathcal S(\mathbb R^n) \hat \otimes_\pi \mathcal S(\mathbb R^m) \simeq \mathcal S(\mathbb R^{n+m})$?
Grothendieck called the $\varepsilon$ tensor topology injective. The inductive topology is still finer and even for nuclear spaces it may be different from the injective one.