bio | website | math.uni-trier.de/~wengenroth |
---|---|---|
location | Universität Trier | |
age | 47 | |
visits | member for | 3 years, 5 months |
seen | yesterday | |
stats | profile views | 1,456 |
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. |