2,297 reputation
725
bio website math.uni-trier.de/~wengenroth
location Universität Trier
age 46
visits member for 2 years, 6 months
seen 6 hours ago
I am Professor for Mathematics at the Unversität Trier (Germany)

6h
comment Left invertible operators of $B(X,Y)$
If $X=Y$ is of finite dimension, injective operators are surjective. One thus needs infinite dimensional spaces to have the set of injectivions not open. Anyway, the "right" question would be not for injections but for monohomomorphisms, i.e., injections with closed range (surjections between Banach spaces are always epihomomorphisms, i.e. open).
1d
comment boundary density of the Von Koch flake
It looks as if $f_r(x)$ tends to the derivative of the measure $\mu(A)= vol(K\cap A)$ w.r.t. Lebesgue measure. Lebegue's theorem (see e.g. Rudin's Real and Complex Analysis, chapter 8) tells you that $\lim f_r = 1_K$ almost everywhere which, unfortunately, does not say anything about the integral w.r.t. the Hausdorff measure on $\partial K$.
2d
comment Power law distribution with support in x=0
What do you mean by the above theorem? All you need to apply thm 2.3.4 in Hoermander's book is that $f_0$ is a distribution with support $\lbrace 0 \rbrace$ (the order is then automatically finite), and for this you would need the convergence of all integrals $\int \varphi(x)f_\epsilon(x)dx$ (thm 2.1.8).
2d
comment In the category of sets epimorphisms are surjective - Constructive Proof?
Is this really different from the argument in my comment?
Aug
18
comment In the category of sets epimorphisms are surjective - Constructive Proof?
Given the epimorphism $f$ define $g:Y\to\lbrace 0,1\rbrace$ by $g(y)=1$ if $y\in f(X)$ and $g(y)=0$ else. For the constant function $h(y)=1$ you have $g\circ f=h\circ f$ so that $g=h$. Hence, every $y\in Y$ belongs to $f(X)$ and $g$ is surjective.
Aug
11
comment Do regular conditional distributions almost surely assign trivial measure to all members of the conditioning $\sigma$-algebra?
I think that this projecteuclid.org/euclid.aop/1015345764 answers your question.
Aug
11
comment Do regular conditional distributions almost surely assign trivial measure to all members of the conditioning $\sigma$-algebra?
I have deleted my answer (which in fact did not answer the question). Meanwhile, I think that the answer to you question is NO. It might be helpful to study (more carefully than I did) the following projecteuclid.org/…
Aug
4
answered Measurable functions lifted onto a space of point measures are measurable
Aug
1
comment Example of a space for which $V \cong Hom(V,V)$
I still have doubts. In particular concerning the isomorphism $(E\otimes E')' = E'\otimes E''$. I think that Remarque 1 on page 47 of Grothendieck's thesis might be relevant here.
Aug
1
comment Example of a space for which $V \cong Hom(V,V)$
@blackburne I think one has to be careful with the "formal manipulations". Isn't your example similar to Andrej Bauer's attempt? If I understand correctly, the space of row-finite matrices is (isomorphic to) $\phi^{\mathbb N}$ and then you have $L(\phi^{\mathbb N},\phi^{\mathbb N})= L(\phi^{\mathbb N},\phi)^{\mathbb N}$. But why is $L(\phi^{\mathbb N},\phi)$ isomorphic to $\phi^{\mathbb N}$?
Jul
4
awarded  Taxonomist
Jul
2
awarded  Curious
Jun
20
comment Closed Graph Theorem and Spaces Of Continuous Functions
There are Springer Lecture Notes of Jean Schmets about locally convex properties of $C(X)$ (SLN 519, Espaces de fonctions continue, 1976).
Jun
17
comment Inductive and projective tensor product
This question might be of some relevance: mathoverflow.net/questions/123879/…
Jun
8
awarded  Notable Question
Jun
5
comment Strong Markov property for Poisson point process
What about Theorem 19.17 in Kallenberg's Foundations?
Jun
4
comment Strong Markov property for Poisson point process
Did you check Kallenberg's Foundation of Modern Probability, Theorem 12.14?
Jun
3
comment Smooth function over a manifold into an algebra
Since $C^\infty(M)$ is a nuclear Frechet space you can choose more or less any of the usual tensor topologies (like the projective $\pi$ or the injective $\varepsilon$) and you get $C^\infty(M,A)\cong C^\infty(M) \tilde{\otimes} A$ (the completed tensor product) for all Banach (or Frechet) spaces $A$.
Jun
3
revised Are functions of moderate growth a bornological space?
References added.
Jun
2
awarded  fa.functional-analysis