bio  website  math.unitrier.de/~wengenroth 

location  Universität Trier  
age  46  
visits  member for  2 years, 6 months 
seen  6 hours ago  
stats  profile views  1,101 
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

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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

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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

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In the category of sets epimorphisms are surjective  Constructive Proof?
Is this really different from the argument in my comment? 
Aug 18 
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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 
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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 
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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 
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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 rowfinite 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 
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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.functionalanalysis 