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# Jochen Wengenroth

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## Registered User

 Name Jochen Wengenroth Member for 1 year Seen 8 hours ago Website Location Universität Trier Age 45
I am Professor for Mathematics at the UnversitÃ¤t Trier (Germany)
 Jun14 answered Complete uniform spaces require complete metrics? Jun13 comment Homeomorphisms and disjoint unionsThe authors also mention that one can modify their construction to obtain an example in $\mathbb R^2$. May17 answered $C_c^{\infty}([0,T];V)$ is dense in $C_c^{1}([0,T];V)$? May17 comment $C_c^{\infty}([0,T];V)$ is dense in $C_c^{1}([0,T];V)$?@MTS: $C_c^\infty(K) = \lbrace f\in C^\infty(\mathbb R): \mathrm{supp}(f) \subseteq K\rbrace$. May13 comment Existence of dominating measure for weak*-compact set of measuresAndy does not have a definition at all because his functionals $L_Z$ make sense only for bounded $Z$. Anyway, weak*-compactness could be obtained from the Banach-Alaoglu theorem. Mar21 comment Smooth function algebra on cartesian product and beyond Just a little remark: That $\otimes_i$ coincides with $\otimes_pi$ in this case is not only because of nuclearity (your remark about the spaces of smooth functions with compact support shows this since $\mathscr D(\mathbb R) \tilde{\otimes}_\pi \mathscr D(\mathbb R) \neq \mathscr D(\mathbb R^2)$).You use first that $\otimes_i = \otimes_\varepsilon$ for Frechet spaces and then nuclearity. Mar19 comment Riesz representation theorem for vector-valued fields@jbc: What do you mean by every Banach space is an inductive limit of its finite dimensional subspaces? In which category? Shouldn't the inductive limit $X=\lim X\alpha$ have the universal property that a linear map $X\to Y$ is continuous (a morphism of the category) iff all restrictions to $X_\alpha$ are continuous? If all $X_\alpha$ are finite dimensional this is no condition. Mar19 comment Weak convergence in measure for negligible sets.For more information about the relation between weak and a kind of "set-wise" convergence look at the so-called Portmanteau theorem. Mar18 comment Uniformly integrable sequence such that a.s. limit and conditional expectation do not commute@Didier: Thanks for editing. Mar16 comment Extending a Hilbert space isometrically@Tom: Even if there is any metric on $F$ inducing the vector space topology such that $f$ is an isometry then $f(H)$ will be already a completely metrizable topological vector space. This not obvious (since the uniformities may be a priori different) but true because of a theorem of Victor Klee (solving a problem of Banach). Therefore, $F=f(H)$ still holds. Mar15 comment Extending a Hilbert space isometrically...normed and $f$ is an isometry then $f(H)$ is complete and hence closed in $X$ (if you assume that $X$ is Hausdorff). Mar15 comment Extending a Hilbert space isometricallyThe obvious idea to embed the separable Hilbert space into a sequentially complete locally convex space is to find a sequence $x_n$ which converges fast to $0$ and to define $f:\ell_2 \to X$ by $f((a_n)_{n\in\mathbb N})= \sum\limits_{n=1}^\infty a_n x_n$. This map will be injective if the sequence is *topologically linearly m-independent$. This is discussed on page 37 in the book Barrelled locally convex spaces of Bonet and Perez-Carreras. The second question is not clear to me: If there is no norm on the Frechet space$F$, what do you mean by isometric? On the other hand, if$F$is ... Mar15 revised Uniformly integrable sequence such that a.s. limit and conditional expectation do not commuteadded 201 characters in body Mar15 answered Uniformly integrable sequence such that a.s. limit and conditional expectation do not commute Mar14 comment stopping time on event A$\lbrace f\in B\rbrace$is a shorthand for$\lbrace \omega\in\Omega: f(\omega)\in B\rbrace$. The conventions for$\cap$and$\cup$are as for$\cdot$and$+$. Mar14 comment stopping time on event AAlthough this is not research level,$\lbrace \sigma' \le t \rbrace = \lbrace \sigma \le t \rbrace \cap A \cup \lbrace \tau \le t \rbrace \cap A^c \in F_t$since$A\in F_\sigma \subseteq F_\tau$. Mar13 comment A possible refinement of a theorem of MalliavinAt first sight it seems to me that in Sotirov's thesis only sequential continuity is considered. Since$\mathscr D(\mathbb R^n)$is very far from being metrizable this can be much weaker than contiunity. Mar12 comment Exponential sums and binary expansionsVery good. Thank you very much. There is a bounty on this question on math.stackexchange.com/questions/324496/… Mar12 comment Exponential sums and binary expansionsYour way to prove the identity is certainly better. Mar11 accepted Inductive tensor product and smooth functions Mar11 comment Inductive tensor product and smooth functions@Allan: You are right, my answer was not correct. I hope that the new answer is better. Mar11 revised Inductive tensor product and smooth functionsadded 35 characters in body Mar11 revised Exponential sums and binary expansionscorreced$n/2-1$to$n/2 +1$Mar11 revised Inductive tensor product and smooth functionsadded 1885 characters in body Mar11 asked Exponential sums and binary expansions Mar7 answered Inductive tensor product and smooth functions Mar1 comment a problem in functional analysis that erdos solved in 2 linesThe transcendence of$\pi$is due to to Bryant? Feb26 comment A question of Allan on infinite divisibilityLooks very good ($x$should also belong to the generators). Thank you very much. Feb26 asked A question of Allan on infinite divisibility Feb24 comment Class of functions that the Fourier inversion holdsThe Fourier transform is an isomorphism also on$\mathscr S'(\mathbb R^d)$, the space of tempered distributions. Feb24 answered spectacular applications of functional analysis in resolutions of apparently unrelated problems Feb21 comment Compactly generated Banach spacesYemon's answer shows that every separable Frechet space is compactly generated. The converse is also true: The compact generator is contained in the closed absolutely convex hull of a sequence converging to$0$, and the countable set of all rational (finite) linear combinations of that sequence is dense. Feb19 comment Expectation of sample varianceThis is not research level. Nevertheless, under your assumptions, the empirical mean and variance are independent so that$E(S^2|\bar{X}=\bar{x})=E(s^2)$. Feb15 comment When is a sequentially closed cone, closed?As for almost all locally convex properties, bornologicity does not reflect properties of single seminorms. The essential point is always the relation between them or how many of them you need. A trivial example: If only countably many seminorms describe the locally convex topology the space is (semi-) metrizable and hence bornological. Feb14 comment When is a sequentially closed cone, closed?I do not understand this question. Barreledness is rather close to bornologicity, for instance, every (locally) complete bornological space is ultrabornological and hence barrelled. This means that barrelledness will not help you very much to conclude closed from sequentially closed. Feb14 revised When is a sequentially closed cone, closed?added 158 characters in body Feb14 revised When is a sequentially closed cone, closed?added 468 characters in body Feb13 comment When is a sequentially closed cone, closed?Lieber Herr Michor, do you agree with the counterexample to your claim that I posted as an answer? Feb12 revised When is a sequentially closed cone, closed?added 1 characters in body Feb12 answered When is a sequentially closed cone, closed? Feb5 comment An interesting summationIt is very easy to calculate the derivative of$f(x)= \sum\limits_{k=1}^n \frac{x^k}{k}$. Feb5 comment Can distribution theory be developed Riemann-free?Of course, if you want a characterization involving all Riemann integrable functions you have to use the Riemann integral. But continuous functions (or, by Weyl's criterion, exponentials) would be enough. Feb1 comment Schwartz kernel theorem for topological spacesThe answers of jbc and Peter Michor are more or less the same as both describe the$\varepsilon$-product$C(K_1) \varepsilon C(K_2)$of Laurent Schwartz. Feb1 awarded ● Yearling Jan30 comment Second differenceIf you only want to have a fixed$x$(as it seems since you accepted Xandi's answer) you can take any odd function$f$with$f(0)=0$. Jan30 comment Second differenceAre you sure that there is a constant$C$that works in this example for all other$x\$? Jan30 answered Metrization of weak convergence of signed measures Jan30 comment An elementary probability questionThe question is probably about good bounds. Jan25 answered Is an additive category a balanced category? Jan9 awarded ● Organizer