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Banach spaces, function spaces, real functions, integral transforms, theory of distributions, measure theory.
14
votes
Accepted
Between compact and locally uniform: What is the name of this convergence?
This notion of convergence is not often refered to; I think mostly because it does not come from a topology. But this an excellent notion of convergence, probably the best we can put on the space of ( …
10
votes
Accepted
A C*-algebra enjoying some different C*-norms
No that's not possible (except the trivial case). Any $*$-homomorphism between $C^*$-algebras is automatically contractive, and if it is injective then it is isometric. You can apply this to the ident …
10
votes
What are the 'wonderful consequences' following from the existence of a minimal dense subspace?
I like to call this result the localic Baire category theorem, and it plays essentially the same role as Baire category theorem: it lets you "construct" object by showing that some spaces are non-empt …
7
votes
3
answers
737
views
Duality between Banach spaces and compact convex spaces
I always had the impression that there was a duality (i.e. a contravariant equivalence of categories) between Banach spaces and certain notion of pointed compact convex set (something like algebras fo …
6
votes
Accepted
Continuous maps on compact topological spaces which induce compact (Fredholm) operators
The image of $f$ is an interval $[a,b] \subset [0,1]$.
$T_f$ induce an isometric inclusion of $C([a,b])$ in $C([0,1])$.
If $T_f$ was compact then weak convergence of a bounded sequence in $C([a,b])$ …
6
votes
Accepted
The Gelfand duality for pro-$C^*$-algebras
The answer is No. Rougly, because it is not a good idea to look at continuous $\mathbb{C}$ valued function on a space which is not completely Haussdorff as completely haussdorf is exactly the hypothes …
6
votes
Accepted
What is the Gelfand dual of an open surjection?
After more thought, I think the correct statement is the following:
Theorem: Let $\pi : X \to Y$ be a continuous map between compact Hausdorff spaces. Then the following condition are equivalents:
(a) …
5
votes
Accepted
Counterexample to Riesz representation for Hilbert modules
Take $A= \mathcal{C}([0,1])$ and $H$ the ideal of $A$ of functions that vanish at $0$.
$H$ is a Hilbert $A$ module (as any ideal, with the natural multiplication of $A$ and the scalar product $(x,y)= …
5
votes
Relation between norm of any element of $C^*$-algebra in terms of self adjoint elements
The norm of the self adjoint part is not sufficient: Even if $b$ and $c$ commutes (which should be the easiest case) then it corresponds (by looking at the commutative algebra generated by $b$ and $c$ …
4
votes
No Hilbert space can have countable Hamel basis without using Baire's Category theorem
Nam-Kiu Tsing argument indeed become way much simpler for Hilbert spaces. In fact I wouldn't be surprise if his argument was inspired from the (very easy) case of Hilbert spaces:
If you have a counta …
4
votes
Why $A^*A =A$ implies that $A$ is a C$^*$ algebra (Proposition 5.2.8 of An Invitation to Qua...
I assume that $A^* A$ denotes the algebra generated by product $a^* b$ with $a, b \in A$, but the following argument also applies to some others interpretation of $A^* A$.
If $A^* A =A$ then for any …
4
votes
Constructive proof of existence of non-separable normed space
AS I said, it depends way to much on your framework to give a definitive answer ! here are some exemples that works in some cases:
Take $E$ to be the free $\mathbb{Q}$-vector space on a set $S$, and …
4
votes
2
answers
353
views
On the descent homomorphsim of Kasparov equivariant KK theory
Hello,
I have recently read about the construction of the descent map in Kasparov KK theory, which, for a group $G$ and two $G$-equivariant $C^*$ algebra $A$ and $B$ send $KK_i^G(A,B)$ to $KK_i(A \rt …
4
votes
Accepted
$H^{*}$ algebras as a generalization of $C^{*}$ algebras
Let assume that you consider unital algebra only (one can still study non unital algebra by unitarizing them, but notion of spectrum is always a little annoying when one want to consider non unital al …
4
votes
Do Hausdorff locally convex inductive limits always exist?
The limit always exists whether or not $H$ is closed: The inductive limit in the category of hausdorff lc vector space will be the quotient of $F$ by the closure of of $H$: a cone for the inductive li …