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Banach spaces, function spaces, real functions, integral transforms, theory of distributions, measure theory.

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 …
Simon Henry's user avatar
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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) …
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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 …
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2 votes

Homomorphism to multiplier algebra of groupoid $C^\ast$-algebra

This is not a complete answer, but that was way too long for a comment: First I started almost sure that this sort of things would be in the literature, but it seems harder than I thought to find som …
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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 …
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2 votes

Does every integer map generate a von Neumann algebra of type I?

I think I have an example: Precisely, I will construct an integer function $m$ such that $M$ is bounded and the algebra $\mathcal{M}$ contains a corner which is the von Neuman algebra completion of a …
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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 …
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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)= …
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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 $c$ …
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1 vote
Accepted

On the second dual of $C[0,1]$

It is easy to see that $\int \psi d\delta_t$ where $\delta_t$ is the Dirac mass at $t$ is $\psi(\{t\})$ So you are starting from a $T \in C([0,1])^{**}$ and you are attaching to it the function $t \m …
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2 votes
Accepted

Stone-Weierstrass, uniform convergence, and sums

As I said in the comment, this is clearly not true in general: the algebra of polynomial function on $[0,1]$ is dense among all continuous functions and generated by the $x^i$ but only functions that …
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1 vote

Comparison between spectra

If $G$ is normal (and you don't care it has compact resolvent or not), then $G_0 = G +P$ is $f(G)$ where $f$ is the measurable function that send $0$ to $1$ and is the identity on other value. If I r …
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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 …
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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 …
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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 …
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