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

6 votes
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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 $C([a,b])$ …
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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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2 votes
Accepted

Infinite Determinant between different Hilbert Spaces

It seems this implies that $V^{-1}U$ can be diagonalised isn't it ? Indeed: $V^{-1}U$ is a unitary, let just call it $W$. If $1-W$ is trace class it is in particular compact, moreover it is normal (a …
Simon Henry's user avatar
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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 …
Simon Henry's user avatar
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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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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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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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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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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 …
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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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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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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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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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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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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 …
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