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Just when I thought I was out, they pull me back in
31m

revised 
Is there a name for this space?
edited tags 
31m

comment 
Is there a name for this space?
Sergei's comment is incorrect as FanZheng points out. HausdorffYoung is not a characterization. I have never seen the space of functions whose FTs are pintegrable given any special name, so I would suggest just inventing some ad hoc notation within your paper and sticking to it. 
47m

comment 
Is the space of all borel measures on $\mathbb R^n$ isomorphic to the tensor product of spaces of borel measures on $\mathbb R$?
The tensor product of $L_2({\bf R})$ with itself as a vector space is merely dense in $L_2({\bf R}^2)$... 
51m

revised 
Determining primitive ideal space of C*algebra
replaced needlessly specific tag with toplevel OA tag 
52m

comment 
Determining primitive ideal space of C*algebra
The question is too openended at the moment. What classes of Cstar algebras are you studying? Moreover, I think you will have problems if your Cstar algebra is not "postliminal" 
54m

comment 
Are “most” operators on an infinitedimensional complex Banach space “diagonalizable”?
Just a comment, given the answer below and the ensuing comments: you will always have problems as soon as there exists something like a onesided shift, i.e. a map which is injective with closed range but not surjective  for such operators, there are points in the spectrum that are not approximate eigenvalues, and these are an obstruction to approximation by operators where every point is an approximate eigenvalue (this class including diagonalizable ones, I think) 
57m

comment 
Is the space of all borel measures on $\mathbb R^n$ isomorphic to the tensor product of spaces of borel measures on $\mathbb R$?
1 until the OP clarifies whether this is the tensor product of Banach spaces, vector spaces, Hilbert spaces, Waelbroeck spaces, etc... 
58m

comment 
Is the space of all borel measures on $\mathbb R^n$ isomorphic to the tensor product of spaces of borel measures on $\mathbb R$?
Just to clarify for anyone else reading: the tensor product completion Gerard describes above is not in general the same as the projective tensor product of Banach spaces 
1h

comment 
Zero divisors and boundary elements of $A^{1}$
Have you tried looking in the short section on Banach algebras in Rudin's FA? Or the book of Rickart? Or the book of Bonsall and Duncan? These should all treat the topics you mention, which are part of the core material one usually covers en route to the basic properties of the spectrum 
19h

comment 
Reference request about the representations of the group $PSL_2(\mathbb{F}_q)$
@FanZheng I believe following the link would answer your question 
20h

answered  Reference request about the representations of the group $PSL_2(\mathbb{F}_q)$ 
1d

comment 
Looking for an example of a contour integral with matrix entries
I mean, what you have is a meromorphic function whose coefficients are powers of A, and diagonalizing A simultaneously diagonalizes all powers of A... 
1d

comment 
Looking for an example of a contour integral with matrix entries
To illustrate the idea consider the case Q(z,A)= zIA, and assume A is diagonal. Then you can explicitly write down an expression for Q(z,A)^{1}, and P will also be diagonal, so aren't you just integrating a function taking values in the diagonal matrices, hence just doing a bunch of scalar valued integrals? What am I missing? 
1d

comment 
Looking for an example of a contour integral with matrix entries
I am not sure where you are stuck. At what point do you not see what to do? 
1d

comment 
Looking for an example of a contour integral with matrix entries
Well conceptually this certainly makes sense, you just want to do a contour integral of a matrixvalued function. If A is symmetric then you can diagonalize the whole expression and then this should reduce down to a bunch of usual scalarvalued integrals? 
Apr 16 
accepted  For discrete groups, does the Haagerup property imply the AP of HaagerupKraus? 
Apr 16 
comment 
For discrete groups, does the Haagerup property imply the AP of HaagerupKraus?
Thanks! Just to make sure I have understood (and for others reading): Osajda's examples have the Haagerup property but do not have Property A, and for discrete groups the AP of HaagerupKraus implies Property A. Is that so? 
Apr 15 
reviewed  Leave Open Does “equalisers always closed” imply $T_2$? 
Apr 15 
revised 
For discrete groups, does the Haagerup property imply the AP of HaagerupKraus?
fixed a misstatement 
Apr 15 
comment 
Decomposition space of $\mathbb{C}$ by concentric circles
Thanks. Firstly, my comment was trivial (as the OP acknowledges). Secondly, my comment is now no longer an answer because the OP has written a new and more substantial question. Thirdly, I can't add an answer even if I wanted to, because the question is currently closed. 