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I don't know about Q1 off the top of my head, but I think that for Q2 you are unlikely to get a good answer. Of course I may have a different view from you as to what "mild" adjectives are reasonable …

A useful test case for RKHS (which is not like the interesting examples, but does satisfy the definitions) is $\Omega={\mathbb N}$ and $H=\ell^2({\mathbb N})$. Note that $\hat{k_n}$ is just the usual …

Fell absorption works for all locally compact groups and all strongly continuous unitary representations; the intertwining map that you give in the discrete case can be generalized to $$ W: L^2(G)\ot …

Iosif Pinelis has given a direct argument, but if you are studying approximate complementation in more general Banach spaces then the following argument might be of interest. In Zhang's original pape …

In a different direction from Yves Cornulier's suggestion: how about something based on Cauchy kernels as elements of Hardy space? I'm writing this off the top of my head so I might not be taking the …

The following is based on a copy of some old handwritten notes; I haven't had time to refresh my memory on all the details. I will try to fill in these details in the next day or so. $\newcommand{\Nat …

As requested, I've moved my comments to an answer. The question is equivalent to the following: Suppose that a doubly-infinite matrix $A=(A_{ij})_{i,j\in {\bf Z}}$ represents a bounded hermitian …

This is true if $0$ belongs to the boundary of the spectrum of $T$ (more or less from the definition of the spectrum) but fails in general. The unilateral left shift $S: \ell^2({\mathbb N})\to\ell^2({ …

Assuming that we normalize so that the area of the unit disc is $\pi$, take $f_n(z) = \sqrt{\frac{n+1}{\pi}} z^n$ for $n=0,1,2,\dots$ to get an orthonormal basis.

The Fourier transform on $L^2({\mathbb R})$ has an complete set of eigenvectors (that is, there is an o.n. basis of $L^2({\mathbb R})$ consisting of eigenfunctions for the FT, and they are all in the …

Firstly, a Hilbert space is complete: the $l^1$-norm is defined on a dense subspace of $l^2$ but not on the whole space. Secondly, if you can equip $l^2$ with another norm, then either this norm is in …

I agree with Eric's belief, albeit for a "cheat" reason: if the abelianization were non-trivial then you'd get a notion of determinant for GL, and that seems unlikely to me. For what it's worth, acco …

This looks a bit like it could be a H/W exercise, but having started to type something up I might as well give most of it. I'm assuming all your functions are real-valued, since otherwise your form is …