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A Hilbert space $H$ is a real or complex vector space endowed with an inner product such that $H$ is a complete metric space when endowed with the norm induced by this inner product.
1
vote
Range of a Certain Linear Operator
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 …
2
votes
Is there a use for a Hilbert space that uses a different norm than the one induced by the in...
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 …
8
votes
Accepted
is a non-invertible operator a boundary point of the group of invertible operators?
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({ …
3
votes
Bounded operators leaving dense subspace invariant
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 …
1
vote
Simultaneous time-frequency concentration of orthonormal sequences?
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 …
1
vote
Abelianization of GL(H)
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 …
4
votes
A homeomorphism between the unit interval $[0,1]$ and a linearly independent subset of a Hil...
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 …
7
votes
Accepted
Fell's trick for Lie groups
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 …
2
votes
Accepted
A reproducing kernel Hilbert space
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 …
7
votes
Accepted
When are finite-dimensional representations on Hilbert spaces completely reducible?
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 …
8
votes
Matrices Representing Bounded Operators and Absolute Values
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 …
6
votes
Accepted
What is the orthonormal basis for the Bergman space on the disk?
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.
3
votes
Approximately complemented subspaces
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 …