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Results tagged with hilbert-spaces
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user 2554
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.
5
votes
Accepted
When is Euclidean distortion finitely determined?
The problem is that you misdefined "Euclidean distortion". The infimum is over all maps into arbitrary Hilbert spaces, not just $\ell_2$. Of course, you can use only $\ell_2$ if the metric space is se …
- 31.5k
2
votes
Accepted
Representing an $L^2$-functional by a non-$L^2$-function on a dense subspace - Part II
Gro-Tsen's answer to your previous question provides a counterexample if you define $D$ to be all vectors in $\ell_2$ that are of the form $\sum_n a_n f_n$, where
$f_n = e_n + e_{n+1}$, $(e_n)$ is the …
- 31.5k
7
votes
Accepted
Quasinilpotent , non-compact operators
On the Argyros-Haydon space every operator is a compact perturbation of a scalar multiple of the identity, and hence every quasinilpotent operator is compact.
- 31.5k
4
votes
Accepted
Compact images of nowhere dense closed convex sets in a Hilbert space
Your revised assumption is that the norm (rather than semi-norm after the revision) on $\ell_2$ given by sup-ing against vectors in $B$ is not equivalent to the usual norm on any finite codimensional …
- 31.5k
5
votes
Characterisation of compact operators
No. Consider $H = (\sum_n \ell_2^{2^n})_2$ with the unit vector basis. In $\ell_2^{2^n}$, let $x_n$ be the sum of the unit vector basis. For the subspace take the closed linear span of $(x_n)$.
- 31.5k
2
votes
Accepted
Geometry of the Hilbert sphere
I think yes. Your balls are of the form $X\cap S(x,r)$ where $x\in X$ and $S(s,r)$ is the slice
$\{y: \|y\|\le 1 \ \text{and} \ \langle x,y\rangle \ge r\}$ of the unit ball. Note that if $y$ is in …
- 31.5k
2
votes
Accepted
Are these ideals in rings of operators on Hilbert space unique?
No proper ideal satisfies your condition. Take the Hilbert space to be $ \ell_2\oplus \ell_2$ and define $f(x,y)=(o,x)$; $g(x,y)= (x,0)$.
- 31.5k
5
votes
Set of invertible operators in B(H) is connected. Is it true? Is there a reference?
Kuiper's theorem gives the contractibility of the invertible operators in $B(H)$. Start with
http://en.wikipedia.org/wiki/Kuiper's_theorem
Check out papers of Boris Mitjagin [Mitiagin] in the 1970 …
- 31.5k
1
vote