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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.
8
votes
0
answers
193
views
Hilbert spaces over the semi-field $\mathbb R_+$
Let $\mathbb R_+$ be the semi-field of non-negative real numbers.
Definition (preliminary): A Hilbert space over $\mathbb R_+$ is a pair $(H,P)$, where $H$ is a complex Hilbert space, and $P\subset H …
64
votes
6
answers
10k
views
Are dagger categories truly evil?
Recall that a dagger category is a category equipped with an involution $*:Hom(x,y)\to Hom(y,x)$ that satisfies $f^{**}=f$ and $f^* g^*=(gf)^*$. A prominent example of a dagger category is the categor …
18
votes
2
answers
1k
views
compact-open topology on $B(H)$
In topology, it is common to use the compact-open topology on the set of continuous maps between two given topological spaces.
Let now $H$ be a Hilbert space and $B(H)$ the set of continuous linear m …
22
votes
2
answers
3k
views
Does it make sense to talk about smooth bundles of Hilbert spaces?
Is there a notion of "smooth bundle of Hilbert spaces" (the base is a smooth finite dimensional manifold, and the fibers are Hilbert spaces) such that:
1• A smooth bundle of Hilbert spaces over a poi …
9
votes
2
answers
470
views
why is this a sufficient condition for a domain to be a core of an unbounded operator?
Let $\alpha:\mathbb R\to U(H)$ be a strongly continuous action of the reals on some Hilbert space, and let $A=-i\frac d{dt}\alpha(t)|_{t=0}$ be its infinitesimal generator, so that $\alpha(t)=e^{itA}$ …
18
votes
2
answers
1k
views
Complex structure on $L^2(\mathbb R)$ generalizing the Hilbert transform
The Hilbert transform on the real Hilbert space $L^2(\mathbb R)$ is the singular integral operator
$$
\mathcal H(f)(x) := \frac{1}{\pi} \int_{-\infty}^\infty \frac{1}{x-y} f(y) dy.
$$
It satisfies $\ …
11
votes
0
answers
526
views
Contraction semigroup on Hilbert space
I'd like to know whether a certain unbounded operator on a Hilbert space is the generator of a strongly continuous contraction semigroup.
(Such operators are known as maximally dissipative operators.) …
10
votes
3
answers
1k
views
ordered exponential of unbounded operators
Let $H$ be a Hilbert space,
and let $A_t$ be a family of unbounded positive (self-adjoint) operators on $H$ parametrized by $\mathbb t\in R_{\ge 0}$. Consider the ordinary differential equation
$$
\qq …
18
votes
0
answers
368
views
Can Rep(G) tell us whether G is discrete?
Given a locally compact group $G$, let $$\mathrm{Rep}(G)$$ be its category of unitary representations.
The objects of that category are strongly continuous unitary representations of $G$ on Hilbert sp …