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3 votes
0 answers
95 views

Commutator of $A\otimes I$ and $I \otimes B$ vanishes?

Consider two Hilbert spaces $H_1$ and $H_2$, and $A$, $B$ unbounded operators on $H_1$, $H_2$ respectively. $(A \otimes I)$ is classically defined as the closure of the operator defined on the set of ...
Hugo's user avatar
  • 31
5 votes
2 answers
149 views

Showing an operator is (or not) closed on $L^2(\mathbb{R})$

I am linearizing nonlinear waves and get operators of the form below. Everything is considered in $L^2(\mathbb{R})$. Consider the operator $L_1=\frac{d}{dx}$. The domain is $H^1(\mathbb{R})$ and it is ...
Gateau au fromage's user avatar
0 votes
0 answers
35 views

Operator-form correspondence without lower semiboundedness

When dealing with a normal unbounded operator $A$, it is often useful to frame questions about the operator in terms of questions about the associated form $\omega,$ which has domain $D(|A|^{1/2})$ ...
user_35's user avatar
  • 109
2 votes
1 answer
244 views

If $A$ is a closed operator, is $A^k$ closed?

Let $A$ be a closed (densely defined) operator on a Hilbert space $H$. We define for a natural number $k$, the operator $A^k$ with its natural domain. Is $A^k$ closed?
Andromeda's user avatar
  • 175
2 votes
1 answer
233 views

Identity for spectral resolution: $dE_{\xi, \xi}= |g|^2 dE_{\eta, \eta}$

Let $(\Omega, \mathcal{F})$ be a measurable space. Let $E: \mathcal{F}\to B(H)$ be a regular resolution of the identity on the Hilbert space $H$, see e.g. Rudin's functional analysis book. Suppose ...
Andromeda's user avatar
  • 175
3 votes
1 answer
332 views

Takesaki II Lemma 1.13: stuck in proof

Consider the following fragment from Takesaki's book "Theory of operator algebras II" (Lemma 1.13 on p8, in chapter VI "Left Hilbert algebras"): Here, we associate with an ...
Andromeda's user avatar
  • 175
0 votes
1 answer
138 views

Antilinear unbounded operator has closed graph

Let $H$ and $K$ be Hilbert spaces and $D(T)$ a vector subspace of $H$. Let $T: D(T) \to K$ be a densely defined antilinear operator. Its adjoint $T^*: D(T^*)\to K$ is defined by the relation $$\langle ...
Andromeda's user avatar
  • 175
1 vote
1 answer
335 views

A consequence of the Min-Max Principle for self-adjoint operators

Let $H=(H, (\cdot, \cdot))$ be a Hilbert space. Let $T_1,T_2:D \subset H \longrightarrow H$ be a self-adjoint operators (not necessarily bounded). It's well-know that the spectrum $\sigma(T_i)$ of $...
Guilherme's user avatar
  • 205
6 votes
2 answers
514 views

Convergence criterion in the domain of an unbounded operator

Cross-post from math.sx. My question is somewhat close to this one, but the counterexamples given there do not apply here. Setup. Given a Hilbert space $\mathcal H$, a closed operator $A$ and a ...
Benjamin's user avatar
  • 245
6 votes
2 answers
665 views

Unbounded Fredholms operators

Motivated by the situation of bounded Fredholm operators, I have the following question about "unbounded Fredholm operators". Let $\mathcal{H}_1$ and $\mathcal{H}_2$ be two Hilbert spaces, and $$ D: ...
Dave Shulman's user avatar
5 votes
2 answers
2k views

On the domains and extensions of unbounded operators

I am not an expert in functional analysis but I was studying some, motivated from some mathematical physics considerations. I am not quite sure whether this is research-level, but let me state some ...
Konstantinos Kanakoglou's user avatar
0 votes
1 answer
54 views

Simplify the expression of $ T^+$ for an unbounded operator $T$?

For a negative unbounded operator $T$, what equals the operator $$ T^+ = \left[\frac{1}{2}(|T| + T) \right]^{**},$$ where $|T|= (T^2)^{1/2}$ and $A^{**} $ is the minimal closed extension of an ...
Z. Alfata's user avatar
  • 650
-3 votes
1 answer
320 views

Is :$\frac{\Bbb d}{\Bbb d x}$ a chaotic operator in infinite-dimensional Hilbert space? [closed]

I proposed this question in SE but no answer ,may I have a problem in my question, I would like to know when $\frac{\Bbb d}{\Bbb d x}$ does chaotic operator in Hilbert space ? Let $H$=$L^2(\mathbb R)$...
zeraoulia rafik's user avatar
1 vote
1 answer
2k views

Operator theory of the Hessian

How can I learn more about the operator theory of the Hessian? The Hessian of a function $u : \Omega \rightarrow \mathbb R$ over a domain $\Omega \subseteq \mathbb R^n$ is the matrix of second ...
shuhalo's user avatar
  • 5,327