All Questions
7 questions
8
votes
1
answer
656
views
When is the adjoint of a hypoelliptic operator also hypoelliptic?
Suppose that $M$ is a smooth manifold with a measure $\mu$ and let $L^2(M, \mu)$ be a space of all square-integrable functions on $M$.
Recall that $L$ is a hypoelliptic differential operator if for ...
5
votes
1
answer
660
views
Hormander's bracket condition for the adjoint of an operator
Let $X_0, X_1, \dots, X_k$ be smooth vector fields over ${\mathbb R}^n$, and let us consider the operator
$$
L = \sum_{i=1}^k X_i^2 + X_0~.
$$
Here, I assume that Hörmander's bracket condition is ...
4
votes
1
answer
165
views
Scattering of relativistic particle by long-range potential
Let
$\mathcal{H}=L^2(\mathbb{R}^3)$,
$H_0=\sqrt{-\Delta+M^2}$, ($M$ is a positive constant, $\Delta$ is the laplacian)
and
$H=H_0+V(\vec{x})$
(where $V(\vec{x})$ is the operator of ...
2
votes
0
answers
654
views
Convergence of operator in norm resolvent sense and their eigenvectors
Let $\{T_n\}_{n=1}^\infty$ and $T$ be (unbounded) self-adjoint operators and $T_n\to T$ in norm resolvent sense, that is, for some $z\in \mathbb{C} \setminus \mathbb{R}$, $\|(zI- T_n)^{-1}- (zI- T)^{-...
2
votes
0
answers
242
views
Core of divergence form operator with unbounded coefficient
Consider the unbounded operator $L$ on $L^2(\mathbb{R^d})$ to be the self-adjoint extension of
$$Lf := \nabla \cdot \left(a(x) \nabla f(x) \right)$$ on $C^2_c(\mathbb{R^d})$.
I also assume that $a(x)...
1
vote
1
answer
275
views
Laplacian on space of measures
Let $X$ be a compact Riemannian manifold and let $\mathcal{M}(X)$ be the space of regular finite Borel measures with the total variation as norm.
The Laplace-Belrami-Operator $\Delta$ on $X$ with ...
1
vote
0
answers
153
views
Maximal domain of an unbounded linear operator in a weighted Hilbert-space
Let's consider the following (unbounded) linear operator. (So called transport operator in some context.)
$$ \mathrm{T}: \mathcal{H} \supset \mathcal{D}(\mathrm{T}) \to \mathcal{H} , f \mapsto \mathrm{...