8

In other words, if $A$ is the infinitesimal generator of $T$, the mild solution of the abstract inhomogeneous Cauchy problem
$$\begin{cases}\dot v =A v +f\\v(0)=0\end{cases}$$
needs not to be $W^{1,1}_{loc}(\mathbb{R}_+, X)$.
For instance an $f\in C^0(\mathbb{R}_+,X)$ of the form $f(t):=T(t)x$ for some $x=x(\theta)\in X$, gives $v(t)=tT(t)x=tf(t)$ ...

answered Sep 19 '18 at 20:27

Pietro Majer

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8

Let me comment on what I know in an open set $\Omega$, trying to control $C$. First of all, the heat semigroup can be expressed through a kernel $p$ which is pointwise dominated by the heat kernel in $\mathbb{R}^n$, written by Bazin, but this is not sufficient, because of $\nabla_x$. One can extend the Gaussian bound to complex times thus getting
$$|p(z,x,y)...

8

The problem is discussed in a more general setting (operator ideals in Banach spaces) for the so-called analytic semigroups (parabolic problems) in
Blunck, S.; Weis, L., Operator theoretic properties of differences of semigroups in terms of their generators, Arch. Math. 79, No. 2, 109-118 (2002). ZBL1006.47036.
The paper seems to be freely accessible. The ...

7

This cannot happen, if, e.g., $Y$ is a countable union of finite dimensional subspaces (by the Baire category theorem). Suitable examples are given by the spaces of finite sequences in $\ell^p$.
Edit in response to questioner's comment. If $Y$ is the domain of a closed operator into a Banach space, then the graph topology makes it into a Banach space ...

7

One can prove a stronger estimate in fact. Suppose $u$ and $v$ are positive with $v>u$ say. Note that
$$
|G(u,x)-G(v,x)| \le \int_{u}^v \Big| \frac{d}{dt} G(t,x)\Big| dt =\int_{u}^{v}
\frac{e^{-x^2/2t}}{\sqrt{2\pi t}} \Big|\frac{x^2}{2t^2}-\frac{1}{2t}\Big| dt.
$$
Integrating this over $x\in {\Bbb R}$ we obtain
$$
\int_{-\infty}^{\infty} |G(u,x)-G(...

fa.functional-analysis ap.analysis-of-pdes stochastic-processes parabolic-pde semigroups-of-operators

7

Following ideas of John von Neumann,
J. von Neumann, Über einen Satz von Herrn M. H. Stone, Ann. Math. (2) 33, 567-573 (1932). ZBL0005.16402,
it can be shown that if a function satisfies the semigroup law and is measurable, then it is already continuous. Hence your identity is only true if the generator is bounded.
See also
Engel, Klaus-Jochen; Nagel, ...

7

This question was considered by Functional Analysis specialists around 1960 or earlier. In the case of Banach Algebras $C(X)$ (for Hausdorff compact $X$) this gets reduced often to studying the auto-homeomorphisms of $X$, and it is extra interesting when the compact space is nice. In this context, in 1961, I have rediscovered that orientation preserving ...

reference-request fa.functional-analysis operator-theory fractional-calculus semigroups-of-operators

7

The purpose of this answer is to extend Christian Remling's answer to dimension $d = 3$. There are two steps. (N.B. below the cut I show how to replace part 1 by a different argument that works in all dimensions, and so this should answer the question posed.)
We assume that we have a solution $\phi$ to the Schrodinger equation such that it vanishes on a (...

6

### Does the generator of a 1-parameter group of Banach space isometries know which elements are entire?

If I am not mistaken, then this is connected to the notion of analytic vectors and related object.
If I understand your question correctly, it is answered in this paper of Chernoff.
Further, for group generators on Banach spaces the analytic vectors are dense and they determine the generator $D$, see Exercise II.3.12(2) in Engel-Nagel.

6

If I understand your question correctly, this would mean that the function $t\mapsto T(t)x$ is Lipschitz continuous, which is equivalent for $x$ to be in the Favard space $\text{Fav}(A)$$. See for example Defintion 8.2. in these lecture notes.
If your space $X$ is reflexive, then the Favard space is exactly $D(A)$, the domain of $A$, see Corollary II.5.21 ...

6

There is a one-dimensional counterexample: Consider the analytic semigroup $z \mapsto e^{iz}$. This semigroup is bounded on the non-negative real line, but it is not bounded on any sector $\Delta_\delta$.
EDIT in response to the comments: One can "modify" each analytic semigroup to obtain the following boundedness property:
Let $0 < \delta' < \delta ...

6

Too long for a comment. Actually I do not think that it is written anywhere but these kind of counterexamples are usually provided by the Ornstein-Uhlenbeck semigroup, generated by $\Delta+Bx \cdot \nabla$, where $B$ is a matrix. Assuming that all eigenvalues of $B$ have negative real parts, then an invariant measure $\mu$ exists (and is given by a Gaussian ...

6

For the record, the result requested in the question is given in Theorem 1.3.3 in: E.B. Davies, Heat Kernels and Spectral Theory, DOI:10.1017/CBO9780511566158.
The assumptions are:
$L$ is a positive definite self-adjoint operator given by a quadratic form $Q$ on the Hilbert space $L^2(\Omega)$, where $\Omega$ is a $\sigma$-finite measure space;
if $u$ is in ...

answered Oct 2 '20 at 17:34

Mateusz Kwaśnicki

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6

Probabilistic approach
I have to think about the right reference, but things are very much different in the non-local case.
The solution $u = u^\epsilon$ can be written in probabilistic terms as
$$ u(x) = \mathbb E^x e^{-k \tau_D} ,$$
where $k = \lambda / \epsilon$, $\tau_D$ is the first exit time from the open set $D = \Omega$ (when probability enters, I ...

fa.functional-analysis pr.probability ap.analysis-of-pdes ca.classical-analysis-and-odes semigroups-of-operators

answered Apr 9 '21 at 21:22

Mateusz Kwaśnicki

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5

Take $K = \{0,1\}^{\mathbf{Z}^2}$ and take for $T_t$ the Glauber dynamic for the Ising model below the critical temperature. Then $T_t$ is Feller and irreducible, but it has two distinct ergodic invariant measures.
If however you know that $T_t$ is strong Feller (or asymptotically strong Feller) then irreducibility does imply uniqueness of the invariant ...

5

An operator $A$ is a group generator if and only if both $A$ and $-A$ are semigroup generators. So under your assumptions, if we denote by $\tilde{A}$ the generator of the extension $\{\tilde{T}(t),t\ge 0\}$ of $\{T(t),t\ge 0\}$ to $Y$, then $-\tilde{A}$ would be a semigroup generator, too.
If by "extend" you mean the usual thing (viz, that $X$ is left ...

5

You remarked in comments that $T$ should be an unbounded operator from $X$ to $X$.
If $X$ is separable then $D(T)$ has to be a Borel set in $X$.
Note that $X \times X$ is a separable Banach space under a norm such as $\|(x_1, x_2)\| = \|x_1\| + \|x_2\|$. The graph $\Gamma(T) = \{(x, Tx) : x \in D(T)\}$ is by assumption a closed linear subspace of $X \...

answered Sep 27 '15 at 15:26

Nate Eldredge

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5

As you write it, this is just a bounded perturbation of the sectorial operator
$$\begin{pmatrix} 0 & 0 & 0 \\ 0 & \Delta & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$
It is quite standard that bounded perturbation does not destroy this property.

5

First, the classical Calderón–Zygmund estimate gives
$$\left\|D^2 u\right\|_{L^p(\mathbb{R}^d)}\leq C(p,d) \left\|\Delta u\right\|_{L^p(\mathbb{R}^d)}.$$
By interpolation, $\left\|u\right\|_{W^{2,p}(\mathbb{R})}$ is equivalent to
$$\left\|\Delta^2 u\right\|_{L^p(\mathbb{R}^d)}+\left\|u\right\|_{L^p(\mathbb{R}^d)}.$$
For detail proof you can look up ...

5

Your conditions is for contraction semigroups equivalent to have uniform exponential stability, i.e., to have growth bound less than zero, see Proposition V.1.7. in
Engel, Klaus-Jochen; Nagel, Rainer, One-parameter semigroups for linear evolution equations, Graduate Texts in Mathematics. 194. Berlin: Springer. xxi, 586 p. (2000). ZBL0952.47036.
Assuming you ...

5

To complete Piero's answer, let me give a definition of dispersiveness. To stay at a rather simple level, let me consider an linear evolution PDE with constant coefficients,
$$P(\partial_t,\nabla_x)u=0.$$
Hereabove, $P$ is a polynomial. Let me assume a property of homogeneity, $$P(\lambda^\alpha\tau,\lambda\xi)\equiv\lambda^m P\tau,\xi).$$
It is met by the ...

5

The estimate the OP is looking for is called an ultracontractivity estimate. A characterisation of semigroups that satisfy such an estimate can be found in the Theorem on page 65, Subsection 7.3.2, of the following survey article:
Wolfgang Arendt: Semigroups and Evolution Equations: Functional Calculus, Regularity and Kernel Estimates in the Handbook of ...

4

This does not answer your question only adds something to your list of examples: two parameter semigroups, as investigated by
Slocinski, M. Unitary dilation of two-parameter semi-groups of
contractions. (English) [J] Bull. Acad. Pol. Sci., Sér. Sci. Math.
Astron. Phys. 22, 1011-1014 (1974)
and many others later on. They are significantly different ...

fa.functional-analysis ct.category-theory operator-theory semigroups-and-monoids semigroups-of-operators

4

Just to expand the answer by Alex, let me mention that the internet seminar of last year:
http://isem17.unisa.it/w/index.php/Phase_1:_The_lectures
was intended as a gentle introduction where a lot of time was spent on finite dimensional theory to motivate the ideas in a simpler situation. Though it stresses the positivity aspects of spectral theory, I am ...

4

Apart from the fact that I do not understand some parts of your question, self-adjoint elements with positive spectrum define a positive cone in your $C^\ast$ algebra. Positivity-preserving semigroups in $C^\ast$ and von Neumann algebras were extensively studies, you should consult the chapter written by Ulrich Groh in the book
W. Arendt, A. Grabosch, G. ...

4

This is a symbol which reminds that the solution of linear system of ODEs is represented as an exponential function of a matrix. This symbol represents the solution family of the heat equation, and indeed called a one-parameter semigroup (or semi-dynamical system).
Great introduction in the subject are the books of Engel and Nagel, the original one and a ...

4

Toric varieties in Algebraic Geometry!! Indeed, the category of normal toric varieties is equivalent with the dual of the category of finitely generated, integral semigroups.

4

### Does the generator of a 1-parameter group of Banach space isometries know which elements are entire?

It is well known in the theory of operator semigroups that the Cauchy problem for the generator $D$ of a strongly continuous SEMIgroup is always UNIQUELY solvable for initial data in the domain of $D$, that is $[0,\infty)\to X$, $t\mapsto \sigma_t(x)$ is the unique solution of $f'(t)=D(f(t))$ for all $t\ge 0$ and $f(0)=x$. Therefore, if you find by some ...

answered Jan 30 '14 at 14:42

Jochen Wengenroth

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4

This is standard, but the argument is short enough to fit in an answer. It is not restrictive to assume that $0\in\Omega$. Denote by $Q(t,x,y)$ the heat kernel associated with the Dirichlet Laplacian on $\Omega$ (this is simply the solution to the heat equation on $\Omega$ with Dirichlet b.c. and the delta function as initial data) and by $P(t,x,y)=ct^{-n/2}\...

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