pr.probability – Show that the adjoint of this Markov semigroup eventually preserves a Wasserstein space

Let $E$ be a separable $mathbb R$-Banach space, $mathcal M_1$ denote the set of probability measures on $(E,mathcal B(E))$, $(kappa_t)_{tge0}$ be a Markov semigroup on $(E,mathcal B(E))$ and $rho$ be a metric on $E$ with $$(mukappa_totimesdelta_0)rhole cint v^{lambda(t)}:{rm d}mutag1$$ for all $tge0$ and $muinmathcal M_1$ for some $cge0$, some continuous $v:Eto(1,infty)$ and some nonincreasing $lambda:(0,infty)to(0,1)$ with $lambda(t)xrightarrow{ttoinfty}0$.

Let $$mathcal S^1:={muinmathcal M_1:exists yin E:(muotimesdelta_y)rho<infty}$$ denote the Wasserstein space associated to $rho$. Are we able to conclude that, if $(kappa_t)_{tge0}$ has an invariant measure, it must belong to $mathcal S^1$?

My idea is to show that there is a $t_0ge0$ such that the adjoint$^1$ $kappa_t^ast$ maps $mathcal M_1$ to $mathcal S^1$ for all $tge t_0$.

I’m only able to prove something weaker: Let $muinmathcal M_1$. By the monotone convergence theorem, $$(mukappa_totimesdelta_0)rhoxrightarrow{ttoinfty}ctag2.$$ So, there is a $t_0ge0$ such that $(mukappa_totimesdelta_0)rho<infty$, but since this $t_0$ clearly depends on $mu$, I’m not sure if this is sufficient for the desired conclusion.


$^1$ $kappa_t^astmu:=mukappa_t$.

Construction of $R$-graph semigroup from a directed graph

In the paper https://arxiv.org/pdf/1209.2578.pdf, Krieger is simply defining his notion of $mathcal{R}$-graph. It is an object $mathcal{G}_{mathcal{R}}(mathfrak{P},mathcal{E}^-,mathcal{E}^+)$, where $mathfrak{P}$ is the vertex set of a finite directed graph with edge set $mathcal{E}$, ${mathcal{E}^-,mathcal{E}^+}$ is a partition of $mathcal{E}$, the graph satisfies certain conditions, and $mathcal{R}$ is a relation on the edges that also satisfies certain conditions.

Specifically, for $mathfrak{q},mathfrak{r}inmathfrak{P}$ we define $mathcal{E}^-(mathfrak{q},mathfrak{r})$ to be the set of edges in $mathcal{E}^-$ from $mathfrak{q}$ to $mathfrak{r}$ and $mathcal{E}^+(mathfrak{q},mathfrak{r})$ to be the set of edges in $mathcal{E}^+$ from $mathfrak{r}$ to $mathfrak{q}$. For each $mathfrak{q},mathfrak{r}inmathfrak{P}$ we require that either both of these sets be empty or both be non-empty, and we further require that the directed graph $langlemathfrak{P},mathcal{E}^-rangle$ (and hence also ($langlemathfrak{P},mathcal{E}^+rangle$) be strongly connected.

The relation $mathcal{R}$ is the union of relations $mathcal{R}(mathfrak{q},mathfrak{r})subseteqmathcal{E}^-(mathfrak{q},mathfrak{r})timesmathcal{E}^+(mathfrak{q},mathfrak{r})$ for $mathfrak{q},mathfrak{r}inmathfrak{P}$.

Further, Kreiger construct a semigroup (with zero) $mathcal S(mathcal{G}_{mathcal{R}}(mathfrak{P},mathcal{E}^-,mathcal{E}^+))$ from an $R$-graph $mathcal{G}_{mathcal{R}}(mathfrak{P},mathcal{E}^-,mathcal{E}^+)$. The semigroup $mathcal S(mathcal{G}_{mathcal{R}}(mathfrak{P},mathcal{E}^-,mathcal{E}^+))$ contains an idempotent $1_{mathfrak p}, mathfrak p in mathfrak{P}$ and $mathcal E$ as a generating set. Besides $1_{mathfrak p}^2 = 1_{mathfrak p}, mathfrak p in mathfrak{P}$, the defining relations are
$$f^-g^+ = left{ begin{array}{cl}
1_{mathfrak{q}} & text{if} ; f^{-} in mathcal{E^{-}}(mathfrak{q}, mathfrak{r}), ; g^{+} in mathcal{E^{+}}(mathfrak{q}, mathfrak{r}), ; (f^{-}, g^{+}) in mathcal{R}(mathfrak{q}, mathfrak{r}), ; mathfrak q, mathfrak{r} in mathfrak{P},\
0 & text{if} ; f^{-} in mathcal{E^{-}}(mathfrak{q}, mathfrak{r}), ; g^{+} in mathcal{E^{+}}(mathfrak{q}, mathfrak{r}), ; (f^{-}, g^{+}) notin mathcal{R}(mathfrak{q}, mathfrak{r}), ; mathfrak q, mathfrak{r} in mathfrak{P},\
0 & text{if} ; f^{-} in mathcal{E^{-}}(mathfrak{q}, mathfrak{r}), ; g^{+} in mathcal{E^{+}}(mathfrak{q}, mathfrak{r}), ; mathfrak{q}, ; mathfrak{q’}, mathfrak{r} in mathfrak{P}, mathfrak{q} ne mathfrak{q’},\
end{array}right.$$

and

begin{array}{cl}
1_{mathfrak{q}}e^{-}= e^{-} 1_{mathfrak{r}} = e^{-}, & e^{-}in mathcal{E}(mathfrak{q}, mathfrak{r}) \
1_{mathfrak{r}}e^{+}= e^{+} 1_{mathfrak{q}} = e^{+}, & e^{+}in mathcal{E}(mathfrak{q}, mathfrak{r}), & mathfrak q, mathfrak{r} in mathfrak{P},
end{array}

and

begin{array}{cl}
1_{mathfrak{q}}
1_{mathfrak{r}} = 0 & mathfrak q, mathfrak{r} in mathfrak{P}, & mathfrak q ne mathfrak{r}.
end{array}

My question is

What is the value of $g^{+}f^{-}$, where $f^{-} in mathcal{E^{-}}(mathfrak{q}, mathfrak{r}), g^{+} in mathcal{E^{+}}(mathfrak{q}, mathfrak{r})$? I think it should be $1_{mathfrak{r}}$ but I don’t know how?

Solving partial differential equations according to the molding method (semigroup theory). Which form (a, j) does the operator A have?

We have the equations

$$ rho (x) frac { partial ^ 2u} { partial ^ 2} (t, x) = div (A (x) degrees u + B (x) grad frac { partial u} { partial t}) (t, x) -div ( gamma (x) v (t, x)) $$

$$ beta (x) frac { partial v} { partial t} (t, x) = div (C (x) degree v) (t, x) – gamma (x) .grad frac { partial u} { partial t} (t, x) $$

This leads to an operator

$$ A = begin {pmatrix} 0 &, I &, 0 \ frac {1} { rho (x)} div A (x) grad &, frac {1} { rho (x)} div B (x) grad &, frac {-1} { rho (x)} div gamma (x) \ 0 &, frac {-1} { beta (x)} gamma (x) grad &, frac {1} { beta (x)} div C (x) grad end {pmatrix} $$

Now A continues to work $ (u_1, u_2, u_3) ^ T $How do I find the associated form of A?
Help Thanks

Analysis of pdes – A question of Schrödinginger Semigroup – By B. Simon

The question comes from the newspaper: B. Simon, Schrödinger Semigroups, Bull A.M., (1982) Vol. 7 (3).

Theorem C.1.2 (underestimation estimate) of the paper states: If $ Hu = 0 $, from where $ H = – Delta + V $ for a limited continuous function $ V $, Then
$$ | u (x) | leq C int_ {B_r (x)} | u (y) | dy, $$
from where $ C $ depends on $ r $ and norm of $ V $ and $ B_r (x) $ takes place within the domain of $ u $,

On page 499 it writes: If $ e ^ {a | x |} u in L ^ 2 $ for all $ | a | <M $then from the Subsolution estimate $ e ^ {a | x |} u in L ^ infty $,

Q How can the result be deduced? $ e ^ {a | x |} u in L ^ 2 $ for all $ | a | <M $, then $ u in L ^ infty $,

PS: What I do not understand is that $ e ^ {a | x |} u $ Does not the equation satisfy, is there a result to solve this problem?

Monoids – Can the nilpotent relationships in the semigroup $ ( wp (X ^ 2), circ) $ be generated by elements of $ 3 $?

In every finite sentence $ X $ the set of relationships $ wp (X times X) = {R: R subseteq X times X } $ forms a semigroup in relation to the relationship composition $ circ $, With this statement, every nilpotent in $ wp (X times X) $ be generated by $ 3 $ Elements of $ wp (X times X) $? I can prove it is possible with only four, though I think that's possible 3% $ Elements. If someone could correct me or quote relevant articles, I would be happy.


Edit: To repeat a relationship $ R subseteq X times X $ is nilpotent iff $ exists n in mathbb {N}: underscore {R circ R circ cdots circ R} _ { emptyset} = emptyset $

fa.functional analysis – convolution with an analytical semigroup

To let $ e ^ {At} $ denote an analytic semigroup in Hilbert space $ X $ generate by $ A: D (A) to X $, Leave too $ f in L ^ 1 (0, tau; X) $, I want to show that folding
$$ g (t) = int_0 ^ t e ^ {A (t-s)} f (s) ds $$
belongs $ W ^ {1,1} (0, tau; X) cap L ^ 1 (0, tau; D (A)) $, When I sit down $ f (s) = e ^ {As} x $ from where $ x in X $ then $ g (t) = te ^ {At} x $, Because of the analyticity of $ e ^ {At} $ we have
$$ sup_ {t> 0} | tAe ^ {At} | _ {L (X)} < infty $$
thus $ g (t) in L ^ { infty} (0, tau; D (A)) $ and $ dot {g} (t) = e ^ {At} x + tAe ^ {At} x in L ^ { infty} (0, tau; X) $, Does that prove that?