## 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 $$mathcal S^1:={muinmathcal M_1:exists yin E:(muotimesdelta_y)rho denote the Wasserstein space associated to $$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 $$(mukappa_totimesdelta_0)rho, but since this $$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?

## The quotient semigroup is an inverse semigroup if the semigroup is not

With an infinite semigroup $$S$$ and $$rho$$ Congruence relationship $$S$$. Could you help me give the necessary conditions for this? $$S / rho$$ is an inverse semigroup if and only if $$S$$ is not an inverse semigroup? I started thinking that $$S$$ must be commutative, but I can't find the other condition.

## 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

## Abstract algebra – In a semigroup, the product of two subgroups is always a subgroup

In a semigroup, the product of two subgroups is always a subgroup.
To let $$A$$.$$B$$ be two subset of semigroup G.
$$AB = {ab: a in A, b inB }$$

I try to prove that result, but I have no idea. So please give me a hint. Please do not give me any answers.

## 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 $$| a | then from the Subsolution estimate $$e ^ {a | x |} u in L ^ infty$$,

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