By Feng-Yu Wang
Stochastic research on Riemannian manifolds with no boundary has been good tested. even though, the research for reflecting diffusion procedures and sub-elliptic diffusion approaches is much from entire. This publication comprises fresh advances during this course in addition to new principles and effective arguments, that are the most important for extra advancements. Many effects contained the following (for instance, the formulation of the curvature utilizing derivatives of the semigroup) are new between present monographs even within the case with out boundary.
Readership: Graduate scholars, researchers and execs in chance thought, differential geometry and partial differential equations.
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Additional info for Analysis for Diffusion Processes on Riemannian Manifolds : Advanced Series on Statistical Science and Applied Probability
Example text
10) for some x, e ∈ E and constant CΦ (x, e) ≥ 0. Obviously, if Φ(r) = rp for some p > 1 then this inequality reduces to the shift Harnack inequality with power p, while when Φ(r) = er it becomes the shift log-Harnack inequality. 5. 10) for all x, e ∈ E := Rd and some non-negative measurable function CΦ on Rd × Rd . Then Φ(P f )(x) ≤ sup f ∈Bb+ (Rd ), Rd Φ◦f (x)dx≤1 1 , x ∈ Rd . t. the Lebesgue measure such that 1 p(x, y)Φ−1 (p(x, y))dy ≤ Φ−1 Rd Rd e−CΦ (x,e) de . 12) (2) If Φ(r) = rp for some p > 1, then 1 p p(x, y) p−1 dy ≤ ( Rd Proof.
For P (x, · − e)-null set A, let f = 1A . 10) implies that Φ(P (x, A)) ≤ 0; hence P (x, A) = 0 since Φ(r) > 0 for r > 0. t. P (x, · − e). Next, let Φ(r) = rΨ(r). 10) for f (y) = Ψ(n ∧ p(x, e; y)) and noting that Ψ(n ∧ p(x, e; y)) P (x, dy) P f (x) = E ≥ Φ(n ∧ p(x, e; y))P (x, dy − e), E we obtain Φ(n ∧ p(x, e; y))P (x, dy − e) ≤ Ψ−1 eCΦ (x,e) . E Then the proof is completed by letting n → ∞. 5 Harnack inequality and derivative estimate In this section, we consider the relationship between Harnack inequalities and derivative estimates of Markov operators on a geodesic space.
T. µ. t. the invariant probability measure is strictly positive. t. µ satisfies p(x, ·)Φ−1 E p(x, ·) dµ ≤ Φ−1 (eΨ(x,y) ), x, y ∈ E, p(y, ·) −1 where Φ (∞) := ∞ by convention. t. µ satisfies p(x, ·)p(y, ·)dµ ≥ e−Ψ(x,y) , x, y ∈ E. E (6) If µ is an invariant probability measure of P , then sup f ∈Bb+ (E),µ(Φ(f ))≤1 P f (x) ≤ 1 , x ∈ E. −Ψ(x,y) µ(dy) e E Proof. Since (6) is obvious, below we prove (1)-(5) respectively. (1) Let f ∈ Bb (E) be positive. 1) to 1 + εf in place of f for ε > 0, we have Φ(1 + εP f (x)) ≤ {P Φ(1 + εf )(y)}eΨ(x,y) , x, y ∈ E, ε > 0.