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Branching capacity and Brownian snake capacity

Tianyi Bai, Jean-François Delmas, Yueyun Hu

Abstract

The branching capacity has been introduced by [Zhu 2016] as the limit of the hitting probability of a symmetric branching random walk in $\mathbb Z^d$, $d\ge 5$. Similarly, we define the Brownian snake capacity in $\mathbb R^d$, as the scaling limit of the hitting probability by the Brownian snake starting from afar. Then, we prove our main result on the vague convergence of the rescaled branching capacity towards this Brownian snake capacity. Our proof relies on a precise convergence rate for the approximation of the branching capacity by hitting probabilities.

Branching capacity and Brownian snake capacity

Abstract

The branching capacity has been introduced by [Zhu 2016] as the limit of the hitting probability of a symmetric branching random walk in , . Similarly, we define the Brownian snake capacity in , as the scaling limit of the hitting probability by the Brownian snake starting from afar. Then, we prove our main result on the vague convergence of the rescaled branching capacity towards this Brownian snake capacity. Our proof relies on a precise convergence rate for the approximation of the branching capacity by hitting probabilities.
Paper Structure (12 sections, 20 theorems, 193 equations, 4 figures)

This paper contains 12 sections, 20 theorems, 193 equations, 4 figures.

Key Result

Theorem 1.1

Let $d\ge 5$ and $\lambda>1$. Assume hyp-tree and hyp-brw. There exists a positive constant $C=C(d,\lambda)$ such that uniformly in $r\ge 1$, $K \subset {\rm B}(0,r)$ nonempty and $x\in \mathbb Z^d$ with $|x|\ge \lambda r$, we have, with $\alpha=(d-4)/2(d-1)$,

Figures (4)

  • Figure 1: An illustration of $\mathcal{T}_{\rm I} =\mathcal{T}_{-}\cup\mathcal{T}_{\text{adj}}^0$.
  • Figure 2: In this picture, we have a random walk starting at $x$ and exiting $B$ at the point $a$. On each point we attach an adjoint branching random walk, illustrated with the small triangles. The quantity $\sum_{a\not\in B}H^{B}_{K}(x,a)$ is the probability that the shaded and solid triangles do not hit $K$, ${\bf e}_K (x)=1-{\bf p}_K^{(-)}(x)=\mathbf P_x({\mathscr R}_-\cap K=\emptyset)$ is the probability that the solid and hollow triangles do not hit $K$, $1-{\bf p}_K^{({\mathrm{adj}})}(x)=\mathbf P_x({\mathscr R}_{\text{adj}} \cap K=\emptyset)$ is the probability that the shaded triangle does not hit $K$, and $1-{\bf p}_K^{(I)}(x)=\mathbf P_x({\mathscr R}_{\rm I}\cap K=\emptyset)$ is the probability that no triangle hits $K$.
  • Figure 3: Relative positions of $a,x,y$ and $B$ in the proof of Lemma \ref{['lem:true_rs']}.
  • Figure 4: An illustration for the case $b\not\in (2B)$. Note that by definition of $H^B_K(b,a)$ in \ref{['def-HBK0']}, $S^\kappa_1$ makes a large jump (in blue) from $b\not\in (2B)$ to some point $z\in B$, so that $H^B_K(b,a)\le \sum_{z\in B} \theta(z-b)G_K(z,a)$.

Theorems & Definitions (44)

  • Theorem 1.1
  • Theorem 1.2
  • Proposition 1.3
  • Theorem 1.4
  • Remark 1.5
  • Remark 2.1
  • Theorem 2.2: dellacherie-meyer
  • Remark 2.3
  • Remark 2.4
  • Lemma 2.5
  • ...and 34 more