The first order correction to harmonic measure for random walks of rotationally invariant step distribution

Abstract

Let $D\subset\mathbb{R}^{d}\ (d\geq2)$ be an open simply-connected bounded domain with smooth boundary $\partial D$ and $\mathbf{0}=(0,\ldots,0)\in D$. Fix any rotationally invariant probability $\mu$ on closed unit ball $\{z\in\mathbb{R}^{d}:\vert z\vert\leq1\}$ with $\mu(\{\mathbf{0}\})<1$. Let $\{S_{n}^{\mu}\}_{n=0}^{\infty}$ be the random walk with step-distribution $\mu$ starting at $\mathbf{0}$. Denote by $\omega_{\delta}(\mathbf{0},\mathrm{d}z;D)$ the discrete harmonic measure for $\{\delta S_{n}^{\mu}\}_{n=0}^{\infty}\ (\delta>0)$ exiting from $D$, which is viewed as a probability on $\partial D$ by projecting suitably the first exiting point to $\partial D$. Denote by $\omega(\mathbf{0},\mathrm{d}z;D)$ the harmonic measure for the $d$-dimensional standard Brownian motion exiting from $D$. Then in the weak convergence topology, \begin{equation*}\lim_{\delta\rightarrow0}\frac{1}{\delta}\bigl[\omega_{\delta}(\mathbf{0} ,\mathrm{d}z;D)-\omega(\mathbf{0},\mathrm{d}z;D)\bigr]=c_{\mu}\rho_{D}(z)\,\vert \mathrm{d}z\vert ,\end{equation*} where $\rho_{D}(\cdot)$ is a smooth function depending on $D$ but not on $\mu$, $c_{\mu}$ is a constant depending only on $\mu$, and $|\mathrm{d}z|$ is the Lebesgue measure with respect to $\partial D$. Additionally, $\rho_{D}(z)$ is determined by the following equation: For any smooth function $g$ on $\partial D$, \begin{equation*}\int_{\partial D}g(z)\rho_{D}(z)\,\vert \mathrm{d}z\vert =\int_{\partial D}\frac{\partial f}{\partial\mathbf{n}_{z}}(z)H_{D}(\mathbf{0},z)\,\vert \mathrm{d}z\vert ,\end{equation*} where $f$ is the harmonic function in $D$ with boundary values given by $g$, $H_{D}(\mathbf{0},z)$ is the Poisson kernel and derivative $\frac{\partial f}{\partial\mathbf{n}_{z}}$ is with respect to the inward unit normal $\mathbf{n}_{z}$ at $z\in\partial D$.