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I have a question about harmonic functions and hitting probabilities.

Let $d \ge 3$ be an integer.

Let $D=\mathbb{R}^d \times (-1,1)$ and denote points $z \in D$ by $z=(r,\theta,y),$ where $(r,\theta)$ denotes the polar coordinates in $\mathbb{R}^d$ and $y \in (-1,1)$. Let $D_0=\{(r,\theta,y) \in D \mid r<1\}$ denote the unit cylinder in $D$.

Let $X=(X_t,P_z)$ be the (normally) reflecting Brownian motion on $\bar{D}$. We define $\sigma=\inf \{t \ge 0 \mid X_t \in D_0\}$. Then, $P_{z}(\sigma<\infty)$ is a harmonic function on $D \setminus D_0$ and satisfies the Neumann boundary condition on $\partial D \setminus \partial D_0$.

There are some harmonic functions on $D \setminus D_0$ and satisfies the Neumann boundary condition on $\partial D \setminus \partial D_0$. For example, if we define $h : \bar{D} \mapsto \mathbb{R}$ by \begin{align*} h(z)&=r^{-d+2}, \quad (z=(r,\theta,y) \in \bar{D},\ r\ge 1), \\ h(z)&=1,\quad (z=(r,\theta,y) \in \bar{D},\ r<1), \end{align*} $h$ is a desired one.

The following should hold: \begin{equation*} (1)\quad P_z(\sigma<\infty)=h(z), \, z \in \bar{D}. \end{equation*}

My question

Is there a theorem which justifies (1)?

What are sufficient conditions for uniqueness of harmonic functions?

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    $\begingroup$ I cannot give a reference at the moment, but it is one of the fundamental links between probability and potential theory that the hitting probability $f(z) = P_z(\sigma < \infty)$ is the equilibrium potential of $\overline{D}$. Knowing this, one can use the standard machinery of potential theory: in order to prove that $f = h$ it is for example sufficient to show that $h$ is bounded by some potential, continuous, harmonic in the complement of $\overline{D}$ and equal to one in $\overline{D}$. (1/2) $\endgroup$ Commented Dec 26, 2018 at 23:01
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    $\begingroup$ Standard references are: Doob's Classical Potential Theory and Its Probabilistic Counterpart for the Brownian motion case (which I do not know well, but it is certainly a good read) and Bliedtner–Hansen Potential Theory for the general case (which I know reasonably well and which is certainly not a good read, although I like it anyway). (2/2) $\endgroup$ Commented Dec 26, 2018 at 23:05
  • $\begingroup$ Thank you for always teaching me carefully. I will check the above references. $\endgroup$
    – sharpe
    Commented Dec 27, 2018 at 11:27

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