Let $U\subseteq\mathbb R^d$ be a bounded domain.

A continuous function $u:\overline U\to\mathbb R$ is called harmonic $:\Leftrightarrow$ $$u(x)=\frac 1{|\partial B_r(x)|}\int_{\partial B_r(x)} u\;do\;\;\;\text{for any open ball }B_r(x)\subset\subset U\;.$$

Now, let

  • $(\Omega,\mathcal A)$ be a measurable space
  • $\mathbb F=\left(\mathcal F_t\right)_{t\ge 0}$ be a filtration on $(\Omega,\mathcal A)$
  • $\left(\operatorname P_x\right)_{x\in\mathbb R^d}$ be a family of probability measures on $(\Omega,\mathcal A)$
  • $(B_t-x)_{t\ge 0}$ be a $d$-dimensional standard Brownian motion on $(\Omega,\mathcal A,\operatorname P_x)$ with respect to $\mathbb F$, for all $x\in\mathbb R^d$

Let $$\tau:=\inf\left\{t\ge 0:B_t\in\partial U\right\}$$ and $\varphi:\partial U\to\mathbb R$ be Borel-measurable. How can we show, that $$u(x):=\operatorname E_x\left[1_{\left\{\tau<\infty\right\}}\varphi\left(B_\tau\right)\right]\;\;\;\text{for }x\in U$$ is harmonic?

Let $B_r(x)\subset\subset U$ and $$\sigma:=\inf\left\{t>0:B_t\notin B_r(x)\right\}\;.$$ The strong Markov property implies, that \begin{equation} \begin{split} u(x)&=\operatorname E_x\left[\operatorname E_x\left[1_{\left\{\tau<\infty\right\}}\varphi\left(B_\tau\right)\mid\mathcal F_{\sigma}\right]\right]\\ &=\operatorname E_x\left[u\left(B_{\sigma}\right]\right] \end{split} \end{equation}

for all $x\in U$. In Brownian Motion by Peter Mörters and Yuval Peres they state in Theorem 3.8, that the last expression would be equal to $$\int_{\partial B_r(x)}u\;d\varpi\;,$$ where $\varpi$ denotes the uniform distribution on $\partial B_r(x)$. I absolutely don't understand why this is true and why this yields the desired statement.

  • 1
    $\begingroup$ Because the isotropy of Brownian motion implies that, under $P_x$, $B_\sigma$ is uniformly distributed on $\partial B_r(x)$ (as most probably the authors explain). $\endgroup$ – Did Oct 20 '15 at 22:39
  • $\begingroup$ Do you understand the step where they use the strong Markov property to conclude $E_x[E_x[\varphi(B_\tau) \mid \mathcal{F}_\sigma]]=E_x[u(B_\sigma)]$? The rest is actually fairly simple: the final integral is equal to the integral at the top more or less by definition, and the final integral is equal to $E_x[u(B_\sigma)]$ because $B$ is rotationally symmetric. $\endgroup$ – Ian Oct 20 '15 at 22:39
  • $\begingroup$ (A side remark: enforcing $\tau<\infty$ with the indicator function is not at all necessary, because Brownian motion escapes any bounded domain in finite time with probability $1$.) $\endgroup$ – Ian Oct 20 '15 at 22:40

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