[proofplan] We show $u(x) \to \varphi(z)$ as $x \to z$ by splitting $\mathbb{E}_x[|\varphi(B_\tau) - \varphi(z)|]$ according to whether $B_\tau$ exits near $z$ or far from $z$. The first contribution is controlled by the continuity of $\varphi$. For the second, the cone condition supplies a positive lower bound on the probability that Brownian motion hits the exterior cone before exiting a ball centred at $z$. We establish this bound via the Poisson kernel for the ball and a domain monotonicity argument: any continuous path from a point outside the cone to the spherical cap cut out by the cone must cross the cone boundary, so the harmonic measure of the cone boundary in the excised region $B(z,\rho) \setminus \overline{C_z}$ is at least the harmonic measure of the cap in the full ball. The Poisson kernel then yields an explicit lower bound $c(\theta, d) > 0$ independent of the ball radius. The [Strong Markov Property](/theorems/1180) converts this one-step bound into geometric decay as $x \to z$. [/proofplan] [step:Split the error $|u(x) - \varphi(z)|$ into near-boundary and far-boundary contributions] Fix $z \in \partial D$ and $\varepsilon > 0$. By continuity of $\varphi$ at $z$, choose $\delta > 0$ such that $|\varphi(y) - \varphi(z)| < \varepsilon$ for all $y \in \partial D \cap B(z, \delta)$. For $x \in D$, write \begin{align*} |u(x) - \varphi(z)| &= |\mathbb{E}_x[\varphi(B_\tau) - \varphi(z)]| \leq \mathbb{E}_x[|\varphi(B_\tau) - \varphi(z)|] \\ &\leq \varepsilon \cdot \mathbb{P}_x(B_\tau \in B(z, \delta)) + 2\|\varphi\|_\infty \cdot \mathbb{P}_x(B_\tau \notin B(z, \delta)). \end{align*} The first term contributes at most $\varepsilon$ (uniformly in $x$). It remains to show that $\mathbb{P}_x(B_\tau \notin B(z, \delta)) \to 0$ as $x \to z$. [/step] [step:Establish a positive lower bound $c(\theta, d) > 0$ on the probability of hitting the cone before exiting a ball] By hypothesis, there exists a cone $C_z$ with vertex $z$, half-angle $\theta \in (0, \pi/2)$, axis direction $e \in \partial B(0,1)$, and radius $r > 0$ such that $C_z \cap B(z, r) \subset \mathbb{R}^d \setminus D$. Explicitly, \begin{align*} C_z = \{y \in \mathbb{R}^d : (y - z) \cdot e > |y - z| \cos\theta\}. \end{align*} Fix $\rho \in (0, \min(\delta/2, r/2)]$ and consider $x \in D$ with $|x - z| \leq \rho/2$. Define the open region $G := B(z, \rho) \setminus \overline{C_z}$. Since $D \cap B(z, \rho) \subset G$ (the cone portion is exterior to $D$), any Brownian path starting at $x \in D \cap B(z, \rho)$ that exits $D$ within $B(z, \rho)$ must first exit $G$. We decompose the boundary $\partial G = S_{\text{sphere}} \cup S_{\text{cone}}$ where \begin{align*} S_{\text{sphere}} &= \partial B(z, \rho) \setminus \overline{C_z}, \\ S_{\text{cone}} &= \partial C_z \cap B(z, \rho). \end{align*} Let $\sigma_G := \inf\{t \geq 0 : B_t \notin G\}$ and $\sigma_{B(z,\rho)} := \inf\{t \geq 0 : B_t \notin B(z, \rho)\}$. Since $G \subset B(z, \rho)$, we have $\sigma_G \leq \sigma_{B(z,\rho)}$. The cone $C_z$ intersects $\partial B(z, \rho)$ in the spherical cap \begin{align*} \Sigma_\theta := \{y \in \partial B(z, \rho) : (y - z) \cdot e > \rho\cos\theta\}. \end{align*} We use a **domain monotonicity** argument. On the event $\{B_{\sigma_{B(z,\rho)}} \in \Sigma_\theta\}$, the Brownian path starts at $x \notin \overline{C_z}$ and reaches a point in $\Sigma_\theta \subset \overline{C_z}$. By continuity of the path, it must cross $\partial C_z$ at some time $t^* < \sigma_{B(z,\rho)}$. At time $t^*$, the path is at a point of $\partial C_z \cap B(z, \rho) = S_{\text{cone}}$, so $B$ has already exited $G$ through $S_{\text{cone}}$ by time $t^* \leq \sigma_G$. Therefore \begin{align*} \mathbb{P}_x(B_{\sigma_G} \in S_{\text{cone}}) \geq \mathbb{P}_x(B_{\sigma_{B(z,\rho)}} \in \Sigma_\theta). \end{align*} It remains to bound the right-hand side from below using the Poisson kernel. For $x \in B(z, \rho)$ and any Borel [set](/page/Set) $A \subset \partial B(z, \rho)$, the exit [distribution](/page/Distribution) from the ball is \begin{align*} \mathbb{P}_x(B_{\sigma_{B(z,\rho)}} \in A) = \int_A P(x, y) \, d\mathcal{H}^{d-1}(y), \quad P(x, y) = \frac{\rho^2 - |x - z|^2}{\omega_d \, \rho \, |x - y|^d}, \end{align*} where $\omega_d = \mathcal{H}^{d-1}(\partial B(0,1))$ is the surface area of the unit sphere in $\mathbb{R}^d$. For $|x - z| \leq \rho/2$ and $y \in \partial B(z, \rho)$, the triangle inequality gives $|x - y| \leq |x - z| + |y - z| \leq 3\rho/2$, so \begin{align*} P(x, y) \geq \frac{\rho^2 - \rho^2/4}{\omega_d \, \rho \, (3\rho/2)^d} = \frac{3}{4\omega_d} \cdot \frac{2^d}{3^d \, \rho^{d-1}} = \frac{2^d}{4 \cdot 3^{d-1} \cdot \omega_d \, \rho^{d-1}}. \end{align*} The $(d-1)$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) of the cap $\Sigma_\theta$ is computed by scaling from the unit sphere: \begin{align*} \mathcal{H}^{d-1}(\Sigma_\theta) = \rho^{d-1} \cdot \omega_d(\theta), \quad \text{where } \omega_d(\theta) := \int_{\{\xi \in \partial B(0,1) : \xi \cdot e > \cos\theta\}} d\mathcal{H}^{d-1}(\xi) > 0. \end{align*} The quantity $\omega_d(\theta)$ is the surface area of a spherical cap of half-angle $\theta$ on the unit sphere in $\mathbb{R}^d$, which is strictly positive for $\theta \in (0, \pi/2)$. Integrating the lower bound on $P(x,y)$ over $\Sigma_\theta$, \begin{align*} \mathbb{P}_x(B_{\sigma_{B(z,\rho)}} \in \Sigma_\theta) \geq \frac{2^d}{4 \cdot 3^{d-1} \cdot \omega_d \, \rho^{d-1}} \cdot \rho^{d-1} \, \omega_d(\theta) = \frac{2^d \, \omega_d(\theta)}{4 \cdot 3^{d-1} \cdot \omega_d} =: c(\theta, d). \end{align*} The constant $c(\theta, d) > 0$ depends only on the half-angle $\theta$ and the dimension $d$. The factors of $\rho^{d-1}$ cancel, so the bound is **independent of $\rho$ and of the starting point $x$** (provided $|x - z| \leq \rho/2$). This scale-invariance reflects the [scaling property](/theorems/1175)(ii) of Brownian motion: the exit distribution from $B(z, \rho)$, when viewed in the rescaled coordinate $\tilde{x} = (x - z)/\rho$, depends only on the normalised starting position $\tilde{x} \in B(0,1)$. Since $S_{\text{cone}} \subset \partial C_z \subset \mathbb{R}^d \setminus D$, hitting $S_{\text{cone}}$ forces exit from $D$ within $B(z, \rho) \subset B(z, \delta)$. Combining, \begin{align*} \mathbb{P}_x(B_\tau \in B(z, \delta)) \geq \mathbb{P}_x(B_{\sigma_G} \in S_{\text{cone}}) \geq c(\theta, d) > 0 \end{align*} for all $x \in D$ with $|x - z| \leq \rho/2$. [/step] [step:Iterate via the Strong Markov Property to obtain geometric decay] The previous step gives, for any $\rho \leq \min(\delta/2, r/2)$ and any $x \in D$ with $|x - z| \leq \rho/2$, \begin{align*} \mathbb{P}_x(B_\tau \notin B(z, \delta)) \leq 1 - c(\theta, d). \end{align*} Fix $\rho_0 := \min(\delta/2, r/2)$ and define the dyadic radii $\rho_k := 2^{-k} \rho_0$ for $k \geq 0$. Consider $x \in D$ with $|x - z| < \rho_0/2$, and define the stopping times \begin{align*} T_0 &:= 0, \\ T_{k+1} &:= \inf\{t > T_k : B_t \notin B(z, \rho_k)\} \quad \text{for } k \geq 0. \end{align*} Each $T_k$ is finite a.s. since $B(z, \rho_k)$ is bounded. If $B$ has not exited $D$ by time $T_k$ (i.e., $\tau > T_k$), then $B_{T_k} \in D \cap \partial B(z, \rho_k)$, so $|B_{T_k} - z| = \rho_k = \rho_{k-1}/2$. The cone $C_z \cap B(z, \rho_k) \subset \mathbb{R}^d \setminus D$ since $\rho_k \leq \rho_0 \leq r/2 < r$. By the [Strong Markov Property](/theorems/1180), the process $(B_{T_k + t} - B_{T_k})_{t \geq 0}$ is a standard Brownian motion independent of $\mathcal{F}_{T_k}^+$. Since $|B_{T_k} - z| = \rho_{k-1}/2$, the estimate from the previous step (applied with radius $\rho_{k-1}$ in place of $\rho$) gives \begin{align*} \mathbb{P}(B \text{ exits } D \text{ within } B(z, \delta) \text{ during } [T_k, T_{k+1}] \mid \mathcal{F}_{T_k}^+,\, \tau > T_k) \geq c(\theta, d). \end{align*} Since the successive opportunities are conditionally independent (by the Strong Markov Property), the probability that $B$ fails to exit $D$ near $z$ during the first $N$ opportunities satisfies \begin{align*} \mathbb{P}_x(\tau > T_N, \, B_\tau \notin B(z, \delta)) \leq (1 - c(\theta, d))^N. \end{align*} Define $N(x) := \max\{k \geq 0 : |x - z| \leq \rho_k/2\}$. Since $\rho_k = 2^{-k}\rho_0$, we have $N(x) \geq k$ whenever $|x - z| \leq 2^{-(k+1)}\rho_0$, so $N(x) \to \infty$ as $x \to z$. For the Brownian path started at $x$, either $B$ exits $D$ within $B(z, \delta)$ during one of the first $N(x)$ opportunities, or $B$ exits $B(z, \rho_0)$ while still in $D$. In either case, \begin{align*} \mathbb{P}_x(B_\tau \notin B(z, \delta)) \leq (1 - c(\theta, d))^{N(x)} \xrightarrow{x \to z} 0, \end{align*} since $N(x) \to \infty$ as $|x - z| \to 0$ and $0 < 1 - c(\theta, d) < 1$. [/step] [step:Conclude boundary continuity] Combining the two contributions: for $|x - z|$ sufficiently small, \begin{align*} |u(x) - \varphi(z)| \leq \varepsilon + 2\|\varphi\|_\infty \cdot \mathbb{P}_x(B_\tau \notin B(z, \delta)). \end{align*} Taking $\limsup_{x \to z}$, the second term vanishes by the geometric decay established in the previous step, giving $\limsup_{x \to z} |u(x) - \varphi(z)| \leq \varepsilon$. Since $\varepsilon > 0$ was arbitrary, $u(x) \to \varphi(z)$ as $x \to z$. Therefore $u$ extends continuously to $\overline{D}$ with $u(z) = \varphi(z)$ for all $z \in \partial D$. [/step]