[proofplan] The proof uses harmonic [functions](/page/Function) as [test functions](/page/Test%20Function) for optional stopping. In dimension $d = 2$, the function $y \mapsto \log|y|$ is harmonic in an annulus; optional stopping for the martingale $\log|B_{t \wedge H}|$ at the exit time $H$ from the annulus gives the hitting probability of the inner sphere, which tends to $1$ as the outer radius $R \to \infty$. In dimension $d \geq 3$, the function $y \mapsto |y|^{2-d}$ plays the analogous role and yields a hitting probability that vanishes as $R \to \infty$, establishing transience. Dimension $d = 1$ follows from point recurrence via the strong Markov property. [/proofplan] [step:Establish neighbourhood recurrence in $d = 2$ using the harmonic function $\log|\cdot|$] Fix $\varepsilon > 0$ and $R > |x| > \varepsilon$. The function $f : \mathbb{R}^2 \setminus \{0\} \to \mathbb{R}$ defined by $f(y) = \log|y|$ satisfies $\Delta f = 0$ on $\mathbb{R}^2 \setminus \{0\}$ (verified by direct computation in polar coordinates: $\Delta f = \partial_r^2 f + r^{-1}\partial_r f + r^{-2}\partial_\theta^2 f = -1/r^2 + 1/r^2 + 0 = 0$). Define the stopping times $T_\varepsilon = \inf\{t \geq 0 : |B_t| = \varepsilon\}$ and $T_R = \inf\{t \geq 0 : |B_t| = R\}$. [Set](/page/Set) $H = \min(T_\varepsilon, T_R)$, the first exit time of $B$ from the annulus $\{y : \varepsilon < |y| < R\}$. Since $B$ is a continuous process and the annulus is bounded, $H < \infty$ $\mathbb{P}_x$-almost surely. The process $f(B_{t \wedge H}) = \log|B_{t \wedge H}|$ is a bounded martingale (bounded between $\log \varepsilon$ and $\log R$), so by the [Optional Stopping Theorem](/theorems/1153), \begin{align*} \log|x| = \mathbb{E}_x[\log|B_H|] = (\log \varepsilon) \, \mathbb{P}_x(T_\varepsilon < T_R) + (\log R) \, \mathbb{P}_x(T_R < T_\varepsilon). \end{align*} Using $\mathbb{P}_x(T_R < T_\varepsilon) = 1 - \mathbb{P}_x(T_\varepsilon < T_R)$ and solving: \begin{align*} \mathbb{P}_x(T_\varepsilon < T_R) = \frac{\log R - \log|x|}{\log R - \log \varepsilon}. \end{align*} Letting $R \to \infty$: $\mathbb{P}_x(T_\varepsilon < \infty) = \lim_{R \to \infty} \mathbb{P}_x(T_\varepsilon < T_R) = 1$. So $B$ hits the ball $B(0, \varepsilon)$ almost surely from any starting point $x$. To show the set $\{t \geq 0 : |B_t| \leq \varepsilon\}$ is unbounded: by the [Strong Markov Property](/theorems/1180) applied at $T_\varepsilon$, the process $(B_{T_\varepsilon + t} - B_{T_\varepsilon})_{t \geq 0}$ is a standard Brownian motion. Starting from $B_{T_\varepsilon}$ with $|B_{T_\varepsilon}| = \varepsilon$, the same argument shows $B$ returns to $B(0, \varepsilon)$ almost surely. By induction, $B$ returns infinitely often. [/step] [step:Show that two-dimensional Brownian motion does not hit individual points] For $x \neq 0$ in $\mathbb{R}^2$, let $\varepsilon \to 0$ in the hitting probability formula: $\mathbb{P}_x(T_\varepsilon < T_R) = (\log R - \log|x|)/(\log R - \log \varepsilon) \to 0$ as $\varepsilon \to 0$ (with $R$ fixed), since the denominator $\log R - \log \varepsilon \to +\infty$. Therefore $\mathbb{P}_x(T_0 < T_R) = 0$ for every $R > |x|$, and taking $R \to \infty$ gives $\mathbb{P}_x(\exists\, t > 0 : B_t = 0) = 0$. For the case $x = 0$: fix $a > 0$. By the Markov property at time $a$, \begin{align*} \mathbb{P}_0(\exists\, t > a : B_t = 0) = \int_{\mathbb{R}^2} \mathbb{P}_y(\exists\, t > 0 : B_t = 0) \, p_a(y) \, d\mathcal{L}^2(y) = 0, \end{align*} where $p_a : \mathbb{R}^2 \to (0, \infty)$ is the heat kernel at time $a$, defined by $p_a(y) = (2\pi a)^{-1} \exp(-|y|^2/(2a))$. The integrand vanishes $\mathcal{L}^2$-almost everywhere since $\mathbb{P}_y(\exists\, t > 0 : B_t = 0) = 0$ for all $y \neq 0$, and $\{0\}$ has $\mathcal{L}^2$-measure zero. Since this holds for every $a > 0$, taking $a \downarrow 0$ gives $\mathbb{P}_0(\exists\, t > 0 : B_t = 0) = 0$. [/step] [step:Establish transience in $d \geq 3$ using the harmonic function $|\cdot|^{2-d}$] The function $g : \mathbb{R}^d \setminus \{0\} \to \mathbb{R}$ defined by $g(y) = |y|^{2-d}$ satisfies $\Delta g = 0$ on $\mathbb{R}^d \setminus \{0\}$ for $d \geq 3$ (this is the fundamental solution of [Laplace's equation](/page/Laplace's%20Equation)). With $T_\varepsilon$, $T_R$, and $H = \min(T_\varepsilon, T_R)$ as before, the same optional stopping argument gives \begin{align*} |x|^{2-d} = \varepsilon^{2-d} \, \mathbb{P}_x(T_\varepsilon < T_R) + R^{2-d} \, \mathbb{P}_x(T_R < T_\varepsilon). \end{align*} Solving: \begin{align*} \mathbb{P}_x(T_\varepsilon < T_R) = \frac{|x|^{2-d} - R^{2-d}}{\varepsilon^{2-d} - R^{2-d}}. \end{align*} As $R \to \infty$: since $d \geq 3$, $R^{2-d} \to 0$, so $\mathbb{P}_x(T_\varepsilon < \infty) = (|x|/\varepsilon)^{2-d} = (\varepsilon/|x|)^{d-2}$. For $|x| > \varepsilon$, this probability is strictly less than $1$. In particular, $B$ does not return to $B(0, \varepsilon)$ with probability $1$. To establish $|B_t| \to \infty$: define $A_n = \{|B_t| > n \text{ for all } t \geq T_{n^3}\}$ where $T_{n^3} = \inf\{t \geq 0 : |B_t| = n^3\}$. By the [Strong Markov Property](/theorems/1180) at $T_{n^3}$, the probability that $B$ returns to $B(0, n)$ after reaching $|B| = n^3$ is $(n/n^3)^{d-2} = n^{-2(d-2)}$. Since $d \geq 3$, $\sum_{n=1}^\infty n^{-2(d-2)} < \infty$, and the [Borel--Cantelli Lemma](/theorems/507) gives that $A_n^c$ occurs only finitely often. Therefore $|B_t| \to \infty$ almost surely. [/step] [step:Establish point recurrence in $d = 1$] For $d = 1$ and $x, z \in \mathbb{R}$, consider $B$ started at $x$ and the target $z$. By translation (replacing $B$ with $B - z$), it suffices to show that $B$ started at $x - z$ hits $0$ almost surely. The [Strong Markov Property](/theorems/1180) then gives infinitely many returns. The case $x = z$ follows from [Immediate Return to Zero](/theorems/1179): $\inf\{t > 0 : B_t = 0\} = 0$ almost surely when $B_0 = 0$. For $x \neq z$, consider $T_0 = \inf\{t \geq 0 : B_t = z\}$ with $B_0 = x$. In dimension $d = 1$, $B_t \sim \mathcal{N}(x, t)$, so $\mathbb{P}_x(B_t > z) \to 1/2$ as $t \to \infty$ (and similarly $\mathbb{P}_x(B_t < z) \to 1/2$). By path [continuity](/page/Continuity), $B$ must cross level $z$ to reach the other side, so $T_0 < \infty$ almost surely. The [Strong Markov Property](/theorems/1180) at successive return times gives that $\{t \geq 0 : B_t = z\}$ is unbounded. [/step]