Harmonic functions of planar Brownian motion

The two functions are space-time harmonic: by [F2] both have vanishing Laplacian and no time dependence, so the lifetime-local assertion [F1] applies with the relatively open set $U=[0,\mathrm{\infty })\times {\mathbb{R}}^2$, whose lifetime is infinity. Hence $B_t^1{B}_t^2=p(B_t)$ and $(B_t^1{)}^2-(B_t^2{)}^2=q(B_t)$ are continuous local martingales.

Time-capped stopping: fix $R>0$ and for each integer $n\ge 1$ set $K_n=[0,n]\times \{y:\mathrm{\mid }y\mathrm{\mid }\le R\}$. This is a compact subset of $U=[0,\mathrm{\infty })\times {\mathbb{R}}^2$ containing $(0,0)$ in its relative interior; the exit time in [F1] is exactly ${\tau }_{K_n}=n\wedge {\tau }_R$. The harmonic theorem makes this a stopping time and supplies its stopped integral identity and square-integrable martingale. Also $\{{\tau }_R\le t\}=\{{\tau }_{K_n}\le t\}$ whenever $n>t$, so ${\tau }_R$ is a stopping time. Given any finite horizon $T$, choose an integer $n>T$; then $t\wedge {\tau }_{K_n}=t\wedge {\tau }_R$ for every $0\le t\le T$. Thus the compact-stopped process and integral from [F1] coincide with the disc-stopped ones throughout that horizon. This proves the disc-stopped martingale property for every pair of finite times without treating a spatial disc as compact space-time.

Explicit form of the stopped integrals: from [F1] and the stopping identity, $p(B_{t\wedge {\tau }_R})={\int }_0^t{1}_{[0,{\tau }_R]}(s)(B_s^{2}d{B}_s^1+B_s^{1}d{B}_s^2),$ $q(B_{t\wedge {\tau }_R})={\int }_0^t{1}_{[0,{\tau }_R]}(s)(2B_s^{1}d{B}_s^1-2B_s^{2}d{B}_s^2)\mathrm{.}$ On each finite horizon the integrands agree with the bounded predictable compact-stopped integrands from step 2.1, including the endpoint indicator; their squared Euclidean norms are bounded by $R^2$ and $4R^2$. Thus each scalar component has finite expected energy and the finite sums are square-integrable martingales. The identities hold up to indistinguishability: intersect the probability-one identities for integer horizons.

Boundary and consistency cases: at $t=0$ both processes start at $0$; as integer $R\to \mathrm{\infty }$, every continuous path is bounded on each compact time interval, so ${\tau }_R\to \mathrm{\infty }$ and the stopped processes eventually equal the unstopped ones on that interval; the proof here asserts square-integrability after disc stopping using bounded gradients; unboundedness on the plane alone is not an obstruction to a true martingale, and no such obstruction is claimed; for $B$ starting at the origin the disc contains the starting point for every $R>0$; and AC enters only through [F4].