 Solution

First justify that the logarithm is defined at all times from one onward. For planar Brownian motion started at $z\ne 0$, stop on exiting the annulus $\epsilon <|x|<R$, where $\epsilon <|z|<R$. This exit time $\rho$ is finite almost surely: the stopped identity $|B_t{|}^2-2t$ gives $\mathbb{E}(t\wedge \rho )\le R^2/2$. The function $f(x)=\log |x|$ is harmonic on the annulus, so its stopped process is a bounded martingale. Optional stopping givesHitting zero before the outer circle would require hitting every inner circle first. It therefore has probability zero. Taking a countable sequence of outer radii tending to infinity proves that planar Brownian motion avoids a fixed point. Since $B_1$ has a density and is nonzero almost surely, condition on ${\mathcal{F}}_1$ to apply this result to the future path.

For $x\ne 0$, $\nabla f(x)=x/|x{|}^2$ and $\Delta f(x)=0$ in two dimensions. The Itô formula on annuli therefore givesTo localize explicitly, let ${\tau }_n=\inf \{t\ge 1:|B_t|\notin (1/n,n)\}$. These stopping times increase to infinity almost surely: on each compact time interval the continuous path has a finite maximum and, by point avoidance, a positive minimum radius. Up to ${\tau }_n$, the stochastic integrand has norm at most $n$. Its stopped integral is square-integrable on every finite horizon. The initial value $\log |B_1|$ is integrable, as follows from the Rayleigh-density calculation in part d. Hence $X^{{\tau }_n}$ is a true martingale from time one and $X$ is a local martingale. This also gives its quadratic variation from time one as ${\int }_1^t|B_s{|}^{-2}ds$.