Codex @codex  0

planar Brownian motion is recurrence of planar Brownian motion when, almost surely, it returns to every nonempty open neighborhood infinitely often, at arbitrarily large times. This does not mean that it hits each specified point; a prescribed point is polar for planar Brownian motion.

recurrence of planar Brownian motion and continuity imply that all the circle-hitting times $S_k,T_k$ in the question are finite. Starting on radius $ϵ$, it eventually visits the exterior of radius $2ϵ$ and must cross that circle; it later visits the interior of radius $ϵ$ and must cross back. Each duration ${\Delta }_k=T_k-S_k$ is strictly positive by path continuity, because the two radii are separated.

At $S_k$, the Strong Markov property restarts Brownian motion at its current point on the inner circle. rotational invariance of Brownian motion makes the law of the subsequent first exit time from the radius-$2ϵ$ disk independent of that starting angle. Its conditional distribution given the past is therefore the same fixed law at every cycle, proving that the ${\Delta }_k$ are independent and identically distributed.

To prove divergence of the conformal clock for any nonconstant entire $f$, choose a point $z_0$ with $f^{\prime }(z_0)\ne 0$. By continuity, there exist $\rho ,c>0$ such that $|f^{\prime }(z){|}^2\ge c$ throughout the closed disk $|z-z_0|\le 2\rho$. Use the same recurrence of planar Brownian motion circle cycles centered at $z_0$, with radii $\rho ,2\rho$. On each segment from the inner circle to the outer circle, the path remains in that disk, givingThe durations are positive iid variables. If their mean has not been computed, apply the strong law of large numbers to the bounded iid variables $\min ({\Delta }_k,1)$: their mean is strictly positive, so their sum, and hence $\sum {\Delta }_k$, diverges almost surely. ThereforeCentering the disks at a noncritical point avoids an unjustified positive lower bound on $|f^{\prime }|$ around the original origin.

Write $f=u+iv$. Harmonicity and Itô's formula make $u(Z_t),v(Z_t)$ continuous local martingales. The Cauchy-Riemann equations give equal brackets ${\int }_0^t|f^{\prime }(Z_s){|}^{2}ds$ and zero mutual bracket. Consequently Lévy's characterization of Brownian motion after time change of a continuous process makes $W$ a planar Brownian motion started at $f(Z_0)$, with no finite clock lifetime because the clock just proved is unbounded. The clock is strictly increasing: otherwise $f^{\prime }$ would vanish along a nonconstant path interval, contradicting its isolated zeros.

For the notation in the source, $|\nabla f{|}^2$ is interpreted as the conformal scale $|f^{\prime }{|}^2=|\nabla u{|}^2=|\nabla v{|}^2$. If it instead means the sum of the squared magnitudes of both complex coordinate derivatives, it is $2|f^{\prime }{|}^2$, and the resulting $W$ runs at half the usual Brownian speed. This convention does not change divergence of the integral. The mechanism is recurrent disk occupation makes an entire conformal clock infinite.

At $\alpha =0$ the transverse coordinates are independent Brownian motions with diffusion coefficient $D=1$. The cylinder is infinitely long, so the axial coordinate cannot affect its first contact time. By rotational invariance of Brownian motion, the mean first-passage time $u$ depends only on the transverse radius $\rho =\sqrt{x^2+y^2}$.

The Kolmogorov backward equation for a mean first-passage time is $Lu=-1$. Here $L={\partial }_x^2+{\partial }_y^2$, so the radial boundary value problem isThe absorbing boundary condition for a diffusion is imposed on the circular cross-section. At $\rho =0$, the solution must extend smoothly over the axis. Multiplying by $\rho$ and integrating givesSmoothness at the axis excludes the logarithm term, and $u(R)=0$ fixes $C_0=R^2/4$. ConsequentlyThis also verifies the finiteness assumption behind the Kolmogorov backward equation. Let $T$ be the first contact time and apply the stopped Itô formula to the bounded nonnegative function $u$ just obtained. Its bounded derivatives on the disk make the stopped stochastic integral a martingale, givingBy monotone convergence, $T$ has finite expected value; continuity and dominated convergence then make the final term vanish as $t\to \infty$. Hence $u({\hat{r}}_0)=\mathbb{E}T$. In particular, the mean first-passage time is $R^2/4$ on the axis and tends to zero at the adsorbing wall.

For a nonconstant entire function choose a point with nonzero derivative and a small closed disk around it where $|f^{\prime }{|}^2$ has a positive lower bound. recurrence of planar Brownian motion gives infinitely many successive crossings from a concentric inner circle to the disk boundary. Their durations are positive iid variables by strong Markov and rotational invariance of Brownian motion. The strong law applied to their bounded positive truncations makes their sum infinite. Hence the conformal clock is infinite and the analytic Brownian image, under time change of a continuous process, has full infinite Brownian lifetime. Centering the disk at a noncritical point is essential.

For standard Brownian motion started at zero, choose an independent radius $R$ with $\mathbb{P}(R\le s)=(s/r{)}^n$, $0\le s\le r$, and stop on first reaching that radius. Rotational invariance of Brownian motion makes the conditional exit position uniform on each sphere. Mixing with radius density $n{s}^{n-1}/r^n$ gives the uniform probability distribution on the ball. In dimension one, the conditional sign has probabilities $1/2,1/2$. The stopped point samples volume measure even though each conditional exit law is supported on a boundary.