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For a nonconstant entire function, its derivative is nonzero somewhere and therefore bounded away from zero on a small disc. Recurrence of planar Brownian motion and the Strong Markov property imply infinite total occupation time in that disc: repeat entrances into a smaller concentric disc after exits from the larger one. On each entrance there is a uniformly positive chance of remaining inside the larger disc for a fixed positive duration. Iterating the conditional failure bound gives infinitely many such sojourns almost surely.

The displayed integral is bounded below by a positive constant times this occupation time, so it diverges. The same conclusion follows by comparing with the divergence of a positive planar Brownian occupation integral. At every finite time the clock is finite, because the Brownian path has compact range and $f^{\prime }$ is continuous. Thus the conformal Brownian clock covers every nonnegative image time; conformal invariance of planar Brownian motion then yields a Brownian image with no missing finite or infinite lifetime.

Write $Z_t=z+B_t^1+i{B}_t^2$, where the coordinates are independent standard Brownian motions, and write $f=u+iv$. The Cauchy-Riemann equations imply $\Delta u=\Delta v=0$. Itô's formula therefore makescontinuous local martingales. Their quadratic variations and quadratic covariation areIndeed the two gradient norms are equal to $|f^{\prime }|$, and their dot product vanishes by the Cauchy-Riemann equations.

The clock is finite at every finite time, since a continuous Brownian path has compact range there and $f^{\prime }$ is continuous. To show its lifetime is infinite, choose a small disc on which $|f^{\prime }{|}^2\ge c>0$; such a disc exists because the entire function is nonconstant. A supporting recurrence result is that planar Brownian motion spends infinite total time in every nonempty open disc almost surely. One proof combines recurrent returns to a smaller disc with the strong Markov property: each visit has a uniformly positive chance of lasting a fixed positive time in the larger disc. Equivalently, use the divergence of a positive planar Brownian occupation integral. ThereforeFinally the common-clock time change of orthogonal local martingales states that two continuous local martingales starting at zero with common bracket $A$, zero cross bracket, and $A_{\infty }=\infty$ are the coordinates of a planar Brownian motion evaluated at $A_t$. Applying it gives a planar Brownian motion $\tilde{Z}$ starting at zero withThis proves conformal invariance of planar Brownian motion in the entire-function setting, including the divergence of the conformal Brownian clock for an entire function.

A finite original exit time alone does not imply that the integral defining the conformal Brownian clock is finite. To establish this, let $\psi ={\varphi }^{-1}$. On the event $\{\tilde{T}=\infty \}$, the Brownian path of part (a) lives forever in $D^{\prime }$, while its inverse clock satisfiesChoose a closed disc $C$ compactly contained in $D^{\prime }$. Its radius may be decreased so that $|{\psi }^{\prime }{|}^2\ge c>0$ on $C$. Recurrence of planar Brownian motion, together with the Strong Markov property, gives infinite total Brownian occupation time in $C$: return repeatedly to a smaller concentric disc, and use the fixed positive probability of remaining in $C$ for a fixed positive duration. The successive trials imply infinitely many such durations. This is the same mechanism as the divergence of a positive planar Brownian occupation integral.

Hence, on any infinite-lifetime Brownian path,This contradicts $\tau (t)<T<\infty$. Thus $\tilde{T}$ is finite almost surely, and equality in law from part (a) gives finite exit from a conformal image of a bounded planar domain:This asserts almost-sure finiteness, not finiteness of the expected exit time.

Use first that planar Brownian motion avoids a fixed point, so the path starting at $1$ never reaches zero. Its continuous logarithmic lift, initialized at zero, has real part $U_t=\log |Z_t|$ and imaginary part the continuous unwrapped angle ${\Theta }_t$. Local branches of the complex logarithm, patched along the path, and the Itô formula show that both coordinates are continuous local martingales withThe clock is finite on every compact time interval, since a continuous nonvanishing path stays a positive distance from zero there. It is continuous and strictly increasing. Moreover $\tau (\infty )=\infty$: by recurrence, planar Brownian motion spends infinite total time in a fixed disk, and inside the unit disk the clock integrand is at least one. This is the divergence of a positive planar Brownian occupation integral.

The common-clock time change of orthogonal local martingales, using the Lévy characterization of multidimensional Brownian motion, supplies independent Brownian coordinates ${\zeta }^1,{\zeta }^2$ such that $U_t+i{\Theta }_t={\zeta }_{\tau (t)}$. Exponentiating givesTo identify the inverse clock, write $A={\tau }^{-1}$. The derivative of an inverse function and the representation just obtained giveConsequentlyThe strict inequality in this generalized inverse has the same infimum as the unique inverse value, since $A$ is continuous and strictly increasing onto the half-line.