I am currently reading this article, page 7 (enumerated 889).
Let $g_t(z)$ be the solution for $t\in\mathbb{R}$ for
$$\partial_tg_t(z)=\frac{2}{ g_t(z)-\xi(t)},\quad g_0(z)=z.$$
Where $\xi$ is a brownian motion with variation parameter $\kappa$. Define $T(u)=\{t\in\mathbb{R}\vert im\; g_t(z)=e^u \}$ define also $\bar{\xi}(t)$ as $\max \vert \xi(s)\vert$ for $s\in [0,t]$.
I don't understand their argument for why $\vert T(u)\vert$ should be finite a.s, located at the bottom of page 7.
why does the inequality $\vert g_t(z)\vert<\vert z\vert+\bar{\xi}(t)+2\sqrt{t}$ hold for all $t<\tau(z)$ ? And what does it mean when $t$ is negative?
It is concluded that$...\geq 2(\vert z\vert+2\bar{\xi}(t)+2\sqrt{t})^{-2} $ to which they add that "...the right-hand-side is not integrable over $[0,\infty)$ nor $(-\infty,0]$..." What is ment by that? As far as i can see, two of the variables are not even defined for $t<0$.